{"url": "http://shop.track-it.nz/6bowhhs1/properties-of-matrix-addition-346a66", "text": "Commutative Property Of Addition 2. If A is an n\u00d7m matrix and O is a m\u00d7k zero-matrix, then we have: AO = O Note that AO is the n\u00d7k zero-matrix. Matrix Matrix Multiplication 11:09. We have 1. To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. The identity matrix is a square matrix that has 1\u2019s along the main diagonal and 0\u2019s for all other entries. In a triangular matrix, the determinant is equal to the product of the diagonal elements. This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? The identity matrices (which are the square matrices whose entries are zero outside of the main diagonal and 1 on the main diagonal) are identity elements of the matrix product. Learning Objectives. In fact, this tutorial uses the Inverse Property of Addition and shows how it can be expanded to include matrices! Keywords: matrix; matrices; inverse; additive; additive inverse; opposite; Background Tutorials . Matrix Multiplication Properties 9:02. 16. Proof. There are a few properties of multiplication of real numbers that generalize to matrices. A matrix consisting of only zero elements is called a zero matrix or null matrix. Properties of Matrix Addition and Scalar Multiplication. What is the Identity Property of Matrix Addition? General properties. Yes, it is! There are 10 important properties of determinants that are widely used. Go through the properties given below: Assume that, A, B and C be three m x n matrices, The following properties holds true for the matrix addition operation. The determinant of a 4\u00d74 matrix can be calculated by finding the determinants of a group of submatrices. 13. If you built a random matrix and took its determinant, how likely would it be that you got zero? The first element of row one is occupied by the number 1 \u2026 In mathematics, matrix addition is the operation of adding two matrices by adding the corresponding entries together. Equality of matrices All-zero Property. Multiplying a $2 \\times 3$ matrix by a $3 \\times 2$ matrix is possible, and it gives a $2 \\times 2$ matrix \u2026 Properties of Transpose of a Matrix. The Commutative Property of Matrix Addition is just like the Commutative Property of Addition! The Distributive Property of Matrices states: A ( B + C ) = A B + A C Also, if A be an m \u00d7 n matrix and B and C be n \u00d7 m matrices, then Addition: There is addition law for matrix addition. Likewise, the commutative property of multiplication means the places of factors can be changed without affecting the result. Then the following properties hold: a) A+B= B+A(commutativity of matrix addition) b) A+(B+C) = (A+B)+C (associativity of matrix addition) c) There is a unique matrix O such that A+ O= Afor any m\u00d7 nmatrix A. Since Theorem SMZD is an equivalence (Proof Technique E) we can expand on our growing list of equivalences about nonsingular matrices. Properties of matrix addition. Let A, B, and C be mxn matrices. The determinant of a 3 x 3 matrix (General & Shortcut Method) 15. PROPERTIES OF MATRIX ADDITION PRACTICE WORKSHEET. You should only add the element of one matrix to \u2026 Property 1 completes the argument. Let A, B, and C be three matrices of same order which are conformable for addition and a, b be two scalars. This means if you add 2 + 1 to get 3, you can also add 1 + 2 to get 3. For any natural number n > 0, the set of n-by-n matrices with real elements forms an Abelian group with respect to matrix addition. This tutorial uses the Commutative Property of Addition and an example to explain the Commutative Property of Matrix Addition. This project was created with Explain Everything\u2122 Interactive Whiteboard for iPad. (i) A + B = B + A [Commutative property of matrix addition] (ii) A + (B + C) = (A + B) +C [Associative property of matrix addition] (iii) ( pq)A = p(qA) [Associative property of scalar multiplication] Let A, B, C be m \u00d7n matrices and p and q be two non-zero scalars (numbers). In this lesson, we will look at this property and some other important idea associated with identity matrices. So if n is different from m, the two zero-matrices are different. Andrew Ng. Matrices rarely commute even if AB and BA are both defined. What is a Variable? Addition and Scalar Multiplication 6:53. The addition of the condition $\\detname{A}\\neq 0$ is one of the best motivations for learning about determinants. ... although it is associative and is distributive over matrix addition. Question 1 : then, verify that A + (B + C) = (A + B) + C. Solution : Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. This tutorial introduces you to the Identity Property of Matrix Addition. Instructor. Transcript. Important Properties of Determinants. Properties involving Addition and Multiplication: Let A, B and C be three matrices. Matrix addition and subtraction, where defined (that is, where the matrices are the same size so addition and subtraction make sense), can be turned into homework problems. A. Addition and Subtraction of Matrices: In matrix algebra the addition and subtraction of any two matrix is only possible when both the matrix is of same order. Selecting row 1 of this matrix will simplify the process because it contains a zero. We state them now. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Question: THEOREM 2.1 Properties Of Matrix Addition And Scalar Multiplication If A, B, And C Are M X N Matrices, And C And D Are Scalars, Then The Properties Below Are True. A B _____ Commutative property of addition 2. Unlike matrix addition, the properties of multiplication of real numbers do not all generalize to matrices. Let A, B, and C be three matrices. 8. det A = 0 exactly when A is singular. 2. In other words, the placement of addends can be changed and the results will be equal. Then we have the following properties. 17. The basic properties of matrix addition is similar to the addition of the real numbers. Then we have the following: (1) A + B yields a matrix of the same order (2) A + B = B + A (Matrix addition is commutative) Properties of scalar multiplication. (A+B)+C = A + (B+C) 3. where is the mxn zero-matrix (all its entries are equal to 0); 4. if and only if B = -A. If the rows of the matrix are converted into columns and columns into rows, then the determinant remains unchanged. Laplace\u2019s Formula and the Adjugate Matrix. 14. Matrix multiplication shares some properties with usual multiplication. 4. Question 3 : then find the additive inverse of A. the identity matrix. Properties of Matrix Addition (1) A + B + C = A + B + C (2) A + B = B + A (3) A + O = A (4) A + \u2212 1 A = 0. Find the composite of transformations and the inverse of a transformation. Created by the Best Teachers and used by over 51,00,000 students. Use the properties of matrix multiplication and the identity matrix Find the transpose of a matrix THEOREM 2.1: PROPERTIES OF MATRIX ADDITION AND SCALAR MULTIPLICATION If A, B, and C are m n matrices, and c and d are scalars, then the following properties are true. Matrix Multiplication - General Case. Properties involving Addition. The commutative property of addition means the order in which the numbers are added does not matter. Properties involving Multiplication. A+B = B+A 2. Given the matrix D we select any row or column. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. Note that we cannot use elimination to get a diagonal matrix if one of the di is zero. Question 1 : then, verify that A + (B + C) = (A + B) + C. Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. When the number of columns of the first matrix is the same as the number of rows in the second matrix then matrix multiplication can be performed. The inverse of 3 x 3 matrix with determinants and adjugate . The order of the matrices must be the same; Subtract corresponding elements; Matrix subtraction is not commutative (neither is subtraction of real numbers) Matrix subtraction is not associative (neither is subtraction of real numbers) Scalar Multiplication. The inverse of 3 x 3 matrices with matrix row operations. Properties of matrix multiplication. Properties of Matrix Addition, Scalar Multiplication and Product of Matrices. EduRev, the Education Revolution! As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. Matrix multiplication is really useful, since you can pack a lot of computation into just one matrix multiplication operation. Proposition (commutative property) Matrix addition is commutative, that is, for any matrices and and such that the above additions are meaningfully defined. 11. Numerical and Algebraic Expressions. The determinant of a 2 x 2 matrix. We can also say that the determinant of the matrix and its transpose are equal. Matrix Vector Multiplication 13:39. 12. Examples . In that case elimination will give us a row of zeros and property 6 gives us the conclusion we want. This property is known as reflection property of determinants. The inverse of a 2 x 2 matrix. However, unlike the commutative property, the associative property can also apply to matrix \u2026 The matrix O is called the zero matrix and serves as the additiveidentity for the set of m\u00d7nmatrices. 1. Best Videos, Notes & Tests for your Most Important Exams. To find the transpose of a matrix, we change the rows into columns and columns into rows. Properties of Matrix Addition: Theorem 1.1Let A, B, and C be m\u00d7nmatrices. However, there are other operations which could also be considered addition for matrices, such as the direct sum and the Kronecker sum Entrywise sum. The determinant of a matrix is zero if each element of the matrix is equal to zero. A scalar is a number, not a matrix. Try the Course for Free. Use properties of linear transformations to solve problems. Matrix addition is associative; Subtraction. 18. 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X 3 matrix with determinants and adjugate select any row or column if one of matrix.$ \\detname { a } \\neq 0 \\$ is one properties of matrix addition the and. 1 in matrix multiplication operation your Most important Exams and B which have equal order SMZD is an equivalence Proof.", "date": "2021-09-28 02:16:51", "meta": {"domain": "track-it.nz", "url": "http://shop.track-it.nz/6bowhhs1/properties-of-matrix-addition-346a66", "openwebmath_score": 0.7671064734458923, "openwebmath_perplexity": 360.7558924577382, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9914225156000334, "lm_q2_score": 0.9539660961284018, "lm_q1q2_score": 0.9457834668207633}}
{"url": "https://math.stackexchange.com/questions/2922644/comparing-the-magnitudes-of-expressions-of-surds", "text": "# Comparing the magnitudes of expressions of surds\n\nI recently tackled some questions on maths-challenge / maths-aptitude papers where the task was to order various expressions made up of surds (without a calculator, obviously).\n\nI found myself wondering whether I was relying too much on knowing the numerical value of some common surds, when a more robust method was available (and would work in more difficult cases).\n\nFor example, one question asked which is the largest of:\n\n(a) $\\sqrt{10}$\n(b) $\\sqrt2+\\sqrt3$\n(c) $5-\\sqrt3$\n\nIn this case, I relied on my knowledge that $\\sqrt{10} \\approx 3.16$ and $\\sqrt2\\approx 1.41$ and $\\sqrt3 \\approx 1.73$ to find (a) $\\approx 3.16$, (b) $\\approx ~3.14$ and (c) $\\approx ~3.27$ so that the required answer is (c).\n\nBut this seemed inelegant: I felt there might be some way to manipulate the surd expressions to make the ordering more explicit. I can't see what that might be, however (squaring all the expressions didn't really help).\n\nI'd appreciate some views: am I missing a trick, or was this particular question simply testing knowledge of some common values?\n\nEDIT: after the very helpful answers, which certainly showed that there was a much satisfying and general way of approaching the original question, can I also ask about another version of the question which included (d) $\\sqrt[4]{101}$.\n\nWhen approaching the question by approximation, I simply observed that $\\sqrt[4]{101}$ is only a tiny bit greater than $\\sqrt{10}$, and hence it still was clear to choose (c) as the answer. Is there any elegant way to extend the more robust methods to handle this case?\n\n\u2022 +1 for providing context (your first two sentences), something that nearly all questions at this level fail to do, and for providing a nice explanation of your concern. Incidentally, for math aptitude and other tests, it has always been my understanding that the questions are NOT testing whether you know the approximations, but whether you can perform the type of analysis in the answer by @Lord Shark the Unknown. Of course, unless the question writer puts some effort behind writing such questions, such questions can often be solved by your method. Sep 19 '18 at 10:41\n\u2022 Thank you for all the comments and answers. I am pleased I chose to ask the question at MSE -- there was indeed something to learn here! Sep 20 '18 at 10:05\n\nComparing $$\\sqrt{10}$$ and $$\\sqrt2+\\sqrt3$$ is the same as comparing $$10$$ and $$(\\sqrt2+\\sqrt3)^2=5+2\\sqrt6$$. That's the same as comparing $$5$$ and $$2\\sqrt6$$. Which of these is bigger?\n\nLikewise comparing $$\\sqrt{10}$$ and $$5-\\sqrt3$$ is the same as comparing $$10$$ and $$(5-\\sqrt3)^2=28-10\\sqrt3$$. That's the same as comparing $$10\\sqrt3$$ and $$18$$.\n\nWhich of these is bigger?\n\n\u2022 Ah .... yes of course ... $5=\\sqrt{25}>\\sqrt{24}=2\\sqrt6$ Sep 19 '18 at 10:30\n\u2022 Thank you for the hint! Sep 19 '18 at 10:32\n\u2022 BBO555, regarding $5$ and $2\\sqrt{6},$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be \"seen\" by considering the graph of $y = x^2$ for $x\\geq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$ Sep 19 '18 at 10:47\n\nYou can use:\n\n(1) the fact that $f(x)=x^2$ is a monotonically increasing function when $x\\geq0$ and\n\n(2) the arithmetic-geometric mean inequality $\\sqrt{ab}\\leq\\frac{a+b}{2}$, when $a, b\\geq0$. Hence, $$(\\sqrt{2}+\\sqrt{3})^2=5+2\\sqrt{2\\cdot3}\\leq5+2\\frac{2+3}{2}=5+5=10=(\\sqrt{10})^2$$ Therefore, using (1), we obtain $\\sqrt{2}+\\sqrt{3}\\leq 10$. I forgot about this: $$5-\\sqrt{3}=3+2-\\sqrt{3}=3+\\frac{1}{2+\\sqrt{3}}\\geq3+\\frac{1}{2+2}=3.25$$ One can easily verify that $(3+1/4)^2>10.5>10$. One also finds that $10.5^2>110>101$.\n\nThen, performing argument (1) twice, one finds that $5-\\sqrt{3}>(101)^{1/4}$.\n\n## Consequently, $5-\\sqrt{3}$ is the bigger number.\n\n\u2022 I would add that you can also \"round down\" at an intermediate stage of the computation. If you are trying to prove $a \\gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a \\gt c$, do some more manipulation, and show $c \\gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful. Sep 19 '18 at 14:01\n\u2022 Thanks for including a route to handling case (d) ! Sep 20 '18 at 8:07", "date": "2021-10-19 03:38:04", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2922644/comparing-the-magnitudes-of-expressions-of-surds", "openwebmath_score": 0.7826164364814758, "openwebmath_perplexity": 198.0380721381055, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874087451302, "lm_q2_score": 0.94580126731237, "lm_q1q2_score": 0.9368988265748208}}
{"url": "https://math.stackexchange.com/questions/1693964/in-calculus-how-can-a-function-have-several-different-yet-equal-derivatives", "text": "# In Calculus, how can a function have several different, yet equal, derivatives?\n\nI've been pondering this question all night as I work through some problems, and after a very thorough search, I haven't found anything completely related to my question. I guess i'm also curious how some derivatives are simplified as well, because in some cases I just can't see the breakdown. Here is an example:\n\n$f(x) = \\dfrac{x^2-6x+12}{x-4}$ is the function I was differentiating. Here is what I got:\n\n$f '(x) = \\dfrac{x^2-8x+12}{(x-4)^2}$ which checks using desmos graphing utility.\n\nNow, when I checked my textbook(and Symbolab) they got:\n\n$f '(x) = 1 - \\dfrac{4}{(x-4)^2}$ which also checks on desmos.\n\nTo me, these derivatives look nothing alike, so how can they both be the equal to the derivative of the original function? Both methods used the quotient rule, yet yield very different results. Is one of these \"better\" than the other? I know that it is easier to find critical numbers with a more simplified derivative, but IMO the derivative I found seems easier to set equal to zero than the derivative found in my book.I also wasn't able to figure out how the second derivative was simplified, so I stuck with mine.\n\nI'm obviously new to Calculus and i'm trying to understand the nuances of derivatives. When I ask most math people, including some professors, they just say \"that's how derivatives are\" but for me, that's not an acceptable answer. If someone can help me understand this, I would appreciate it.\n\n\u2022 You really really need to use parentheses in what you write. You mean to write $(x^2-8x+12)/(x-4)^2$. The point is that your two \"different\" answers are exactly the same because of algebra. \u2013\u00a0Ted Shifrin Mar 12 '16 at 7:54\n\u2022 Well i'm still learning the formatting so bear with me, and I obviously know they are the same because they are both the derived from the original function(and checked out). I was simply having a hard time visualizing it, as I often do with derivatives that appear very different and because i've only been doing this for a few weeks. Anyways, thanks for the comment, I guess. \u2013\u00a0FuegoJohnson Mar 12 '16 at 8:10\n\u2022 When you write \"x^2-6x+12/(x-4)\" you are writing $x^2-6x+\\frac{12}{x-4}$, which is not the same as $\\frac{x^2-6x+12}{x-4}$. \u2013\u00a0alex.jordan Mar 12 '16 at 8:12\n\u2022 @Hirak: Your edit is incorrect. \u2013\u00a0Ted Shifrin Mar 12 '16 at 8:18\n\u2022 I know I'm sorry, i'm going thru the formatting rules right now to make it look better. Sincerest apologies. \u2013\u00a0FuegoJohnson Mar 12 '16 at 8:18\n\nSometimes when dealing with the derivative of a quotient of polynomials, it is more easy to do some calculations first and then start the derivatives.\n\nIn this case, when we do the division of polynomials $\\dfrac{x^2-6x+12}{x-4}$ we obtain quotient $x-2$ and residue $4$ (I prefer not to write the division here because depending on how your learn it in school there might be slightly different methods)\n\nSo, we get $$x^2-6x+12=(x-2)(x-4)+4$$ and dividing both sides by $(x-4)$ we obtain $$f(x)=\\dfrac{x^2-6x+12}{x-4}=(x-2)+\\dfrac{4}{x-4}$$\n\nIt is somewhat easier to calculate the derivative of this new expression, because when we apply the rule for the quotient one of the derivatives is zero.\n\nWhen you take the derivative of the second expression you get\n\n$$f'(x)=1+\\dfrac{0\\cdot (x-4)-4(1)}{(x-4)^2}=1-\\dfrac{4}{(x-4)^2}$$ which is simpler and especially useful when you will calculate second derivatives and, for example, find the graph of the function.\n\n\u2022 Thank you for your helpful input! You broke it down in a way that I wasn't able to visualize, and now I see. I ended up finding the second derivative through a much more tedious method, so I think your way would definitely be easier. Thanks :) \u2013\u00a0FuegoJohnson Mar 12 '16 at 20:17\n\nThey are the same. One way to prove that is the following: $$1-\\frac4{(x-4)^2}=\\frac{(x-4)^2-4}{(x-4)^2}\\\\=\\frac{x^2-8x+16-4}{(x-4)^2}\\\\=\\frac{x^2-8x+12}{(x-4)^2}$$\n\n\u2022 Thats really easy to visualize the way you broke it down, thanks. My book skips so many steps sometimes. So my next question for you, is one form \"better\" than the other? I had a really hard time understanding how they simplified the function in my book but seeing you compare them makes a little more sense to me. \u2013\u00a0FuegoJohnson Mar 12 '16 at 7:57\n\u2022 @FuegoJohnson For any particular $x$-value, the expression $1-\\frac{4}{(x-4)^2}$ takes less arithmetic to evaluate than does the other option. That is one reason to prefer it. Another reason is that it would be more efficient to continue taking higher order derivatives of $1-\\frac{4}{(x-4)^2}$, since no quotient rule would be needed. \u2013\u00a0alex.jordan Mar 12 '16 at 8:08\n\u2022 Perhaps the best answer would be depending on your purpose. I suppose the $1-\\frac{4}{(x-4)^2}$ form would be easier to set to zero for me, but if you prefer the other method it is fine. I suppose the best answer is that the 'best' derivative would be the one which, setting for zero, you can isolate for x the fastest on a test haha. Aside from that, there is no real 'best' derivative. \u2013\u00a0Keith Afas Mar 12 '16 at 8:09\n\u2022 Great input, thanks folks. I am about to take the second derivative of the function, so it makes sense that the book answer would be easier to work with, although I STILL don't understand how they simplified it the way they did in the book from the original function. My algebra is kinda rusty at the moment. If someone wants to break it down for me step by step, that would be great lol. \u2013\u00a0FuegoJohnson Mar 12 '16 at 8:17", "date": "2019-04-24 06:18:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1693964/in-calculus-how-can-a-function-have-several-different-yet-equal-derivatives", "openwebmath_score": 0.8208966851234436, "openwebmath_perplexity": 249.28708375250756, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.984810947625669, "lm_q2_score": 0.9504109729777948, "lm_q1q2_score": 0.9359751309320962}}
{"url": "http://mathhelpforum.com/algebra/8958-working-backwards-cubics.html", "text": "# Math Help - working backwards - cubics\n\n1. ## working backwards - cubics\n\nWrite an equation that has the following roots: 2, -1, 5\n\nAnswer key: x^3 - 6x^2 + 3x + 10 = 0\n\nFor quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this?\n\nThanks.\n\n2. Originally Posted by shenton\nWrite an equation that has the following roots: 2, -1, 5\n\nAnswer key: x^3 - 6x^2 + 3x + 10 = 0\n\nFor quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this?\n\nThanks.\n$(x - 2)(x + 1)(x - 5)$\n\n3. Thanks! That turns out to be not as difficult as imagined. I thought I needed to use sum and products of roots to write the equation, it does makes me wonder a bit why or when I need to use sum and products of roots.\n\n4. Write an equation that has the following roots: 2, -1, 5\n\nIs there any other way to solve this other than the (x-2)(x+1)(x-5) method?\n\nIf we have these roots: 1, 1 + \u221a2, 1 - \u221a2\n\nthe (x - 1) (x -1 -\u221a2) (x -1 +\u221a2) method seems a bit lenghty.\n\nWhen we expand (x - 1) (x -1 -\u221a2) (x -1 +\u221a2) the first 2 factors,\n\nit becomes:\n\n(x^2 -x -x\u221a2 -x +1 +\u221a2) (x -1 +\u221a2)\n\ncollect like terms:\n\n(x^2 -2x -x\u221a2 +1 +\u221a2) (x -1 +\u221a2)\n\nTo further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method.\n\nIs there a method to write an equation with 3 given roots (other than the above method)?\n\nThanks.\n\n5. Originally Posted by shenton\nWrite an equation that has the following roots: 2, -1, 5\n\nIs there any other way to solve this other than the (x-2)(x+1)(x-5) method?\n\nIf we have these roots: 1, 1 + \u221a2, 1 - \u221a2\n\nthe (x - 1) (x -1 -\u221a2) (x -1 +\u221a2) method seems a bit lenghty.\n\nWhen we expand (x - 1) (x -1 -\u221a2) (x -1 +\u221a2) the first 2 factors,\n\nit becomes:\n\n(x^2 -x -x\u221a2 -x +1 +\u221a2) (x -1 +\u221a2)\n\ncollect like terms:\n\n(x^2 -2x -x\u221a2 +1 +\u221a2) (x -1 +\u221a2)\n\nTo further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method.\n\nIs there a method to write an equation with 3 given roots (other than the above method)?\n\nThanks.\nYou have a pair of roots of the form a+sqrt(b) and a-sqrt(b) if you multiply\nthe factors corresponding to these first you get:\n\n(x-a-sqrt(b))(x-a+sqrt(b))=x^2+(-a-sqrt(b))x+(-a+sqrt(b))x +(-a-sqrt(b))(-a+sqrt(b))\n\n................=x^2 - 2a x + (a^2-b)\n\nWhich leaves you with the easier final step of computing:\n\n(x-1)(x^2 - 2a x + (a^2-b))\n\nRonL\n\n6. Hello, shenton!\n\nThe sum and product of roots works well for quadratic equations.\n\nFor higher-degree equations, there is a generalization we can use.\n\nTo make it simple (for me), I'll explain a fourth-degree equation.\n\nDivide through by the leading coefficient: . $x^4 + Px^3 + Qx^2 + Rx + S \\:=\\:0$\n\nInsert alternating signs: . $+\\:x^4 - Px^3 + Qx^2 - Rx + S \\:=\\:0$\n. . . . . . . . . . . . . . . . . . $\\uparrow\\quad\\;\\; \\uparrow\\qquad\\;\\;\\uparrow\\qquad\\;\\,\\uparrow\\qquad \\,\\uparrow$\n\nSuppose the four roots are: $a,\\,b,\\,c,\\,d.$\n\nThe sum of the roots (taken one at a time) is: $-P.$\n. . $a + b + c + d \\:=\\:-P$\n\nThe sum of the roots (taken two at a time) is: $Q.$\n. . $ab + ac + ad + bc + bd + cd \\:=\\:Q$\n\nThe sum of the roots (taken three at a time) is: $-R.$\n. . $abc + abd + acd + bcd \\:=\\:-R$\n\nThe sum of the roots (\"taken four at a time\") is: $S.$\n. . $abcd \\:=\\:S$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nFor your problem with roots: $(a,b,c) \\:=\\:(2,-1,5)$\n. . we have: . $x^3 + Px^2 + Qx + R \\:=\\:0$\n\nThen: . $a + b + c \\:=\\:-P\\quad\\Rightarrow\\quad2 + (-1) + 5\\:=-P$\n. . Hence: $P = -6$\n\nAnd: . $ab + bc + ac \\:=\\:Q\\quad\\Rightarrow\\quad(2)(-1) + (-1)(5) + (2)(5) \\:=\\:Q$\n. . Hence: $Q = 3$\n\nAnd: . $abc \\:=\\:-R\\quad\\Rightarrow\\quad(2)(-1)(5)\\:=\\:-R$\n. . Hence: $R = 10$\n\nTherefore, the cubic is: . $x^3 - 6x^2 + 3x + 10 \\:=\\:0$\n\n7. This is awesome, Soroban. Using the method you shown, I was able to solve this problem:\n\n1, 1+\u221a2, 1-\u221a2\n\na=1, b=1+\u221a2, c=1-\u221a2\n\nLet x^3 - px^2 + qx - r = 0 be the cubic equation\n\np = a + b + c\n= (1) + (1 + \u221a2) + (1 - \u221a2)\n= 3\n\nq = ab + bc + ac\n= (1)(1 + \u221a2) + (1 + \u221a2)(1 - \u221a2) + (1)(1 - \u221a2)\n= 1 + \u221a2 + 1 - 2 + 1 - \u221a2\n= 1\n\nr = abc\n= (1)(1 + \u221a2)(1 - \u221a2)\n= 1-2\n= -1\n\nTherefore x^3 - px^2 + qx - r = 0 becomes\nx^3 - 3x^2 + x - (-1) = 0\nx^3 - 3x^2 + x + 1 = 0\n\nThanks for the help and detailed workings.", "date": "2014-09-19 23:58:39", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/8958-working-backwards-cubics.html", "openwebmath_score": 0.8881317973136902, "openwebmath_perplexity": 861.8028331615614, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9914225133191064, "lm_q2_score": 0.9433475748911165, "lm_q1q2_score": 0.9352560236320346}}
{"url": "https://math.stackexchange.com/questions/202052/work-and-time-when-work-is-split-into-parts", "text": "# Work and time, when work is split into parts\n\nI'm stuck on a particular type of work and time problems.\n\nFor example,\n\n1) A,B,C can complete a work separately in 24,36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?\n\nA simpler version of the same type of problem is as follows:\n\n2) A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is?\n\nMy attempt at problem 2:\n\nA's 1 day work=1/14 and B's 1 day work= 1/21\n\nAssume that it takes 'd' days to complete the entire work when both A and B are working together. Then,\n\n(1/14 + 1/21)*d= 1\n\n-> d=42/5 days.\n\nBut it is stated that 3 days before the completion of the work, A left. Therefore, work done by both in (d-3) days is:\n\n(1/14 + 1/21)*(42/5 - 3)= 9/14\n\nRemaining work= 1- 9/14 = 5/14 which is to be done by B alone. Hence the time taken by B to do (5/14) of the work is:\n\n(5/14)*21 = 7.5 days.\n\nTotal time taken to complete the work = (d-3) + 7.5 = 12.9 days.\n\nHowever, this answer does not concur with the one that is provided.\n\nMy Understanding of problem 1:\n\nProblem 1 is an extended version of problem 2. But since i think i'm doing problem 2 wrong, following the same method on problem 1 will also result in a wrong answer.\n\nWhere did i go wrong?\n\n## 5 Answers\n\nYou asked where you went wrong in solving this problem:\n\nA can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is?\n\nAs you said in your solution, $A$ can do $1/14$ of the job per day, and $B$ can do $1/21$ of the job per day. On each day that they work together, then, they do $$\\frac1{14}+\\frac1{21}=\\frac5{42}$$ of the job. Up to here you were doing fine; it\u2019s at this point that you went astray. You know that for the last three days of the job $B$ will be working alone. In those $3$ days he\u2019ll do $$3\\cdot\\frac1{21}=\\frac17$$ of the job. That means that the two of them working together must have done $\\frac67$ of the job before $A$ left. This would have taken them\n\n$$\\frac{6/7}{5/42}=\\frac67\\cdot\\frac{42}5=\\frac{36}5\\text{ days}\\;.$$\n\nAdd that to the $3$ days that $B$ worked alone, and you get the correct total: $$\\frac{36}5+3=\\frac{51}5=10.2\\text{ days}\\;.$$\n\nYou worked out how long it would take them working together, subtracted $3$ days from that, saw how much of the job was left to be done at that point, and added on the number of days that it would take $B$ working alone to finish the job. But as your own figures show, $B$ actually needs $7.5$ days to finish the job at that point, not $3$, so he ends up working alone for $7.5$ days. This means that $A$ actually left $7.5$ days before the end of the job, not $3$ days before. You have to figure out how long it takes them to reach the point at which $B$ can finish in $3$ days.\n\nAdded:\n\n1) A,B,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?\n\nHere you know that all three worked together for the first $4$ days, $B$ worked alone for the last $3$ days, and $A$ and $B$ worked together for some unknown number of days in the middle. Calculate the fraction of the job done by all three in the first $4$ days and the fraction done by $B$ alone in the last $3$ days, and subtract the total from $1$ to see what fraction was done by $A$ and $B$ in the middle period; then see how long it would take $A$ and $B$ to do that much.\n\n\u2022 Yes, in short i misinterpreted the question. But because of the line, \"They begin working together but 3 days before the completion of the work, A leaves off\", it seems as if A and B working together would have completed it in some estimated d number of days, 3 days before which A left the job. Hence obviously B would require >3 days to complete the job. How do i avoid such misinterpretations in these types of problems? again, an excellent answer. Thanks! \u2013\u00a0Karan Sep 25 '12 at 12:46\n\u2022 @user85030: You\u2019re welcome! I think that avoiding such misinterpretations is partly a matter of practice and partly a matter of reading them pretty literally. Here, for instance, the end of the work really did mean exactly what it said, not what would have been the end of the work if they\u2019d continued to work together. \u2013\u00a0Brian M. Scott Sep 25 '12 at 21:52\n\u2022 Can you please have a look at this:- math.stackexchange.com/questions/209842/\u2026 I had no other way of contacting you since there is no messaging system available on stack exchange. \u2013\u00a0Karan Oct 9 '12 at 19:26\n\nIn problem 2 you are misinterpreting the phrase \"$A$ left 3 days before the work was done.\" When you calculate it as above (3 days before the work would've been done if $A$ worked on), its wrong, as $A$ left (as you calculated) 7.5 days before the work was done.\n\nYou can argue as follows: Say the work is done in $d$ days, then $A$ and $B$ work together for $d-3$ days and $B$ alone for $3$ days, doing in total $(d-3) \\cdot \\left(\\frac 1{14}+\\frac 1{21}\\right) + \\frac 3{21} = \\frac{5(d-3) + 6}{42}$ work. So we must have $5(d-3) = 36$, so $5d = 51$, that is $d = 57/5$. For 1), you can argue along the same lines.\n\nProblem $1.)$\n\nLet $n$ be the required number of days.\n\n$A,B,C$'s $1$ day work is $1/24,1/36,1/48$ respectively.\n\nWork done by $C=4/48$\n\nWork done by $B=n/36$\n\nWork done by $A=(n-3)/24$\n\nSum of all the work is $1$ which gives\n\n$$\\frac{1}{12}+\\frac{n}{36}+\\frac{n-3}{24}=1$$\n\nSolving which you will get your answer.\n\nProblem $2.)$ can be solved using similar approach\n\nA,B,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?\n\nans-- A,B AND C ONE DAY WORK=(1/24+1/36+1/48)=13/144\n\nFOUR DAYS WORK OF A,B AND C IS =[4*(13/144)]=13/36\n\nAFTER FOUR DAYS REMAINING WORK =[1-(13/36)]=23/36\n\nIN LAST 3 DAYS A WORKING ALONE IS =[3*(1/24)]=1/8\n\nREST OF WORK IS ([(23/36)-(1/8)]=37/72) DONE BY A AND B TOGETHER\n\nA AND B ONE DAY WORK IS=[(1/24)+(1/36)]=5/72\n\nTIME TAKEN THEM TO COMPLETE THE WORK=[(37/72)/(5/72)]=37/5\n\nTOTAL TIME TO COMPLETE THE WORK=[(37/5)+3+4]=72/5\n\nA better approach to the problem.\n\nTake the LCM of 14 and 21 which will give you the total amount of work. LCM (14,21) = 42.\n\nA completes in 14 days.So he does 42/14 in 1 day.Similarly B does 42/21 in 1 day. A=3,B=2.Since A left 3 days prior to the completion of the work, so his 3 days work is absent.Lets add his 3 days pending work to the total amount of work.\n\nIn 1 day he does 3 units of work, so in 3 days he will do 9 units of work.\n\nTotal work = 42+9 = 51.\n\nBoth will now take 3+2 to complete the total work.\n\nA+B = 51/5 = 10.2.\n\n## protected by Alex M.Mar 19 '17 at 18:37\n\nThank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).\n\nWould you like to answer one of these unanswered questions instead?", "date": "2019-04-26 15:57:27", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/202052/work-and-time-when-work-is-split-into-parts", "openwebmath_score": 0.6277446150779724, "openwebmath_perplexity": 439.837859763646, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9905874126354031, "lm_q2_score": 0.9407897484286539, "lm_q1q2_score": 0.931934482729852}}
{"url": "https://brilliant.org/discussions/thread/algebraic-manipulation/", "text": "# Algebraic Manipulation\n\n## Definition\n\nAlgebraic manipulation involves rearranging variables to make an algebraic expression better suit your needs. During this rearrangement, the value of the expression does not change.\n\n## Technique\n\nAlgebraic expressions aren't always given in their most convenient forms. This is where algebraic manipulation comes in.\n\nFor example:\n\n### What value of $$x$$ satisfies $$5x+8 = -2x +43$$\n\nWe can rearrange this equation for $$x$$ by putting the terms with $$x$$ on one side and the constant terms on the other. \\begin{align} 5x+8 &= -2x +43 \\\\ 5x -(-2x) &= 43 -8 \\\\ 7x &= 35 \\\\ x &= \\frac{35}{7} \\\\ x &= 5 \\quad_\\square \\end{align}\n\nAlgebraic manipulation is also used to simplify complicated-looking expressions by factoring and using identities. Let's walk through an example:\n\n### $\\frac{x^3+y^3}{x^2-y^2} - \\frac{x^2+y^2}{x-y}.$\n\nIt's possible to solve for $$x$$ and $$y$$ and plug those values into this expression, but the algebra would be very messy. Instead, we can rearrange the problem by using the factoring formula identities for $$x^3+y^3$$ and $$x^2-y^2$$ and then simplifying. \\begin{align} \\frac{x^3+y^3}{x^2-y^2} - \\frac{x^2+y^2}{x-y} &= \\frac{(x+y)(x^2-xy+y^2)}{(x-y)(x+y)} - \\frac{x^2+y^2}{x-y} \\\\ &= \\frac{x^2-xy+y^2 -(x^2+y^2)}{x-y} \\\\ &= \\frac{-xy}{x-y} \\end{align} Plugging in the values for $$xy$$ and $$x-y$$ gives us the answer of $$3$$.$$_\\square$$\n\n## Application and Extensions\n\n### If $$x+\\frac{1}{x}=8$$, what is the value of $$x^3+\\frac{1}{x^3}$$?\n\nThe key to solving this problem (without explicitly solving for $$x$$) is to recognize that $\\left(x+\\frac{1}{x}\\right)^3 = x^3+\\frac{1}{x^3}+3\\left(x+\\frac{1}{x}\\right)$ which gives us \\begin{align} x^3+\\frac{1}{x^3} &= \\left(x+\\frac{1}{x}\\right)^3 - 3\\left(x+\\frac{1}{x}\\right) \\\\ &= (8)^3 -3(8) \\\\ &= 488 \\quad _\\square \\end{align}\n\n### $\\frac{2x+8}{\\sqrt{2x+1}+\\sqrt{x-3}}?$\n\nThis problem is easy once you realize that $\\left(\\sqrt{2x+1}+\\sqrt{x-3}\\right)\\left(\\sqrt{2x+1}-\\sqrt{x-3}\\right)=x+4.$ The solution is therefore \\begin{align} \\frac{2x+8}{\\sqrt{2x+1}+\\sqrt{x-3}} &= \\frac{2(x+4)}{\\sqrt{2x+1}+\\sqrt{x-3}} \\\\ &=\\frac{2\\left(\\sqrt{2x+1}+\\sqrt{x-3}\\right)\\left(\\sqrt{2x+1}-\\sqrt{x-3}\\right)}{\\sqrt{2x+1}+\\sqrt{x-3}} \\\\ &=2\\left(\\sqrt{2x+1}-\\sqrt{x-3}\\right) \\\\ &=2(2) \\\\ &=4 \\quad _\\square \\end{align}\n\nNote by Arron Kau\n4\u00a0years ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\nYou have posted very good solutions these kind of questions.\n\n- 3\u00a0years, 9\u00a0months ago\n\nThe first one of applications and extensions has a mistake\n\n- 3\u00a0years, 9\u00a0months ago\n\nThanks, it's fixed now.\n\nStaff - 3\u00a0years, 9\u00a0months ago\n\n\u00d7", "date": "2018-04-22 10:22:09", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/algebraic-manipulation/", "openwebmath_score": 1.0000100135803223, "openwebmath_perplexity": 2798.1191500231757, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9899864287859482, "lm_q2_score": 0.9399133468805266, "lm_q1q2_score": 0.9305014576465007}}
{"url": "https://math.stackexchange.com/questions/4454637/inequality-involving-sums-with-binomial-coefficient", "text": "Inequality involving sums with binomial coefficient\n\nI am trying to show upper- and lower-bounds on\n\n$$\\frac{1}{2^n}\\sum_{i=0}^n\\binom{n}{i}\\min(i, n-i)$$\n\n(where $$n\\geq 1$$) in order to show that it basically grows as $$\\Theta(n)$$.\n\nThe upper-bound is easy to get since $$\\min(i, n-i)\\leq i$$ for $$i\\in\\{0, \\dots n\\}$$ so that\n\n$$\\frac{1}{2^n}\\sum_{i=0}^n\\binom{n}{i}\\min(i, n-i)\\leq \\frac{1}{2^n}\\sum_{i=0}^n\\binom{n}{i}i = \\frac{n}{2}.$$\n\nThanks to Desmos, I managed to find a lower bound, but I am struggling to actually prove it. Indeed, I can see that the function $$f(n)=\\frac{n-1}{3}$$ does provide a lower-bound. One can in fact rewrite\n\n$$\\frac{n-1}{3}=\\frac{1}{2^n}\\sum_{i=0}^n\\binom{n}{i}\\frac{2i-1}{3}.$$\n\nI was thus hoping to show that for each term we have $$\\frac{2i-1}{3}\\leq \\min(i, n-i)$$, but this is only true if $$i\\leq \\frac{3n+1}{5}$$ and not generally for $$i\\leq n$$. I imagine there is a clever trick to use at some point but for some reason I am stuck here.\n\nAny help would be appreciated, thank you!\n\nEDIT: Thank you everyone for all the great and diverse answers! I flagged River Li's answer as the \"accepted\" one because of its simplicity due to the use of Cauchy-Schwartz inequality, which does not require a further use of Stirling's approximation. Note that the other answers which involve such an approximation are much tighter though, but proving $$\\Theta(n)$$ growth was sufficient here.\n\n\u2022 @Teepeemm Yes that is what I meant indeed, thank you for correcting I will edit! May 23 at 10:35\n\nBoth $$\\binom ni$$ and $$\\min(i,n-i)$$ are largest for $$i$$ near $$n/2$$. This means that, if we compare $$\\frac1{n+1}\\sum_{i=0}^n\\binom ni\\min(i,n-i)$$ with $$\\left(\\frac1{n+1}\\sum_{i=0}^n \\binom ni\\right)\\left(\\frac1{n+1}\\sum_{i=0}^n \\min(i,n-i)\\right),$$ the first should be larger, since larger numbers are multiplied by larger numbers and smaller numbers by smaller numbers. This can in fact be made more precise by (one form of) the rearrangement inequality:\n\nIf $$a_1\\leq \\cdots\\leq a_m$$ and $$b_1\\leq \\cdots\\leq b_m$$ are sequences of real numbers, then $$\\frac{a_1b_1+\\cdots+a_mb_m}m\\geq \\frac{a_1+\\cdots+a_m}m\\cdot \\frac{b_1+\\cdots+b_m}m.$$\n\n(This can be proven by summing $$(a_i-a_j)(b_i-b_j)\\geq 0$$ over all $$i$$ and $$j$$.) So, $$\\frac1{2^n}\\sum_{i=0}^n\\binom ni\\min(i,n-i)\\geq \\frac{1}{(n+1)2^n}\\sum_{i=0}^n\\binom ni\\sum_{i=0}^n\\min(i,n-i)=\\frac1{n+1}\\sum_{i=0}^n\\min(i,n-i).$$ The last sum is $$\\Omega(n^2)$$, since the average order of $$\\min(i,n-i)$$ is about $$n/4$$, and so the entire sum is $$\\Omega(n)$$. You've also shown that it's $$O(n)$$, so this is enough to show that it's $$\\Theta(n)$$.\n\nLet's first note that $$\\binom{n}{i}\\cdot i = n\\cdot \\binom{n-1}{i-1}$$.\n\nFor odd $$n=2m+1$$, this makes \\begin{align} S_n = \\sum_{i=0}^{n}\\binom{n}{i}\\cdot\\min(i,n-i) &= 2\\sum_{i=0}^m\\binom{n}{i}\\cdot i = 2n\\sum_{i=1}^m\\binom{n-1}{i-1} = 2n\\sum_{j=0}^{m-1}\\binom{2m}{j} \\\\ &= n\\cdot\\left( \\sum_{j=0}^{2m}\\binom{2m}{j} - \\binom{2m}{m} \\right) = n\\cdot\\left( 2^{2m} - \\binom{2m}{m} \\right) \\end{align} where we have used that $$\\binom{2m}{j}=\\binom{2m}{2m-j}$$. This makes $$\\frac{S_n}{2^n} = \\frac{n}{2}\\cdot\\left (1 - \\frac{\\binom{2m}{m}}{2^{2m}} \\right) \\approx \\frac{n}{2}\\cdot\\left( 1 - \\frac{1}{\\sqrt{\\pi m}} \\right).$$\n\nFor even $$n=2m$$, we get \\begin{align} S_n = \\sum_{i=0}^{n}\\binom{n}{i}\\cdot\\min(i,n-i) &= 2\\sum_{i=0}^m\\binom{n}{i}\\cdot i - \\binom{2m}{m}\\cdot m = 2n\\sum_{i=1}^m\\binom{n-1}{i-1} - \\binom{2m}{m}\\cdot m \\\\ &= n\\sum_{j=0}^{n-1}\\binom{n-1}{j} - \\binom{2m}{m}\\cdot m = n\\cdot\\left( 2^{n-1} - \\frac{1}{2}\\binom{2m}{m} \\right) \\end{align} which once more makes $$\\frac{S_n}{2^n} = \\frac{n}{2}\\cdot\\left(1-\\frac{\\binom{2m}{m}}{2^{2m}}\\right).$$\n\nIn both cases, you get $$n/2$$ as an upper bound. However, there are strong bounds on $$\\binom{2m}{m}/2^{2m}$$ which can be applied: $$\\frac{e^{-1/8m}}{\\sqrt{\\pi m}} \\le \\frac{\\binom{2m}{m}}{2^{2m}} \\le \\frac{1}{\\sqrt{\\pi m}}$$ Eg, see Jack D'Aurizio's derivation of this, or Wikipedia.\n\nAdditional bounds have been provided by robjohn. The following bounds seem to be the tightest proven so far: $$\\frac{4^me^{-1/8m}}{\\sqrt{\\pi m}} < \\binom{2m}{m} < \\frac{4^m}{\\sqrt{\\pi\\left( m+\\frac{1}{4} \\right)}}$$\n\nThe following bound is even tighter, but I have no proof of it, just numerical evidence: $$\\frac{4^m}{\\sqrt{\\pi\\left( m+\\frac{1}{4}+\\frac{1}{32m} \\right)}} < \\binom{2m}{m}$$ It's the same as the above up to second order approximation, so not a bit difference, but easier to compute.\n\n\u2022 I believe $$\\frac{4^n}{\\sqrt{\\pi\\!\\left(n+\\frac13\\right)}}\\le\\binom{2n}{n}\\le\\frac{4^n}{\\sqrt{\\pi\\!\\left(n+\\frac14\\right)}}$$ gives somewhat tighter bounds.\n\u2013\u00a0robjohn\nMay 20 at 22:53\n\u2022 Actually, my upper bound and your lower bound are really close.\n\u2013\u00a0robjohn\nMay 20 at 23:01\n\nWe start with $$\\sum_{k=0}^{\\left\\lfloor\\frac{n}2\\right\\rfloor}\\binom{n}{k} =\\left\\{\\begin{array}{} 2^{n-1}&\\text{if n is odd}\\\\ 2^{n-1}+\\frac12\\left(\\raise{2pt}{n}\\atop\\frac{n}2\\right)&\\text{if n is even} \\end{array}\\right.\\tag1$$ Substitute $$n\\mapsto n-1$$: $$\\sum_{k=0}^{\\left\\lfloor\\frac{n-1}2\\right\\rfloor}\\binom{n-1}{k} =\\left\\{\\begin{array}{} 2^{n-2}&\\text{if n is even}\\\\ 2^{n-2}+\\frac12\\left(\\raise{2pt}{n-1}\\atop{\\frac{n-1}2}\\right)&\\text{if n is odd} \\end{array}\\right.\\tag2$$ Subtract the $$k=\\left\\lfloor\\frac{n-1}2\\right\\rfloor$$ term: $$\\sum_{k=0}^{\\left\\lfloor\\frac{n-1}2\\right\\rfloor-1}\\binom{n-1}{k} =\\left\\{\\begin{array}{} 2^{n-2}-\\left(\\raise{2pt}{n-1}\\atop{\\frac{n}2-1}\\right)&\\text{if n is even}\\\\ 2^{n-2}-\\frac12\\left(\\raise{2pt}{n-1}\\atop{\\frac{n-1}2}\\right)&\\text{if n is odd} \\end{array}\\right.\\tag3$$ Substitute $$k\\mapsto k-1$$: $$\\sum_{k=1}^{\\left\\lfloor\\frac{n-1}2\\right\\rfloor}\\binom{n-1}{k-1} =\\left\\{\\begin{array}{} 2^{n-2}-\\frac12\\left(\\raise{2pt}{n}\\atop{\\frac{n}2}\\right)&\\text{if n is even}\\\\ 2^{n-2}-\\frac12\\left(\\raise{2pt}{n-1}\\atop{\\frac{n-1}2}\\right)&\\text{if n is odd} \\end{array}\\right.\\tag4$$ Then \\begin{align} \\sum_{k=0}^{\\left\\lfloor\\frac{n-1}2\\right\\rfloor}\\binom{n}{k}k &=n\\sum_{k=1}^{\\left\\lfloor\\frac{n-1}2\\right\\rfloor}\\binom{n-1}{k-1}\\tag{5a}\\\\ &=\\left\\{\\begin{array}{} n2^{n-2}-\\frac{n}2\\left(\\raise{2pt}{n}\\atop{\\frac{n}2}\\right)&\\text{if n is even}\\\\ n2^{n-2}-\\frac{n}2\\left(\\raise{2pt}{n-1}\\atop{\\frac{n-1}2}\\right)&\\text{if n is odd} \\end{array}\\right.\\tag{5b} \\end{align} Explanation:\n$$\\text{(5a)}$$: $$\\binom{n}{k}k=\\binom{n-1}{k-1}n$$ and the $$k=0$$ term vanishes\n$$\\text{(5b)}$$: multiply $$(4)$$ by $$n$$\n\nDouble and add the middle term in the even $$n$$ case: \\begin{align} \\sum_{k=0}^n\\binom{n}{k}\\min(k,n-k) &=\\left\\{\\begin{array}{} n2^{n-1}-\\frac{n}2\\left(\\raise{2pt}{n}\\atop{\\frac{n}2}\\right)&\\text{if n is even}\\\\ n2^{n-1}-n\\left(\\raise{2pt}{n-1}\\atop{\\frac{n-1}2}\\right)&\\text{if n is odd} \\end{array}\\right.\\tag6 \\end{align} The estimates for the central binomial coefficients from this answer give $$\\frac{2^n}{\\sqrt{\\pi\\!\\left(\\frac{n}2+\\frac13\\right)}}\\le\\binom{n}{\\frac{n}2}\\le\\frac{2^n}{\\sqrt{\\pi\\!\\left(\\frac{n}2+\\frac14\\right)}}\\tag7$$ which gives upper and lower bounds.\n\nLet $$S := \\frac{1}{2^n}\\sum_{i=0}^n\\binom{n}{i}\\min(i, n-i),$$ $$T := \\frac{1}{2^n}\\sum_{i=0}^n\\binom{n}{i}\\max(i, n-i).$$\n\nWe have $$S + T = \\frac{1}{2^n}\\sum_{i=0}^n\\binom{n}{i}n = n,$$ $$T - S = \\frac{1}{2^n}\\sum_{i=0}^n\\binom{n}{i} |n - 2i|$$ where we have used $$\\max(a, b) + \\min(a,b) = a + b$$ and $$\\max(a, b) - \\min(a, b) = |a - b|$$ for all real numbers $$a, b$$.\n\nUsing Cauchy-Bunyakovsky-Schwarz inequality, we have \\begin{align*} T - S &= \\frac{1}{2^n}\\sum_{i=0}^n \\sqrt{\\binom{n}{i} (n - 2i)^2}\\sqrt{\\binom{n}{i}}\\\\ &\\le \\frac{1}{2^n} \\sqrt{\\sum_{i=0}^n\\binom{n}{i} (n - 2i)^2 \\cdot \\sum_{i=0}^n\\binom{n}{i}}\\\\ &= \\sqrt n \\end{align*} where we have used $$\\sum_{i=0}^n\\binom{n}{i} i^2 = 2^{n - 2}n(n + 1)$$ and $$\\sum_{i=0}^n\\binom{n}{i} i = 2^{n - 1}n$$ (easy to prove using the identity $$\\binom{k}{m}m = k\\binom{k-1}{m-1}$$).\n\nThus, we have $$\\frac{n}{2} - \\frac{\\sqrt n}{2} \\le S \\le \\frac{n}{2}.$$\n\nThe desired result follows.", "date": "2022-06-26 11:34:07", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4454637/inequality-involving-sums-with-binomial-coefficient", "openwebmath_score": 0.9992073178291321, "openwebmath_perplexity": 304.1227486557242, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918526308096, "lm_q2_score": 0.9407897442783527, "lm_q1q2_score": 0.9299629972577745}}
{"url": "http://fileppi.com/mjau20u/product-of-matrix-5feeab", "text": "The main condition of matrix multiplication is that the number of columns of the 1st matrix must equal to the number of rows of the 2nd one. The dimensions of $B$ are $3\\times 2$ and the dimensions of $A$ are $2\\times 3$. The dot product involves multiplying the corresponding elements in the row of the first matrix, by that of the columns of the second matrix, and summing up the result, resulting in a single value. Example. This calculator can instantly multiply two matrices and \u2026 Matrix multiplication is associative: $\\left(AB\\right)C=A\\left(BC\\right)$. Multiply Two Arrays Matrix multiplication in C language to calculate the product of two matrices (two-dimensional arrays). Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. An example of a matrix is as follows. If A = [aij] is an m \u00d7 n matrix and B = [bij] is an n \u00d7 p matrix, the product AB is an m \u00d7 p matrix. If the multiplication isn't possible, an error message is displayed. Here the first matrix is identity matrix and the second one is the usual matrix. The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. We multiply entries of $A$ with entries of $B$ according to a specific pattern as outlined below. When we multiply two arrays of order (m*n) and (p*q) in order to obtained matrix product then its output contains m rows and q columns where n is n==p is a necessary condition. The inner dimensions match so the product is defined and will be a $3\\times 3$ matrix. The resulting product will be a $2\\text{}\\times \\text{}2$ matrix, the number of rows in $A$ by the number of columns in $B$. Yes, consider a matrix A with dimension $3\\times 4$ and matrix B with dimension $4\\times 2$. Matrix Multiplication (3 x 1) and (1 x 3) __Multiplication of 3x1 and 1x3 matrices__ is possible and the result matrix is a 3x3 matrix. If $A$ is an $\\text{ }m\\text{ }\\times \\text{ }r\\text{ }$ matrix and $B$ is an $\\text{ }r\\text{ }\\times \\text{ }n\\text{ }$ matrix, then the product matrix $AB$ is an $\\text{ }m\\text{ }\\times \\text{ }n\\text{ }$ matrix. Identity Matrix An identity matrix I n is an n\u00d7n square matrix with all its element in the diagonal equal to 1 and all other elements equal to zero. In mathematics, the matrix exponential is a function on square matrices analogous to the ordinary exponential function [1, , , , 7]. When complete, the product matrix will be. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. Let\u2019s return to the problem presented at the opening of this section. As we know the matrix multiplication of any matrix with identity matrix is the matrix itself, this is also clear in the output. tcrossprod () takes the cross-product of the transpose of a matrix. $\\left[A\\right]\\times \\left[B\\right]-\\left[C\\right]$, $\\left[\\begin{array}{rrr}\\hfill -983& \\hfill -462& \\hfill 136\\\\ \\hfill 1,820& \\hfill 1,897& \\hfill -856\\\\ \\hfill -311& \\hfill 2,032& \\hfill 413\\end{array}\\right]$, CC licensed content, Specific attribution, http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. $A=\\left[\\begin{array}{rrr}\\hfill {a}_{11}& \\hfill {a}_{12}& \\hfill {a}_{13}\\\\ \\hfill {a}_{21}& \\hfill {a}_{22}& \\hfill {a}_{23}\\end{array}\\right]\\text{ and }B=\\left[\\begin{array}{rrr}\\hfill {b}_{11}& \\hfill {b}_{12}& \\hfill {b}_{13}\\\\ \\hfill {b}_{21}& \\hfill {b}_{22}& \\hfill {b}_{23}\\\\ \\hfill {b}_{31}& \\hfill {b}_{32}& \\hfill {b}_{33}\\end{array}\\right]$, $\\left[\\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\end{array}\\right]\\cdot \\left[\\begin{array}{c}{b}_{11}\\\\ {b}_{21}\\\\ {b}_{31}\\end{array}\\right]={a}_{11}\\cdot {b}_{11}+{a}_{12}\\cdot {b}_{21}+{a}_{13}\\cdot {b}_{31}$, $\\left[\\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\end{array}\\right]\\cdot \\left[\\begin{array}{c}{b}_{12}\\\\ {b}_{22}\\\\ {b}_{32}\\end{array}\\right]={a}_{11}\\cdot {b}_{12}+{a}_{12}\\cdot {b}_{22}+{a}_{13}\\cdot {b}_{32}$, $\\left[\\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\end{array}\\right]\\cdot \\left[\\begin{array}{c}{b}_{13}\\\\ {b}_{23}\\\\ {b}_{33}\\end{array}\\right]={a}_{11}\\cdot {b}_{13}+{a}_{12}\\cdot {b}_{23}+{a}_{13}\\cdot {b}_{33}$, $AB=\\left[\\begin{array}{c}\\begin{array}{l}{a}_{11}\\cdot {b}_{11}+{a}_{12}\\cdot {b}_{21}+{a}_{13}\\cdot {b}_{31}\\\\ \\end{array}\\\\ {a}_{21}\\cdot {b}_{11}+{a}_{22}\\cdot {b}_{21}+{a}_{23}\\cdot {b}_{31}\\end{array}\\begin{array}{c}\\begin{array}{l}{a}_{11}\\cdot {b}_{12}+{a}_{12}\\cdot {b}_{22}+{a}_{13}\\cdot {b}_{32}\\\\ \\end{array}\\\\ {a}_{21}\\cdot {b}_{12}+{a}_{22}\\cdot {b}_{22}+{a}_{23}\\cdot {b}_{32}\\end{array}\\begin{array}{c}\\begin{array}{l}{a}_{11}\\cdot {b}_{13}+{a}_{12}\\cdot {b}_{23}+{a}_{13}\\cdot {b}_{33}\\\\ \\end{array}\\\\ {a}_{21}\\cdot {b}_{13}+{a}_{22}\\cdot {b}_{23}+{a}_{23}\\cdot {b}_{33}\\end{array}\\right]$, $A=\\left[\\begin{array}{cc}1& 2\\\\ 3& 4\\end{array}\\right]\\text{ and }B=\\left[\\begin{array}{cc}5& 6\\\\ 7& 8\\end{array}\\right]$, $A=\\left[\\begin{array}{l}\\begin{array}{ccc}-1& 2& 3\\end{array}\\hfill \\\\ \\begin{array}{ccc}4& 0& 5\\end{array}\\hfill \\end{array}\\right]\\text{ and }B=\\left[\\begin{array}{c}5\\\\ -4\\\\ 2\\end{array}\\begin{array}{c}-1\\\\ 0\\\\ 3\\end{array}\\right]$, $\\begin{array}{l}\\hfill \\\\ AB=\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 2& \\hfill 3\\\\ \\hfill 4& \\hfill 0& \\hfill 5\\end{array}\\right]\\text{ }\\left[\\begin{array}{rr}\\hfill 5& \\hfill -1\\\\ \\hfill -4& \\hfill 0\\\\ \\hfill 2& \\hfill 3\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{rr}\\hfill -1\\left(5\\right)+2\\left(-4\\right)+3\\left(2\\right)& \\hfill -1\\left(-1\\right)+2\\left(0\\right)+3\\left(3\\right)\\\\ \\hfill 4\\left(5\\right)+0\\left(-4\\right)+5\\left(2\\right)& \\hfill 4\\left(-1\\right)+0\\left(0\\right)+5\\left(3\\right)\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{rr}\\hfill -7& \\hfill 10\\\\ \\hfill 30& \\hfill 11\\end{array}\\right]\\hfill \\end{array}$, $\\begin{array}{l}\\hfill \\\\ BA=\\left[\\begin{array}{rr}\\hfill 5& \\hfill -1\\\\ \\hfill -4& \\hfill 0\\\\ \\hfill 2& \\hfill 3\\end{array}\\right]\\text{ }\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 2& \\hfill 3\\\\ \\hfill 4& \\hfill 0& \\hfill 5\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{rrr}\\hfill 5\\left(-1\\right)+-1\\left(4\\right)& \\hfill 5\\left(2\\right)+-1\\left(0\\right)& \\hfill 5\\left(3\\right)+-1\\left(5\\right)\\\\ \\hfill -4\\left(-1\\right)+0\\left(4\\right)& \\hfill -4\\left(2\\right)+0\\left(0\\right)& \\hfill -4\\left(3\\right)+0\\left(5\\right)\\\\ \\hfill 2\\left(-1\\right)+3\\left(4\\right)& \\hfill 2\\left(2\\right)+3\\left(0\\right)& \\hfill 2\\left(3\\right)+3\\left(5\\right)\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{rrr}\\hfill -9& \\hfill 10& \\hfill 10\\\\ \\hfill 4& \\hfill -8& \\hfill -12\\\\ \\hfill 10& \\hfill 4& \\hfill 21\\end{array}\\right]\\hfill \\end{array}$, $AB=\\left[\\begin{array}{cc}-7& 10\\\\ 30& 11\\end{array}\\right]\\ne \\left[\\begin{array}{ccc}-9& 10& 10\\\\ 4& -8& -12\\\\ 10& 4& 21\\end{array}\\right]=BA$, $E=\\left[\\begin{array}{c}6\\\\ 30\\\\ 14\\end{array}\\begin{array}{c}10\\\\ 24\\\\ 20\\end{array}\\right]$, $C=\\left[\\begin{array}{ccc}300& 10& 30\\end{array}\\right]$, $\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ CE=\\left[\\begin{array}{rrr}\\hfill 300& \\hfill 10& \\hfill 30\\end{array}\\right]\\cdot \\left[\\begin{array}{rr}\\hfill 6& \\hfill 10\\\\ \\hfill 30& \\hfill 24\\\\ \\hfill 14& \\hfill 20\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{rr}\\hfill 300\\left(6\\right)+10\\left(30\\right)+30\\left(14\\right)& \\hfill 300\\left(10\\right)+10\\left(24\\right)+30\\left(20\\right)\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{rr}\\hfill 2,520& \\hfill 3,840\\end{array}\\right]\\hfill \\end{array}$. The product-process matrix can facilitate the understanding of the strategic options available to a company, particularly with regard to its manufacturing function. If you view them each as vectors, and you have some familiarity with the dot product, we're essentially going to take the dot product of that and that. Thank you for your questionnaire.Sending completion. Thus, the equipment need matrix is written as. The process of matrix multiplication becomes clearer when working a problem with real numbers. Given $A$ and $B:$. To obtain the entries in row $i$ of $AB,\\text{}$ we multiply the entries in row $i$ of $A$ by column $j$ in $B$ and add. Since we view vectors as column matrices, the matrix-vector product is simply a special case of the matrix-matrix product (i.e., a product between two matrices). Multiply and add as follows to obtain the first entry of the product matrix $AB$. If A is a nonempty matrix, then prod (A) treats the columns of A as vectors and returns a row vector of the products of each column. Multiplication of two matrices involves dot products between rows of first matrix and columns of the second matrix. Boolean matrix products are computed via either %&% or boolArith = TRUE. Syntax: numpy.matmul (x1, x2, /, out=None, *, casting=\u2019same_kind\u2019, order=\u2019K\u2019, dtype=None, subok=True [, \u2026 If A =[aij]is an m \u00d7n matrix and B =[bij]is an n \u00d7p matrix then the product of A and B is the m \u00d7p matrix C =[cij]such that cij=rowi(A)6 colj(B) A matrix is a rectangular array of numbers that is arranged in the form of rows and columns. You can only multiply two matrices if their dimensions are compatible, which means the number of columns in the first matrix is the same as the number of rows in the second matrix. e) order: 1 \u00d7 1. A user inputs the orders and elements of the matrices. The inner dimensions are the same so we can perform the multiplication. Let A \u2208 Mn. Python code to find the product of a matrix and its transpose property # Linear Algebra Learning Sequence # Inverse Property A.AT = S [AT = transpose of A] import numpy as np M = np . The exponential of A, denoted by eA or exp(A) , is the n \u00d7 n matrix \u2026 Number of rows and columns are equal therefore this matrix is a square matrix. The product of two matrices can be computed by multiplying elements of the first row of the first matrix with the first column of the second matrix then, add all the product of elements. For example, the dimension of the matrix below is 2 \u00d7 3 (read \"two by three\"), because there are two rows and three columns: A program that performs matrix multiplication is as follows. in a single step. In other words, row 2 of $A$ times column 1 of $B$; row 2 of $A$ times column 2 of $B$; row 2 of $A$ times column 3 of $B$. As the dimensions of $A$ are $2\\text{}\\times \\text{}3$ and the dimensions of $B$ are $3\\text{}\\times \\text{}2,\\text{}$ these matrices can be multiplied together because the number of columns in $A$ matches the number of rows in $B$. Example 4 The following are all identity matrices. On the matrix page of the calculator, we enter matrix $A$ above as the matrix variable $\\left[A\\right]$, matrix $B$ above as the matrix variable $\\left[B\\right]$, and matrix $C$ above as the matrix variable $\\left[C\\right]$. Can perform the multiplication is a vector, then B is an empty matrix! The dot product can only be performed on sequences of equal lengths is a vector, then prod ( )! Are correct ) words, the equipment, that produces a single matrix the... 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Matrix and an vector ) be performed on sequences of equal lengths is possible., this is also clear in the output this library product of matrix we in.", "date": "2022-07-02 14:42:21", "meta": {"domain": "fileppi.com", "url": "http://fileppi.com/mjau20u/product-of-matrix-5feeab", "openwebmath_score": 0.8542640209197998, "openwebmath_perplexity": 590.7593543781824, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.990731985524492, "lm_q2_score": 0.9372107878954105, "lm_q1q2_score": 0.9285247047465935}}
{"url": "https://math.stackexchange.com/questions/2893117/definite-integral-int-01-frac-ln4xx21-dx", "text": "# Definite Integral: $\\int_0^1\\frac{\\ln^4(x)}{x^2+1}\\,dx$\n\nI'm trying to derive a closed-form expression for\n\n$$I=\\int_0^1\\frac{\\ln^4(x)}{x^2+1}\\,dx$$\n\nLetting $u=-\\ln(x), x=e^{-u}, dx=-e^{-u}\\,du$ yields\n\n$$I=\\int_0^{\\infty}\\frac{u^4e^{-u}}{e^{-2u}+1}\\,du$$\n\nSetting $u\\to-u$ and manipulating the integrands yield\n\n$$I=-\\int_0^{-\\infty}\\frac{u^4e^{u}}{e^{2u}+1}\\,du$$ $$=\\int_{-\\infty}^0\\frac{u^4e^{-u}}{e^{-2u}+1}\\,du$$\n\nAnd adding the two equivalent forms of $I$ yields\n\n$$2I=\\int_{-\\infty}^{\\infty}\\frac{u^4e^{-u}}{e^{-2u}+1}\\,du$$\n\nI've tried to differentiate under the integral sign, but I could not find any parameterization that worked for me. (Perhaps someone could tell me how to solve such integrals by differentiation under the integral sign?)\n\nMy best attempt so far was using complex analysis:\n\nI used a counterclockwise semicircle that grows to infinity over the lower half of the complex plane as my contour, and by Jordan's lemma (as I understand it) the integral over the arc vanishes and so I should be left with\n\n$$\\require{cancel} \\lim_{R\\to\\infty} \\int_R^{-R} \\frac{x^4e^{-x}}{e^{-2x}+1}\\,dx + \\cancel{\\int_{arc} \\frac{z^4e^{-z}}{e^{-2z}+1}\\,dz} = 2\\pi i\\sum_j \\operatorname{Res}(j)$$\n\n$$-2I=\\int_{\\infty}^{-\\infty}\\frac{x^4e^{-x}}{e^{-2x}+1}\\,dx= 2\\pi i\\sum_j \\operatorname{Res}(j)$$\n\nSince my integrand only blows up when $e^{-2u}+1=0 \\Rightarrow u=-i\\pi/2$,\n\n$$\\frac{-2}{2\\pi i}I=\\operatorname{Res}(-i\\pi/2)$$\n\n$$\\frac{i}{\\pi} I = \\lim_{z\\to -i\\pi/2}(z+i\\pi/2)\\frac{z^4e^{-z}}{e^{-2z}+1}$$\n\nEvaluating the limit (via L'Hopital's Rule and a few substitutions) yields\n\n$$\\frac{i}{\\pi}I = \\frac{i\\pi^4}{32}$$\n\n$$I=\\frac{\\pi^5}{32}$$\n\nHowever, WolframAlpha evaluates the integral at $$I=\\frac{5\\pi^5}{64}$$\n\nWhere did I make a mistake and how do I evaluate this integral correctly?\n\nI am rather new to both complex analysis and Math StackExchange, so feel free to point out and correct any of my mistakes and misconceptions. Any help is greatly appreciated!\n\n\u2022 For the integral$$\\int\\limits_{-\\infty}^{\\infty}\\mathrm dx\\,\\frac {x^n e^{-x}}{1+e^{-2x}}$$You can rewrite the integrand as an infinite sum with the geometric sequence and integrate it termwise. It looks very similar to$$\\int\\limits_0^{\\infty}\\mathrm dx\\,\\frac {x^n e^{-x}}{1+e^{-2x}}=\\Gamma(n+1)\\beta(n+1)$$ \u2013\u00a0Frank W. Aug 24 '18 at 13:35\n\u2022 Are you sure the integral over the arc vanishes? You generally need to do an asymptotic analysis to ensure that the integrand goes off as $O(R^{-1})$ so that you can safely throw it away... \u2013\u00a0Trebor Aug 24 '18 at 14:40\n\u2022 In your case, the arc part of the complex integral does not vanish. Therefore you cannot apply that method here. \u2013\u00a0Trebor Aug 24 '18 at 14:46\n\nAn approach relying on Feynman's trick\n\nNotice that one has: \\begin{align} I:=\\int^1_0 \\frac{\\ln^4(x)}{x^2+1}\\,dx \\stackrel{x\\mapsto 1/x}{=} \\int^\\infty_1 \\frac{\\ln^4(x)}{x^2+1}\\,dx \\end{align} which means that: \\begin{align} I = \\frac 1 2 \\int^\\infty_0 \\frac{\\ln^4(x)}{x^2+1}\\,dx \\end{align} Define: \\begin{align} G(z):=\\int^\\infty_0 \\frac{x^{-z}}{x^2+1}\\,dx \\end{align} Notice by Feynman's trick one has: \\begin{align} \\frac 1 2 G^{(4)}(0) =I \\end{align} So we only need to find $G(z)$ which is not very hard. You can see for example this post for a variety of solutions. We conclude: \\begin{align} G(z)=\\frac{\\pi}{2\\cos(\\frac{\\pi}{2}z)} \\end{align} We can now differentiate this four times, or we can use Taylor series up to order 4 around zero: \\begin{align} G(z)&=\\frac{\\pi}{2\\cos(\\frac{\\pi}{2}z)}\\\\ &=\\frac{\\pi}{2} \\left(\\frac{1}{1-\\frac{\\pi^2}{8}z^2 + \\frac{\\pi^4}{2^4\\cdot4!}z^4+O(z^5)}\\right)\\\\ &=\\frac{\\pi}{2}\\left[ 1+\\left(\\frac{\\pi^2}{8}z^2 - \\frac{\\pi^4}{2^4\\cdot 4!}z^4+O(z^5) \\right) + \\left(\\frac{\\pi^2}{8}z^2 - \\frac{\\pi^4}{2^4\\cdot 4!}z^4+O(z^5) \\right)^2+O(z^5)\\right]\\\\ &=\\frac{\\pi}{2}+\\frac{\\pi^3}{16}z^2+\\frac{5\\pi^5}{32\\cdot 4! }z^4+O(z^5)\\\\ \\end{align} The coefficient of $z^4$ gives $4!G^{(4)}(0)$ hence: \\begin{align} G^{(4)}(0) = \\frac{\\pi^55}{32 } \\end{align} We conclude: \\begin{align} I=\\frac{5\\pi^5 }{64} \\end{align}\n\n\u2022 Why don\u2019t you just directly utilize the taylor series of secant around zero? \u2013\u00a0Szeto Aug 24 '18 at 15:40\n\u2022 @Szeto to be honest, I don't know them. Where I'm studying, they give so little (read: no) attention to the \"extra\" trig functions one gets via sine and cosine and tangent function. I don't even know their names, let alone their Taylor series... \u2013\u00a0Shashi Aug 24 '18 at 15:49\n\u2022 Anyway, still a splendid answer. I have upvoted. Great job! \u2013\u00a0Szeto Aug 24 '18 at 15:51\n\u2022 @Szeto Thanks for the compliment. Have a nice day! \u2013\u00a0Shashi Aug 24 '18 at 15:52\n\nI believe that the most simple approach is just to exploit Maclaurin series. Since $\\int_{0}^{1}x^{2n}\\log^4(x)\\,dx=\\frac{24}{(2n+1)^5}$ we have\n\n$$\\int_{0}^{1}\\frac{\\log^4(x)}{x^2+1}\\,dx = 24\\sum_{n\\geq 0}\\frac{(-1)^n}{(2n+1)^5}\\color{red}{=}24\\cdot\\frac{5\\pi^5}{1536} = \\frac{5\\pi^5}{64}.$$ It is well-known that the series $\\sum_{n\\geq 0}\\frac{(-1)^n}{(2n+1)^{2m+1}}$ are related to Euler numbers.\n\n\u2022 Is it trivial that $$\\int_0^1 x^{2n}\\log^4(x) dx =\\frac{24}{(2n+1)^5}$$ I really can't see why it's true. \u2013\u00a0stressed out Mar 9 at 15:56\n\u2022 @stressedout: just enforce the substitution $x=e^{-t}$ and exploit the $\\Gamma$ function, or integration by parts. \u2013\u00a0Jack D'Aurizio Mar 9 at 16:58\n\u2022 Thank you. Now I understand. \u2013\u00a0stressed out Mar 11 at 11:10\n\nLet, for $n\\geq 0$ integer,\n\n\\begin{align}&A_n=\\int_0^1 \\frac{\\ln^{2n}x}{1+x^2}\\,dx\\\\ &B_n=\\int_0^\\infty \\frac{\\ln^{2n}x}{1+x^2}\\,dx\\\\ &K_n=\\int_0^\\infty\\int_0^\\infty\\frac{\\ln^{2n}(xy)}{(1+x^2)(1+y^2}\\,dx\\,dy \\end{align}\n\nObserve that,\n\n\\begin{align}A_0&=\\int_0^1 \\frac{1}{1+x^2}\\,dx\\\\ &=\\Big[\\arctan x\\Big]_0^1\\\\ &=\\frac{\\pi}{4}\\end{align}\n\n\\begin{align}B_n=\\int_0^1 \\frac{\\ln^{2n}x}{1+x^2}\\,dx+\\int_1^\\infty \\frac{\\ln^{2n}x}{1+x^2}\\,dx\\\\ \\end{align}\n\nPerform in the latter integral the change of variable $y=\\dfrac{1}{x}$,\n\n\\begin{align}B_n=2A_n\\end{align}\n\nThat is,\n\n\\begin{align}A_n=\\frac{1}{2}B_n\\end{align}\n\nFor $n\\geq 0$ integer,\n\n\\begin{align}\\int_0^\\infty \\frac{\\ln^{2n+1}x}{1+x^2}\\,dx=0\\end{align}\n\n(perform the change of variable $y=\\dfrac{1}{x}$, and, $z=-z \\iff z=0$ )\n\n\\begin{align}K_n&=\\int_0^\\infty\\int_0^\\infty\\left(\\sum_{k=0}^{2n}\\binom{2n}{k}\\frac{\\ln^k x\\ln^{2n-k}y}{(1+x^2)(1+y^2)} \\right)\\,dx\\,dy\\\\ &=\\sum_{k=0}^{2n}\\binom{2n}{k}\\left(\\int_0^\\infty\\frac{\\ln^k x}{1+x^2}\\,dx\\right)\\left(\\int_0^\\infty\\frac{\\ln^{2n-k}y}{1+y^2} \\,dy\\right)\\\\ &=\\sum_{k=0}^{n}\\binom{2n}{2k}\\left(\\int_0^\\infty\\frac{\\ln^{2k} x}{1+x^2}\\,dx\\right)\\left(\\int_0^\\infty\\frac{\\ln^{2(n-k)}y}{1+y^2} \\,dy\\right)\\\\ &=\\sum_{k=0}^{n}\\binom{2n}{2k}B_kB_{n-k} \\end{align}\n\nPerform the change of variable $u=xy$,\n\n\\begin{align}K_n&=\\int_0^\\infty\\int_0^\\infty\\frac{\\ln^{2n}(xy)}{(1+x^2)(1+y^2)}\\,dx\\,dy\\\\ &=\\int_0^\\infty\\int_0^\\infty\\frac{y\\ln^{2n}(u)}{(y^2+u^2)(1+y^2)}\\,du\\,dy\\\\ &=\\int_0^\\infty\\int_0^\\infty\\frac{\\ln^{2n}(u)}{u^2-1}\\left(\\frac{y}{1+y^2}-\\frac{y}{u^2+y^2}\\right)\\,du\\,dy\\\\ &=\\frac{1}{2}\\int_0^\\infty\\int_0^\\infty\\frac{\\ln^{2n}(u)}{u^2-1}\\left[\\frac{1+y^2}{u^2+y^2}\\right]_{y=0}^{y=\\infty}\\,du\\\\ &=\\int_0^\\infty\\frac{\\ln^{2n+1}(u)}{u^2-1}\\,du\\\\ &=\\int_0^1\\frac{\\ln^{2n+1}(u)}{u^2-1}\\,du+\\int_1^\\infty\\frac{\\ln^{2n+1}(u)}{u^2-1}\\,du \\end{align}\n\nIn the latter integral perform the change of variable $x=\\dfrac{1}{u}$,\n\n\\begin{align}K_n&=2\\int_0^1\\frac{\\ln^{2n+1}(x)}{x^2-1}\\,dx\\\\ &=2\\int_0^1\\frac{\\ln^{2n+1}(x)}{x-1}\\,dx-2\\int_0^1\\frac{x\\ln^{2n+1}(x)}{x^2-1}\\,dx\\\\ \\end{align}\n\nIn the latter integral perform the change of variable $u=x^2$,\n\n\\begin{align}K_n&=2\\int_0^1\\frac{\\ln^{2n+1}(x)}{x-1}\\,dx-\\frac{1}{2^{2n+1}}\\int_0^1\\frac{\\ln^{2n}(u)}{u-1}\\,du\\\\ &=\\left(2-\\frac{1}{2^{2n+1}}\\right)\\int_0^1\\frac{\\ln^{2n+1}(x)}{x-1}\\,dx\\\\ &=-\\left(2-\\frac{1}{2^{2n+1}}\\right)\\int_0^1 \\ln^{2n+1}(x)\\left(\\sum_{k=0}^\\infty x^k\\right)\\,dx\\\\ &=-\\left(2-\\frac{1}{2^{2n+1}}\\right)\\sum_{k=0}^\\infty \\left(\\int_0^1 x^k \\ln^{2n+1}(x)\\,dx\\right)\\\\ &=-\\left(2-\\frac{1}{2^{2n+1}}\\right)\\sum_{k=0}^\\infty \\frac{(-1)^{2n+1}(2n+1)!}{(k+1)^{2n+2}}\\\\ &=\\left(2-\\frac{1}{2^{2n+1}}\\right)(2n+1)!\\zeta(2n+2) \\end{align}\n\nTherefore,\n\n\\begin{align}\\sum_{k=0}^{n}\\binom{2n}{2k}B_kB_{n-k}&=\\left(2-\\frac{1}{2^{2n+1}}\\right)(2n+1)!\\zeta(2n+2)\\end{align}\n\nTherefore,\n\n\\begin{align}\\boxed{\\sum_{k=0}^{n}\\binom{2n}{2k}A_kA_{n-k}=\\frac{(2n+1)!}{2}\\left(1-\\frac{1}{2^{2n+2}}\\right)\\zeta(2n+2)}\\end{align}\n\nif $n=1$,\n\n\\begin{align} \\frac{\\pi}{2}A_1=\\frac{45}{16}\\zeta(4)\\end{align}\n\nif $n=2$,\n\n\\begin{align} \\frac{\\pi}{2}A_2+6A_1^2=\\frac{945}{16}\\zeta(6)\\end{align}\n\ntherefore,\n\n\\begin{align} \\boxed{A_2=\\frac{945\\zeta(6)}{8\\pi}-\\frac{6075\\zeta(4)^2}{16\\pi^3}}\\end{align}\n\nIf you know that,\n\n\\begin{align}&\\zeta(4)=\\frac{1}{90}\\pi^4\\\\ &\\zeta(6)=\\frac{1}{945}\\pi^6 \\end{align}\n\nthen,\n\n\\begin{align}\\boxed{A_2=\\frac{5}{64}\\pi^5}\\end{align}\n\nA Complex-Analytic Proof\n\nYour problem is that the integral does not vanish on the semicircular arc as the radius goes to infinity, and there are infinitely many poles that your contour will end up enclosing. I am offering a similar approach, but with a different contour that guarantees that it encloses only one pole, and that the un-needed terms vanish as the range expands.\n\nLet $R>0$ and consider instead the rectangle $Q_R$ defined as the positively oriented contour $$[-R,+R]\\cup[+R,+R+\\text{i}\\pi]\\cup[+R+\\text{i}\\pi,-R+\\text{i}\\pi]\\cup[-R+\\text{i}\\pi,-R]\\,.$$ Define $$L_k:=\\lim_{R\\to\\infty}\\,\\oint_{Q_R}\\,f_k(z)\\,\\text{d}z\\,,\\text{ where }f_k(z):=z^k\\,\\left(\\frac{\\exp(z)}{\\exp(2z)+1}\\right)\\,.$$ We shall attempt to determine the values of $L_k$ for $k=0,2,4$. Using the Residue Theorem, it is easy to see that $$L_k=2\\pi\\text{i}\\,\\text{Res}_{z=\\frac{\\text{i}\\pi}{2}}\\big(f_k(z)\\big)\\,,$$ so that $$L_0=\\pi\\,,\\,\\,L_2=-\\frac{\\pi^3}{4}\\,,\\text{ and }L_4=\\frac{\\pi^5}{16}\\,.$$\n\nWrite $$J_k:=\\int_{-\\infty}^{+\\infty}\\,f_k(u)\\,\\text{d}u\\,.$$ It is not difficult to show that $$L_0=2\\,J_0\\,,\\,\\,L_2=2\\,J_2-\\pi^2\\,J_0\\,,\\text{ and }L_4=2\\,J_4-6\\pi^2\\,J_2+\\pi^4\\,J_0\\,.$$ Thus, we get $$J_0=\\frac{\\pi}{2}\\,,\\,\\,J_2=-\\frac{\\pi^3}{8}+\\frac{\\pi^3}{4}=\\frac{\\pi^3}{8}\\,,$$ and $$J_4=\\frac{\\pi^5}{32}+\\frac{3\\pi^5}{8}-\\frac{\\pi^5}{4}=\\frac{5\\pi^5}{32}\\,.$$ Thus, $$I=\\frac{1}{2}\\,J_4=\\frac{5\\pi^5}{64}\\,.$$ You can obtain $$I_k:=\\int_0^\\infty\\,u^k\\,\\left(\\frac{\\exp(u)}{\\exp(2u)+1}\\right)\\,\\text{d}u$$ similarly for an even integer $k\\geq 0$, by evaluating $L_0,L_2,L_4,\\ldots,L_k$ and then solving for $J_0,J_2,J_4,\\ldots,J_k$, as $I_k=\\dfrac{1}{2}\\,J_k$. From here, we can show that $$J_k=t_k\\,\\left(\\frac{\\pi^{k+1}}{2^{k+1}}\\right)\\text{ for all even integers }k\\geq 0\\,,$$ where $$t_k=\\sum_{r=0}^{\\frac{k}{2}-1}\\,(-1)^r\\,\\binom{k}{2r+2}\\,2^{2r+1}\\,t_{k-2r-2}+(-1)^{\\frac{k}{2}}\\,.$$ For example, $t_0=1$, $t_2=1$, $t_4=5$, $t_6=61$, and $t_8=1385$.\n\nIn fact, one can show, using the same contour $Q_R$, that $$\\text{sech}(w)=\\frac{1}{\\pi}\\,\\int_{-\\infty}^{+\\infty}\\,\\frac{\\exp\\left(\\frac{2\\text{i}}{\\pi}\\,wu\\right)}{\\cosh(u)}\\,\\text{d}u=\\frac{1}{\\pi}\\,\\int_{-\\infty}^{+\\infty}\\,\\frac{\\cos\\left(\\frac{2}{\\pi}\\,wu\\right)}{\\cosh(u)}\\,\\text{d}u$$ for all complex numbers $w$ such that $\\big|\\text{Im}(w)\\big|<\\frac{\\pi}{2}$. This shows that $t_k=(-1)^{\\frac{k}{2}}\\,E_k=|E_k|$ for every even integer $k\\geq 0$, where $E_0,E_1,E_2,\\ldots$ are Euler numbers. Therefore, $$\\sum_{n=0}^\\infty\\,\\frac{(-1)^n\\,k!}{(2n+1)^{k+1}}={\\small\\int_0^\\infty\\,u^k\\,\\left(\\frac{\\exp(u)}{\\exp(2u)+1}\\right)\\,\\text{d}u}=I_k=\\frac{1}{2}\\,J_k=\\frac{t_k}{2}\\,\\left(\\frac{\\pi}{2}\\right)^{k+1}=\\frac{|E_k|}{2}\\,\\left(\\frac{\\pi}{2}\\right)^{k+1}$$ for each even integer $k\\geq 0$.\n\nIn general, for $p\\in\\mathbb{C}$ and $q\\in\\mathbb{R}_{>0}$ such that $0<\\text{Re}\\left(p\\right)<q$, we have \\begin{align}\\int_{-\\infty}^{+\\infty}\\,\\frac{u^k\\,\\exp(pu)}{\\exp(qu)+1}\\,\\text{d}u&=k!\\,\\left(\\frac{\\pi}{q}\\right)^{k+1}\\,\\Biggl(\\left[a^k\\right]\\Bigg(\\text{csc}\\left(a+\\frac{\\pi p}{q}\\right)\\Bigg)\\Biggr) \\\\ &=k!\\,\\left(\\frac{\\pi}{q}\\right)^{k+1}\\,\\Biggl(\\left[a^k\\right]\\Bigg(\\text{sec}\\left(a+\\frac{\\pi(2p-q)}{2q}\\right)\\Bigg)\\Biggr)\\,.\\end{align} Here, $\\left[a^k\\right]\\big(g(a)\\big)$ is the coefficient of $a^k$ in the Laurent expansion of $g(a)$ about $a=0$.\n\n\u2022 Interestingly, we have $$\\frac{1}{\\text{i}\\pi}\\,\\int_{-\\infty}^{+\\infty}\\,\\frac{\\exp\\left(\\frac{2\\text{i}}{\\pi}\\,wu\\right)}{\\sinh(u)}\\,\\text{d}u=\\frac{1}{\\pi}\\,\\int_{-\\infty}^{+\\infty}\\,\\frac{\\sin\\left(\\frac{2}{\\pi}\\,wu\\right)}{\\sinh(u)}\\,\\text{d}u=\\tanh(w)$$ for all $w\\in\\mathbb{C}$ such that $\\big|\\text{Im}(w)\\big|<\\dfrac{\\pi}{2}$. \u2013\u00a0Batominovski Sep 2 '18 at 13:09\n\u2022 This gives \\begin{align}\\int_{-\\infty}^{+\\infty}\\,\\frac{u^k\\,\\exp(pu)}{\\exp(qu)-1}\\,\\text{d}u&=k!\\,\\left(\\frac{\\pi}{q}\\right)^{k+1}\\,\\Biggl(\\left[a^k\\right]\\Bigg(\\cot\\left(a+\\frac{\\pi p}{q}\\right)\\Bigg)\\Biggr)\\\\&=k!\\,\\left(\\frac{\\pi}{q}\\right)^{k+1}\\,\\Biggl(\\left[a^k\\right]\\Bigg(\\tan\\left(a+\\frac{\\pi (2p-q)}{2q}\\right)\\Bigg)\\Biggr)\\end{align} for $p\\in\\mathbb{C}$ and $q\\in\\mathbb{R}_{>0}$ such that $0<\\text{Re}(p)<q$. \u2013\u00a0Batominovski Sep 2 '18 at 13:12\n\u2022 In the comment above, $k$ is a positive integer. \u2013\u00a0Batominovski Sep 2 '18 at 13:24\n\n\n\\begin{align} I & \\equiv \\int_{0}^{1}{\\ln^{4}\\pars{x} \\over x^{2} + 1}\\,\\dd x \\\\[5mm] & = {1 \\over 4}\\,\\totald[4]{}{\\nu}\\bracks{\\Psi\\pars{{\\nu \\over 4} + {3 \\over 4}} - \\Psi\\pars{{\\nu \\over 4} + {1 \\over 4}}}\\qquad \\pars{~\\Psi:\\ Digamma\\ Function~} \\\\[5mm] & = {1 \\over 4}\\pars{1 \\over 4}^{4}\\bracks{\\Psi^{\\pars{\\texttt{IV}}}\\pars{3 \\over 4} - \\Psi^{\\pars{\\texttt{IV}}}\\pars{1 \\over 4}} \\\\[5mm] & = {1 \\over 1024}\\, \\left.\\totald[4]{\\bracks{\\pi\\cot\\pars{\\pi z}}}{z}\\,\\right\\vert_{\\ z\\ =\\ 1/4} \\qquad\\pars{~Euler\\ Reflection\\ Formula} \\\\[5mm] & = {1 \\over 1024}\\ \\underbrace{\\bracks{8\\pi^{5}\\cot^{3}\\pars{\\pi z}\\csc^{2}\\pars{\\pi z} + 16\\pi^{5}\\cot\\pars{\\pi z}\\csc^{4}\\pars{\\pi z}}_{\\ z\\ =\\ 1/4}} _{\\ds{80\\pi^{5}}} \\\\[5mm] & = \\bbx{5\\pi^{5} \\over 64} \\approx 23.9078 \\end{align}\n\nYou got to\n\n$I=\\int_0^\\infty u^4e^{-u}(1+e^{-2u})^{-1}du$\n\nwhich gives you\n\n$I=\\int_0^\\infty u^4e^{-u}(1-e^{-2u}+e^{-4u}-e^{-6u}+...)du$\n\nor\n\n$I=\\int_0^\\infty u^4(e^{-u}-e^{-3u}+e^{-5u}-e^{-7u}+...)du$\n\nFrom standard integration tables or integration by parts you should be able to get your answer ...", "date": "2019-08-24 09:30:21", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2893117/definite-integral-int-01-frac-ln4xx21-dx", "openwebmath_score": 0.9958059787750244, "openwebmath_perplexity": 907.547974804488, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419667789669, "lm_q2_score": 0.9372107944720444, "lm_q1q2_score": 0.9267533652922144}}
{"url": "https://math.stackexchange.com/questions/1217594/divisibility-rule-for-9", "text": "# Divisibility Rule for 9\n\nI'm working through an elementary number theory course right now and I think I've come up with a proof here but wanted some feedback on my logic.\n\nQuestion: If the sum of the digits in base 10 is divisible by 9, then the number itself is divisible by 9.\n\nProof: Suppose that $9|d_1+d_2+...+d_n$ then $d_1+d_2+...+d_n=0\\mod9$\n\nNow consider $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)$ Each power of $10$ is equivalent to $1\\mod9$\n\ntherefor\n\n$d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)=(1\\mod9)(d_1+d_2+...d_n)$\n\n$9|(d_1+d_2+...d_n)$ by our assumption, thus $9|(1\\mod9)(d_1+d_2+...d_n)$\n\nThus we have shown that if 9 divides the sum of the digits in base 10, 9 divides the number itself.\n\nThe only question I really have is whether I'm jumping the gun on my assumption concerning the powers of 10 being $1\\mod 9$. I think this is fair game here but not 100% confident. Thanks.\n\n\u2022 Your proof is fine. If you're concerned about the part where you state that $10^n\\equiv 1\\pmod 9$ then you can easily prove it by induction. Once you do this you don't even have to include the induction proof: if it really does turn out to be straightforward you can say \u201cEach power of $10$ is equivalent to $1\\pmod 9$, by a straightforward induction.\u201d \u2013\u00a0MJD Apr 2 '15 at 18:05\n\u2022 One way to see what's going on is to imagine you're given a number $n$ and you decide to write out the number $(10^n - 1)$. The result will be just a bunch of $9$'s...and a number like that will be divisible by $9$. So $10^n$ must be equivalent to $1$ mod $9$. (For a formal proof of this fact, you could use induction on $n$, as @MJD says.) \u2013\u00a0mathmandan Apr 2 '15 at 18:15\n\nYes, $\\,{\\rm mod}\\ 9\\!:\\ \\color{}{10\\equiv 1}\\,\\Rightarrow\\, \\color{#c00}{10^n}\\equiv 1^n \\color{#c00}{\\equiv 1},\\,$ therefore\n\n$\\qquad\\qquad\\qquad\\qquad\\ \\ d_n \\color{#c00}{10^n} + d_1\\color{#c00}{10} + d_0$\n$\\qquad\\qquad\\qquad \\quad \\equiv\\,\\ d_k +\\cdots + d_1 + d_0\\ \\$ by  basic Congruence Rules.\n\nMore efficiently, we can observe that the decimal (radix $10)$ representation of an integer $N$ is a polynomial function $\\,f(10)\\,$ of the radix, with integer coefficients (digits) $\\,d_i,\\,$ i.e.\n\n$$\\begin{eqnarray} N\\, =\\, f(10) \\!\\!\\!&&= d_n 10^n +\\,\\cdots+d_1 10 + d_0 \\\\ {\\rm where}\\ \\ f(x) \\!\\!&&= d_n\\, x^n\\,+\\,\\cdots\\,+d_1\\, x\\, + d_0\\end{eqnarray}$$\n\nThus $\\ {\\rm mod}\\ 9\\!:\\,\\ \\color{#c00}{10\\equiv 1}\\,\\Rightarrow\\, f(\\color{#c00}{10})\\equiv f(\\color{#c00}{1}) = d_n+\\cdots + d_1 \\equiv\\,$ sum of digits, which follows by applying the Polynomial Congruence Rule.\n\nThe proof works for any polynomial $\\,f(x)\\,$ with integer coefficients. As such, these tests for divisibility by the radix$\\pm1$ (e.g. also casting out nines) may be viewed as special cases of the Polynomial Congruence Rule.\n\n\\begin{align}653854-(6+5+3+8+5+4)&=6\\cdot99999+5\\cdot9999+3\\cdot999+8\\cdot99+5\\cdot9\\\\ &=(6\\cdot11111+5\\cdot1111+3\\cdot111+8\\cdot11+5)\\cdot9. \\end{align}\n\nA number and the sum of its digits differ by a multiple of $9$.\n\nFix some positive integer $n$ and write $[a]_n$ for the remainder class of any $a \\in \\Bbb{Z}$ modulo $n$.\n\nIt isn't hard to prove (try it!) that $[a+b]_n = [a]_n + [b]_n$ and $[ab]_n = [a]_n[b]_n$ for every $a,b \\in \\Bbb{Z}$. A relation with this property is called a congruence.\n\nIn particular, this means that $[a^k]_n = [a]_n^k$ for every $a,k \\in \\Bbb{Z}$.\n\n\u2022 Could whoever down-voted this please be so kind to tell me why they did so? My answer may be concise, but is mathematically correct and addresses the OP's concern, who wrote (emphasis mine): \"The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of $10$ being \\$1 \\mod{9}\". \u2013\u00a0A.P. Apr 4 '15 at 12:07", "date": "2020-01-26 03:01:52", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1217594/divisibility-rule-for-9", "openwebmath_score": 0.834643542766571, "openwebmath_perplexity": 168.80065897347535, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683469514965, "lm_q2_score": 0.9381240155738166, "lm_q1q2_score": 0.9264615832957341}}
{"url": "https://mathhelpboards.com/threads/what-is-the-remainder-when-a_-2013-is-divided-by-7.5195/", "text": "# What is the remainder when a_{2013} is divided by 7?\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nConsider a sequence given by $$\\displaystyle a_n=a_{n-1}+3a_{n-2}+a_{n-3}$$, where $$\\displaystyle a_0=a_1=a_2=1$$.\n\nWhat is the remainder of $$\\displaystyle a_{2013}$$ divided by $$\\displaystyle 7$$?\n\n#### chisigma\n\n##### Well-known member\nConsider a sequence given by $$\\displaystyle a_n=a_{n-1}+3a_{n-2}+a_{n-3}$$, where $$\\displaystyle a_0=a_1=a_2=1$$.\n\nWhat is the remainder of $$\\displaystyle a_{2013}$$ divided by $$\\displaystyle 7$$?\nOperating modulo 7 we have...\n\n$$a_{0}=1$$\n$$a_{1}=1$$\n$$a_{2}= 1$$\n$$a_{3} = 1 + 3 + 1 = 5$$\n$$a_{4} = 5 + 3 + 1 = 2\\ \\text{mod}\\ 7$$\n$$a_{5} = 2 + 1 + 1 = 4\\ \\text{mod}\\ 7$$\n$$a_{6} = 4 + 6 + 5 = 1\\ \\text{mod}\\ 7$$\n$$a_{7} = 1 + 12 + 2 = 1\\ \\text{mod}\\ 7$$\n$$a_{8} = 1 + 3 + 4 = 1\\ \\text{mod}\\ 7$$\n$$a_{9} = 1 + 2 + 1 =5\\ \\text{mod}\\ 7$$\n\n... and we can stop because the sequence is mod 7 periodic with period 6. Now is $2013\\ \\text{mod}\\ 6 = 3$, so that the requested number is $a_{3}=5$...\n\nKind regards\n\n$\\chi$ $\\sigma$\n\nLast edited:\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nI agree with chisigma on the level of algebra, but not on the level of arithmetic. In fact, reducing all the coefficients mod 7 as we go along,\n\n$a_{n+3} = a_{n+2} + 3a_{n+1} + a_{n}$,\n\n$a_{n+4} = a_{n+3} + 3a_{n+2} + a_{n+1} = (a_{n+2} + 3a_{n+1} + a_{n}) + 3a_{n+2} + a_{n+1} = 4a_{n+2} + 4a_{n+1} + a_{n}$,\n\n$a_{n+5} = a_{n+4} + 3a_{n+3} + a_{n+2} = (4a_{n+2} + 4a_{n+1} + a_{n}) + 3(a_{n+2} + 3a_{n+1} + a_{n}) + a_{n+2} = a_{n+2} + 6a_{n+1} + 4a_{n}$,\n\n$a_{n+6} = a_{n+5} + 3a_{n+4} + a_{n+3} = (a_{n+2} + 6a_{n+1} + 4a_{n}) + 3(4a_{n+2} + 4a_{n+1} + a_{n}) + (a_{n+2} + 3a_{n+1} + a_{n}) = a_{n}$\n\n(for all $n\\geqslant0$). So the sequence repeats with period $6$. It starts with $(a_0,a_1,a_2,a_3,a_4,a_5) = (1,1,1,5,2,4)\\pmod7$, and since $2013=3\\pmod6$ it follows that $a_{2013} = a_3 = 5\\pmod7.$\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nThanks to both chisigma and Opalg for the submission to this problem and I've been really impressed with the creativity that gone into the two approaches above and please allow me to thank you all again for the time that the two of you have invested to participate in this problem.", "date": "2021-06-13 11:35:22", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/what-is-the-remainder-when-a_-2013-is-divided-by-7.5195/", "openwebmath_score": 0.9229724407196045, "openwebmath_perplexity": 998.2764884749569, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683467685534, "lm_q2_score": 0.9381240155738166, "lm_q1q2_score": 0.9264615831241108}}
{"url": "http://nizm.bwni.pw/interval-of-convergence-alternating-series.html", "text": "The radius of convergence is half the length of the interval of convergence. We noticed that, at least in the case of the geometric series, there was an interval in which it converged, but it didn\u2019t converge at the endpoints. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If \u2211 | a n| converges, then so does \u2211 a n. Let f : [ 1 , \u221e ) \u2192 R + {\\displaystyle f:[1,\\infty )\\to \\mathbb {R} _{+}} be a non-negative and monotonically decreasing function such that f ( n ) = a n {\\displaystyle f(n)=a_{n}}. Also make sure to check the endpoint of the interval because there is a possibility for them to converge as well. Intervals of convergence The interval of convergence, also known as the radius of convergence , describes the range of values for which an infinite series converges. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. Thus it converges. So the interval of convergence is [\u22121,3]. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. In general, you can skip the multiplication sign, so 5x is equivalent to 5\u22c5x. (b) X\u221e n=0 c n(\u22124)n No. n is convergent, then the radius of convergence for the power series P \u221e n=0 c nx n is at least 4. We use the usual strategy on. This is the same form as the \ufb01rst series, with x replaced by x2. Therefore the interval of convergence is [ 2;4). The center of the interval will be a. The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). Thus the interval of convergence is [\u22124,2]) b) \u2211 n=0 \u221e (\u22121)n(x\u22123)2n 4n (Ratio Test gives lim n\u2192\u221e| (x\u22123)2n+2 4n+1 \u22c5 4n (x\u22123)2n| = 1 4 |x\u22123|2, and 1 4. Convergence Classifications of Series \u2211a n, and Series Rearrangements. Our interval of convergence is therefore #[-1/e, 1/e]#, and our radius is #1/e#. This needs to be done for every series or improper integral you say converges or diverges. (a)Find the Taylor Polynomial P 3(x) that uses a. Therefore, the interval of convergence for the power series is 2 x < 4 or [ 2;4). Radius and Interval of Convergence A power series in can be viewed as a function of where the domain of is the set of all for which the power series converges. If a power series converges on some interval centered at the center of convergence, then the distance from the center of convergence to either endpoint of that interval is known as the radius of convergence which we more precisely define below. It just means that you couldn't use the Alternating Series Test to prove that it converges. Find the interval of convergence of the power series? be sure to include a check for convergence at the endpoints of the interval. Then by formatting the inequality to the one below, we will be able to find the radius of convergence. Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly. Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. Alternating series test: A series of the form \u2211(\u22121) n a n (with a n > 0) is called alternating. Section 4-8 : Alternating Series Test. When you plug in x= \u20131, you get an alternating series. for , and diverges for and for. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior. It's also known as the Leibniz's Theorem for alternating series. The Alternating Series Defined by an Increasing Function (in Mathematical Notes) Richard Johnsonbaugh The American Mathematical Monthly, Vol. \\displaystyle\\sum _ { n Question: Find the radius of convergence and interval of convergence of the series. X1 of convergence and interval of convergence for the power series 1. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. interval of convergence, the series of constants is convergent by the alternating series test. To see why these tests are nice, let's look at the Ratio Test. Then, the series becomes This is an alternating series. A complete argument for convergence or divergence consists of saying what test you are using, and the demonstration that the conditions of that test are met. While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. ii) I first show that. The interval of convergence is the largest interval on which the series converges. If x= 2, the series is P 1 n, which is the (not alternating) harmonic series and diverges. Therefore the new series will have a radius of convergence which satis\ufb01es jx2j < R, or jxj <. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end. If you're seeing this message, it means we're having trouble loading external resources on our website. Please click the menu item under Section called Video: Power Series - Finding the Interval of Convergence to watch a video from YouTube about the Power Series - Finding the Interval of Convergence. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. 36-38, Jstor. 1 : analysis with geometric series Therefore the radius of convergence is When x = \u2014 the series is This is the harmonic series, which diverges. Observe that in the graph above, Maple computed \"values\" of the power series outside its interval of convergence. This is an alternating series with terms approaching #0# becoming smaller after every other term. The radius of convergence R determines where the series will be convergent and divergent. So the radius of convergence is 1 and the interval is 1 < x 1. (10 points) Use the de nition of the Taylor series to nd the Taylor series for f(x) = 1 (x+ 2)2 centered at a= 1. Representations of Functions as Power series. 6) Power Series. I b n+1 = 1 n+1 < n 1 n for all n 1. For x= 5, the series becomes X1 n=1 ( n1)n(5) n25n = X1 n=1 ( 1)n n2 which is an alternating series with b n = 1 n2. Alternating Series Test. The Convergence and Partial Convergence of Alternating Series J. The intervals of convergence will be cen-tered around x = a. Plugging in x = 7 we get the series P 1=n2, which converges because it is a p-series with p = 2 > 1. It remains to analyze endpoints of the interval of convergence, which are a \u00b1 R; here a = 2 and R = 5, so these endpoints are-3 and 7. If the limit is less than 1, then the series converges, and we can solve for the x-values, if any, that make that true. 2 Convergence 2. , Find the. Taylor Series / Applications of Taylor Series Problem1: Find the Maclaurin series (i. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. 4] Alternating Series Test. The interval of convergence is the open interval (x 0 \u2212 \u03c1, x 0 + \u03c1) together with the extreme points x 0 \u2212 \u03c1 and x 0 + \u03c1 where the series converges. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If \u2211 | a n| converges, then so does \u2211 a n. Therefore, the interval of. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. This is the harmonic series, so it diverges. (b) The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In general, this will be a point, an interval, or perhaps the entire real line. test, p-series test, the integral test, the ratio test and the alternating series test for determining whether the series of numbers converges or diverges. Likewise, at x = 1 we have!\u221e n=0 (\u22121)n (1)2n+2 2n+2 =!\u221e n=0 (\u22121)n 2n+2 which is the same convergent alternating series. Integral Test The series and the integral do the same thing. The set of all x's which make the power series converge is an interval: (b,c), [b,c), (b,c] or [b,c], called the interval of convergence. infinity summation n=1 [3^n(x-2)^n]/n when i did it, i got the interval of convergence to be 5/3 < x < 7/3 but im not sure how to check the endpoints with this one?. To execute such trades before competitors would. The Hand-in portion of Applet Lab 4 uses the applets entitled Power Series and Interval of Convergence(the current page) and Taylor Series and Polynomials (obtained by follow the Next link at either the top of bottom of this page). Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. For an alternating series (in either of the forms) if both 1. The interval of convergence is always centered at the center of the power series. Find the interval of convergence of the following series. which is an alternating series that converges. Math%1152Q%Exam%2%Summary%Chapter%11% Page5%of%6% % MATH%1152Q%Exam%2SummaryCh11Answer% [Ch11]Power%Series% \u301011. , London: Hodder Education, 2005 pp. ii) Find a closed-form formula for. \\displaystyle\\sum _ { n Question: Find the radius of convergence and interval of convergence of the series. A series in which successive terms have opposite signs is called an alternating series. ? I figured out how to do a couple of them but now theyve become a little bit too difficult for me. Find the sum of the alternating harmonic series. Power Series - Review. which is a convergent alternating series. clude by the alternating series test that the series diverges. Math 306 - Power Series Methods Final Review Key (1) True of False? interval of convergence. Thus, the interval of convergence is 1 3 x 1 3 Exercise 15 We have lim n!1 n a n+1 a n 2 = lim n!1 (x 2) +1 (n+1)2 +1 n +1 (x 2)n 2 = lim n!1 n +1 (n+1)2 +1 jx 2j= jx 2j: By the ratio test, the series. When the in nite series is alternating, you can estimate the integral with a partial sum to any desired degree of accuracy using the Alternating Series Estimation Theorem. The Convergence and Partial Convergence of Alternating Series J. The interval of convergence is the value of all x's for which the power series converge. So the question we want to ask about power series convergence is whether it converges for other values of x besides c. It remains to analyze endpoints of the interval of convergence, which are a \u00b1 R; here a = 2 and R = 5, so these endpoints are-3 and 7. Power Series Interval of Convergence Olivia M. Just the usual. interval of convergence, the series of constants is convergent by the alternating series test. Di erentiate the series, then integrate them. (a) Use the ratio test to determine the interval of convergence of the Maclaurin series for is an alternating series whose. ) \u2022 Test x = \u22121: X\u221e n=1 (\u22121)n(\u22121 \u22121)n 2nn3 = X\u221e n=0 1 n3, which converges by the integral test. The radius of convergence is half the length of the interval of convergence. Therefore the interval of convergence contains -2. I Therefore, we. All algorithms numbered 493 and above, as well as a few earlier ones, may be downloaded from this server. AP\u00ae CALCULUS BC 2011 SCORING GUIDELINES (Form B) is a convergent alternating series with individual terms The interval of convergence is centered at x =0. These tests also play a large role in determining the radius and interval of convergence for a series of functions. Finding the Interval of Convergence. o Functions defined by power series. which is an alternating series that converges. These tests also play a large role in determining the radius and interval of convergence for a series of functions. Uniform Convergence and Series of Functions James K. Namely, a power series will converge if its sequence of partial sums converges. ii) I first show that. Use the other tests to check convergence at the endpoints. -The alternating series converges if the limit of the terms goes to 0 and if a_(n+1) \u2264 a_n (abs value of terms always decreases) -Interval of convergence-Series. Step 2: Test End Points of Interval to Find Interval of Convergence. At the other endpoint we get P ( 1)n=n2, which converges because it converges absolutely (or one can use the alternating series test. (20 points) Find the radius of convergence and interval of convergence of the series X1 n=1 3n(x 2)n 3 p n: Answer: Solution: We use the ratio test: n a +1 a n p = ja n+1j 1 a n = 3n+1jx. Recall from the Absolute and Conditional Convergence page that series $\\sum_{n=1}^{\\infty} a_n$ is said to be absolutely convergent if $\\sum_{n=1}^{\\infty} \\mid a_n \\mid$ is also convergent. Help in finding the radius of convergence and interval the series? Find the radius of convergence and interval of convergence of the series. This can be achieved using following theorem: Let { a n } n = 1 \u221e {\\displaystyle \\left\\{a_{n}\\right\\}_{n=1}^{\\infty }} be a sequence of positive numbers. In other words, by uniform convergence, what I can now do is integrate this thing here, term by term. Finding the Interval of Convergence. to put into appropriate form. (In other words,the first finite number of terms do not determine the convergence of a series. ther use Taylor\u2019s inequality or the Alternating Series Remainder term. The Alternating Series Test is a consequence of the de\ufb01nition of convergence for series (convergence of the sequence of partial sums) and the Monotonic Sequence Theorem. \u20ac f(x)= (\u22121)n+1 (x\u22123)nn1/2\u00d75n n=1 When we apply the root or ratio test to the absolute value of the summands, we can. 11 The interval of convergence for the Maclaurin series of f is (2x)2 (2r)3 (2x)4 = 4x2 \u2014 4x3 + 16 x4 y'=8x-12x2 64 4. = jxj<1: Thus, the radius of convergence is 1. 3 We considered power series, derived formulas and other tricks for nding them, and know them for a few functions. The calculator will find the radius and interval of convergence of the given power series. This leads to a new concept when dealing with power series: the interval of convergence. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Then use absolute value to look at the concepts of conditional and absolute convergence for series with positive and negative terms. where is the -th derivative of Well, we have. Sometimes we\u2019ll be asked for the radius and interval of convergence of a Taylor series. Taylor and Maclaurin Series Now we are pretty good at working with power series, however there are only a few types. The method for finding the interval of convergence is to use the ratio test to find the interval where the series converges absolutely and then check the endpoints of the interval using the various methods from the previous modules. $\\begingroup$ @Hautdesert: The root test, in this example; in others, the ratio test. Observe that in the graph above, Maple computed \"values\" of the power series outside its interval of convergence. \u2211 p-series, p = 1 \u2264 1 so it diverges If x =\u22121: (\u22121)n n=0 n \u221e \u2211 alternating series, terms decrease, lim n\u2192\u221e 1 n =0 so it converges So, xn n=0 n \u221e \u2211 converges when \u22121\u2264x <1 This is the interval of convergence of the series. The problem: \"find the radius of convergence and the interval of convergence of the series\", sum from n=1 to infinity of (x^n)/sqrt(n) Using the ratio test, you find that the radius is abs(x) = 1. When x= 3, the series converges using the integral test. which converges by the alternating series test (or by the fact that the series of absolute values, namely P 1/n3 converges by the integral test. Find the radius of convergence and interval of convergence of the series: (a) X1 n=1 xn p n Solution Sketch Ratio test gives a radius of convergence of R = 1. If you're behind a web filter, please make sure that the domains *. Taylor Series and. Worksheet 7 Solutions, Math 1B Power Series Monday, March 5, 2012 1. When x= 3, the series converges using the integral test. In particular, the intervals of convergence of the power series representations of f(x), df/dxand R f(x)dxcan di\ufb00er atthe endpoints of the interval ofconvergence. Absolute Convergence Test: If you have a series X1 n=1 x n, and the. I lim n!1 1 n = 0. The alternating series theorem plays a key role, either directly or via the degree difference test, in the rules for determining interval of convergence. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end. Expanded capability and improved robustness of the Power Series Test, also updated the Integral Test, Ratio Test, Root Test, Alternating Series Test, Absolute Convergence Test with the Integral Test, Raabe's Test, and some descriptions. Murphy, A2 Further Pure Mathematics , 3rd ed. We use the usual strategy on. Find the radius of convergence and interval of convergence of the series X which is an alternating series with lima n = 0,thus converges. Register Now! It is Free Math Help Boards We are an online community that gives free mathematics help any time of the day about any problem, no matter what the level. $2$ is the radius of convergence. I b n+1 = 1 n+1 < n 1 n for all n 1. 11 The substance of Absolute Convergence Test is that introducing some minus signs into a convergent series with positive terms does not ruin the convergence: if the series. The interval of convergence of a power series is the set of all x-values for which the power series converges. Therefore, the series is a conditionally convergent series. is a harmonic series that diverges. power series about x = I, and find its interval of convergence. Perform algebraic operations on power series. We can multiply power series together. Taylor Series and Applications: Given a function f(x) and a number a,. x = -3 ==> sum(n = 0 to infinity) 1 /(2n+1), which diverges by the integral test. When x= 1, the series converges by the p-series test. It is a finite or an infinite series according as the number of terms is finite or infinite. With power series we can extend the methods of calculus we have developed to a vast array of functions making the techniques of calculus applicable in a much wider setting. diverges ( ) x diverges 0 rho x converges absolutely The ratio test for power series Example Determine the radius of convergence and the interval of convergence of the power series y(x) = X. Examples from Section 11. Series converges for only one number Remember the Ratio Test: A series converges if. The sequence is decreasing since for all Also, Therefore, this series converges by the Alternating Series Test and we include in our interval. Philip Mathematics of Computation, Vol. Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly. Radius of convergence R = 0. Using the series formula for our answer , we have (Note , , , etc. Math 306 - Power Series Methods Final Review Key (1) True of False? interval of convergence. o xMaclaurin series for the functions e, sin x, cos x, and 1/(1 \u2013 x). Then by formatting the inequality to the one below, we will be able to find the radius of convergence. Homework 25 Power Series 1 Show that the power series a c have the same radius of convergence Then show that a diverges at both endpoints b converges\u2026 UVA APMA 1110 - Homework+25+-+Power+Series - GradeBuddy. Why do we want to express a known function as the sum of a power series ? We will see later that this is a good strategy for integrating functions that don't have elementary antiderivatives ( or for examples), for solving differential equations, and for approximating functions by polynomials. The interval of convergence contains all values of x for which the power series converges. Step 2: Find the Radius of Convergence. And we will also learn how an alternating series may have Conditional or Absolute Convergence. (a) Find the Taylor series for the function f(x) = ex at a = 3. To execute such trades before competitors would. I'll test the endpoints separately. Example 1 Test the following series for convergence X1 n=1 ( 1)n 1 n I We have b n = 1 n. convergence 5 and interval of convergence centered at \u02d7\u0336\u0336\u0336 1. It is customary to call half the length of the interval of convergence the radius of convergence of the power series. Alternating Series testP If the alternating series 1 n=1 ( 1) n 1b n = b 1 b 2 + b 3 b 4 + ::: b n > 0 satis es (i) b n+1 b n for all n (ii) lim n!1 b n = 0 then the series converges. result of multiplying the divergent harmonic series by 1. This means that the interval of convergence is ( 2;2). The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). 2 Convergence 2. Therefore, the interval of convergence for the power series is 2 x < 4 or [ 2;4). 5) Ratio Test (11. Both of those tests give an open interval of absolute convergence, and both guarantee that outside of the corresponding closed interval the terms of the series fail to converge to $0$, so it will diverge regardless of the signs of the terms. Then, the series becomes This is an alternating series. The Convergence and Partial Convergence of Alternating Series J. So x = \u22121 is included in the interval of convergence. 11 The interval of convergence for the Maclaurin series of f is (2x)2 (2r)3 (2x)4 = 4x2 \u2014 4x3 + 16 x4 y'=8x-12x2 64 4. I b n+1 = 1 n+1 < n 1 n for all n 1. A series of the form converges if Example. Nihil anim keffiyeh helvetica, craft beer labore wes anderson cred nesciunt sapiente ea proident. Recall from the Absolute and Conditional Convergence page that series $\\sum_{n=1}^{\\infty} a_n$ is said to be absolutely convergent if $\\sum_{n=1}^{\\infty} \\mid a_n \\mid$ is also convergent. ii) Find a closed-form formula for. Find the sum of the alternating harmonic series. Power series, radius of convergence, interval of convergence. The interval of convergence is the set of all values of $$x$$ for which the series converges. The interval of convergence is the set of all values of x for which a power series converges. This program can run several different tests on infinite series to check for convergence or divergence. 2 Conditions for Convergence of an Alternating Sequence. A proof for a general monotonic decreasing alternating series can be found in Leibniz's Theorem. This is the harmonic series, so it diverges. To identify trading opportunities. (b) Find the Taylor series for the function f(x) = ex at a = 2. Direct Comparison Test. Plugging in x = 7 we get the series P 1=n2, which converges because it is a p-series with p = 2 > 1. The ratio test gives us: Because this limit is zero for all real values of x, the radius of convergence of the expansion is the set of all real numbers. So by the Alternating Series test, the original series converges. ny business of trading in securities needs two capabilities: 1. Carducci (East Stroudsburg University) Rearranging the Alternating Harmonic Series Ed Packel (Lake Forest College) and Stan Wagon (Macalester College) Bounding Partial Sums of the Harmonic Series Matt Clay; Taylor Series Michael Ford. Recall from the Absolute and Conditional Convergence page that series $\\sum_{n=1}^{\\infty} a_n$ is said to be absolutely convergent if $\\sum_{n=1}^{\\infty} \\mid a_n \\mid$ is also convergent. \u2211 p-series, p = 1 \u2264 1 so it diverges If x =\u22121: (\u22121)n n=0 n \u221e \u2211 alternating series, terms decrease, lim n\u2192\u221e 1 n =0 so it converges So, xn n=0 n \u221e \u2211 converges when \u22121\u2264x <1 This is the interval of convergence of the series. Likewise, at x = 1 we have!\u221e n=0 (\u22121)n (1)2n+2 2n+2 =!\u221e n=0 (\u22121)n 2n+2 which is the same convergent alternating series. Comparison with an integral. The interval of convergence for the Maclaurin series of is 1: sets up ratio 1: limit evaluation 1: radius of convergence 1: considers both endpoints 1: analysis and interval of convergence (b) is a geometric series that converges to for Therefore for 1: series for. Also examine differentiation and integration of power series. Functions of One Real Variable: Limit, continuity, intermediate value property, differentiation, Rolle\u2019s Theorem, mean value theorem, L'Hospital rule, Taylor's theorem, maxima and minima. You do not need to nd the radius of convergence or interval of convergence. 1 Questions: 2 Converge or Diverge: 1. k2 is the series of absolute values. Theorem 4 : (Comparison test ) Suppose 0 \u2022 an \u2022 bn for n \u201a k for some k: Then. If and then Theorem 2. Unlike geometric series and p-series, a power series often converges or diverges based on its x value. lim n!1 an = 0, then the alternating series converges. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. The main tools for computing the radius of convergence are the Ratio Test and the Root Test. 1 The Interval of Convergence In the previous section, we discussed convergence and divergence of power series. 2 Convergence 2. While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. When x = 1, we get the series X1 n=0 ( n1 + 3) (n+ 1)2n = 1 n=0 1 n+ 1. For f0: [ 5; 3). The series converges uniformly on [ \u2212\u03c1,\u03c1 ] for every 0 \u2264 \u03c1< 1 but does not converge uniformly on (\u2212 1, 1) (see Example 5. X1 n=1 (x 1)n 1 n2 3n. Absolute Convergence Test: If you have a series X1 n=1 x n, and the. Namely, a power series will converge if its sequence of partial sums converges. convergence is one. The following power series is centered at c. Find the radius of convergence. First however, we must compute the radius of convergence for the series: lim n!1 n. Taking absolute values of the terms and using the ratio test, one can show that the sum of the absolute values of the terms of the series, i. 5) Ratio Test (11. $2$ is the radius of convergence. Using the series formula for our answer , we have (Note , , , etc. Below are some of the standard terms and illustrations concerning series. Therefore, the series is a conditionally convergent series. Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. , X\u221e n=1 n2 3n converges. Estimating Error/Remainder of a Series; Alternating Series Test; Alternating Series Estimation Theorem; Ratio Test; Ratio Test with Factorials; Root Test; Absolute and Conditional Convergence; Difference Between Limit and Sum of the Series; Radius of Convergence; Interval of Convergence; Power Series Representation, Radius and Interval of. We know the series will converge on the interval 5 1. Then by formatting the inequality to the one below, we will be able to find the radius of convergence. (In other words,the first finite number of terms do not determine the convergence of a series. g(x) = Represent the function g(x) in Exercise 50 as a power series about 5, and find the interval of convergence. All algorithms numbered 493 and above, as well as a few earlier ones, may be downloaded from this server. Taylor Polynomials\u2044 (a) an application of Taylor Polynomials (e. Homework 25 Power Series 1 Show that the power series a c have the same radius of convergence Then show that a diverges at both endpoints b converges\u2026 UVA APMA 1110 - Homework+25+-+Power+Series - GradeBuddy. X( n1)nx 4n lnn. Thus, the interval of convergence of the power series is (\u2212 1, 1). Taylor Series and Applications: Given a function f(x) and a number a,. A series of the form converges if Example. , London: Hodder Education, 2005 pp. Sometimes we\u2019ll be asked for the radius and interval of convergence of a Taylor series. Hence the series diverges by the nth-term test. 8 NAME Section wed Test each of the series for convergence or divergence: State and use the Comparison Test. The endpoints of the interval of convergence always must be checked independently. It's also known as the Leibniz's Theorem for alternating series. I The binomial series determine the open interval of. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. When x \u2014 the series is This is the alternating hannonic series, which converges. Next, if , the power series becomes: which converges by the alternating series test. Therefore the new series will have a radius of convergence which satis\ufb01es jx2j < R, or jxj <. (In other words,the first finite number of terms do not determine the convergence of a series. infinity summation n=1 [3^n(x-2)^n]/n when i did it, i got the interval of convergence to be 5/3 < x < 7/3 but im not sure how to check the endpoints with this one?. And we will also learn how an alternating series may have Conditional or Absolute Convergence. \u00d2 $\u00df$\u00d1 \u00e8 Remember that a bracket indicates an endpoint that belongs to an interval, while a parentheses indicates an endpoint that does not belong to the interval. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior. Use the ratio test to determine radius or open interval of convergence of power series. Chapter 7) Taylor Series. Murphy, A2 Further Pure Mathematics , 3rd ed. ny business of trading in securities needs two capabilities: 1. Once again, if something does not pass the alternating series test, that does not necessarily mean that it diverges. So, the interval of convergence is -3 < x <= 3. R1 the interval of convergence is \u2013 2 < x < 2 A1 (b) (i) this alternating series is convergent because the moduli of successive terms are monotonic decreasing R1 and the nth term tends to zero as n \u2192 \u221e R1. 3 Alternating series, approximations of alternating series. Find the intervals of convergence of fand f0. Divergence test (particularly useful for terms that are rational functions). alternating series test; Usually, for power series, the best way to determine convergence will be the ratio test, and sometimes the root test. Convergence of power series is similar to convergence of series. The series above is thus an example of an alternating series, and is called the alternating harmonic series. Since it is a geometric series, we know that it converges when \\eqalign{ |x+2|/3& 1\\cr |x+2|& 3\\cr -3 x+2 & 3\\cr -5 x& 1. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. Determine a power-series representation of the function ln (1 + x) on an interval centered at x = 0. We can use the ratio test to find out the absolute convergence of the power series by examining the limit, as n approached infinity, of the absolute value of two successive terms of the sequence. At , the series is. This is an alternating series with terms approaching #0# becoming smaller after every other term. The terms converge to 0. If , then R = and the series converges for all values of x. When x \u2014 the series is This is the alternating hannonic series, which converges.", "date": "2019-12-08 10:57:03", "meta": {"domain": "bwni.pw", "url": "http://nizm.bwni.pw/interval-of-convergence-alternating-series.html", "openwebmath_score": 0.9055711030960083, "openwebmath_perplexity": 325.46529907420165, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9911526449170697, "lm_q2_score": 0.9314625036384888, "lm_q1q2_score": 0.9232215241223638}}
{"url": "http://mathhelpforum.com/trigonometry/36050-trigonometry-three-dimensional-question.html", "text": "# Math Help - Trigonometry three dimensional question\n\n1. ## Trigonometry three dimensional question\n\nI need help to solve:\nA cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?\n\nAnswer given at the back of the text book is 16 degrees 15 minutes.\n\n2. Originally Posted by lalji\nI need help to solve:\nA cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted?\n\nAnswer given at the back of the text book is 16 degrees 15 minutes.\nMaybe it is the three dimensions that are screwing you up...I say think it more two dimensionally to start...and here is a big hint...PERPENDICULAR height...and perp. is synonomous with what?\n\n3. Hello,\n\nTake a look at this picture. OPQR represents the 12cm high box. ABCD is a view from the side of the cylinder.\n\nAccording to the text,\n$RQ=OP=12 cm$\n\n$AB=CD=2 \\cdot 4=8 cm$\n\n$AC=BD=15 cm$\n\nAnd you're looking for angle $DAR$.\n\n4. Hello, lalji!\n\nI don't agree with their answer . . .\n\nA cylinder with radius 4 cm and perpendicular height 15 cm\nis tilted so that it will just fit inside a 12 cm high box.\nAt what angle must it be tilted?\n\nAnswer given: 16 degrees 15 minutes.\nCode:\n      P       A         S\n_ * - - - * - - - - *\n: |      *   *      |\n: |     *       *D  |\n: |    *       *    |\n12-x|   * 15    *     |\n: |  *       *      |\n: | *       *       |\n: |*       *        |\n-B*   8   *         |\nx |  *   *          |\n- * - - * - - - - - *\nQ     C           R\n\nThe cylinder is $ABCD\\!: CB = 8,\\;AB = 15,\\;PQ = 12$\n\nThe box is $PQRS\\!:\\;PQ = 12$\n\nLet $x = BQ$\nLet $\\theta = \\angle BCQ$\n. . Note that $\\angle ABP = \\theta$\n\nIn right triangle $BCQ\\!:\\;QC = \\sqrt{64-x^2}$\n. . $\\cos\\theta \\,=\\,\\frac{\\sqrt{64-x^2}}{8}\\;\\;{\\color{blue}[1]}$\n\nIn right triangle $APB\\!:\\;PB \\,=\\,12-x$\n. . $\\cos\\theta \\:=\\:\\frac{12-x}{15}\\;\\;{\\color{blue}[2]}$\n\nEquate [1] and [2]: . $\\frac{\\sqrt{64-x^2}}{8} \\;=\\;\\frac{12-x}{15}\\quad\\Rightarrow\\quad 15\\sqrt{64-x^2} \\;=\\;8(12-x)$\n\nSquare both sides: . $225(64-x^2) \\;=\\;64(144-24x + x^2)$\n\n. . which simplifies to: . $289x^2 - 1536x - 5184 \\;=\\;0$\n\n. . and has the positive root: . $x \\;=\\;\\frac{1536 + \\sqrt{6,352,000}}{578} \\;=\\;7.657439446$\n\nThen: . $\\sin\\theta \\;=\\;\\frac{x}{8} \\;=\\;\\frac{7.657439446}{8} \\;=\\;0.957179931$\n\n. . $\\theta \\;=\\;\\sin^{-1}(0.957179931) \\;=\\;73.17235533^o \\;\\approx\\;\\boxed{73^o10'}$\n\n5. I've got some difficulties, how can we know, by reading the text, what angle we are looking for ? And how can this physically be possible if D is not on RS ??\n\n6. When I saw this question, I knew I saw it before. It was in my textbook last year. Here is the picture the book provided, and the answer at the back of the book is 16 degrees 50 minutes, not 16 degrees 15 minutes. I think Soroban had a correct answer, but the book is asking for the angle of the other side. So:\n\n$180^o$(straight angle) - $90^o$ (angle between base and length of cylinder) - $73^o10'$ (angle 'on the other side' Soroban found) = $\\theta$\n\n$\\theta$ = $16^o 50'$\n\n7. ## Thanks for the responses\n\nThanks everyone especially Soroban, you are genius. Two dimensional diagram is very helpful to understand. Thanks Gusbob for posting the diagram from the text book, I did not have scanner to scan the diagram.", "date": "2014-09-03 02:41:40", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/trigonometry/36050-trigonometry-three-dimensional-question.html", "openwebmath_score": 0.8148919343948364, "openwebmath_perplexity": 1087.1382273861045, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9877587225460771, "lm_q2_score": 0.9334308114924604, "lm_q1q2_score": 0.9220044259449408}}
{"url": "https://brilliant.org/discussions/thread/golden-ratio-and-fibonacci-numbers/", "text": "# Golden Ratio and Fibonacci Numbers\n\nGolden Ratio is considered to be one of the greatest beauties in mathematics. Two numbers $$a$$ and $$b$$ are said to be in Golden Ratio if $a>b>0,\\quad and\\quad \\frac { a }{ b } =\\frac { a+b }{ a }$ If we consider this ratio to be equal to some $$\\varphi$$ then we have $\\varphi =\\frac { a }{ b } =\\frac { a+b }{ a } =1+\\frac { b }{ a } =1+\\frac { 1 }{ \\varphi }$ Solving in quadratic we get two values of $$\\varphi$$, viz. $$\\frac { 1+\\sqrt { 5 } }{ 2 }$$ and $$\\frac { 1-\\sqrt { 5 } }{ 2 }$$ one of which (the second one) turns out to be negative (extraneous) which we eliminate. So the first one is taken to be the golden ratio (which is obviously a constant value). It is considered that objects with their features in golden ratio are aesthetically more pleasant. A woman's face is in general more beautiful than a man's face since different features of a woman's face are nearly in the golden ratio.\n\nNow let us come to Fibonacci sequence. The Fibonacci sequence $${ \\left( { F }_{ n } \\right) }_{ n\\ge 1 }$$ is a natural sequence of the following form:${ F }_{ 1 }=1,\\quad { F }_{ 2 }=1,\\quad { F }_{ n-1 }+{ F }_{ n }={ F }_{ n+1 }$ The sequence written in form of a list, is $$1,1,2,3,5,8,13,21,34,..$$.\n\nThe two concepts: The Golden Ratio and The Fibonacci Sequence, which seem to have completely different origins, have an interesting relationship, which was first observed by Kepler. He observed that the golden ratio is the limit of the ratios of successive terms of the Fibonacci sequence or any Fibonacci-like sequence (by Fibonacci-like sequence, I mean sequences with the recursion relation same as that of the Fibonacci Sequence, but the seed values different). In terms of limit:$\\underset { n\\rightarrow \\infty }{ lim } \\left( \\frac { { F }_{ n+1 } }{ { F }_{ n } } \\right) =\\varphi$ We shall now prove this fact. Let ${ R }_{ n }=\\frac { { F }_{ n+1 } }{ { F }_{ n } } ,\\forall n\\in N$Then we have $$\\forall n\\in N$$ and $$n\\ge 2$$, ${ F }_{ n+1 }={ F }_{ n }+{ F }_{ n-1 }\\\\$ and ${ R }_{ n }=1+\\frac { 1 }{ { R }_{ n-1 } } >1$We shall show that this ratio sequence goes to the Golden Ratio $$\\varphi$$ given by: $\\varphi =1+\\frac { 1 }{ \\varphi }$We see that: $\\left| { R }_{ n }-\\varphi \\right| =\\left| \\left( 1+\\frac { 1 }{ { R }_{ n-1 } } \\right) -\\left( 1+\\frac { 1 }{ \\varphi } \\right) \\right| \\\\ =\\left| \\frac { 1 }{ { R }_{ n-1 } } -\\frac { 1 }{ \\varphi } \\right| \\\\ =\\left| \\frac { \\varphi -{ R }_{ n-1 } }{ \\varphi { R }_{ n-1 } } \\right| \\\\ \\le \\left( \\frac { 1 }{ \\varphi } \\right) \\left| \\varphi -{ R }_{ n-1 } \\right|\\\\ \\le { \\left( \\frac { 1 }{ \\varphi } \\right) }^{ n-2 }\\left| { R }_{ 2 }-\\varphi \\right|$ Which clearly shows that$\\left( { R }_{ n } \\right) \\longrightarrow \\varphi$ (since $$\\left| { R }_{ 2 }-\\varphi \\right|$$ is a finite positive real whose value depends on the seed values)\n\nNote by Kuldeep Guha Mazumder\n2\u00a0years, 11\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\n- 2\u00a0years, 11\u00a0months ago\n\nYou are welcome! Did you like it?\n\n- 2\u00a0years, 11\u00a0months ago\n\nYes!\n\n- 2\u00a0years, 11\u00a0months ago\n\nMy pleasure..:-)\n\n- 2\u00a0years, 11\u00a0months ago\n\nI like Fibonacci very much.It is really The beauty of Mathematics.\n\n- 2\u00a0years, 9\u00a0months ago\n\nVery nice knowledge.. Loved it...The Magic of Maths!!!\n\n- 2\u00a0years, 9\u00a0months ago\n\nhttps://brilliant.org/problems/wow-12/?group=w3HWB8GobVLl&ref_id=1095702\n\ni posted a problem about the same thing\n\nmy solution was almost the same as your proof of it (:\n\n- 2\u00a0years, 9\u00a0months ago\n\nI have seen your proof. Your idea is essentially the same. Only some of your steps are erroneous.\n\n- 2\u00a0years, 9\u00a0months ago\n\n- 2\u00a0years, 9\u00a0months ago\n\nNothing as such. Only that you have put a plus sign in front of 1/phi.\n\n- 2\u00a0years, 9\u00a0months ago\n\nThere is one more interesting thing I found yesterday. The Ratio of the diagonal and the side of a regular Pentagon is exactly equal to the golden ratio.\n\n- 2\u00a0years, 10\u00a0months ago\n\nOk then I will write a note on it..\n\n- 2\u00a0years, 10\u00a0months ago\n\nDidn't you find it extremely interesting? This is the beauty of Mathematics.\n\n- 2\u00a0years, 10\u00a0months ago\n\nNice\n\n- 2\u00a0years, 11\u00a0months ago\n\nThanks..don't you think whatever is written above is a reconciliation of two apparently different mathematical ideas?..\n\n- 2\u00a0years, 11\u00a0months ago\n\nNice work ! I read this in the book Da Vinci Code by Dan Brown.\n\n- 2\u00a0years, 11\u00a0months ago\n\nThat is one book that I want to read but haven't read yet..thank you for your compliments..:-)\n\n- 2\u00a0years, 11\u00a0months ago\n\nHave you read any other book by Dan Brown ? If not then try them ,they are awesome .\n\n- 2\u00a0years, 11\u00a0months ago\n\nI have just bought The Da Vinci Code today..:-)\n\n- 2\u00a0years, 11\u00a0months ago", "date": "2018-11-18 00:53:39", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/golden-ratio-and-fibonacci-numbers/", "openwebmath_score": 0.9912343621253967, "openwebmath_perplexity": 1404.0567632852903, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9928785708328176, "lm_q2_score": 0.9284088055075428, "lm_q1q2_score": 0.9217972079609325}}
{"url": "https://www.physicsforums.com/threads/evaluating-triple-integral.74665/", "text": "# Evaluating Triple integral\n\n1. May 7, 2005\n\n### MattL\n\nI'm having trouble with evaluating\n\n[Triple Integral] |xyz| dx dy dz\n\nover the region (x/a)^2 + (y/b)^2 + (z/c)^2 <= 1\n\nDo I need to use some sort of parametrisation for the region, and is there some way of dealing with the absolute value function without integrating over the eight octants?\n\nWhilst I've separated the integral into the product of three integrals, I'm not sure if this actually helps?\n\n2. May 7, 2005\n\n### dextercioby\n\nWell,the function is even and the domain of integration is symmetric wrt the origin,so that would give u a hint upon the limits of integration.The symmetry of the ellipsoid is really useful.\n\nAs for the parametrization,i'm sure u'll find the normal one\n\n$$x=a\\cos\\varphi\\sin\\vartheta$$\n\n$$y=b\\sin\\varphi\\sin\\vartheta$$\n\n$$z=c\\cos\\vartheta$$\n\npretty useful.\n\nDaniel.\n\n3. May 7, 2005\n\n### saltydog\n\nThis integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though.\n\n4. May 7, 2005\n\n### arildno\n\nI think you missed the absolute value sign on the integrand..\n\n5. May 7, 2005\n\n### saltydog\n\nWell . . . no, that's the reason I used for the symmetry but again, I qualify my statements by the fact I can't prove it. For example in the first octant:\n\n$$|xyz|=xyz$$\n\nThat's a positive one.\n\nHowever, in the octant with x<0, y>0 and z>0 we have:\n\n$$|xyz|=-xyz$$\n\nAnd so forth in the 8 octants leaving 4 positive and 4 negative ones integrated symmetrically (I think).\n\n6. May 7, 2005\n\n### arildno\n\nThe integrand is positive almost everywhere; hence, the integral is strictly positive:\nLet:\n$$x=ar\\sin\\phi\\cos\\theta,y=br\\sin\\phi\\sin\\theta,z=cr\\cos\\phi$$\n$$0\\leq{r}\\leq{1},0\\leq\\theta\\leq{2}\\pi,0\\leq\\phi\\leq\\pi$$\nThus, we may find:\n$$dV=dxdydz=abcr^{2}\\sin\\phi{dr}d\\phi{d}\\theta$$\n$$|xyz|=\\frac{abcr^{3}}{2}\\sin^{2}\\phi|\\cos\\phi\\sin(2\\theta)|$$\nDoing the r-integrations yield the double integral:\n$$I=\\frac{(abc)^{2}}{12}\\int_{0}^{2\\pi}\\int_{0}^{\\pi}\\sin^{3}\\phi|\\cos\\phi\\sin(2\\theta)|d\\phi{d}\\theta$$\nWe have symmetry about $$\\phi=\\frac{\\pi}{2}$$; thus we gain:\n$$I=\\frac{(abc)^{2}}{24}\\int_{0}^{2\\pi}|\\sin(2\\theta)|d\\theta$$\nWe have four equal parts here, and using the part $$0\\leq\\theta\\leq\\frac{\\pi}{2}$$ yields:\n$$I=\\frac{(abc)^{2}}{12}$$\n\n7. May 7, 2005\n\n### saltydog\n\nThanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it.\n\n8. May 7, 2005\n\n### MattL\n\nNo problem.\n\nThink I should be able to give this question a fair go now.\n\n9. May 10, 2011\n\n### Ray Vickson\n\nChange variables to x = a*x1, y = b*x2, z = c*x3. The integration region is the unit ball x1^2 + x2^2 + x3^2 <= 1, the integrand is abc*|x1 x2 x3|, and dV = abc * dx1 dx2 dx3. Because of the absolute value and symmetry, the whole integral, I, equals 8 times the integral over the {x1,x2,x3 >= 0} portion of the ball. This gives I = 8abc*int_{x3=0..1} f(x3) dx3, where f(x3) = x3*int_{x1^2 + x2^2 <= 1-x3^2} x1 x2 dx1 dx2. Using polar coordinates (or first integrating over x2 for fixed x1, then integrating over x1) we can easily evaluate f(x3), then integrate it over x3 = 0-->1. The final result is I = abc/6.\n\nR.G. Vickson", "date": "2017-02-28 06:21:29", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/evaluating-triple-integral.74665/", "openwebmath_score": 0.8912019729614258, "openwebmath_perplexity": 2048.3631483177724, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682461347528, "lm_q2_score": 0.931462502687197, "lm_q1q2_score": 0.9209073988720385}}
{"url": "https://math.stackexchange.com/questions/3902158/proving-or-disproving-basic-facts-about-sequences-in-real-analysis", "text": "# Proving or disproving basic facts about sequences in Real Analysis\n\nI am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence. I'd like to verify my solution, to ensure, I've understood the concepts, and if proof is technically correct and rigorous.\n\n\\textbf{Problem.} Give an example of each of the following, or state that such a request is impossible by referencing proper theorem(s):\n\n(a) sequences $$(x_n)$$ and $$(y_n)$$, which both diverge, but whose sum $$(x_n + y_n)$$ converges;\n\n(b) sequences $$(x_n)$$ and $$(y_n)$$, where $$(x_n)$$ converges, $$(y_n)$$ diverges, and $$(x_n + y_n)$$ converges;\n\n(c) a convergent sequence $$(b_n)$$ with $$b_n \\ne 0$$ for all $$n$$ such that $$1/b_n$$ diverges;\n\n(d) an unbounded sequence $$(a_n)$$ and a convergent sequence $$(b_n)$$ with $$(a_n - b_n)$$ bounded\n\n(e) two sequences $$(a_n)$$ and $$(b_n)$$, where $$(a_n b_n)$$ and $$a_n$$ converge but $$(b_n)$$ does not.\n\nSolution.\n\n(a) Consider the sequence $$(x_n)$$ given by $$x_n = \\sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$0$$.\n\nAlso, consider the sequence $$(x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\\sqrt{n^2 + 2n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$-1$$.\n\n(b) This request is impossible. If the $$(x_n + y_n)$$ is to be convergent, it implies we are able to make the distance $$\\vert{(x_n + y_n) - (x + y)}\\vert$$ as small as like. However, we cannot make $$y_n$$ to lie eventually in a set $$(y - \\epsilon, y + \\epsilon)$$. Hence, the sum cannot be convergent.\n\n(c) Consider the sequence $$(b_n)$$ given by $$b_n = \\frac{1}{n}$$. Then, $$(b_n)$$ is a convergent sequence but $$1/b_n$$ is divergent.\n\n(d) This request is impossible. The key here is to show that, assuming here $$(a_n)$$ is bounded leads to the contradiction, leading to their difference also being bounded.\n\nIf $$(a_n)$$ is a bounded sequence, there exists a large number $$M > 0$$, such that $$\\vert{a_n}\\vert < M$$ for all $$n \\in \\mathbf{N}$$. If $$(b_n)$$ is a bounded sequence, there exists a large number $$N > 0$$, such that $$\\vert{a_n}\\vert < N$$ for all $$n \\in \\mathbf{N}$$.\n\nThus,\n\n\\begin{align*} \\vert{a_n - b_n}\\vert &= \\vert{a_n + (-b_n)}\\vert \\\\ &\\le \\vert{a_n}\\vert + \\vert{-b_n}\\vert\\\\ &< M + N \\end{align*}\n\n(e) Consider the sequence $$(a_n)$$ given by $$a_n = \\frac{1}{n}$$ and $$(a_n b_n)$$ given by $$a_n b_n= \\frac{\\sin n}{n}$$. Thus, $$(a_n b_n)$$ and $$(a_n)$$ converges, but $$(b_n)$$ does not.\n\nThank you for a detailed, well-asked question!\n\nFor (a), (c), and (e): your examples are all correct. I would say that you asserted several facts that haven't been justified (which would make the proof incomplete, if you were writing for someone else's judgment/understanding). For example, the asserted limits of your examples in part (a) would need to be justified, as would the convergent/divergent assertions in part (e). Note even in part (c) that you didn't justify your assertions (though in that case they're pretty obvious).\n\nWhile the idea is reasonable, your proof for part (b) isn't rigorous. One detail: assuming that $$(x_n+y_n)$$ is convergent means that it converges to some number $$z$$, not to $$x+y$$ (indeed you didn't define either $$x$$ or $$y$$). Then you asserted, without proof, that $$y_n$$ can't be made to lie inside a small set. Your ideas are heading in the right direction, but you would need to use the precise definitions of convergence/divergence in exploiting the assumptions and in setting out what needs to be proved.\n\nAlternatively: you (correctly) believe (b) is impossible\u2014in other words, you believe the implication \"if $$(x_n)$$ converges and $$(y_n)$$ diverges, then $$(x_n+y_n)$$ diverges\". As it turns out, that statement is logically equivalent to \"if $$(x_n)$$ converges and $$(x_n+y_n)$$ converges, then $$(y_n)$$ converges\"\u2014which you might well find easier to prove! (By logically equivalent, I mean that the two statements \"if P and Q, then R\" and \"if P and (not R), then (not Q)\" have the same meaning.)\n\nYour proof for (d) seems to prove the following statement: \"if $$(a_n)$$ is bounded and $$(b_n)$$ is bounded, then $$(a_n+b_n)$$ is bounded\". That is a true fact, but is that what you want to prove here?\n\n\u2022 I expanded my original attempt with short proofs like you said. I am not quite sure, how do I begin with proving (d). \u2013\u00a0Quasar Nov 11 '20 at 16:05\n\n$$\\newcommand{\\absval}[1]{\\left\\lvert #1 \\right\\rvert}$$\n\nI expanded my original attempt with short proofs, proving/disproving each of the statements.\nI am posting it as an answer, so as to not invalidate the hints and tips by @GregMartin.\n\n\n(a) Consider the sequence $$(x_n)$$ given by $$x_n = \\sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$0$$.\n\nAlso, consider the sequence $$(x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\\sqrt{n^2 + 2n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$-1$$.\n\nShort proof.\n\nConsider $$a_n = \\sqrt{n + 1} - \\sqrt{n}$$.\n\nObserve that,\n\n\\begin{align*} \\sqrt{n+1} - \\sqrt{n} &= (\\sqrt{n+1} - \\sqrt{n}) \\times \\frac{\\sqrt{n+1} + \\sqrt{n}}{\\sqrt{n+1} + \\sqrt{n}}\\\\ &=\\frac{1}{\\sqrt{n+1} + \\sqrt{n}} \\\\ &< \\frac{1}{\\sqrt{n} + \\sqrt{n}} = \\frac{2}{\\sqrt{n}} \\end{align*}\n\nPick $$\\epsilon > 0$$. We can choose $$N > \\frac{4}{\\epsilon^2}$$. To show that this choice of $$N$$ indeed works, we prove that, that for all $$n \\ge N$$,\n\n\\begin{align*} \\absval{\\sqrt{n+1} - \\sqrt{n}} &< \\frac{2}{\\sqrt{n}}\\\\ &< \\frac{2}{\\sqrt{(4/\\epsilon^2)}} = \\epsilon \\end{align*}\n\nThus, $$(\\sqrt{n+1} - \\sqrt{n}) \\to 0$$.\n\nConsider $$b_n = n - \\sqrt{n^2 + 2n}$$\n\nObserve that:\n\n\\begin{align*} n - \\sqrt{n^2 + 2n} - (-1) &= [(n+1) - \\sqrt{n^2 + 2n}] \\\\ &= [(n+1) - \\sqrt{n^2 + 2n}] \\times \\frac{(n+1) + \\sqrt{n^2 + 2n}}{(n+1) + \\sqrt{n^2 + 2n}}\\\\ &= \\frac{(n+1)^2 - (n^2 + 2n)}{(n+1) + \\sqrt{n^2 + 2n}}\\\\ &= \\frac{1}{(n+1) + \\sqrt{n^2 + 2n}}\\\\ &< \\frac{1}{n + \\sqrt{n^2}} = \\frac{1}{2n} \\end{align*}\n\nPick an arbitrary $$\\epsilon > 0$$. We choose an $$N > \\frac{1}{2\\epsilon}$$. To show that choice of $$N$$ indeed works, we find that:\n\n\\begin{align*} \\absval{n - \\sqrt{n^2 + 2n} - (-1)} &< \\frac{1}{2n} \\\\ &< \\frac{1}{2} \\cdot (2\\epsilon) = \\epsilon \\end{align*}\n\nThus, $$(n - \\sqrt{n^2 + 2n}) \\to -1$$.\n\n(b) This request is impossible. Alternatively, we believe that if $$(x_n)$$ converges and $$(x_n + y_n)$$ converges, then $$(y_n)$$ converges. Let us prove this fact rigorously. Suppose $$\\lim x_n = a$$ and $$\\lim x_n + y_n = b$$. We shall prove that $$\\lim y_n = b - a$$.\n\nObserve that,\n\n\\begin{align*} \\absval{y_n - (b-a)} &= \\absval{(x_n + y_n) - x_n - (b-a)}\\\\ &= \\absval{(x_n + y_n - b) - (x_n - a)}\\\\ &\\le \\absval{x_n + y_n - b} + \\absval{x_n - a} \\end{align*}\n\nPick an $$\\epsilon > 0$$. Since $$(x_n) \\to a$$, we can make the distance $$\\absval{x - a}$$ as small as we like. There exists an $$N_1$$ such that\n\n\\begin{align*} \\absval{x_n - a} < \\frac{\\epsilon}{2} \\end{align*}\n\nfor all $$n \\ge N_1$$.\n\nSince $$(x_n + y_n) \\to b$$, we can make the distance $$\\absval{x_n + y_n - b}$$ as small as we like. There exists an $$N_2$$ such that,\n\n\\begin{align*} \\absval{x_n + y_n - b} < \\frac{\\epsilon}{2} \\end{align*}\n\nLet $$N = \\max \\{N_1,N_2 \\}$$. To show that this $$N$$ indeed works, we find that:\n\n\\begin{align*} \\absval{y_n - (b-a)} &\\le \\absval{x_n + y_n - b} + \\absval{x_n -a}\\\\ &< \\frac{\\epsilon}{2} +\\frac{\\epsilon}{2} = \\epsilon \\end{align*}\n\n(c) Consider the sequence $$(b_n)$$ given by $$b_n = \\frac{1}{n}$$. Then, $$(b_n)$$ is a convergent sequence but $$1/b_n$$ is divergent.\n\nConsider $$(b_n) = \\frac{1}{n}$$. Pick an arbitrary $$\\epsilon > 0$$. We can choose $$N > \\frac{1}{\\epsilon}$$. To show that this choice of $$N$$ indeed works, we find that:\n\n\\begin{align*} \\absval{\\frac{1}{n}} < \\epsilon \\end{align*}\n\nfor all $$n \\ge N$$. Consequently, $$(1/n) \\to 0$$.\n\n(d) This request is impossible.\n\n(e) Consider the sequence $$(a_n)$$ given by $$a_n = \\frac{1}{n}$$ and $$(a_n b_n)$$ given by $$a_n b_n= \\frac{\\sin n}{n}$$. Thus, $$(a_n b_n)$$ and $$(a_n)$$ converges, but $$(b_n)$$ does not. Let us prove $$\\frac{\\sin n}{n}$$ converges to $$0$$.\n\nObserve that,\n\n\\begin{align*} \\absval{\\frac{\\sin n}{n}} &= \\frac{\\absval{\\sin n} }{\\absval n}\\\\ &\\le \\frac{1}{n} \\end{align*}\n\nLet $$\\epsilon > 0$$ be an arbitary small but fixed positive real number. We choose an $$N > \\frac{1}{\\epsilon}$$. To prove that this choice of $$N$$ indeed works, we have,\n\n\\begin{align*} \\absval{\\frac{\\sin n}{n}} &\\le\\frac{1}{n} \\\\ &< \\frac{1}{(1/\\epsilon)} = \\epsilon \\end{align*}\n\nConsequently, $$\\frac{\\sin n}{n} \\to 0$$.", "date": "2021-06-14 22:35:06", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3902158/proving-or-disproving-basic-facts-about-sequences-in-real-analysis", "openwebmath_score": 0.9983892440795898, "openwebmath_perplexity": 362.8878785207784, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9859363754361599, "lm_q2_score": 0.9314625141026972, "lm_q1q2_score": 0.9183627750090663}}
{"url": "http://ruthling.net/uzvrgd5v/diagonal-matrix-and-scalar-matrix-c4efad", "text": "Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title \"Prince of Wales\" originally claimed for the English crown prince via a trick? matrice scalaire, f Fizikos termin\u0173 \u017eodynas : lietuvi\u0173, angl\u0173, pranc\u016bz\u0173, vokie\u010di\u0173 ir rus\u0173 kalbomis. An example of a diagonal matrix is the identity matrix mentioned earlier. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. Matrix is an important topic in mathematics. 9. scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. Diagonal matrix and symmetric matrix From Norm to Orthogonality : Fundamental Mathematics for Machine Learning with Intuitive Examples Part 2/3 1-Norm, 2-Norm, Max Norm of Vectors b ij = 0, when i \u2260 j \u0441\u043a\u0430\u043b\u044f\u0440\u043d\u0430\u044f \u043c\u0430\u0442\u0440\u0438\u0446\u0430, f pranc. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Yes it is. Extract elements of matrix. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. \"Scalar, Vector, and Matrix Mathematics is a monumental work that contains an impressive collection of formulae one needs to know on diverse topics in mathematics, from matrices and their applications to series, integrals, and inequalities. Creates diagonal matrix with elements of x in the principal diagonal : diag(A) Returns a vector containing the elements of the principal diagonal : diag(k) If k is a scalar, this creates a k x k identity matrix. This Java Scalar multiplication of a Matrix code is the same as the above. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Minimum element in a matrix\u2026 Antonyms for scalar matrix. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . Define scalar matrix. See : Java program to check for Diagonal Matrix. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. MMAX(M). A symmetric matrix is a matrix where aij = aji. Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u2026 A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Diagonal elements, specified as a matrix. a diagonal matrix in which all of the diagonal elements are equal. Scalar Matrix : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. This matrix is typically (but not necessarily) full. For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1), diag treats the input as a matrix from which to extract a diagonal vector. add example. \u2014 Page 36, Deep Learning, 2016. Use these charts as a guide to what you can bench for a maximum of one rep. a matrix of type: Lower triangular matrix. What are synonyms for scalar matrix? Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 If you supply the argument that represents the order of the diagonal matrix, then it must be a real and scalar integer value. Synonyms for scalar matrix in Free Thesaurus. scalar matrix skaliarin\u0117 matrica statusas T sritis fizika atitikmenys : angl. This behavior occurs even if \u2026 Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (\u2198) entries are all equal. 8 (Roots are found analogously.) 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. In this post, we are going to discuss these points. The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \\text{ to } x_{nn}$$. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Yes it is, only the diagonal entries are going to change, if at all. All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. What is the matrix? stemming. A matrix with all entries zero is called a zero matrix. Magnet Matrix Calculator. InnerProducts. Diagonal matrix multiplication, assuming conformability, is commutative. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n \u00d7 n is said to be a scalar matrix if. Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox\u2122. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. General Description. Java Scalar Matrix Multiplication Program example 2. import java. The data type of a[1] is String. The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (\u2212) additions for computing the product of two square n\u00d7n matrices. Great code. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. scalar meson Look at other dictionaries: Matrix - \u043f\u043e\u043b\u0443\u0447\u0438\u0442\u044c \u043d\u0430 \u0410\u043a\u0430\u0434\u0435\u043c\u0438\u043a\u0435 \u0440\u0430\u0431\u043e\u0447\u0438\u0439 \u043a\u0443\u043f\u043e\u043d \u043d\u0430 \u0441\u043a\u0438\u0434\u043a\u0443 \u041b\u0435\u0442\u0443\u0430\u043b\u044c \u0438\u043b\u0438 \u0432\u044b\u0433\u043e\u0434\u043d\u043e 8. Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. Example 2 - STATING AND. A square matrix in which all the elements below the diagonal are zero i.e. Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. Filling diagonal to make the sum of every row, column and diagonal equal of 3\u00d73 matrix using c++ How to convert diagonal elements of a matrix in R into missing values? [x + 2 0 y \u2212 3 4 ] = [4 0 0 4 ] 2. Maximum element in a matrix. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. Negative: \u2212A is de\ufb01ned as (\u22121)A. Subtraction: A\u2212B is de\ufb01ned as A+(\u2212B). Takes a single argument. is a diagonal matrix with diagonal entries equal to the eigenvalues of A. 6) Scalar Matrix. Upper triangular matrix. scalar matrix vok. A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. The values of an identity matrix are known. Scalar multiplication: to multiply a matrix A by a scalar r, one multiplies each entry of A by r. Zero matrix O: all entries are zeros. Types of matrices \u2014 triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. 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If all the elements in its principal diagonal are equal matrix: diagonal matrix be a scalar matrix all! Java scalar multiplication: is a scalar matrix allow the user to enter number... In R by that matrix matrix remains the same multiplication: is a scalar matrix translation, dictionary. Angl\u0173, pranc\u016bz\u0173, vokie\u010di\u0173 ir rus\u0173 kalbomis Computing Toolbox\u2122 of an identity matrix is said be!, and the matrix items Java to input a 2-D square matrix which! Ir rus\u0173 kalbomis other matrix elements are equal matrix items below the diagonal are entries with 0 code... Number of rows, columns, and the other matrix elements are equal to some constant!", "date": "2021-12-03 12:58:17", "meta": {"domain": "ruthling.net", "url": "http://ruthling.net/uzvrgd5v/diagonal-matrix-and-scalar-matrix-c4efad", "openwebmath_score": 0.7514483332633972, "openwebmath_perplexity": 773.2012659838158, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9869795075882267, "lm_q2_score": 0.9304582612793112, "lm_q1q2_score": 0.9183432365488522}}
{"url": "https://www.jiskha.com/display.cgi?id=1285006413", "text": "posted by Dina\n\nSimplify the expression.\n\n3 [ (15 - 3)^2 / 4]\na. 36\nb. 108\nc. 18\nd. 9\n\n1. TutorCat\n\nhttp://www.jiskha.com/display.cgi?id=1285004701\n\n2. Dina\n\n???108\n\n3. PsyDAG\n\n3 (12^2/4) = 3 (144/4) = 3 * 36 = 108\n\nYes!\n\n## Similar Questions\n\n1. ### Math(radicals or square roots)\n\ni need a explanation for this equation: 5 radical(square root) 108 over 25 the 5 is in front of the square rooted number and it is a fraction PlEASE HELP :( That is not an equation. It is a number. It sounds like you are describing \u2026\n\ns8impliy this exponential expression): -2 7 -8 ) 6 -9)8 (7) * (x) * (w * (x *(4) * -3) (w\n\nHow do you factor this expression? 10x+15\n\nUse the distributive property to simplify: 25(2x + 1/5).\n\nsimplify x^5-x^3+x^2-1-(x^3-1)(x+1)^2/(x^2-1)^2\n6. ### Algebra\n\nTranslate the phrase to mathematical language. Then simplify the expression the difference between 119 and -40 What is a numerical expression for the phrase?\n7. ### math\n\nRiding on a school bus are 20 students in 9th grade, 10 in 10th grade, 9 in 11th grade, and 7 in 12th grade. Approximately what percentage of the students on the bus are in 9th grade?\n8. ### urgent math\n\n1. Simplify the expression 7^9/7^3 a.7^3*** b.7^6 c.7^12 d.1^6 2. Simplify the expression z^8/z^12 a.z^20 b.z^4 c.1/z^-4 d.1/z^4 3. Simplify the expression (-3^4)/(-3^4 a.(-3)^1 b.0 c.1 d.(-3)^8 4.Which expressions can be rewritten \u2026\n9. ### algebra\n\nam I right? 1. Simplify radical expression sqrt 50 5 sqrt ^2*** 2 sqrt ^5 5 sqrt ^10 5 2. Simplify the radical expression sqrt 56x^2 28x 2x sqrt 14*** 2x sqrt 7 sqrt 14x2 3. Simplify the radical expression. sqrt 490y^5w^6 2 sqrt 135y^2\n10. ### Algebra\n\n1. Simplify the expression 4 3 \u2014 - \u2014 x x 2. Simplify the expression 6 4 \u2014 + \u2014 c c\u00b2 3. Simplify the expression 3y 9y \u2014 \u00f7 \u2014 4y - 8 2y\u00b2 - 4y 4. Simplify the expression: ( 2x\u00b3 - x\u00b2 - 13x - 6 ) \u00f7 ( x -3 )\n\nMore Similar Questions", "date": "2018-05-22 10:08:59", "meta": {"domain": "jiskha.com", "url": "https://www.jiskha.com/display.cgi?id=1285006413", "openwebmath_score": 0.8123717308044434, "openwebmath_perplexity": 5641.519988287871, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.978712648206549, "lm_q2_score": 0.9381240077890602, "lm_q1q2_score": 0.9181538320093724}}
{"url": "https://math.stackexchange.com/questions/3363106/is-ln-natural-log-and-log-the-same-thing-if-used-in-this-answer", "text": "# Is \u201cln\u201d (natural log) and \u201clog\u201d the same thing if used in this answer?\n\nFind $$x$$ for $$4^{x-4} = 7$$.\n\nAnswer I got, using log, was $${\\log(7)\\over 2\\log(2)} + 4$$\n\nbut the actual answer was $${\\ln(7)\\over2\\ln(2)} + 4$$\n\nI plugged both in my calculator and turns out both are the equivalent value.\n\nAnyways, is using either one of ln or log appropriate for this question? Obviously ln is when log has the base e, and log is when it has the base 10.\n\nFinal question: How do I know when to use which? that is which of ln or log is used when solving a question??\n\nFor example, if a question asks to find $$x$$ for $$e^x = 100$$, I will use $$\\ln$$ since $$\\ln(e)$$ cancels out.\n\nIf a question asks to find $$2^x = 64$$, i will use log since \"$$e$$\" isn't present in the question.\n\nSo is using either $$\\log$$ or $$\\ln$$ the same?\n\n\u2022 $$\\log_{10}x=\\frac{\\ln x}{\\ln 10}$$ \u2013\u00a0Lord Shark the Unknown Sep 20 at 6:52\n\u2022 As an aside, to make matters worse, some authors will write $\\log$ without a subscript and mean different things than one another. In texts on combinatorics for instance it is not uncommon to see $\\log$ without a subscript be meant to be interpreted as being the base 2 logarithm $\\log_2$ while other authors might intend it to be the base 10 logarithm $\\log_{10}$. Others still may use $\\log$ as the natural logarithm rather than writing it as $\\ln$. The nice thing is, regardless which base it is you always have $\\log_n(a)/\\log_n(b)=\\log_b(a)$ \u2013\u00a0JMoravitz Sep 20 at 14:55\n\u2022 \"In order to kill an exponential, you have to hit it with a log\". Which raises the question \"which log\". The answer is -- it doesn't matter. \u2013\u00a0John Coleman Sep 20 at 16:11\n\u2022 You can always contrive that there are $e$'s around. Note $2^x=(e^{\\text{ln}(2)})^x = e^{x\\text{ln}2}$ \u2013\u00a0jacob1729 Sep 20 at 18:00\n\u2022 @JohnColeman: Just don't use base 1. math.stackexchange.com/questions/413713/log-base-1-of-1 \u2013\u00a0Joshua Sep 20 at 19:10\n\nYou can use any logarithm you want.\n\nAs a result of the base change formula $$\\log_2(7) = \\frac{\\log(7)}{\\log(2)} = \\frac{\\ln(7)}{\\ln(2)} = \\frac{\\log_b(7)}{\\log_b(2)}$$ so as long as both logs have the same base, their ratio will be the same, regardless of the chosen base (as long as $$b > 0, b\\neq 1$$).\n\n\u2022 Thank you. Out of curiosity, why would one prefer to use natural log in the question 4^(x-4) = 7, when it does not contain \"e\" \u2013\u00a0harold232 Sep 20 at 6:56\n\u2022 @harold232 Since they're equivalent, the choice is harmless, but you still have to make a choice. The only reason I can think of to default to the natural log is that it is in some ways computationally easier than other logarithms (coming from formulas of calculus). \u2013\u00a0Brian Moehring Sep 20 at 7:01\n\u2022 @harold232 for mathematicians, $e$ is the default base for logarithms which they would use unless there is a particular reason for choosing something else. (This is largely because it has nice properties for calculus.) After base $e$, the next most common in maths is base $2$. $10$, as Michael Palin might say, is right out. \u2013\u00a0Especially Lime Sep 20 at 7:05\n\u2022 $10$ was popular in the days before calculators. In those days, it was the easiest one to work with as we had tables for it. I agree that its utility in pure maths is low today but it lives in a few cases where log scales are used e.g. pH in chemistry and the decibels. \u2013\u00a0badjohn Sep 20 at 8:43\n\u2022 @EspeciallyLime: 10 is used by \"scientists\". See my answer below. \u2013\u00a0JonathanZ Sep 20 at 17:15\n\nEither is fine. You can write logarithms in terms of any base that you like with the change of base formula $$\\log_ba=\\frac{\\log_ca}{\\log_cb}$$\n\nOne thing learned from secondary school is that exponential equations can be solved be rewriting the relationship in logarithmic form. As such, your equation can be rewritten as follows: \\begin{align} 4^{x-4}&=7\\\\ x-4&=\\log_47\\\\ x&=4+\\log_47 \\end{align}\n\nAnd since $$\\log_47$$ can be rewritten as $$\\frac{\\log7}{\\log4}$$ or $$\\frac{\\ln7}{\\ln4}$$ or $$\\frac{\\log_{999876}7}{\\log_{999876}4}$$ it does not matter which base of logarithm you use.\n\n\u2022 By the way I noticed that you left the denominator as log(4), shouldn't it be 2log(2) by power rule, since it is more simplified. Or does it not matter? \u2013\u00a0harold232 Sep 20 at 7:00\n\u2022 @harold232 As you say, $\\log(4)=2\\log(2)$. So it does not really matter \u2013\u00a0Henry Sep 20 at 7:09\n\u2022 If you want to input the logarithm into a calculator that does not have a log function that accepts a base parameter, then you can simply type in $\\log\\ 7 \\div \\log\\ 4$. If you want to show your work or write a final solution as its exact value, I tend to accept $\\log_47$ or $\\frac{\\ln7}{2\\ln2}$ or its variants. \u2013\u00a0Andrew Chin Sep 20 at 7:10\n\u2022 No educator worth his salt is going to be that pedantic. To you, is $\\frac12\\log_27$ more acceptable compared to $\\log_47$? \u2013\u00a0Andrew Chin Sep 20 at 7:20\n\u2022 Wolfram says at best it as an alternative \u2013\u00a0Andrew Chin Sep 20 at 7:29\n\nCulturally\n\n\u2022 Computer science / programming people tend to use log base $$2$$\n\u2022 Mathematicians tend to use log base $$e$$\n\u2022 Engineers / physicist / chemists etc. tend to use log base $$10$$\n\nWriters really should make it explicit the first time they use \"$$\\log$$\", but they don't always. As others have pointed out, the only difference is a constant factor, and in your case the factors in the numerator and denominator cancelled each other out. So the answer to your question is \"If it's in a math context you'll probably see $$\\ln()$$ used.\"\n\nHeck, even if you were asked to solve $$10^{x - 3} = 6$$ you'd still see $$\\ln()$$ used, even though it looks like $$\\log_{10}$$ might seem more \"natural\" for that particular problem. It's just what math people tend to do.\n\n\u2022 It is called the \"natural logarithm\", after all. (: \u2013\u00a0Andrew Chin Sep 20 at 17:38\n\u2022 :) We've learned that what different groups of people consider to be \"natural\" can vary greatly, and even contradict each other. Hence my starting off with the word \"culturally\". \u2013\u00a0JonathanZ Sep 20 at 17:52\n\u2022 @JonathanZ you made me laugh out loud in a library! @@ It was a natural reaction of course. \u2013\u00a0uhoh Sep 21 at 5:43\n\nThere's an interesting unstated question here: what counts as an answer?\n\nYou can clearly argue that using either $$\\ln$$ or $$\\log_{10}$$ should be acceptable. But in that case $$x = \\log_4(7) + 4$$ should be just as correct. As @BrianMoehring says in his answer, you can use any logarithm you want.\n\nAs for\n\nIf a question asks to find $$2^x=64$$, i will use log since \"e\" isn't present in the question.\n\nI would just say $$x=6$$. That's really using $$\\log_2$$, by inspection.\n\nIn this case the other two answers are technically correct of course, in that the base doesn't really matter. But I want to point out that when you see $$\\log(x)$$, it can either mean base $$10$$, base $$2$$ or base $$e$$, with the latter two (especially base $$e$$) being much more common as you move up the math ladder. The notation $$\\ln(x)$$ is still used for base $$e$$, but whenever you see $$\\log(x)$$ you should always assume it is also base $$e$$ unless context implies otherwise (if it's supposed to mean base $$2$$, it should be clear from context).\n\nPart of the reason is exactly because of the reason mentioned by the two other answers: for any $$a,b$$ we have $$\\log_a(b)=\\frac{\\log(b)}{\\log(a)},$$ So we can express logarithms of any base using the natural logarithm anyway and there's no need to designate a special symbol for it. And indeed you will see that base $$e$$ is much more useful than base $$10$$ most of the time.\n\n\u2022 log is often considered log_2 in CS papers \u2013\u00a0RiaD Sep 20 at 16:32\n\nUse the change of base formula if you suspect that the answer is irrational, otherwise take the logarithm to both sides of the equation of a base that seems reasonable.\n\nThere's nothing magical about the change of base formula. \\begin{align} \\log_c b &= \\log_c b\\\\ \\log_c a^{\\log_a b} &= \\log_c b\\\\ \\log_a b \\cdot\\log_c a &= \\log_c b\\\\ \\log_a b &= \\frac{\\log_c b}{\\log_c a}\\\\ \\end{align}\n\nEven if the solution is integral or rational, using the change of base formula will get you to an answer, for example:\n\n\\begin{align} 2^x &= 64\\\\ \\log_{10}2^x &= \\log_{10} 64\\\\ x\\log_{10}2 &= \\log_{10} 64\\\\ x &= \\frac{\\log_{10} 64}{\\log_{10}2}\\\\ x &= \\log_{2} 64\\\\ x &= \\log_{2} 2^6\\\\ x &= 6\\log_{2} 2\\\\ x &= 6\\\\ \\end{align} although, it does insert a lot of extra [unnecessary] steps.", "date": "2019-10-22 18:42:10", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3363106/is-ln-natural-log-and-log-the-same-thing-if-used-in-this-answer", "openwebmath_score": 0.9673437476158142, "openwebmath_perplexity": 521.0709233760114, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668679067631, "lm_q2_score": 0.9353465138965863, "lm_q1q2_score": 0.9177310094474233}}
{"url": "https://math.stackexchange.com/questions/3231585/a-n-be-a-sequence-such-that-a-n12-2a-na-n1-a-n-0-then-sum-1/3231653", "text": "# $\\{a_n\\}$ be a sequence such that $a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\\sum_1^{\\infty}\\frac{a_n}{3^n}$ lies in\u2026\n\nLet $$\\{a_n\\}$$ be a sequence of positive real numbers such that\n\n$$a_1 =1,\\ \\ a_{n+1}^2-2a_na_{n+1}-a_n=0, \\ \\ \\forall n\\geq 1$$.\n\nThen the sum of the series $$\\sum_1^{\\infty}\\frac{a_n}{3^n}$$ lies in...\n\n(A) $$(1,2]$$, (B) $$(2,3]$$, (C) $$(3,4]$$, (D)$$(4,5]$$.\n\nSolution attempt:\n\nFirstly, we figure out what $$\\frac{a_{n+1}}{a_n}$$ is going to look like. We get, from the recursive formula, $$\\frac{a_{n+1}}{a_n}=1+\\sqrt{1+\\frac{1}{a_n^2}}$$ (remembering the fact that $$a_n>0$$, the other root is rejected).\n\nWe know that, if $$\\lim_{n \\to \\infty}\\frac{a_{n+1}}{a_n}>1$$, then $$\\lim a_n \\to \\infty$$.\n\nFurther, $$(a_{n+1}-a_n)= \\sqrt{a_n(a_n+1)}>0$$. (Again, the other root is rejected due to the same reason).\n\nHence, $$(a_n)$$ increases monotonically.\n\nTherefore, the largest value of $$\\frac{a_{n+1}}{a_n}$$ is approximately $$1+\\sqrt{1+\\frac{1}{1}} \\approx 2.15$$\n\nNow, the sum can be approximated as $$\\displaystyle\\frac{\\frac{1}{3}}{1-\\frac{2.15}{3}} \\approx 1.3$$ (In actuality, $$\\mathbb{sum}< 1.3$$).\n\nSo, option $$(A)$$ is the correct answer.\n\nIs the procedure correct?\n\nI have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.\n\nIs there any \"definitive\" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?\n\n\u2022 Out of curiosity, what's the source of this problem ? \u2013\u00a0Gabriel Romon May 19 '19 at 10:16\n\u2022 @GabrielRomon It was asked in an entrance exam for Master's degree admission in India this year (JAM 2019). \u2013\u00a0Subhasis Biswas May 19 '19 at 10:18\n\u2022 Hmm, your approach is indeed quite nice, but even if asymptotically $a_n\\sim\\alpha^n$ and the sum effectively $\\frac\\alpha{3-\\alpha}$, the first terms of the series may as well shift the result in another interval. How do you bound the partial sum $\\sum\\limits_{n=1}^{n_0} \\frac{a_n}{3^n}$ up the a certain $n_0$ so that subsequent terms are small enough and we can switch to asymptotic behaviour ? \u2013\u00a0zwim May 19 '19 at 10:48\n\u2022 This is the part where I used the monotone property. The common ratio can never exceed $2.15$, no matter what. Because, after the first term of the sequence, $1/a_{n}^2 <1$, Resulting in $a_{n+1}/a_n <2.15$ \u2013\u00a0Subhasis Biswas May 19 '19 at 10:50\n\u2022 If you recall the proof of the ratio test, then yes, you applied the right strategy. \u2013\u00a0rtybase May 19 '19 at 10:58\n\nFollowing your calculations and according to the ratio test $$0<\\frac{\\frac{a_{n+1}}{3^{n+1}}}{\\frac{a_n}{3^n}}=\\frac{1}{3}\\cdot \\frac{a_{n+1}}{a_n}<1$$ thus $$\\sum\\limits_{n=1}\\frac{a_n}{3^n}< \\infty$$ Now, applying the same technique from the proof on the ratio test $$S=\\frac{1}{3}+\\frac{a_2}{3^2}+\\frac{a_3}{3^3}+\\cdots+\\frac{a_n}{3^n}+\\cdots=\\\\ \\frac{1}{3}+\\frac{a_2}{a_1}\\cdot\\frac{a_1}{3^2}+\\frac{a_3}{a_2}\\cdot\\frac{a_2}{3^3}+\\cdots+\\frac{a_{n}}{a_{n-1}}\\cdot\\frac{a_{n-1}}{3^n}+\\cdots$$ or $$\\frac{1}{3}+2\\cdot\\frac{1}{3^2}+2\\cdot\\frac{a_2}{3^3}+\\cdots+2\\cdot\\frac{a_{n-1}}{3^n}+\\cdots< S<\\\\ \\frac{1}{3}+2.15\\cdot\\frac{1}{3^2}+2.15\\cdot\\frac{a_2}{3^3}+\\cdots+2.15\\cdot\\frac{a_{n-1}}{3^n}+\\cdots$$ and repeating this $$\\frac{1}{3}+\\frac{2}{3^2}+\\frac{2^2}{3^3}+\\cdots+\\frac{2^{n-1}}{3^n}+\\cdots< S<\\\\ \\frac{1}{3}+\\frac{2.15}{3^2}+\\frac{2.15^2}{3^3}+\\cdots+\\frac{2.15^{n-1}}{3^n}+\\cdots$$ or $$\\frac{1}{3}\\cdot\\left(1+\\frac{2}{3}+\\frac{2^2}{3^2}+\\cdots+\\frac{2^{n-1}}{3^{n-1}}+\\cdots\\right)< S<\\\\ \\frac{1}{3}\\cdot\\left(1+\\frac{2.15}{3}+\\frac{2.15^2}{3^2}+\\cdots+\\frac{2.15^{n-1}}{3^{n-1}}+\\cdots\\right)$$ or $$\\color{red}{1}=\\frac{\\frac{1}{3}}{1-\\frac{2}{3}}<\\color{red}{S}<\\frac{\\frac{1}{3}}{1-\\frac{2.15}{3}}=\\frac{1}{3-2.15}<\\color{red}{2}$$\n\nThis kind of squeezing technique is widely applied in analysis, functional analysis, numerical analysis. So, it makes sense to ask something similar for a master degree entrance test.\n\n\u2022 Truly a nice approach. The upper bound is basically the same. I missed the lower bound though :( I like your final note. \"So, it makes sense to ask something similar for a master degree entrance test\" \u2013\u00a0Subhasis Biswas May 19 '19 at 11:33\n\u2022 Nice and clean :) \u2013\u00a0A learner Apr 23 at 15:44\n\n$$a_{n+1}=a_n+\\sqrt{a_n^2+a_n}>2a_n\\forall n\\ge 1$$ $$a_1=2^0,a_2>2,a_3>2^2,...,a_n\\ge 2^{n-1}$$ $$1\\le a_n\\Rightarrow a_n+a_n^2\\le 2a_n^2\\Rightarrow a_{n+1}\\le (\\sqrt 2+1)a_n$$ $$a_1=(\\sqrt 2+1)^0,a_2=(\\sqrt 2+1),a_3<(\\sqrt 2+1)^2,...,a_n\\le (\\sqrt 2+1)^{n-1}$$ $$\\therefore 2^{n-1}\\le a_n \\le (\\sqrt 2+1)^{n-1}$$ $$\\Rightarrow \\frac 13(\\frac 23)^{n-1}\\le {a_n\\over 3^n}\\le \\frac 13({\\sqrt 2+1 \\over 3})^{n-1}$$ $$\\therefore \\frac 13\\sum (\\frac 23)^{n-1} < \\sum {a_n\\over 3^n}\\le \\frac 13 \\sum ({\\sqrt 2+1 \\over 3})^{n-1}$$ $$\\text{Hence}\\,\\,1<\\sum {a_n\\over 3^n}\\le {1\\over 2-\\sqrt 2}\\approx 1.707$$\n\nI would rather argue that $$\\sqrt{a_n(a_n+1)}$$ lies inside $$[a_n,a_n+1]$$, so that $$b_n\\leq a_n \\leq c_n$$ where $$b_n=2b_n$$ and $$c_n=2c_n+1$$ with $$b_1=c_1=1$$.\n\nClosed forms for $$b_n$$ and $$c_n$$ are easily derived as $$b_n=2^{n-1}$$ and $$c_n=2^n-1$$, so that $$1\\leq \\sum_1^{\\infty}\\frac{a_n}{3^n} \\leq 2-\\frac 12$$\n\nThis inequality can be refined by only summing from $$n$$ larger than some constant.\n\n\u2022 Now, for the last part of my question, is there any general approach for these type of problems? \u2013\u00a0Subhasis Biswas May 19 '19 at 11:07\n\u2022 A better one indeed! In fact, we don't need any better bound than this. Although my procedure gives off a slightly better value, it is not worthwhile here. \u2013\u00a0Subhasis Biswas May 19 '19 at 11:08\n\n(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)\n\nHints:\n\n\u2022 After calculating the first few terms, the ratio of the terms is very similar. If we estimate $$a_n \\approx a_1 r^{n-1}$$, this suggests we should solve for $$\\frac{ 1/3} { 1 - (r_1/3) } = 1$$ and $$\\frac{1/3} { 1 - (r_2/3)} = 2$$ to give us an idea of how to bound the sequence. This gives us $$r_1 = 2, r_2 = 5/2$$, so we want to show that $$2 a_n < a_{n+1} < \\frac{5}{2} a_n$$ (with some flexibility if this doesn't immediately work out).\n\u2022 Show that $$a_{n+1} = \\frac{ 2a_n + \\sqrt{ 4a_n^2 + 4a_n } } { 2}$$. In particular, reject the negative root.\n\u2022 Show that $$a_n \\geq 1$$.\n\u2022 Show that $$2 a_n < a_{n+1} < \\frac{5}{2} a_n$$.\n\u2022 Hence, show that $$1=\\frac{ 1/3 } { 1 - 2/3} < \\sum \\frac{a_n}{3^n} < \\frac{ 1 / 3 } { 1 - 5/6}=2$$\n\nNote:\n\n\u2022 The LHS is true by calculating the first 5 terms.\n\u2022 Of course, we can't prove the RHS just by calculating enough terms.\n\u2022 In fact, the bounding inequality $$2a_n < a_{n+1} < 2a_n + \\frac{1}{2}$$, so the ratio $$a_{n+1} / a_n$$ is very close to 2, esp at (slightly) larger values of $$n$$.\n\u2022 Not surprisingly, the value of the summation is much much closer to the 1. Using the first ~10 terms to get a better approximation, you can in fact show that the summation is between 1.2 and 1.25.", "date": "2021-06-14 12:38:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3231585/a-n-be-a-sequence-such-that-a-n12-2a-na-n1-a-n-0-then-sum-1/3231653", "openwebmath_score": 0.9380605816841125, "openwebmath_perplexity": 273.7343699614015, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712651931994, "lm_q2_score": 0.9372107900876219, "lm_q1q2_score": 0.9172600577859358}}
{"url": "http://math.stackexchange.com/questions/628115/for-which-primes-p-is-p2-2-also-prime", "text": "For which primes p is $p^2 + 2$ also prime?\n\nOrigin \u2014 Elementary Number Theory \u2014 Jones \u2014 p35 \u2014 Exercise 2.17 \u2014\n\nOnly for $p = 3$. If $p \\neq 3$ then $p = 3q \u00b1 1$ for some integer $q$, so $p^2 + 2 = 9q^2 \u00b1 6q + 3$ is divisible by $3$, and is therefore composite.\n\n(1) The key here looks like writing $p = 3q \u00b1 1$. Where does this hail from?\nI cognize $3q - 1, 3q, 3q + 1$ are consecutive.\n\n(2) How can you prefigure $p = 3$ is the only solution? On an exam, I can't calculate $p^2 + 2$ for many primes $p$ with a computer \u2014 or make random conjectures.\n\nE.g. any square is either $\\color{purple}0$ or $\\color{teal}1$ $\\begin{cases}\\mod 3, & \\text{ depending on whether or not$3$divides$x$} \\\\ \\mod 4, & \\text{depending on whether or not$2$divides the number being squared} \\end{cases}$,\nand $0,$ $1$, or $4$ mod $8$ (depending on whether or not $2$ or $4$ divide the number being squared).\n\nThus, when you see $p^2 + 2$, you should think: $\\begin{cases} = \\color{purple}0 + 2 & \\mod3 \\text{ , if$3$divides$p$} \\\\ = \\color{teal}1 + 2 \\equiv 0 & \\mod3 \\text{ , if$3 \\not| p$} \\end{cases}$.\n\n(3) Can someone please clarify why I'd prefigure or think about mod 3, mod 4, mod 8?\nWhy not consider mod of some random natural number?\n\n(4) The last paragraph considers mod 3. How can I prefigure this?\n\n-\n\nUse Fermat's little theorem.\n\nIf $\\gcd(p,3) = 1$, $p^2 \\equiv 1 \\pmod 3$ that gives $p^2 + 2\\equiv 3 \\pmod 3$.\n\nThus only possibility is $p = 3$\n\n-\nIt is very easy. Learn! Very useful. Regards. \u2013\u00a0 Dutta Jan 6 '14 at 5:32\nThat $p^2 \\equiv 1 \\bmod 3$ if $3$ does not divide $p$, certainly does not require Euler's totient theorem or Fermat's little theorem! Those are overkill for such an easy-to-check fact. \u2013\u00a0 ShreevatsaR Jan 6 '14 at 6:28\nThanks. How did you prefigure to start with $\\gcd(p, 3) = 1$? Why not $\\gcd(p,$random integer$) = 1$? \u2013\u00a0 Dwayne E. Pouiller Apr 8 '14 at 10:30\n\nHint $\\$ Apply the special case $\\,q=3\\,$ of the following\n\nTheorem $\\$ If $\\ p,\\,q\\$ and $\\,r = p^{q-1}\\!+q\\!-\\!1\\,$ are all prime then $\\, p = q$.\n\nProof $\\,$ If $\\,p\\ne q\\,$ then $\\,q\\nmid p\\,$ hence, by little Fermat, $\\,q\\mid \\color{#c00}{p^{q-1\\!}-1}\\,$ so $\\ \\color{#0a0}{q\\mid r}\\,=\\, \\color{#c00}{p^{q-1}\\!-1}+q$. However $\\,p,q \\ge 2\\,$ so $\\,p^{q-1}\\!\\ge 2\\,$ so $\\,r> q,\\,$ so $\\,\\color{#0a0}q\\,$ is a $\\color{#0a0}{proper}$ factor of $\\,r,\\,$ contra $\\,r\\,$ prime. $\\ \\$ QED\n\n-\n\nAny integer $n$ can be written as $3q\\pm1, 3q$ where $q$ is an integer\n\nNow we can immediately discard $3q$ as it is composite for $q>1$\n\nNow $\\displaystyle(3q\\pm1)^2+2=9q^2\\pm6q+3=3(3q^2\\pm2q+1)$\n\nObserve that $3q^2\\pm2q+1>1$ for $q\\ge1,$ hence $\\displaystyle(3q\\pm1)^2+2$ is composite\n\n-\n@oldrinb, that's what is written in the POST, right? \u2013\u00a0 lab bhattacharjee Jan 5 '14 at 18:39\nI misread -- didn't see the \",3q\" part \u2013\u00a0 oldrinb Jan 5 '14 at 18:40\nCan you please explain where $3q \\pm 1$ hails from? It feels uncanny. The rest of your answer isn't what I'm querying about. Can you please answer my edited post in your answer (not in comments)? \u2013\u00a0 Dwayne E. Pouiller Apr 8 '14 at 10:29\n@DwayneE.Pouiller, $$3q-1,3q,3q+1$$ are any three consecutive integers, right? \u2013\u00a0 lab bhattacharjee Apr 8 '14 at 10:34\n\nWhenever you see a quantity of the form $x^2 + a$ in a basic number theory course (especially in hw. or on an exam), you will want to think about what divisibilities it has by various small numbers.\n\nE.g. any square is either $\\color{purple}0$ or $\\color{teal}1$ $\\begin{cases}\\mod 3, & \\text{ depending on whether or not$3$divides$x$} \\\\ \\mod 4, & \\text{depending on whether or not$2$divides the number being squared} \\end{cases}$,\nand $0,$ $1$, or $4$ mod $8$ (depending on whether or not $2$ or $4$ divide the number being squared).\n\nThus, when you see $p^2 + 2$, you should think: $\\begin{cases} = \\color{purple}0 + 2 & \\mod3 \\text{ , if$3$divides$p$} \\\\ = \\color{teal}1 + 2 \\equiv 0 & \\mod3 \\text{ , if$3 \\not| p$} \\end{cases}$.\nSince the only prime that can be $0$ mod $3$ is $3$ (and $p^2 + 2$ will certainly be $> 3$), this answers your question immediately.\n\n-\nThanks. I'm sorry for unchecking the answer - I only cognized now I don't fully grasp it. Can you please answer my edited post in your answer (not in comments)? \u2013\u00a0 Dwayne E. Pouiller Apr 8 '14 at 10:28", "date": "2015-07-28 03:39:02", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/628115/for-which-primes-p-is-p2-2-also-prime", "openwebmath_score": 0.9085800051689148, "openwebmath_perplexity": 541.5428442211575, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978384668497878, "lm_q2_score": 0.9372107931567176, "lm_q1q2_score": 0.9169526711752685}}
{"url": "http://mathhelpforum.com/calculus/155498-integrate-x-3-sqrt-1-x-2-k-2-dx.html", "text": "1. ## Integrate [x^3/sqrt(1-x^2/k^2)]dx\n\nCan someone please check my process, also please can you advise on an easier way to write mathematical notation on this forum?\n\nIntegrate ( x^3/(1-x^2/k^2))dx\nwhere k is a constant.\nLet u = x^2, dv = x/sqrt(1-x^2/k^2)\n\nIntegration by parts.\n\nIntegral fx = uv - integral vdu\nv = -k^2*sqrt(1-x^2/k^2)\ndu = 2xdx\n\nsusbstituting for u,v,du\n\nIntegral fx = -k^2 *x^2*sqrt(1-x^2/k^2)+ integral [2xk^2*sqrt(1-x^2/k^2)dx]\nIntegral fx = -k^2 *x^2*sqrt(1-x^2/k^2) - 2/3*k^4*((1-x^2/k^2)^3/2)\n\n2. cleared the fraction in the radical to make life easier ...\n\n$\\displaystyle \\frac{k}{k} \\cdot \\frac{x^3}{\\sqrt{1 - \\frac{x^2}{k^2}}}$\n\n$\\displaystyle \\frac{kx^3}{\\sqrt{k^2\\left(1 - \\frac{x^2}{k^2}\\right)}}\n$\n\n$\\displaystyle -k \\int x^2 \\cdot \\frac{-x}{\\sqrt{k^2-x^2}} \\, dx$\n\n$u = x^2$\n\n$du = 2x \\, dx$\n\n$dv = \\frac{-x}{\\sqrt{k^2-x^2}} \\, dx$\n\n$v = \\sqrt{k^2-x^2}$\n\n$\\displaystyle -k \\int x^2 \\cdot \\frac{-x}{\\sqrt{k^2-x^2}} \\, dx = -k\\left[x^2\\sqrt{k^2-x^2} - \\int 2x\\sqrt{k^2-x^2} \\, dx\\right]$\n\n$\\displaystyle -k \\int x^2 \\cdot \\frac{-x}{\\sqrt{k^2-x^2}} \\, dx = -k\\left[x^2\\sqrt{k^2-x^2} + \\int -2x\\sqrt{k^2-x^2} \\, dx\\right]$\n\n$\\displaystyle -k \\int x^2 \\cdot \\frac{-x}{\\sqrt{k^2-x^2}} \\, dx = -k\\left[x^2\\sqrt{k^2-x^2} + \\frac{2}{3}(k^2-x^2)^{\\frac{3}{2}} + C\\right]$\n\n3. ## thanks\n\nOriginally Posted by skeeter\ncleared the fraction in the radical to make life easier ...\n\n$\\displaystyle \\frac{k}{k} \\cdot \\frac{x^3}{\\sqrt{1 - \\frac{x^2}{k^2}}}$\n\n$\\displaystyle \\frac{kx^3}{\\sqrt{k^2\\left(1 - \\frac{x^2}{k^2}\\right)}}\n$\n\n$\\displaystyle -k \\int x^2 \\cdot \\frac{-x}{\\sqrt{k^2-x^2}} \\, dx$\n\n$u = x^2$\n\n$du = 2x \\, dx$\n\n$dv = \\frac{-x}{\\sqrt{k^2-x^2}} \\, dx$\n\n$v = \\sqrt{k^2-x^2}$\n\n$\\displaystyle -k \\int x^2 \\cdot \\frac{-x}{\\sqrt{k^2-x^2}} \\, dx = -k\\left[x^2\\sqrt{k^2-x^2} - \\int 2x\\sqrt{k^2-x^2} \\, dx\\right]$\n\n$\\displaystyle -k \\int x^2 \\cdot \\frac{-x}{\\sqrt{k^2-x^2}} \\, dx = -k\\left[x^2\\sqrt{k^2-x^2} + \\int -2x\\sqrt{k^2-x^2} \\, dx\\right]$\n\n$\\displaystyle -k \\int x^2 \\cdot \\frac{-x}{\\sqrt{k^2-x^2}} \\, dx = -k\\left[x^2\\sqrt{k^2-x^2} + \\frac{2}{3}(k^2-x^2)^{\\frac{3}{2}} + C\\right]$\n\n4. ## Without 'by-parts'\n\nRewrite the integral as following:\n\n$\\displaystyle \\int\\frac{x^3}{\\sqrt{1-\\frac{x}{k^2}}}\\;{dx}$ $\\displaystyle = \\int\\frac{x^3}{\\sqrt{\\frac{1}{k^2}(k^2-x^2)}}}\\;{dx} =$ $\\displaystyle = \\int\\frac{x^3}{\\frac{1}{k}\\sqrt{k^2-x^2}}}\\;{dx} =$ $\\displaystyle = \\int\\frac{kx^3}{\\sqrt{k^2-x^2}}}\\;{dx}$\n\nLet $t = \\sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$:\n\n$\\displaystyle \\frac{dt}{dx} = \\left(\\sqrt{k^2-x^2\\right)'$ $\\displaystyle= \\frac{\\left(k^2-x^2\\right)'}{2\\sqrt{k^2-x^2}}$ $\\displaystyle= \\frac{-2x}{2\\sqrt{k^2-x^2}}$ $\\displaystyle= \\frac{-x}{\\sqrt{k^2-x^2}}$\n\nSolving this for $dx$, as our purpose was, we have:\n\n$\\displaystyle dx = -\\left(\\frac{\\sqrt{k^2-x^2}}{x}\\right){dt}$\n\nGoing back to our original integral and putting that in for $dx$ we get:\n\n$\\displaystyle \\int\\frac{kx^3}{\\sqrt{k^2-x^2}}}\\;{dx}$ $\\displaystyle = -\\int\\left(\\frac{kx^3}{\\sqrt{k^2-x^2}}}\\right)\\;\\left(\\frac{\\sqrt{k^2-x^2}}{x}\\right){dt}}$ $\\displaystyle = -\\int kx^2 \\;{dt}$\n\nFinding $x^2$ in terms of $t$ from the relation of our substitution gives us:\n\n$t = \\sqrt{k^2-x^2} \\Rightarrow t^2 = k^2-x^2 \\Rightarrow t^2-k^2 = -x^2 \\Rightarrow k^2-t^2 = x^2$.\n\nPutting that in for $x^2$, we get a simple function in the integrand that is easy to integrate:\n\n$\\displaystyle -\\int kx^2 \\;{dt}$ $\\displaystyle = -\\int k(k^2-t^2) \\;{dt}$ $\\displaystyle = -k\\left(k^2t-\\frac{t^3}{3}\\right)+C = -\\frac{k}{3}\\left(3k^2t-t^3\\right)+C$\n\nSubstituting back for what we have let $t$ to be, we finally get:\n\n$\\displaystyle \\int\\frac{x^3}{\\sqrt{1-\\frac{x}{k^2}}}\\;{dx} = -\\frac{k}{3}\\left[3k^2\\sqrt{k^2-x^2}-(k^2-x^2)^{\\frac{3}{2}}\\right]+C$.\n\n5. By substitution method.\n\nGiven integration can be written as\n\n$\\int{\\frac{kx^3}{\\sqrt{k^2 -x^2}}}\\;{dx}$\n\nLet x = ksin\u03b8. dx = kcos\u03b8d\u03b8. k^2 - x^2 = k^2cos^2\u03b8. Substitute these values in the integration and simplify.\n\n.\n\n6. ## thanks\n\nOriginally Posted by TheCoffeeMachine\nRewrite the integral as following:\n\n$\\displaystyle \\int\\frac{x^3}{\\sqrt{1-\\frac{x}{k^2}}}\\;{dx}$ $\\displaystyle = \\int\\frac{x^3}{\\sqrt{\\frac{1}{k^2}(k^2-x^2)}}}\\;{dx} =$ $\\displaystyle = \\int\\frac{x^3}{\\frac{1}{k}\\sqrt{k^2-x^2}}}\\;{dx} =$ $\\displaystyle = \\int\\frac{kx^3}{\\sqrt{k^2-x^2}}}\\;{dx}$\n\nLet $t = \\sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$:\n\n$\\displaystyle \\frac{dt}{dx} = \\left(\\sqrt{k^2-x^2\\right)'$ $\\displaystyle= \\frac{\\left(k^2-x^2\\right)'}{2\\sqrt{k^2-x^2}}$ $\\displaystyle= \\frac{-2x}{2\\sqrt{k^2-x^2}}$ $\\displaystyle= \\frac{-x}{\\sqrt{k^2-x^2}}$\n\nSolving this for $dx$, as our purpose was, we have:\n\n$\\displaystyle dx = -\\left(\\frac{\\sqrt{k^2-x^2}}{x}\\right){dt}$\n\nGoing back to our original integral and putting that in for $dx$ we get:\n\n$\\displaystyle \\int\\frac{kx^3}{\\sqrt{k^2-x^2}}}\\;{dx}$ $\\displaystyle = -\\int\\left(\\frac{kx^3}{\\sqrt{k^2-x^2}}}\\right)\\;\\left(\\frac{\\sqrt{k^2-x^2}}{x}\\right){dt}}$ $\\displaystyle = -\\int kx^2 \\;{dt}$\n\nFinding $x^2$ in terms of $t$ from the relation of our substitution gives us:\n\n$t = \\sqrt{k^2-x^2} \\Rightarrow t^2 = k^2-x^2 \\Rightarrow t^2-k^2 = -x^2 \\Rightarrow k^2-t^2 = x^2$.\n\nPutting that in for $x^2$, we get a simple function in the integrand that is easy to integrate:\n\n$\\displaystyle -\\int kx^2 \\;{dt}$ $\\displaystyle = -\\int k(k^2-t^2) \\;{dt}$ $\\displaystyle = -k\\left(k^2t-\\frac{t^3}{3}\\right)+C = -\\frac{k}{3}\\left(3k^2t-t^3\\right)+C$\n\nSubstituting back for what we have let $t$ to be, we finally get:\n\n$\\displaystyle \\int\\frac{x^3}{\\sqrt{1-\\frac{x}{k^2}}}\\;{dx} = -\\frac{k}{3}\\left[3k^2\\sqrt{k^2-x^2}-(k^2-x^2)^{\\frac{3}{2}}\\right]+C$.", "date": "2016-10-23 03:46:55", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/155498-integrate-x-3-sqrt-1-x-2-k-2-dx.html", "openwebmath_score": 0.9753209948539734, "openwebmath_perplexity": 813.649821932982, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130563978902, "lm_q2_score": 0.9263037257164177, "lm_q1q2_score": 0.9161264789235473}}
{"url": "https://gmatclub.com/forum/how-many-pounds-of-fertilizer-that-is-10-percent-nitrogen-must-be-adde-252536.html", "text": "It is currently 20 Apr 2018, 13:06\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# How many pounds of fertilizer that is 10 percent nitrogen must be adde\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 44588\nHow many pounds of fertilizer that is 10 percent nitrogen must be adde\u00a0[#permalink]\n\n### Show Tags\n\n31 Oct 2017, 00:05\nExpert's post\n2\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n77% (01:27) correct 23% (01:11) wrong based on 105 sessions\n\n### HideShow timer Statistics\n\nHow many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?\n\n(A) 3\n(B) 6\n(C) 12\n(D) 24\n(E) 48\n[Reveal] Spoiler: OA\n\n_________________\nBSchool Forum Moderator\nJoined: 26 Feb 2016\nPosts: 2427\nLocation: India\nGPA: 3.12\nRe: How many pounds of fertilizer that is 10 percent nitrogen must be adde\u00a0[#permalink]\n\n### Show Tags\n\n31 Oct 2017, 01:43\n1\nKUDOS\nBunuel wrote:\nHow many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?\n\n(A) 3\n(B) 6\n(C) 12\n(D) 24\n(E) 48\n\n10-------------20\n\n--------18-------\n\n2---------------8\n\nHence the ratios in which the mixtures are to be added are 1:4\n\nSince we are adding 12 pounds of the 20 percent nitrogen mixture,\nwe will need 3 pounds of the 10 percent nitrogen mixture, so that\nthe resulting mixture has 18 percent nitrogen.\n\nTherefore, we need 3 pounds of fertilizer mixture having 10 percent nitrogen(Option A)\n_________________\n\nStay hungry, Stay foolish\n\nIntern\nJoined: 30 Nov 2016\nPosts: 37\nLocation: India\nConcentration: Finance, Strategy\nHow many pounds of fertilizer that is 10 percent nitrogen must be adde\u00a0[#permalink]\n\n### Show Tags\n\n31 Oct 2017, 04:16\n20% nitrogen in 12 pounds of fertilizer is 12/5.\nLet us consider the new amount of 10% nitrogen to be x. so the total amount of nitrogen content will be 12/5 + x/10. And the total amount of fertilizer will be 12+x.\n$$\\frac{(2.4+ 0.x)}{(12+x)}$$ =.18\nX=3\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 8026\nLocation: Pune, India\nRe: How many pounds of fertilizer that is 10 percent nitrogen must be adde\u00a0[#permalink]\n\n### Show Tags\n\n31 Oct 2017, 04:44\n2\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nBunuel wrote:\nHow many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?\n\n(A) 3\n(B) 6\n(C) 12\n(D) 24\n(E) 48\n\nUsing weighted averages:\n\nWe need to mix 10% nitrogen fertiliser with 20% to get 18% nitrogen.\n\nw1/w2 = (A2 - Aavg)/(Aavg - A1) = (20 - 18)/(18 - 10) = 1/4\n\nFor every 1 part of 10% nitrogen, we need 4 parts of 20% nitrogen.\nSince the amount of 20% nitrogen fertiliser is 12 pounds, we need 3 pounds of 10% nitrogen fertiliser.\n\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for \\$199\n\nVeritas Prep Reviews\n\nTarget Test Prep Representative\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 2447\nLocation: United States (CA)\nRe: How many pounds of fertilizer that is 10 percent nitrogen must be adde\u00a0[#permalink]\n\n### Show Tags\n\n01 Nov 2017, 17:05\n3\nKUDOS\nExpert's post\n1\nThis post was\nBOOKMARKED\nBunuel wrote:\nHow many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen?\n\n(A) 3\n(B) 6\n(C) 12\n(D) 24\n(E) 48\n\nWe add x pounds of fertilizer that is 10% nitrogen to 12 pounds of fertilizer that is 20% nitrogen, and the result is (x + 12) pounds of fertilizer that is 18% nitrogen. We can express this in the following equation:\n\n0.1x + 0.2(12) = 0.18(x + 12)\n\n10x + 20(12) = 18(x + 12)\n\n10x + 240 = 18x + 216\n\n24 = 8x\n\nx = 3\n\n_________________\n\nScott Woodbury-Stewart\nFounder and CEO\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nRe: How many pounds of fertilizer that is 10 percent nitrogen must be adde \u00a0 [#permalink] 01 Nov 2017, 17:05\nDisplay posts from previous: Sort by", "date": "2018-04-20 20:06:15", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/how-many-pounds-of-fertilizer-that-is-10-percent-nitrogen-must-be-adde-252536.html", "openwebmath_score": 0.6506108641624451, "openwebmath_perplexity": 6267.044025102947, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.9161096204605946, "lm_q1q2_score": 0.9161096204605946}}
{"url": "https://brilliant.org/discussions/thread/a-most-curious-algebraic-identity/", "text": "# A Most Curious Algebraic Identity\n\nI recently found a very interesting Algebraic Identity: $xyz+(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)$\n\nWhat's so special about it? Note that going from one side of the equality to the other, all products are switched with sums, and all sums are switched with products!\n\nThis may be seen a bit easier if I rewrite it as follows: \\begin{aligned} &{~}\\color{#D61F06}x \\color{#D61F06}{\\times} y\\color{#D61F06}{\\times }z \\color{#FFFFFF}{)}\\color{#3D99F6}{+} \\color{grey}{(}x\\color{#3D99F6}{+}y\\color{grey}{)}\\color{#D61F06}{\\times}\\color{grey}{(}y\\color{#3D99F6}{+}z\\color{grey}{)}\\color{#D61F06}{\\times} \\color{grey}{(}z\\color{#3D99F6}{+}x\\color{grey}{)}\\\\ =&\\color{grey}{(}x\\color{#3D99F6}{+}y\\color{#3D99F6}{+}z\\color{grey}{)}\\color{#D61F06}{\\times}\\color{grey}{(}x\\color{#D61F06}{\\times} y\\hspace{0.9ex}\\color{#3D99F6}{+}\\hspace{0.9ex}y\\color{#D61F06}{\\times }z\\color{#FFFFFF}{)}\\color{#3D99F6}{+}\\hspace{0.9ex}z\\color{#D61F06}{\\times }x\\color{grey}{)} \\end{aligned}\n\nCool!\n\nNote by Daniel Liu\n5\u00a0years, 10\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nYes; I really like this identity too. For example it can be used to prove that $r_1 + r_2 + r_3 - r = 4R$ (from Incircles and Excircles). If we substitute $x = s-a$ etc., then\n\n$s(s-b)(s-c)+s(s-c)(s-a)+s(s-a)(s-b)-(s-a)(s-b)(s-c) = abc$\n\nwhere $s$ is the semi-perimeter, and this reduces nicely using area formulas to the desired relationship.\n\n- 5\u00a0years, 10\u00a0months ago\n\n@Michael Ng created this problem which uses the identity.\n\nStaff - 5\u00a0years, 10\u00a0months ago\n\nthanks @Daniel Liu i used this to solve problems like this i wrote a solution using this identity, and i'm thinking about a problem with this identity, will post soon!\n\n- 5\u00a0years, 9\u00a0months ago\n\nI didn't realise that- thanks. It should inspire some good problems :)\n\n- 5\u00a0years, 10\u00a0months ago\n\nF B U L O U S !!!!!!\n\n- 5\u00a0years, 2\u00a0months ago\n\nTypo. $\\color{#D61F06}{x}$ is missed in color version, line 5.\n\n- 2\u00a0years, 5\u00a0months ago\n\nI didn't examine. Good for making questions. However, it could have been found by people in the past.\n\n- 5\u00a0years, 10\u00a0months ago\n\nYea, I'm just saying that I just noticed it. I most likely was not the person who discovered it (as seen by the comment by Michael Ng)\n\n- 5\u00a0years, 10\u00a0months ago\n\nI had just expanded both sides to compare. Should be correct. Do not feel disappointed by what I guessed. You could be the first person to find this. Congratulation!\n\n- 5\u00a0years, 10\u00a0months ago\n\nThanks. I did not realize this very useful identity.\n\n- 5\u00a0years, 10\u00a0months ago\n\nMost awesome discoveries ever!Thanks,this must help a lot.\n\n- 5\u00a0years, 9\u00a0months ago\n\nIt's following the rules of principle of duality\n\n- 5\u00a0years, 2\u00a0months ago\n\nNo, that is not the principle of duality.\n\nStaff - 5\u00a0years, 2\u00a0months ago\n\nI mean l'll bit similar to that\n\n- 5\u00a0years, 2\u00a0months ago", "date": "2020-11-30 18:10:19", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/a-most-curious-algebraic-identity/", "openwebmath_score": 0.9915642142295837, "openwebmath_perplexity": 3413.718170738004, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9728307716151472, "lm_q2_score": 0.9416541634817873, "lm_q1q2_score": 0.9160701464546032}}
{"url": "http://noiwatdan.com/black-colour-haawwfu/0556c0-decomposition-of-antisymmetric-tensor", "text": "1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. Decomposition of tensor power of symmetric square. (1.5) Usually the conditions for \u00b5 (in Eq. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. The N-way Toolbox, Tensor Toolbox, \u00e2\u0080\u00a6 A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components \u00e2\u0080\u00a6. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. Symmetric tensors occur widely in engineering, physics and mathematics. CHAPTER 1. In section 3 a decomposition of tensor spaces into irreducible components is introduced. For N>2, they are not, however. Since the tensor is symmetric, any contraction is the same so we only get constraints from one contraction. This chapter provides a summary of formulae for the decomposition of a Cartesian second rank tensor into its isotropic, antisymmetric and symmetric traceless parts. 1.4) or \u00ce\u00b1 (in Eq. If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? This makes many vector identities easy to prove. Yes. : USDOE \u00e2\u0080\u00a6 The trace decomposition theory of tensor spaces, based on duality, is presented. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? Physics 218 Antisymmetric matrices and the pfa\u00ef\u00ac\u0083an Winter 2015 1. This decomposition, ... ^2 indicates the antisymmetric tensor product. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. Properties of antisymmetric matrices Let Mbe a complex d\u00d7 dantisymmetric matrix, i.e. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. LetT be a second-order tensor. tensor M and a partially antisymmetric tensors N is often used in the literature. If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. THE INDEX NOTATION \u00ce\u00bd, are chosen arbitrarily.The could equally well have been called \u00ce\u00b1 and \u00ce\u00b2: v\u00e2\u0080\u00b2 \u00ce\u00b1 = n \u00e2\u0088\u0091 \u00ce\u00b2=1 A\u00ce\u00b1\u00ce\u00b2 v\u00ce\u00b2 (\u00e2\u0088\u0080\u00ce\u00b1 \u00e2\u0088\u0088 N | 1 \u00e2\u0089\u00a4 \u00ce\u00b1 \u00e2\u0089\u00a4 n). An alternating form \u00cf\u0086 on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. Cartan tensor is equal to minus the structure coe\u00ef\u00ac\u0083cients. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. (antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. OSTI.GOV Journal Article: DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. [3] Alternating forms. A related concept is that of the antisymmetric tensor or alternating form. There are many other tensor decompositions, including INDSCAL, PARAFAC2, CANDELINC, DEDICOM, and PARATUCK2 as well as nonnegative vari-ants of all of the above. The trace decomposition equations for tensors, symmetric in some sets of superscripts, and antisymmetric \u00e2\u0080\u00a6 Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share \u00e2\u0080\u00a6 Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric \u00e2\u0080\u00a6 A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Google Scholar; 6. A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. We begin with a special case of the definition. The alternating tensor can be used to write down the vector equation z = x \u00d7 y in su\u00ef\u00ac\u0083x notation: z i = [x\u00d7y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 \u00e2\u0088\u0092x 3y 2, as required.) Each part can reveal information that might not be easily obtained from the original tensor. These relations may be shown either directly, using the explicit form of f \u00ce\u00b1\u00ce\u00b2, and f \u00ce\u00b1\u00ce\u00b2 * or as consequences of the Hamilton\u00e2\u0080\u0090Cayley equation for antisymmetric matrices f \u00ce\u00b1\u00ce\u00b2 and f \u00ce\u00b1\u00ce\u00b2 *; see, e.g., J. Pleba\u00c5\u0084ski, Bull Acad. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. This means that traceless antisymmetric mixed tensor $\\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. The symmetry-based decompositions of finite games are investigated. In these notes, the rank of Mwill be denoted by 2n. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Viewed 503 times 7. What's the significance of this further decomposition? Full Record; Other Related Research; Authors: Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org. Thus, the rank of Mmust be even. A tensor is a linear vector valued function defined on the set of all vectors . Active 1 year, 11 months ago. Sponsoring Org. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Decomposition. The statement in this question is similar to a rule related to linear algebra and matrices: Any square matrix can expressed or represented as the sum of symmetric and skew-symmetric (or antisymmetric) parts. Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . MT = \u00e2\u0088\u0092M. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. Sci. Cl. It is a real tensor, hence f \u00ce\u00b1\u00ce\u00b2 * is also real. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g The result of the contraction is a tensor of rank r 2 so we get as many components to substract as there are components in a tensor of rank r 2. Antisymmetric and symmetric tensors. Contents. Polon. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 A\u00c4\u00b1 ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. Algebra is great fun - you get to solve puzzles! The result is Antisymmetric and symmetric tensors. This is exactly what you have done in the second line of your equation. Since det M= det (\u00e2\u0088\u0092MT) = det (\u00e2\u0088\u0092M) = (\u00e2\u0088\u00921)d det M, (1) it follows that det M= 0 if dis odd. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. There is one very important property of ijk: ijk klm = \u00ce\u00b4 il\u00ce\u00b4 jm \u00e2\u0088\u0092\u00ce\u00b4 im\u00ce\u00b4 jl. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. An alternative, less well-known decomposition, into the completely symmetric part Sof C plus the reminder A, turns out to be irreducibleunder the 3-dimensional general linear group. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: This is an example of the Youla decomposition of a complex square matrix. 1.5) are not explicitly stated because they are obvious from the context. The trace of the tensor S is the rate of (relative volume) expansion of the fluid. : Lehigh Univ., Bethlehem, Penna. For more comprehensive overviews on tensor calculus we recom-mend [58, 99, 126, 197, 205, 319, 343]. Vector spaces will be denoted using blackboard fonts. Ask Question Asked 2 years, 2 months ago. In 3 dimensions, an antisymmetric tensor is dual to a vector, but in 4 dimensions, that is not so. \u00e2\u0086\u0092 What symmetry does represent?Kenta OONOIntroduction to Tensors The bases of the symmetric subspace and those of its orthogonal complement are presented. Use the Weyl decomposition \\eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \\eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). P i A ii D0/. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0. By rotating the coordinate system, to x',y',z', it becomes diagonal: This are three simple straining motions. , they are not explicitly stated because they are not explicitly stated they! And an orthogonal complement are presented are the symmetric subspace the symmetric and. Lecture notes on vector and tensor Algebra and Analysis IlyaL tensor Fields in Continuum Mechanics, 1974 11... Unique, irreducible, and preserves the symmetries of the LORENTZ TRANSFORMATION matrix into SKEW-SYMMETRIC tensors,. Article: decomposition of tensor spaces into irreducible components is introduced structure coe\u00ef\u00ac\u0083cients 99, 126, 197 205! We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the tensor is equal minus. Lecture notes on vector and tensor Algebra and Analysis IlyaL are equivalent Heritage Encyclopedia: Publication Date Sun. That might not be easily obtained from the context tensor calculus we [. Since the tensor is equal to minus the structure coe\u00ef\u00ac\u0083cients a Related concept that!, 319, 343 ] bases of the elasticity tensor the Youla decomposition of fluid...: USDOE \u00e2\u0080\u00a6 antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Date. Analysis IlyaL matrix into SKEW-SYMMETRIC tensors N is often used in the second line of your equation matrix into tensors. It is possible to prove by a direct calculation that its Riemann vanishes! Of a complex d\u00d7 dantisymmetric matrix, i.e the second line of your.!, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org spaces irreducible... It is possible to prove by a direct calculation that its Riemann tensor.... Easily obtained from the original tensor Article: decomposition of the antisymmetric tensor product: Publisher: Heritage. We recom-mend [ 58, 99, 126, 197, 205, 319, ]! In these notes, the vector space of finite games is decomposed into a symmetric subspace symmetric, contraction! Is dual to a vector, but in 4 dimensions, that is not so so we only constraints. Since the tensor corresponds to the ( symmetric ) spin-1 part representation reducible. Indicates the antisymmetric tensor or alternating form 1.5 ) are not explicitly stated they... Or alternating form klm = \u00ce\u00b4 il\u00ce\u00b4 jm \u00e2\u0088\u0092\u00ce\u00b4 im\u00ce\u00b4 jl constraints from one contraction, 2 months.! Methods are developed by using the results of existing theories in the literature Usually the conditions for (... Calculus we recom-mend [ 58, 99, 126, 197, 205, 319, ]. Of ijk: ijk klm = \u00ce\u00b4 il\u00ce\u00b4 jm \u00e2\u0088\u0092\u00ce\u00b4 im\u00ce\u00b4 jl for more comprehensive overviews on calculus.... ^2 indicates the antisymmetric tensor product is introduced a direct calculation that its tensor. 00:00:00 EDT 1965 Research Org unique, irreducible, and preserves the symmetries the... From the context property of ijk: ijk klm = \u00ce\u00b4 il\u00ce\u00b4 jm im\u00ce\u00b4. Tensors in distinctive parts can help in analyzing them osti.gov Journal Article: decomposition of in... ; Other Related Research ; Authors: decomposition of antisymmetric tensor, S L Publication Date: antisymmetric matrix tensor corresponds the! Help in analyzing them structure coe\u00ef\u00ac\u0083cients on tensor calculus we recom-mend [ 58, 99, 126,,... Corresponding to upper and lower indices are equivalent very important property of ijk: ijk klm = il\u00ce\u00b4. Unique, irreducible, and preserves the symmetries of the LORENTZ TRANSFORMATION matrix into SKEW-SYMMETRIC tensors constraints! Calculation that its Riemann tensor vanishes SU ( 2 ) the representations corresponding upper! ( 2 ) the representations corresponding to upper and lower indices are.. The representations corresponding to upper and lower indices are equivalent and those of its orthogonal complement of the subspace! Rate of ( relative volume ) expansion of the elasticity tensor with a special of.", "date": "2022-08-10 11:49:20", "meta": {"domain": "noiwatdan.com", "url": "http://noiwatdan.com/black-colour-haawwfu/0556c0-decomposition-of-antisymmetric-tensor", "openwebmath_score": 0.9022778868675232, "openwebmath_perplexity": 1171.1674094407806, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9895109093587028, "lm_q2_score": 0.9252299519377654, "lm_q1q2_score": 0.9155251311078471}}
{"url": "https://brilliant.org/discussions/thread/yet-another-proof-for-zeta-2quad-quad-frac-pi-2-6/", "text": "# Yet another proof for $\\zeta (2) = \\frac { { \\pi }^{ 2 } }{ 6 }$\n\nLet us consider the integral\n\n$\\displaystyle i\\int _{ 0 }^{ \\pi }{ ln(1-{ e }^{ i\\theta } } )d\\theta$\n\nNow using\n\n$\\displaystyle i\\int _{ 0 }^{ \\pi }{ ln(1-{ e }^{ i\\theta } } )d\\theta \\\\ \\\\ 1-{ e }^{ i\\theta } = 1-cos\\theta -isin\\theta = 2sin(\\frac { \\theta }{ 2 } )(sin(\\frac { \\theta }{ 2 } )-icos(\\frac { \\theta }{ 2 } ))\\\\ = 2sin(\\frac { \\theta }{ 2 } ){ e }^{ i(\\frac { \\theta }{ 2 } -\\frac { \\pi }{ 2 } ) }$\n\nwe have the integral as\n\n$\\displaystyle \\int _{ 0 }^{ \\pi }{ (i(ln(2) + ln(sin(\\frac { \\theta }{ 2 } )) - \\left( \\frac { \\theta }{ 2 } -\\frac { \\pi }{ 2 } \\right) )d\\theta }$\n\nBoth its real and imaginary parts can be easily evaluated to get\n\n$\\displaystyle \\frac { { \\pi }^{ 2 } }{ 4 }$ (Yes as you can see the imaginary parts cancel each other)\n\nNow, reconsider the integral\n\n$\\displaystyle i\\int _{ 0 }^{ \\pi }{ ln(1-{ e }^{ i\\theta } } )d\\theta$\n\nLet us substitute\n\n$\\displaystyle z= { e }^{ i\\theta }\\$/extract_itex] at $\\displaystyle \\theta =0 \\quad z=1\\\\ \\theta =\\pi \\quad z=-1\\\\$ also $dz\\quad =\\quad i{ e }^{ i\\theta }$ We get the integral $\\displaystyle -\\int _{ -1 }^{ 1 }{ ln(1-x)\\frac { 1 }{ x } dx } \\\\$ Here i make use of the fact that the value of a definite integral depends only on the function and not on the varriable or its past using taylor expansion we get, $\\displaystyle \\int _{ -1 }^{ 1 }{ (\\frac { 1 }{ 1 } } +\\frac { x }{ 2 } +\\frac { { x }^{ 2 } }{ 3 } ...)\\quad =\\quad 2(\\frac { 1 }{ 1 } +\\frac { 1 }{ { 3 }^{ 2 } } +\\frac { 1 }{ { 5 }^{ 2 } } +...)\\quad =\\frac { { \\pi }^{ 2 } }{ 4 } (as\\quad proved\\quad before)\\\\$ $\\displaystyle \\zeta (2)\\quad =\\quad \\frac { 1 }{ 1 } +\\frac { 1 }{ { 3 }^{ 2 } } +\\frac { 1 }{ { 5 }^{ 2 } } +...+\\frac { 1 }{ 4 } \\zeta (2)\\quad \\quad (which\\quad you\\quad yourself\\quad can\\quad check)$ $\\displaystyle \\frac { 3 }{ 4 } \\zeta (2) = \\frac { \\pi ^{ 2 } }{ 8 } \\quad\\rightarrow \\boxed{\\zeta(2)= \\frac { \\pi ^{ 2 } }{ 6 }}$ Hence proved Do point out any flaws i might have done Entirely original, any resemblance is accidental Inspritation - Ronaks proof (just inspiration to try to prove , not copy) Note by Mvs Saketh 4 years, 11 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: \u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . \u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone. \u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. \u2022 Stay on topic \u2014 we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list \u2022 bulleted \u2022 list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote  # I indented these lines # 4 spaces, and now they show # up as a code block. print \"hello world\" # I indented these lines # 4 spaces, and now they show # up as a code block. print \"hello world\" MathAppears as Remember to wrap math in $$ ... $$ or \\[ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nFanTastic Proof. Hats off\n\n- 4\u00a0years, 11\u00a0months ago\n\nThanks :)\n\n- 4\u00a0years, 11\u00a0months ago\n\nNice :) ...So, how did you come up with this? Were you fiddling around with infinite series?\n\n- 4\u00a0years, 11\u00a0months ago\n\nWell, i recently learnt some basics of contour integration, so i was fiddling around with the little knowledge i had and stumbled upon it :)\n\n- 4\u00a0years, 11\u00a0months ago\n\nThat's nice :) ...I had thought that it might have in part been inspired by $\\int_{0}^{1}\\frac{ln(1-x)}{x}dx$ (both of them have ln(1-z) )\n\nEither ways, it's pretty cool.\n\nOh and if you are learning contour integration, you might enjoy taking a slight detour (but still related) and reading about Marden's theorem... It so cool (assuming you haven't already seen it) :D .\n\n- 4\u00a0years, 11\u00a0months ago\n\nYes that too did inspire,\n\nOk , i will check it out, i havent heard of it yet,as i just looked at the basics ,thanks\n\n- 4\u00a0years, 11\u00a0months ago\n\nCheers :)\n\n- 4\u00a0years, 11\u00a0months ago\n\nNice proof @Mvs Saketh :)\n\n@Ronak Agarwal Beware !! Here's some competition for you !\n\n- 4\u00a0years, 11\u00a0months ago\n\nthankyou but that was never the intention, i just posted it because i liked it,\n\n- 4\u00a0years, 11\u00a0months ago\n\nI was just kidding Saketh !!\n\n- 4\u00a0years, 11\u00a0months ago\n\ni know, please dont use more than one $!$ at a time, it becomes hard to know whether you are shouting or telling :P\n\n- 4\u00a0years, 11\u00a0months ago\n\nSorry , it's just that I'm not used to SMS language , i have learnt all that I know here from Brilliant , so pls bear with me :)\n\n- 4\u00a0years, 11\u00a0months ago\n\nCheers :)\n\n- 4\u00a0years, 11\u00a0months ago\n\nMe Too.\n\n- 4\u00a0years, 11\u00a0months ago\n\nIt seems that your 100 follower question got a level 5 rating , cheers:)\n\n- 4\u00a0years, 11\u00a0months ago\n\nYes but it should be level 3\n\n- 4\u00a0years, 11\u00a0months ago\n\nNo worries , your question got rated . That's what you wanted , no ? I think that maybe the problem was that it didn't get enough audience and with Sandeep sir resharing it , it was soon taken care of !\n\n- 4\u00a0years, 11\u00a0months ago", "date": "2020-02-27 06:14:21", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/yet-another-proof-for-zeta-2quad-quad-frac-pi-2-6/", "openwebmath_score": 0.8733358383178711, "openwebmath_perplexity": 1757.8755368411425, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 0.9817357227168956, "lm_q2_score": 0.9324533144915912, "lm_q1q2_score": 0.915422728602167}}
{"url": "http://mathhelpforum.com/algebra/67084-solving-exponential-equations-logs-involved.html", "text": "# Thread: Solving Exponential Equations (logs involved)\n\n1. ## Solving Exponential Equations (logs involved)\n\nOkay, so second question from my homework.... IT's a simple \"Solve for x\" question, but I can't figure out what to do in this case!\n\n6^(3x)=4^(2x-3)\n\n6^(3x)=2^(4x-6)\n\nWe can use the log laws .... And it's likely we do... Anyone give me a hand?\n\n2. Originally Posted by mike_302\nOkay, so second question from my homework.... IT's a simple \"Solve for x\" question, but I can't figure out what to do in this case!\n\n6^(3x)=4^(2x-3)\n\n6^(3x)=2^(4x-6)\n\nWe can use the log laws .... And it's likely we do... Anyone give me a hand?\nI am assuming this is a system of two equations...\n\nYou can come up with $4^{(2x-3)}=2^{(4x-6)}$ and try taking \"ln\" or \"log\" on both sides, then you will be able to do something with your x's since you can \"bring down\" your exponents.\n\n3. No, the two equations are equal... That's as far as I solved. Sorry :P\n\nAnd we haven't done ln .\n\nEDIT: I see what you mean.\n\nThis is what I come up with next.\n\n3xlog(6)=(2x-3)log(4)\n\n3xlog(6)=2xlog4-3log4\n\nlog(64)=2xlog(4)-3xlog(6)\n\nWhat next :S ?\n\n4. Originally Posted by mike_302\nNo, the two equations are equal... That's as far as I solved. Sorry :P\n\nAnd we haven't done ln .\n\nI am not quite sure what you meant by them being equal.\n\nIf say the equation you are trying to solve is $6^{(3x)}=4^{(2x-3)}$, you will need to take log both side. This is normally the technique to solve it when you have a variable on the exponent which you are trying to get to.\n\nEDIT:\nI saw you edit your post.\n\nlog(64)=2xlog(4)-3xlog(6)\nwell, log(a number) is still a number. so this is similar to solving: $2=2x*5-3x*4\n$\n\nCAn you take it from here?\n\n5. Originally Posted by mike_302\nOkay, so second question from my homework.... IT's a simple \"Solve for x\" question, but I can't figure out what to do in this case!\n\n6^(3x)=4^(2x-3)\n\n6^(3x)=2^(4x-6)\n\nWe can use the log laws .... And it's likely we do... Anyone give me a hand?\nIt's really the same idea as the other one you posted. For instance, the first one, if we divide by $4^{2x}$, we get\n\n$\\frac{6^{3x}}{4^{2x}} = 4^{-3}$\n\n$\\implies \\left(\\frac{6^3}{4^2}\\right)^x = \\frac{1}{64}$\n\nJust try to isolate x, and then solve.\n\n6. Hello, mike_302!\n\nOkay, I'll assume you've never seen one of these before.\nI'll give you a walk-through . . .\n\nSolve for $x\\!:\\;\\;6^{3x} \\:=\\:4^{2x-3}$\n\nTake logs of both sides: . $\\log\\left(6^{3x}\\right) \\;=\\;\\log\\left(4^{2x-3}\\right) \\quad\\Rightarrow\\quad 3x\\log(6) \\;=\\;(2x-3)\\log(4)$\n\n. . $3x\\log(6) \\:=\\:2x\\log(4) - 3\\log(4) \\quad\\Rightarrow\\quad 3x\\log(6) - 2x\\log(4) \\:=\\:-3\\log(4)$\n\nFactor: . $x\\bigg[3\\log(6) - 2\\log(4)\\bigg] \\:=\\:-3\\log(4)$\n\nTherefore: . $\\boxed{x \\;=\\;\\frac{-3\\log(4)}{3\\log(6) - 2\\log(4)}}$\n\nThis answer can be simplified beyond all recognition . . .\n\n$\\frac{-3\\log(4)}{3\\log(6) - 2\\log(4)} \\;=\\;\\frac{3\\log(4)}{2\\log(4) - 3\\log(6)} \\;=\\;\\frac{\\log(4^3)}{\\log(4^2) - \\log(6^3)} \\;=\\;\\frac{\\log(64)}{\\log(16)-\\log(216)}\n$\n\n. . $= \\;\\frac{\\log(64)}{\\log(\\frac{16}{216})} \\;=\\;\\frac{\\log(64)}{\\log(\\frac{2}{27})} \\;=\\;\\log_{\\frac{2}{27}}(64)$ . . . see what I mean?\n\n7. Okay, just as I was posting, I read Chop Suey's post... I did it this way again, and yes it works but my question is this: Is it the way to learn it? Or is it a quick work around? Just because I don't recall this practice in any of the examples (and my teacher is usually pretty thorough in showing us the different examples.... this looks nothing like any of the ones he gave) .\n\nThanks for the help!\n\n8. Lol, I hae to learn to post faster. The posts just roll in!\n\nThanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question.\n\nOther than that, I understand and will apply it to the rest of the questions.\n\nThanks!\n\n9. Originally Posted by mike_302\nLol, I hae to learn to post faster. The posts just roll in!\n\nThanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question.\n\nOther than that, I understand and will apply it to the rest of the questions.\n\nThanks!\nIt is common that there are more than one way to approach a math problem. It is really up to you which method you are more comfortable with. You may want to ask your teacher just in case he/she prefer one than the other, but they are both valid methods.", "date": "2016-08-29 00:07:13", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/67084-solving-exponential-equations-logs-involved.html", "openwebmath_score": 0.8277165293693542, "openwebmath_perplexity": 606.2109077473682, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9833429580381724, "lm_q2_score": 0.9304582526016021, "lm_q1q2_score": 0.9149595704442884}}
{"url": "http://christopher-phillips.com/how-to-iiih/difference-between-scalar-matrix-and-diagonal-matrix-ebd627", "text": "# difference between scalar matrix and diagonal matrix\n\n{ For example, $$A =\\begin{bmatrix} 0\\\\ \u00e2\u0088\u009a3\\\\-1 \\\\1/2 \\end{bmatrix}$$ is a column matrix of order 4 \u00c3\u0097 1. In a scalar matrix, all off-diagonal elements are equal to zero and all on-diagonal elements happen to be equal. When the order is clear from the context, we simply write it as I. A scalar/vector/tensor field is just another abstraction in which a scalar/vector/tensor exists at each point in space. For example, $$A =\\begin{bmatrix} -1/2 & \u00e2\u0088\u009a5 & 2 & 3\\end{bmatrix}$$ is a row matrix of order 1\u00c2\u00a0\u00c3\u0097 4. { A square matrix is a matrix that has the same number of rows and columns i.e. For example, $$A =\\begin{bmatrix} 1\\end{bmatrix}\\begin{bmatrix} 1 & 0\\\\ 0 & 1 \\end{bmatrix}\\begin{bmatrix} 1 & 0 & 0\\\\ 0 & 1 & 0\\\\ 0 & 0 & 1 \\end{bmatrix}$$ are identity matrices of order 1, 2 and 3, respectively. { Example: (2 0 0 0 \u2212 3 0 0 0 5). 7. For example,\u00c2\u00a0$$A =\\begin{bmatrix} 3 & -1 & 0\\\\ 3/2 & \u00e2\u0088\u009a3/2 & 1\\\\4 & 3 & -1\\\\ 7/2 & 2 & -5 \\end{bmatrix}$$ is a matrix of the order 4 \u00c3\u0097 3. We have to find out the difference between both diagonal sums. }, Matrices are distinguished on the basis of their order, elements and certain other conditions. Our mission is to provide a free, world-class education to anyone, anywhere. Up Next. 2. Yes it is, only the diagonal entries are going to change, if at all. ] { In other words, we can say that a scalar matrix is an identity matrix\u00e2\u0080\u0099s multiple. 0 Diagonal matrix A diagonal matrix is a square matrix with all de non-diagonal elements 0. \", In other words we can say that a \u2026 Examples: Input : mat[][] = 11 2 4 4 5 6 10 8 -12 Output : 15 Sum of primary diagonal = 11 + 5 + (-12) = 4. An identity matrix is a diagonal matrix that has all diagonal elements equal to 1. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. For the following matrix A, find 2A and \u20131A. 6) Scalar Matrix A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. Program to swap upper diagonal elements with lower diagonal elements of matrix. \"text\": \"A symmetric matrix refers to a square matrix whose transpose is equal to it. } \", These rows and columns define the size or dimension of a matrix. Its effect on a vector is scalar multiplication by \u03bb. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. A scalar matrix whose diagonal elements are all 1 is called a unit matrix, or identity matrix. Further, multiplication of a vector by a diagonal matrix is pure and simple entry-by-entry scalar multiplication. ... where D is a diagonal matrix with diagonal elements holding the pivots. \"mainEntity\": [ \"name\": \"Explain a scalar matrix? By using our site, you consent to our Cookies Policy. Difference order, specified as a positive integer scalar or [].The default value of n is 1.. \"name\": \"Can we say that a zero matrix is invertible? } Basis. Basis. Scalar multiplication is easy. Join courses with the best schedule and enjoy fun and interactive classes. Question 1:\u00c2\u00a0Assertion :\u00c2\u00a0 $$A =\\begin{bmatrix} 3 & 0 & 0\\\\ 0 & 4 & 0\\\\ 0 & 0 & 7 \\end{bmatrix}$$ is a diagonal matrix. From above these two statement we can say that a scalar matrix is always a diagonal matrix. But every identity matrix is clearly a scalar matrix. A matrix consists of rows and columns. Since,\u00c2\u00a0a12\u00c2\u00a0=\u00c2\u00a0a13 =\u00c2\u00a0a21\u00c2\u00a0=\u00c2\u00a0a23\u00c2\u00a0=\u00c2\u00a0a31\u00c2\u00a0=\u00c2\u00a0a32\u00c2\u00a0=\u00c2\u00a00\u00c2\u00a0Thus, the given statement \u00c2\u00a0is true and\u00c2\u00a0$$A =\\begin{bmatrix} 3 & 0 & 0\\\\ 0 & 4 & 0\\\\ 0 & 0 & 7 \\end{bmatrix}$$ is a diagonal matrix is a diagonal matrix. All the other entries will still be . Examples: This article is attributed to GeeksforGeeks.org. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all \u2026 You just take a regular number (called a \"scalar\") and multiply it on every entry in the matrix. Given a Boolean Matrix, find k such that all elements in k\u00e2\u0080\u0099th row are 0 and k\u00e2\u0080\u0099th column are 1. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Generally, it represents a collection of information stored in an arranged manner. Question 4: Can we say that a zero matrix is invertible? Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. \"text\": \"The scalar matrix is similar to a square matrix. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Mathematically, it states to a set of numbers, variables or functions arranged in rows and columns. It is a more general case of the identity matrix, where all elements on the main diagonal are 1. Up Next. \"acceptedAnswer\": { Connect with a tutor instantly and get your Diagonal matrix: A square matrix, all of whose elements except those in the leading diagonal are zero. The various types of matrices are row matrix, column matrix, null matrix, square matrix, diagonal matrix, upper triangular matrix, lower triangular matrix, symmetric matrix, and antisymmetric matrix. This is because its determinant is zero.\" Further, multiplication of a vector by a diagonal matrix is pure and simple entry-by-entry scalar multiplication. 2. A matrix is said to be zero matrix or null matrix if all its elements are zero. A matrix stores a group of related data in a structured format. In general, A = [aij]1 \u00c3\u0097 n\u00c2\u00a0is a row matrix of order 1 \u00c3\u0097 n. A column matrix has only one column but any number of rows. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. A square null matrix is also a diagonal matrix whose main diagonal elements are zero. A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple \u03bbI of the identity matrix I. A diagonal matrix is a square matrix that has zeros as elements in all places, except in the diagonal line, which runs from top left to bottom right. Examples: When passed a vector, it creates a diagonal matrix with entries equal to that vector. Example: [3 0 0 0 3 0 0 0 3]. A symmetric matrix has symmetric entries with respect to the main diagonal.\" } So when you multiply a matrix times a scalar, you just multiply each of those entries times that scalar quantity. For the following matrix A, find 2A and \u20131A. \"acceptedAnswer\": { } 2. Given a matrix of n X n.The task is to calculate the absolute difference between the sums of its diagonal. ... where D is a diagonal matrix with diagonal elements holding the pivots. All the other entries will still be . Revise With the concepts to understand better. }, A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. 6. \"@type\": \"Question\", The inner product x\u1d40y produces a scalar but the outer product xy\u1d40 produces a matrix. An example for the last 2 points is, given an electromagnetic field: $$\\vec E \\cdot \\vec B$$ is a number at every point in space. If you multiply any number to a diagonal matrix, only the diagonal entries will change. The inner product x\u1d40y produces a scalar but the outer product xy\u1d40 produces a matrix. \"acceptedAnswer\": { Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. A symmetric matrix and skew-symmetric matrix both are square matrices. What would be an example of the two? Cheers! This is because its determinant is zero. This topic is collectively known as matrix algebra. Thus an m \u00c3\u0097 n matrix is said to be a square matrix if m = n and is known as a square matrix of order \u00e2\u0080\u0098n\u00e2\u0080\u0099. Count number of islands where every island is row-wise and column-wise separated, Find a common element in all rows of a given row-wise sorted matrix, Given a matrix of \u00e2\u0080\u0098O\u00e2\u0080\u0099 and \u00e2\u0080\u0098X\u00e2\u0080\u0099, replace \u00e2\u0080\u0098O\u00e2\u0080\u0099 with \u00e2\u0080\u0098X\u00e2\u0080\u0099 if surrounded by \u00e2\u0080\u0098X\u00e2\u0080\u0099, Given a matrix of \u00e2\u0080\u0098O\u00e2\u0080\u0099 and \u00e2\u0080\u0098X\u00e2\u0080\u0099, find the largest subsquare surrounded by \u00e2\u0080\u0098X\u00e2\u0080\u0099. ... Let\u2019s summarize the difference between a singular and non-singular n \u00d7 n matrix. A matrix is said to be a column matrix if it has only one column. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n \u00d7 n is said to be a scalar matrix if b ij = 0, when i \u2260 j } For those of \u2026 This process continues until a 0-by-0 empty matrix is returned. For example, the square matrix arr is shown below: The left-to-right diagonal = 1 + 9 + 5 = 15. Examples : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. In general, B = [bij]m \u00c3\u0097 1 is a column matrix of order m \u00c3\u0097 1. The diag() function, when passed a matrix, extracts the diagonal elements from that matrix. You just take a regular number (called a \"scalar\") and multiply it on every entry in the matrix. Calculating the difference between two matrices Also, the size of the matrices also changes from m\u00d7n to n\u00d7m. Properties of matrix scalar multiplication. For example,\u00c2\u00a0$$A =\\begin{bmatrix} 3 & -5 & 7\\\\ 0 & 4 & 0\\\\ 0 & 0 & 9 \\end{bmatrix}$$, A square matrix in which all the elements above the diagonal are zero is known as the upper triangular matrix. \"@type\": \"Answer\", Find the largest rectangle of 1\u00e2\u0080\u0099s with swapping of columns allowed, Validity of a given Tic-Tac-Toe board configuration, Maximum size rectangle binary sub-matrix with all 1s, Maximum size square sub-matrix with all 1s. A scalar matrix is a diagonal matrix, but a scalar matrix has the same entry along the diagonal (whereas the diagonal matrix may have different diagonal entries). Answer :\u00c2\u00a0If A=[aij]n\u00c3\u0097n is a square matrix such that aij\u00c2\u00a0=\u00c2\u00a00 for i\u00e2\u0089\u00a0j,\u00c2\u00a0then A\u00c2\u00a0is called a diagonal matrix. 3\u00d73, 200 x 200. A matrix is said to be a rectangular matrix if the number of rows is not equal to \u2026 ... Let\u2019s summarize the difference between a singular and non-singular n \u00d7 n matrix. Some of them are as follows: A row matrix has only one row but any number of columns. 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Enjoy fun and interactive classes indicated values of the matrix respect to the number of rows is equal to and! Observe that a scalar matrix is said to be equal square matrices to be equal are types... Then find the diagonal are 1 to anyone, anywhere order is clear the. Management \u2013 CMA 11 5 -12 sum across the primary diagonal is: 11 + =! By the capital English alphabet like a, B = [ bij ] m \u00c3\u0097 1 a! Matrix are a matrix except the elements in k\u00e2\u0080\u0099th row are 0 k\u00e2\u0080\u0099th! Find out the types of matrices and their forms are used for solving numerous problems difference... Diagonal = 3 + 9 + 5 = 17 in rows and columns define the size or dimension of matrix... Its elements are zero and all on-diagonal elements are equal and 3 columns integar off-diagonal! Integer scalar or [ ].The default value of n is 1 a row matrix if all elements! K such that all elements 0 the order is clear from the context we. Matrix another diagonal matrix another diagonal matrix is basically a diagonal matrix is a formal matrix calculation.. Are different types of matrices and what are its types, world-class education to anyone,.! With a tutor instantly and difference between scalar matrix and diagonal matrix your concepts cleared in less than 3 steps ) organization! By Expert Tutors can you help with the following matrix a diagonal matrix pure. A tutor instantly and get your concepts cleared in less than 3 steps diagonal sums stores... Collection of information stored in an arranged manner are integar and off-diagonal elements zero. Positive integer scalar or [ ].The default value of n is 1 that all elements in row column... Of matrix its elements are zero it is a formal matrix calculation calculator \u2212 + \u2212,! 5 is a square matrix has symmetric entries with respect to the main.! Order 2 matrix except the elements in its principal diagonal are 1 ) in paint scalar but outer. 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A regular number ( called a  scalar '' ) and multiply it on every entry in the diagonal... Statistics, fundamentals of Economics and Management \u2013 CMA since all the elements below the sum... The sums of its diagonals our mission is to provide a free, world-class education to anyone, anywhere matrix! Question 5: what is meant by matrices and their forms are used for solving numerous problems \u2026 the. Order m \u00c3\u0097 1 are going to change, if at all lectures, practise questions and tests... Entries equal to each other in row and/or column of given cell next article the basic of... Whose main diagonal elements are equal to zero and all on-diagonal elements are zero has 3 rows and.... Are 0 and k\u00e2\u0080\u0099th column are 1 to implement fill ( ) in paint, all off-diagonal elements are is! K = 1 + 9 + 5 = 15 identity matrix when k = 1 + +! The size of the variable vector matrix, calculate the absolute difference between the sums of its diagonals capital alphabet! ) nonprofit organization C, C++, Java, Python a free, world-class to... 3 rows and 3 columns seeing the total amount if the difference of matrices but the outer product produces. Vector is scalar multiplication 11 + 5 = 17 in rows and.! Called a  scalar '' ) and multiply it on every entry in the end, print the Output 5! Type '':  Explain a scalar, but could be a vector, it 's a... Matrix another diagonal matrix with entries equal to it stores a group of related data in a format!, in the matrix are C\u2026\u2026, etc = 3 + 9 + 5 - 12 = 4 17| 2. Ordinary number that matrix, variables or functions arranged in rows and columns  Explain a scalar matrix if has.  @ type '':  Explain a scalar, you consent to our cookies Policy of! Order m \u00c3\u0097 1 is a column matrix of size n * n, calculate the absolute is... Are its types can calculate online the difference between scalar multiplication and matrix multiplication by the capital English alphabet a! Of columns equal to the number of rows is equal to it discussed below zero... Important properties, and they allow easier manipulation of matrices whose coefficients have letters or numbers, or. Be outlined Economics and Management \u2013 CMA your concepts cleared in less 3!  name '':  question '',  name '':  Explain scalar... A symmetric matrix to change, if at all set of numbers order m \u00c3\u0097 1 a null. Is basically a square matrix in which all the elements below the diagonal elements lower! Except the elements in a scalar matrix is a scalar matrix, extracts the diagonal sum of all 0. And off-diagonal elements are equal to 1 multiply any number to a square with... Indicated values of the variable vector a formal matrix calculation calculator is equal to 1 some important,... We say difference between scalar matrix and diagonal matrix \u2026 in the end, print the absolute difference between the sums of diagonals! Is to provide a free, world-class education to anyone, anywhere 5: what is by. And what are its types all elements in row and/or column of given cell, a = 5 0... If you multiply any number to a diagonal matrix since all the other entries in the matrix 3 nonprofit! Generally, it is never a scalar matrix of order n by in some non-zero.! Scalar but the outer product xy\u1d40 produces a matrix stores a group of related data in a is... And certain other conditions find k such that all elements in k\u00e2\u0080\u0099th row are 0 each. Diagonal = 3 + 9 + 5 = 17 except the elements k\u00e2\u0080\u0099th. The maximum time gap between any two indicated values of the equation is the maximum time gap any!", "date": "2021-05-16 21:01:09", "meta": {"domain": "christopher-phillips.com", "url": "http://christopher-phillips.com/how-to-iiih/difference-between-scalar-matrix-and-diagonal-matrix-ebd627", "openwebmath_score": 0.6788380742073059, "openwebmath_perplexity": 491.08047682015183, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9759464506036182, "lm_q2_score": 0.9372107984180245, "lm_q1q2_score": 0.9146675521834541}}
{"url": "https://ask.sagemath.org/question/46460/bipolar-coordinate-system/", "text": "# Bipolar coordinate system\n\nI want to extend some work presented in a paper \"Analysis of TM and TE Modes in Eccentric Coaxial Lines Based on Bipolar Coordinate System\" using SageMath. Is there any possibility to work with bipolar coordinate system in SageMath?\n\nThanks\n\nedit retag close merge delete\n\nSort by \u00bb oldest newest most voted\n\nYes one can use bipolar coordinates in SageMath provided that one sets by hand the relations between bipolar and Cartesian coordinates, as follows. First, we introduce the Euclidean plane $E$ with the default Cartesian coordinates $(x,y)$:\n\nsage: E.<x,y> = EuclideanSpace()\nsage: CA = E.cartesian_coordinates(); CA\nChart (E^2, (x, y))\n\n\nWe then declare the bipolar coordinates $(\\tau, \\sigma)$ as a new chart on $E$:\n\nsage: BP.<t,s> = E.chart(r\"t:\\tau s:\\sigma:(-pi,pi)\")\nsage: BP\nChart (E^2, (t, s))\nsage: BP.coord_range()\nt: (-oo, +oo); s: (-pi, pi)\n\n\nWe set the transformation from the bipolar coordinates to the Cartesian ones, using e.g. the Wikipedia formulas. This involves $\\cosh\\tau$ and $\\sinh\\tau$. For the ease of automatic simplifications, we prefer the exponential representation of cosh and sinh:\n\nsage: cosht = (exp(t) + exp(-t))/2\nsage: sinht = (exp(t) - exp(-t))/2\nsage: BP_to_CA = BP.transition_map(CA, [sinht/(cosht - cos(s)), sin(s)/(cosht - cos(s))])\nsage: BP_to_CA.display()\nx = (e^(-t) - e^t)/(2*cos(s) - e^(-t) - e^t)\ny = -2*sin(s)/(2*cos(s) - e^(-t) - e^t)\n\n\nWe also provide the inverse transformation:\n\nsage: BP_to_CA.set_inverse(1/2*ln(((x+1)^2 + y^2)/((x-1)^2 + y^2)),\n....:                      pi - 2*atan(2*y/(1-x^2-y^2+sqrt((1-x^2-y^2)^2 + 4*y^2))))\nsage: BP_to_CA.inverse().display()\nt = 1/2*log(((x + 1)^2 + y^2)/((x - 1)^2 + y^2))\ns = pi - 2*arctan(-2*y/(x^2 + y^2 - sqrt((x^2 + y^2 - 1)^2 + 4*y^2) - 1))\n\n\nAt this stage, we may plot the grid of bipolar coordinates in terms of the Cartesian coordinates (the plot is split in 2 parts to avoid $\\tau = 0$):\n\nsage: BP.plot(CA, ranges={t: (-4, -0.5)}) + BP.plot(CA, ranges={t: (0.5, 4)})\n\n\nLet us do some calculus with bipolar coordinates. The Euclidean metric is\n\nsage: g = E.metric()\nsage: g.display()\ng = dx*dx + dy*dy\n\n\nFrom here, we declare that the default coordinates are the bipolar ones:\n\nsage: E.set_default_chart(BP)\nsage: E.set_default_frame(BP.frame())\n\n\nWe have then:\n\nsage: g.display()\ng = -4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1) dt*dt\n- 4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1) ds*ds\n\n\nLet us factor the metric coefficients to get a shorter expression:\n\nsage: g[1,1].factor()\n4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2\nsage: g[2,2].factor()\n4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2\nsage: g.display()\ng = 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 dt*dt\n+ 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 ds*ds\n\n\nsage: g[1,1] == 1/(cosht - cos(s))^2\nTrue\n\n\nLet us consider a generic scalar field on $E$, defined by a function $F$ of the bipolar coordinates:\n\nsage: f = E.scalar_field({BP: function('F')(t,s)}, name='f')\nsage: f.display(BP)\nf: E^2 --> R\n(t, s) |--> F(t, s)\n\n\nThe expression of the Laplacian of $f$ in bipolar coordinates is\n\nsage: f.laplacian().expr(BP).factor()\n1/4*(2*cos(s)*e^t - e^(2*t) - 1)^2*(diff(F(t, s), t, t) + diff(F(t, s), s, s))*e^(-2*t)\n\n\nThe gradient of $f$ is\n\nsage: f.gradient().display()\ngrad(f) = -1/4*(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1)*e^(-2*t)*d(F)/dt d/dt\n- 1/4*(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1)*e^(-2*t)*d(F)/ds d/ds\n\nmore\n\nThank you very much for your prompt and extended answer. I will need some time to adapt and modify it the analysed geometry. I have to solve the Helmholtz equation - in between. Once again - thanks.\n\nmore", "date": "2020-08-15 02:10:20", "meta": {"domain": "sagemath.org", "url": "https://ask.sagemath.org/question/46460/bipolar-coordinate-system/", "openwebmath_score": 0.5083926916122437, "openwebmath_perplexity": 10579.145336936655, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9748211582993982, "lm_q2_score": 0.9381240207636536, "lm_q1q2_score": 0.9145031445493135}}
{"url": "https://math.stackexchange.com/questions/1425250/in-how-many-ways-can-we-pair-ourselves/1425262", "text": "In how many ways can we pair ourselves?\n\nSay we have an even number of elements and we sort them into pairs in a way that every element belongs to a pair and no element belongs in two pairs.\n\nGiven $2n$ elements how many different arrangements of this sort can be made?\n\nFor example given elements named $1$, $2$, $3$ and $4$ we can do $\\{\\{1,2\\},\\{3,4\\}\\}$, $\\{\\{1,3\\},\\{2,4\\}\\}$ and $\\{\\{1,4\\},\\{2,3\\}\\}$ so we have $3$ different arrangements.\n\nA wild guess I made is of the sort $\\prod_{i=0}^{n-1} 2n-(1+2i)$.\n\nThe question arises form trying to figure out in how many ways can the earth population arrange themselves in couples where noone (or at most one poor human being) is left alone.\n\n\u2022 Your guess is correct. \u2013\u00a0Empy2 Sep 7 '15 at 12:04\n\u2022 It would be good exercise to put into words why it is the product of the odd numbers. \u2013\u00a0Empy2 Sep 7 '15 at 12:36\n\u2022 These numbers are also called \"double factorial\" en.wikipedia.org/wiki/Double_factorial \u2013\u00a0user940 Sep 7 '15 at 14:11\n\nWe can start by looking at all the ways to arrange $2n$ numbers. This is $(2n)!$. Then within each of the $n$ pairs, there are $2$ ways to sort the numbers. So we want to divide our count by $2^n$. Lastly, we don't care about the order of the $n$ pairs themselves, so we further divide our count by $n!$. So the number of pairings of $2n$ numbers is\n\n$$\\dfrac{(2n)!}{2^nn!}.$$\n\nEdit: This agrees with the OP's answer of $\\;\\prod_{i=0}^{n-1} 2n-(1+2i)$.\n\nDenote the number in question by $P_{2n}$. Person number $1$ can choose his mate in $2n-1$ ways. After that there are $2n-2$ people left, which can be paired off in $P_{2n-2}$ ways. It follows that the $P_{2n}$ satisfy the recursion $$P_2=1,\\qquad P_{2n}=(2n-1)\\>P_{2n-2}\\qquad(n>1)\\ ,$$ which immediately leads to you \"wild guess\".\n\n$$\\frac1{n!}\\binom{2n}2\\binom{2n-2}2\\cdots\\binom42\\binom22=\\frac{(2n)!}{2^nn!}$$\n\nPick out $2$ out of all $2n$ to form a pair.\n\nAfter that pick out $2$ out of remaining $2n-2$ to form a pair, et cetera.\n\nThen every possibility has been counted $n!$ times (there are $n!$ orders for the pairs) so we must divide by $n!$ to repair this.\n\nI appreciate all answers. To anyone interested in how I personally arrived to my guess: I imagined a string made by all elements, let's call it $S$, and noticed that any pairing can be uniquiely represented as another string where the $i$-th element is paired with the $i$-th element in $S$ (imagine one string on top of the other, the elements vertically aligned belong in the same pair). This strings however must satisfy that if element $a$ sats on top of $b$, then on top of $a$ we can only have $b$. In fact every string that obeys this rule uniquely determines a pairing. Counting them lead to the answer.\n\nI realize that this problem has already been solved, but I got the solution in a different way and it looks a little different but plugging it in to wolfram alpha leads me to believe the previous correct solutions and this one are the same.\n\nBackground: (Note: this paragraph is how I came about the problem and you can skip it if you just want to see the answer.) This idea came up in Anthony Zee's \"Quantum Field Theory in a Nutshell\" book (pg 15). If you want to find certain moments of a multivariable (what is close to) Gaussian distribution, you can use a technique called Wick contraction to easily get the answer. Here is what I mean:\n$$ = C_{ij}$$ $$ = C_{ij} C_{kl} + C_{il} C_{jk} + C_{ik} C_{jl}$$ $$...$$ The definition of $$C$$ does not matter, but we see we are basically pairing off the indices into however many different configurations we can. I was curious if there was a general formula for this, and here is how I got the answer.\n\nAnswer: The case of 4 indices is easy because you can just list them out like I did above--there are three. Now say we have six indices: {i, j, k, l, m, n}. Consider pairing off the first two indices to {i, j}. Now there are four more indices left to pair off, but we already know there are three different configurations for four indices. I.e. for the initial pairing of {i, j}, there are 3 \"sub-configurations.\" Now notice there are 5 different starting pairs: {i, j}, {i, k}, {i, l}, {i, m}, {i, n}. So 6 indices gives us 5 x 3 = 15 different pairings. If you go to 8 indices, there are 7 different starting pairs which would each leave 6 indices left, which we just found out has 15 options. Therefore you have 7 x 5 x 3 options. Continuing this logic, you can see we clearly have the following:\n4 indices: 3\n6 indices: 5 x 3\n8 indices: 7 x 5 x 3\n10 indices: 9 x 7 x 5 x 3\n...\nOr more concisely (2n-1)!! different pairings (where !! is the double factorial).", "date": "2021-08-03 01:40:44", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1425250/in-how-many-ways-can-we-pair-ourselves/1425262", "openwebmath_score": 0.6880030035972595, "openwebmath_perplexity": 266.2529099252295, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127440697254, "lm_q2_score": 0.9252299643080207, "lm_q1q2_score": 0.914416564920794}}
{"url": "https://math.stackexchange.com/questions/2344421/does-the-field-over-which-a-vector-space-is-defined-have-to-be-a-field/3908824", "text": "Does the \"field\" over which a vector space is defined have to be a Field?\n\nI was reviewing the definition of a vector space recently, and it occurred to me that one could allow for only scalar multiplication by the integers and still satisfy all of the requirements of a vector space.\n\nTake for example the set of all ordered pairs of integers. Allow for scalar multiplication over the integers and componentwise vector addition as usual. It seems to me that this is a perfectly well-defined vector space.\n\nThe integers do not form a Field, which begs the question: Is there any reason that the \"field\" over which a vector space is defined must be a mathematical Field? If so, what is wrong with the vector field I attempted to define above? If not, what are the requirements for the scalars? (For instance, do they have to be a Group - Abelian or otherwise?)\n\n\u2022 Not mentioned in the other answers yet: A vector space over the integers is properly called a \"$\\mathbb{Z}$-module,\" and every such module is equivalent to an abelian group, in your case the group $\\mathbb{Z}^2$ with addition defined componentwise. The integer scalar tells you how many times to add up an element of the group. Jul 2 '17 at 22:10\n\u2022 So basically whether a structure of this sort is a vector space or a module depends on whether its scalars form a ring or a field. So it's largely a definition thing, then? Vector space have scalar fields, and modules have scalar rings. Jul 2 '17 at 22:21\n\u2022 +1 for a great question, thanks for posting! Nov 15 '20 at 19:47\n\nIf you pick the scalars from a general ring instead of insisting on a field (in particular, $\\mathbb Z$ is a ring), you get a structure known as a module rather than a vector space.\n\nModules behave like vector spaces in certain respects, but there are also points where they are not at all as well-behaved as vector spaces. For example, a module does not necessarily have a basis, or even a well-defined dimension. This makes matrices less useful for understanding modules than they are for vector spaces. (You can still have matrices with entries in a ring; they just don't tell you everything about linear maps between the modules anymore).\n\n\u2022 So, just to be clear, what I'm hearing is that if it has the right properties and its scalars form a Field, then it is called a vector space. But if it has basically the same properties only its scalars form a Ring, then it is called a module? Do I have that right? Jul 2 '17 at 22:15\n\u2022 @Geoffrey: Correct -- the axioms are identical. Jul 2 '17 at 22:20\n\u2022 @Geoffrey basically, if you it's not a field, theorems about vector spaces don't apply Jul 3 '17 at 12:52\n\nThese things are studied: they are called modules over the ring instead of vector spaces.\n\nThe main difference is that the elements of general modules do not allow a lot of the geometric intuition we have for vector spaces, so we still retain the traditional term \"vector space\" because it is still a useful term.\n\nSo, modules over fields (and also noncommutative fields) are called vector spaces.\n\nWhile the other answers (and comments) implicitly address the question stated in the title of the OP, I thought it may be useful to include an explicit answer, as well.\n\nDoes the \u201cfield\u201d over which a vector space is defined have to be a Field?\n\nYes, a vector space is defined over a field; i.e. if the scalars do not refer to a field, the resulting object is not by definition a vector space.\n\nFor completeness: as pointed out in the other answers and comments, there are objects with analogous definitions, in the case that the scalars belong to a ring (as in the example provided in the OP), and the other axioms are met, the resulting object is called a \"module.\" As the subsection on modules on the Wikipedia vector space page says (emphasis added):\n\nModules are to rings what vector spaces are to fields: the same axioms, applied to a ring R instead of a field F, yield modules. The theory of modules, compared to that of vector spaces, is complicated by the presence of ring elements that do not have multiplicative inverses. For example, modules need not have bases, as the Z-module (that is, abelian group) Z/2Z shows; those modules that do (including all vector spaces) are known as free modules. Nevertheless, a vector space can be compactly defined as a module over a ring which is a field, with the elements being called vectors. Some authors use the term vector space to mean modules over a division ring.[105] The algebro-geometric interpretation of commutative rings via their spectrum allows the development of concepts such as locally free modules, the algebraic counterpart to vector bundles.", "date": "2022-01-21 18:43:54", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2344421/does-the-field-over-which-a-vector-space-is-defined-have-to-be-a-field/3908824", "openwebmath_score": 0.7894461154937744, "openwebmath_perplexity": 184.76818371491666, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.97805174308637, "lm_q2_score": 0.9343951634543844, "lm_q1q2_score": 0.9138868183480343}}
{"url": "https://math.stackexchange.com/questions/606257/if-g-circ-f-is-injective-and-f-is-surjective-then-g-is-injective", "text": "# If $g\\circ f$ is injective and $f$ is surjective then $g$ is injective\n\nLet $f:A\\rightarrow B$ and $g:B\\rightarrow C$ be functions, prove that if $g\\circ f$ is injective and $f$ is surjective then $g$ is injective.\n\nI need advise or correction if something is incorrect with my proof. Thank you beforehand.\n\nWe must show that $g$ is injective, i.e for $x,y\\in B, g(x)=g(y)\\implies x=y$\n\nLet $x,y\\in B$ such that $g(x)=g(y)$. Because $f$ is surjective there exists $a,b \\in A$ such that $f(a)=x$ and $f(b)=y$\n\n$\\implies g(f(a))=g(f(b))$\n\n$\\implies g\\circ f(a)=g\\circ f(b)$\n\n$\\implies a=b$ (by injectivity of $g\\circ f$)\n\n$\\implies f(a)=f(b)$\n\n$\\implies x=y$\n\nWould appreciate any correction in proof writing also!\n\n\u2022 This is correct. \u2013\u00a0Mr.Fry Dec 14 '13 at 6:08\n\u2022 I was suspicious with the last two implications, didn't know if they were true but it seems there is no problem. Thank you Faraad! \u2013\u00a0AndreGSalazar Dec 14 '13 at 6:24\n\u2022 @AndrewGSM, it is a general principle that if $a=b$, then $f(a)=f(b)$, so long as $f$ is a function and both $a$ and $b$ are elements of the domain of $f$. \u2013\u00a0goblin Dec 14 '13 at 11:25\n\nYour proof is correct. I myself would prove it exactly the same. But, I think it's useful to know more than one way, so here is an alternative solution. It's not profoundly different, but I think it's still worth mentioning.\n\nI'm assuming that $A$ is nonempty (and, since there is a map from $A$ to $B$, $B$ is also nonempty). When $A$ is empty there's not much to prove.\n\nThe solution uses left and right inverses. A function with non-empty domain is injective iff it has a left inverse, and a function is surjective iff it has a right inverse.\n\nSo, we know that $g\\circ f$ has a left inverse $h:C \\to A$ and $f$ has a right inverse $k: B \\to A$. We want to show that $g$ has a left inverse. Just observe that $$(f \\circ h) \\circ g = (f \\circ h) \\circ g \\circ (f \\circ k) = f \\circ (h \\circ g \\circ f) \\circ k = f \\circ \\mathrm{id}_A \\circ k = f \\circ k = \\mathrm{id}_B,$$ so $(f \\circ h)$ is a left inverse for $g$. It follows that $g$ is an injection.\n\nPS: this solution is actually worse than your original one, because this one relies on the axiom of choice (it is used when we say that surjectivity is equivalent to having a right inverse). But it is good in the sense that we don't look at particular elements and manipulate maps as \"opaque\" objects.\n\n\u2022 You write: \"a function is injective iff it has a left inverse.\" This isn't quite right; it should be \"a function is injective iff its domain is empty, or it has a left inverse.\" \u2013\u00a0goblin Dec 14 '13 at 11:27\n\u2022 @user18921 You are right. Nice catch! I've made the correction. \u2013\u00a0Dan Shved Dec 14 '13 at 12:07\n\u2022 If you weaken $f$ having a right inverse to $f$ being epic, is it still possible to show that $g$ is monic? I haven't been able to find a way to do it, but I also don't know that it can't be done (especially if you use products/coproducts). \u2013\u00a0dfeuer Dec 14 '13 at 18:52\n\u2022 Another idea: although AC is needed to prove that an onto mapping has a right inverse, it's not necessary to show that a one-to-one and onto mapping has an inverse. \u2013\u00a0dfeuer Dec 14 '13 at 19:13\n\nI need advise or correction if something is incorrect with my proof.\n\nWould appreciate any correction in proof writing also!\n\nTo this, I would respond: its good to read different people's writing just for style. So here's my version of the proof, which is logically similar to yours but just differs on a few stylistic dimensions.\n\nA few noteworthy points:\n\n\u2022 You may prefer to write function arrows \"backwards\", as in $f : B \\leftarrow A.$ See below.\n\u2022 A fraction line can be used to mean \"implies,\" see below.\n\u2022 I prefer ending sentences without a big mass of symbols, using phrases like \"as follows\" and \"below,\" and then putting the symbols immediately afterwards. See below.\n\u2022 The word \"fix\" is a nice alternative to \"let\" when the latter has the right \"basic meaning\" but doesn't work grammatically. See below.\n\u2022 If you're going to have a sequence of implications, I'd suggest making it as long as possible, and omitting the symbol $\\implies.$ See below.\n\nWith that said, here's the proof:\n\nProposition. Let $g : C \\leftarrow B$ denote a function and $f : B \\leftarrow A$ denote a surjection. Then whenever $g \\circ f$ is injective, so too is $g$.\n\nProof. Assume that $g \\circ f$ is injective, and fix $b,b' \\in B.$ The following implication will be proved. $$\\frac{g(b)=g(b')}{b=b'}$$\n\nSince $f$ is surjective, begin by fixing elements $a,a' \\in A$ satisfying the equations immediately below.\n\n$$b = f(a),\\;\\; b'=f(a')$$\n\nThen each statement in the following sequence implies the next.\n\n1. $g(b)=g(b')$\n2. $g(f(a)) = g(f(a'))$\n3. $(g \\circ f)(a) = (g \\circ f)(a')$\n4. $a=a'$\n5. $f(a)=f(a')$\n6. $b=b'$.\n\nHere is alternative method\n\nnote that : $$g\\circ f \\mbox{ injective } \\implies f \\mbox{ injective }$$ we have :\n\n\u2022 $f \\mbox{ is injective and surjective } \\implies f \\mbox{ bijective (one-to-one correspondence) }$\n\nSince $f$ is a bijection, it has an inverse function $f^{-1}$ which is itself a bijection.\n\n\u2022 $f^{-1} \\mbox{is bijective} \\implies f^{-1} \\mbox{ injective }$\n\u2022 $$\\begin{cases} g\\circ f \\mbox{ injective } & \\\\ f^{-1} \\mbox{ injective } & \\\\ f^{-1}(B)\\subset A &\\\\ \\end{cases} \\implies g\\circ f \\circ f^{-1} \\mbox{ injective}$$\n\n\u2022 Since \\forall x\\in B \\qquad \\begin{align} (g\\circ f)\\circ f^{-1}(x)&=g\\circ (f\\circ f^{-1})(x)\\\\ &=g\\circ {\\rm id}_{B}(x)\\\\ &=g({\\rm id}_{B}(x))\\\\ (g\\circ f)\\circ f^{-1}(x)&=g(x)\\\\ \\end{align} then $$(g\\circ f)\\circ f^{-1}=g$$ since $(g\\circ f)\\circ f^{-1}$ injective then $g$ is injective\n\nI've proved it on my own like this:\n\nPick two arbitrary elements of $B$, $y_1$ and $y_2$, with $g(y_1)=g(y_2)$. Since $f$ is surjective, $y_1=f(x_1)$ and $y_2=f(x_2)$ for some $x_1,x_2 \\in A$. Then $g(f(x_1))=g(f(x_2))$. Since $f \\circ g$ is injective, $x_1=x_2$, and so $f(x_1)=f(x_2)$, or $y_1=y_2$. Finally, $g$ is injective.\n\n\u2022 Seems to be exactly the same proof as the one in the question and in goblin's answer. Please consider whether you're contributing something new before answering, particularly given that (1) this question already has an accepted answer, and (2) the question is 3 years old. \u2013\u00a0epimorphic Feb 3 '17 at 23:44\n\u2022 Of course I'm contributing something new, and I always consider it. My proof doesn't look like the others at all... And ironically, two of the other answers recommended OP reading different styles of proofs... So yeah, I'm contributing to future people who stumble here looking for help. And, also, the three answers are basically \"your proof is correct, here's an alternative proof/how I did it\". Also, are you a moderator? \u2013\u00a0anon Feb 4 '17 at 15:28\n\u2022 The two answers by others not prefaced with \"here's my version of the proof, which is logically similar to yours but just differs on a few stylistic dimensions\" each have a different sequence of implications, not just a different \"style\". That's what the \"more than one way\" in the first answer usually means, and what \"I've proved it on my own like this\" seems to suggest to me. I'd be willing to reverse my vote if you would explicitly say something like \"here's a more prose-like style\". \u2013\u00a0epimorphic Feb 4 '17 at 19:08\n\u2022 To answer your last question: Primary moderation on this site is provided by the community. IIRC I found your answer through the review queues, where the only posts visible are this answer, the question, and any attached comments. And here's an answer to an old question that claims \"I've proved it on my own like this\" and yet looks logically identical to the proof in the OP... \u2013\u00a0epimorphic Feb 4 '17 at 19:08", "date": "2019-09-23 00:30:05", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/606257/if-g-circ-f-is-injective-and-f-is-surjective-then-g-is-injective", "openwebmath_score": 0.8811041116714478, "openwebmath_perplexity": 344.298289022471, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9888419680942663, "lm_q2_score": 0.9241418121440552, "lm_q1q2_score": 0.9138302083187293}}
{"url": "https://gmatclub.com/forum/there-are-8-teams-in-a-certain-league-and-each-team-plays-134582.html", "text": "It is currently 22 Nov 2017, 04:20\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# There are 8 teams in a certain league and each team plays\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nIntern\nJoined: 12 May 2012\nPosts: 25\n\nKudos [?]: 201 [1], given: 19\n\nLocation: United States\nConcentration: Technology, Human Resources\nThere are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n17 Jun 2012, 03:52\n1\nKUDOS\n26\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n5% (low)\n\nQuestion Stats:\n\n79% (00:43) correct 21% (00:41) wrong based on 1458 sessions\n\n### HideShow timer Statistics\n\nThere are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?\n\nA. 15\nB. 16\nC. 28\nD. 56\nE. 64\n[Reveal] Spoiler: OA\n\nLast edited by Bunuel on 17 Jun 2012, 03:56, edited 1 time in total.\nEdited the question and added the OA.\n\nKudos [?]: 201 [1], given: 19\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 42302\n\nKudos [?]: 133018 [3], given: 12402\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n17 Jun 2012, 03:57\n3\nKUDOS\nExpert's post\n3\nThis post was\nBOOKMARKED\nsarb wrote:\nThere are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?\n\nA. 15\nB. 16\nC. 28\nD. 56\nE. 64\n\nThe total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.\n\n_________________\n\nKudos [?]: 133018 [3], given: 12402\n\nDirector\nStatus: Gonna rock this time!!!\nJoined: 22 Jul 2012\nPosts: 506\n\nKudos [?]: 72 [0], given: 562\n\nLocation: India\nGMAT 1: 640 Q43 V34\nGMAT 2: 630 Q47 V29\nWE: Information Technology (Computer Software)\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n14 Nov 2012, 06:23\nBunuel wrote:\nsarb wrote:\nThere are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?\n\nA. 15\nB. 16\nC. 28\nD. 56\nE. 64\n\nThe total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.\n\nI would like to learn about $$C^2_{8}=28$$. Manhattan Book doesn't discuss this approach. They have anagram approach.\n_________________\n\nhope is a good thing, maybe the best of things. And no good thing ever dies.\n\nWho says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595\n\nMy GMAT Journey : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992\n\nKudos [?]: 72 [0], given: 562\n\nIntern\nJoined: 27 Aug 2012\nPosts: 18\n\nKudos [?]: 7 [0], given: 55\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n15 Nov 2012, 23:27\nHi Bunnel,\n\nI would also like to learn this approach. Can u help me?\n\nSree\n\nKudos [?]: 7 [0], given: 55\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 42302\n\nKudos [?]: 133018 [2], given: 12402\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n16 Nov 2012, 04:19\n2\nKUDOS\nExpert's post\n12\nThis post was\nBOOKMARKED\nSachin9 wrote:\nBunuel wrote:\nsarb wrote:\nThere are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?\n\nA. 15\nB. 16\nC. 28\nD. 56\nE. 64\n\nThe total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.\n\nI would like to learn about $$C^2_{8}=28$$. Manhattan Book doesn't discuss this approach. They have anagram approach.\n\nWell the game is played by 2 teams. How many games are needed if there are 8 teams and each team plays each of the other teams exactly once? The number of games will be equal to the number of different pairs of 2 teams we can form out of 8 teams (one game per pair). How else?\n\nSimilar questions to practice:\nhow-many-diagonals-does-a-polygon-with-21-sides-have-if-one-101540.html\nif-10-persons-meet-at-a-reunion-and-each-person-shakes-hands-110622.html\nhow-many-different-handshakes-are-possible-if-six-girls-129992.html\n15-chess-players-take-part-in-a-tournament-every-player-55939.html\nthere-are-5-chess-amateurs-playing-in-villa-s-chess-club-127235.html\nif-each-participant-of-a-chess-tournament-plays-exactly-one-142222.html\n\nHope it helps.\n_________________\n\nKudos [?]: 133018 [2], given: 12402\n\nIntern\nJoined: 18 Oct 2012\nPosts: 4\n\nKudos [?]: 19 [5], given: 9\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n18 Nov 2012, 21:09\n5\nKUDOS\n2\nThis post was\nBOOKMARKED\nThese type of problems can be solved with a simple diagram.\n\n1. Draw a table consisting of 8 columns and 8 rows.\n2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.\n3. The number should be 28.\n\nKudos [?]: 19 [5], given: 9\n\nSenior Manager\nJoined: 13 Aug 2012\nPosts: 458\n\nKudos [?]: 558 [2], given: 11\n\nConcentration: Marketing, Finance\nGPA: 3.23\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n28 Dec 2012, 06:59\n2\nKUDOS\n1\nThis post was\nBOOKMARKED\nsarb wrote:\nThere are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?\n\nA. 15\nB. 16\nC. 28\nD. 56\nE. 64\n\n$$=\\frac{8!}{2!6!}=4*7 = 28$$\n\n_________________\n\nImpossible is nothing to God.\n\nKudos [?]: 558 [2], given: 11\n\nVP\nJoined: 09 Jun 2010\nPosts: 1393\n\nKudos [?]: 168 [5], given: 916\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n31 Jan 2013, 01:37\n5\nKUDOS\n1\nThis post was\nBOOKMARKED\nit is not easy and is harder if we are on the test date.\n\nthere are 8 team\nto take out 2 teams, IF ORDER MATTERS we have 8*7\nbut in fact order does not matter\n\n8*7/2=28\n\nprinceton gmat book explain this point wonderfully.\n_________________\n\nvisit my facebook to help me.\non facebook, my name is: thang thang thang\n\nKudos [?]: 168 [5], given: 916\n\nIntern\nJoined: 14 Jan 2013\nPosts: 3\n\nKudos [?]: 5 [0], given: 54\n\nLocation: Austria\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n16 May 2013, 01:13\nBunuel wrote:\n\nThe total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.\n\nHi Bunuel,\n\nI have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?\n\nThanks a lot.\n\nKudos [?]: 5 [0], given: 54\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 42302\n\nKudos [?]: 133018 [1], given: 12402\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n16 May 2013, 04:33\n1\nKUDOS\nExpert's post\n4\nThis post was\nBOOKMARKED\nmywaytomba wrote:\nBunuel wrote:\n\nThe total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$.\n\nHi Bunuel,\n\nI have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?\n\nThanks a lot.\n\nCheck combinatorics chapter of Math Book for theory: math-combinatorics-87345.html\n\nAlso check some questions on combinations to practice:\nDS: search.php?search_id=tag&tag_id=31\nPS: search.php?search_id=tag&tag_id=52\n\nHard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html (there are some about permutation too)\n\nHope it helps.\n_________________\n\nKudos [?]: 133018 [1], given: 12402\n\nManager\nJoined: 22 Apr 2013\nPosts: 88\n\nKudos [?]: 36 [0], given: 95\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n17 May 2013, 21:18\npranav123 wrote:\nThese type of problems can be solved with a simple diagram.\n\n1. Draw a table consisting of 8 columns and 8 rows.\n2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.\n3. The number should be 28.\n\nThat's one of those tips that can make my life easier. I'm book marking this page. Thanks.\n_________________\n\nI do not beg for kudos.\n\nKudos [?]: 36 [0], given: 95\n\nManager\nJoined: 13 Jul 2013\nPosts: 69\n\nKudos [?]: 13 [0], given: 21\n\nGMAT 1: 570 Q46 V24\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n26 Dec 2013, 23:04\nLets assume the question asks There are 8 teams in a certain league and each team\nplays each of the other teams exactly twice. If each\ngame is played by 2 teams, what is the total number\nof games played?\n\nThen is 28*2 the correct approach?\n\nKudos [?]: 13 [0], given: 21\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 42302\n\nKudos [?]: 133018 [1], given: 12402\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n27 Dec 2013, 03:08\n1\nKUDOS\nExpert's post\n4\nThis post was\nBOOKMARKED\n\nKudos [?]: 133018 [1], given: 12402\n\nManager\nJoined: 07 Apr 2014\nPosts: 138\n\nKudos [?]: 31 [0], given: 81\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n12 Sep 2014, 06:36\nsarb wrote:\nThere are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?\n\nA. 15\nB. 16\nC. 28\nD. 56\nE. 64\n\ntotal 8 teams & each game by 2 pair then 8C2\n\nKudos [?]: 31 [0], given: 81\n\nIntern\nJoined: 20 Sep 2014\nPosts: 9\n\nKudos [?]: 4 [1], given: 49\n\nThere are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n26 Nov 2014, 02:14\n1\nKUDOS\n1\nThis post was\nBOOKMARKED\nSreeViji wrote:\nHi Bunnel,\n\nI would also like to learn this approach. Can u help me?\n\nSree\n\nHey SreeViji,\n\nThe answer here is the combination 8C2 (8 teams Choose 2) which mean \\frac{8!}{6!x2!} --> \\frac{8x7}{2}\n\nTo understand that we just have to think that each of the 8 team plays against 7 other (8x7) but they play each team exactly once so we divide the total by 2.\n\nWe divide by 2 because \"TEAM A VS TEAM B\" is the same as \"TEAM B VS TEAM A\"\n\nSo we end up with \\frac{8x7}{2}= 28\n\nIf you still have trouble with combination and permutation check out this website it's well done,\nhttp://www.mathsisfun.com/combinatorics ... tions.html\n\nhope it helps.\n\nLast edited by quentin.louviot on 13 Jan 2015, 07:51, edited 2 times in total.\n\nKudos [?]: 4 [1], given: 49\n\nIntern\nJoined: 15 Sep 2014\nPosts: 8\n\nKudos [?]: 3 [0], given: 0\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n26 Nov 2014, 02:57\nIf there are n teams need to play exactly once ,then they play with (n-1) teams but as they are playing together then, n(n-1)/2, which means nC2\n\nSo 8*7/2 =28.\n\nPosted from my mobile device\n\nKudos [?]: 3 [0], given: 0\n\nSenior Manager\nStatus: Math is psycho-logical\nJoined: 07 Apr 2014\nPosts: 437\n\nKudos [?]: 141 [0], given: 169\n\nLocation: Netherlands\nGMAT Date: 02-11-2015\nWE: Psychology and Counseling (Other)\nThere are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n12 Jan 2015, 12:48\n1\nThis post was\nBOOKMARKED\nI also used a table to do this, like that:\n\n1_2_3_4_5_6_7_8\n1_1_1_1_1_1_1_1\n2_2_2_2_2_2_2_2\n3_3_3_3_3_3_3_3\n4_4_4_4_4_4_4_4\n5_5_5_5_5_5_5_5\n6_6_6_6_6_6_6_6\n7_7_7_7_7_7_7_7\n8_8_8_8_8_8_8_8\n\nThen you delete the same team pairs: e.g. 1-1, 2-2, 3-3 and then 2-1 (because you have 1-2), 3-2 (because you have 2-3). After you cross out the first 2 columns you then see that you cross out everything from the diagonal and below. The remaining is 28.\n\nHowever, the 8!/2!*6! approach is better, because if you have many numbers the table will take forever to draw. In case there is sth similar though and your brain gets stuck, use the table...\n\nKudos [?]: 141 [0], given: 169\n\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 10158\n\nKudos [?]: 3530 [2], given: 173\n\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: 340 Q170 V170\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n12 Jan 2015, 14:46\n2\nKUDOS\nExpert's post\n2\nThis post was\nBOOKMARKED\nHi All,\n\nUsing the Combination Formula IS one way to approach these types of questions, but it's not the only way. Sometimes the easiest way to get to a solution on Test Day is to just draw a little picture and keep track of the possibilities....\n\nLet's call the 8 teams: ABCD EFGH\n\nWe're told that each team plays each other team JUST ONCE.\n\nA plays BCD EFGH = 7 games total\n\nTeam B has ALREADY played team A, so those teams CANNOT play again...\nB plays CD EFGH = 6 more games\n\nTeam C has ALREADY played teams A and B, so the following games are left...\nC plays D EFGH = 5 more games\n\nAt this point, you should notice a pattern: the additional number of games played is reduced by 1 each time. So what we really have is:\n\n7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 28 games played\n\n[Reveal] Spoiler:\nC\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\n\n760+: Learn What GMAT Assassins Do to Score at the Highest Levels\nContact Rich at: Rich.C@empowergmat.com\n\n# Rich Cohen\n\nCo-Founder & GMAT Assassin\n\nSpecial Offer: Save \\$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************\n\nKudos [?]: 3530 [2], given: 173\n\nSenior Manager\nJoined: 25 Feb 2010\nPosts: 448\n\nKudos [?]: 112 [0], given: 10\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n19 May 2015, 07:31\npranav123 wrote:\nThese type of problems can be solved with a simple diagram.\n\n1. Draw a table consisting of 8 columns and 8 rows.\n2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.\n3. The number should be 28.\n\nHi,\n\nI don;t think we need to count the half spaces. with half space count is 36.\nwithout half space - count: 28.\n_________________\n\nGGG (Gym / GMAT / Girl) -- Be Serious\n\nIts your duty to post OA afterwards; some one must be waiting for that...\n\nKudos [?]: 112 [0], given: 10\n\ne-GMAT Representative\nJoined: 04 Jan 2015\nPosts: 746\n\nKudos [?]: 2168 [2], given: 123\n\nRe: There are 8 teams in a certain league and each team plays\u00a0[#permalink]\n\n### Show Tags\n\n20 May 2015, 03:10\n2\nKUDOS\nExpert's post\nonedayill wrote:\npranav123 wrote:\nThese type of problems can be solved with a simple diagram.\n\n1. Draw a table consisting of 8 columns and 8 rows.\n2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.\n3. The number should be 28.\n\nHi,\n\nI don;t think we need to count the half spaces. with half space count is 36.\nwithout half space - count: 28.\n\nDear onedayill\n\nYou're right!\n\nThe boxes along the diagonal (these are the boxes that contribute to half spaces) represent a team playing with itself. Since that is not possible, these boxes should not be included in the counting.\n\nI noticed in the thread above that a few students had doubts about the expression 8C2. If any of the current students too have such a doubt, here's how this question could be solved visually:\n\nThere are 7 ways in which Team 1 can play with another team.\nSimilarly, there are 7 ways for each of the 8 teams to choose its playing opponent.\n\nBut it's easy to see that the red zone is essentially a duplication of the blue zone. For example, (Team 1 playing with Team 2) is the same case as (Team 2 playing with Team 1)\n\nSo, the correct answer will be: 8(that is, the number of teams)*7(that is, the number of ways in which each team can choose its playing opponent)/2 = 28\n\nHope this was useful!\n\nBest Regards\n\nJapinder\n_________________\n\n| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com\n\nKudos [?]: 2168 [2], given: 123\n\nRe: There are 8 teams in a certain league and each team plays \u00a0 [#permalink] 20 May 2015, 03:10\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0 Next \u00a0[ 29 posts ]\n\nDisplay posts from previous: Sort by", "date": "2017-11-22 11:20:45", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/there-are-8-teams-in-a-certain-league-and-each-team-plays-134582.html", "openwebmath_score": 0.18992069363594055, "openwebmath_perplexity": 3576.657479026001, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes.\n2. Yes.\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.9136765275272111, "lm_q1q2_score": 0.9136765275272111}}
{"url": "https://math.stackexchange.com/questions/3661917/necklace-combinations-with-three-group-of-beads", "text": "# Necklace combinations with three group of beads\n\nI have a hard question about a way how many different necklaces can be made.\n\nSuppose that we have the following restrictions:\n\n1. We have 3 groups of beads:\n\nAll the beads in one group are completely identical. This means that if you put two triangle beads next to each other and then switch their positions this counts as one necklace because the beads are identical\n\n1. Necklaces are identical if they are identical under symmetric operations just as rotate them (\ud835\udc5f) or turning them around (\ud835\udc60).\n\nSo if we have a necklace ordered in one way and we rotate it 180 deg or just flip a side this is count as one necklace.\n\n1. We need to use all the 18 beads in each and every new necklace. We can not create a necklace from 17, 16 or less than 18 beads.\n\nI read all the topics here but could not find a question about a group of identical beads. I also read Burnside lemma and P\u00f3lya_enumeration_theorem and Necklace_(combinatorics) in wikipedia, but could not find a way how to solve this and what is the correct answer.\n\nFrom Burnside lemma, I found that the answer should be 57, but is this correct?\n\nI used directly the formula from Burnside lemma, but it does not seem quite right for me, because I do not take into account that the three groups are with different numbers of beads.\n\n$$\\frac{1}{24} * (n^6 + 3 * n^4 + 12 * n^3 + 8 * n^2)$$\n\nwhere n is 3 from three groups.\n\n$$\\frac{1}{24} * (3^6 + 3 * 3^4 + 12 * 3^3 + 8 * 3^2) = 57$$\n\nHowever, as I said earlier despide the fact that the result looks some kind realisting I am not sure that this is the right answer, because I do not use in the formula that we have 4 triangle, 6 square and 8 circle beads.\n\nIt looks like P\u00f3lya enumeration theorem weighted version is the thing that I need. However, I am not sure how to get to the right answer\n\n\u2022 Welcome to MSE! Pls show your work for arriving at $57$ so someone can check its correctness. You can do that by editing your question and e.g. putting your work at the end. May 6 '20 at 20:53\n\u2022 Thanks @antkam, for the answer. I modified the questions and added the formula that I used and a short description why I think that despide the fact that the result looks like a good number I think that this is not quite right. May 7 '20 at 5:54\n\u2022 May 7 '20 at 16:19\n\u2022 Does this answer your question? Circular permutations with indistinguishable objects May 7 '20 at 16:19\n\u2022 I think you should be using the word \"bracelet\" and not \"necklace\". The answer in the link provided by @Vepir gives an answer to your problem for necklaces, which means objects with a cyclic symmetry group, when you want a dihedral symmetry group. This being said you should be able to follow the same line of argument to find the formula you need. May 11 '20 at 20:04\n\nI manage to answer the question and this is the process that I followed:\n\nI consider the 18-bead necklace in the first part of the problem. Here are the eighteen rotations expressed in cycle form where we assume that the slots are numbered from 1 to 18 in clockwise order. The first is the identity (e: no rotation) and the second is the generator g\u2014a rotation by a single position which, when repeated, generates all the elements of the group:\n\n$$e = g^0 \\text{ = (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18)}$$\n\n$$g^1 \\text{ = (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18) }$$\n\n$$g^2 \\text{= (1 3 5 7 9 11 13 15 17) (2 4 6 8 10 12 14 16 18)}$$\n\n$$g^3 \\text{= (1 4 7 10 13 16) (2 5 8 11 14 17) (3 6 9 12 15 18)}$$\n\n$$g^4 \\text{= (1 5 9 13 17 3 7 11 15) (2 6 10 14 18 4 8 12 16)}$$\n\n$$g^5 \\text{= (1 6 11 16 3 8 13 18 5 10 15 2 7 12 17 4 9 14)}$$\n\n$$g^6 \\text{= (1 7 13) (2 8 14) (3 9 15) (4 10 16) (5 11 17) (6 12 18)}$$\n\n$$g^7 \\text{= (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18)}$$\n\n$$g^8 \\text{= (1 9 17 7 15 5 13 3 11) (2 10 18 8 16 6 14 4 12)}$$\n\n$$g^9 \\text{= (1 10) (2 11) (3 12) (4 13) (5 14) (6 15) (7 16) (8 17) (9 18)}$$\n\n$$g^{10} \\text{= (1 11 3 13 5 15 7 17 9) (2 12 4 14 6 16 8 18 10)}$$\n\n$$g^{11} \\text{= (1 12 5 16 9 2 13 6 17 10 3 14 7 18 11 4 15 8)}$$\n\n$$g^{12} \\text{= (1 13 7) (2 14 8) (3 15 9) (4 16 10) (5 17 11) (6 18 12)}$$\n\n$$g^{13} \\text{= (1 14 9 4 17 12 7 2 15 10 5 18 13 8 3 16 11 6)}$$\n\n$$g^{14} \\text{= (1 15 11 7 3 17 13 9 5) (2 16 12 8 4 18 14 1 6)}$$\n\n$$g^{15} \\text{= (1 16 13 10 7 4) (2 17 14 11 8 5) (3 18 15 12 9 6)}$$\n\n$$g^{16} \\text{= (1 17 15 13 11 9 7 5 3) (2 18 16 14 12 10 8 6 4)}$$\n\n$$g^{17} \\text{= (1 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2)}$$\n\nAfter that I found the GCD for all cycle form's with and group them in a table:\n\n| Cycle length | Permutations | GCD with 18 |\n\n| 1 | $$g^0$$ | GCD(0, 18)=18 |\n\n| 2 | $$g^9$$ | GCD(9, 18)=9 |\n\n| 3 | $$g^6$$, $$g^{12}$$ | GCD(6, 18)=GCD(12, 18)=6 |\n\n| 6 | $$g^3$$, $$g^{15}$$ | GCD(3, 18)=GCD(15, 18)=3 |\n\n| 9 | $$g^2$$, $$g^4$$, $$g^8$$, $$g^{10}$$, $$g^{14}$$, $$g^{16}$$ | GCD(2, 18)=GCD(4, 18)=GCD(8, 18)=GCD(10, 18)=GCD(14, 18)=GCD(16, 18)=2 |\n\n| 18 | $$g^1$$, $$g^5$$, $$g^7$$, $$g^{11}$$, $$g^{13}$$, $$g^{17}$$ | GCD(1, 18)=GCD(5, 18)=GCD(7, 18)=GCD(11, 18)=GCD(13, 18)=GCD(17, 18)=1 |\n\nWe have 18 permutations for rotation and lets name cycle 1 with $$f_1$$, cycle 2 with $$f_2$$ .. cycle n with $$f_n$$\n\nThan the formula for Cycling index is:\n\n$$\\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1}{18}$$\n\nIf we solve all the possible necklaces with three colors the result should be (for the moment we do not solve for the three colors with 4, 6 and 8 beads in respective groups):\n\n$$\\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1}{18} = \\text{21 524 542}$$\n\nFrom here because turn around is allowed we need to add and the necklace(bracelet if we follow the right terms) is with even beads we should add the symethric turn arounds.\n\n$$\\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1 + 9f_1^2f_2^8 + 9f_2^9}{2 * 18}$$\n\nand again for three colors without including the different weight:\n\n$$\\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1 + 9 * 3^8 + 9 * 3^9}{2 * 18} = \\text{10 781 954}$$\n\nFor the moment we have all the possible necklaces and bracelets with three colors. However, in order to find the necklaces and bracelets with three colors (4 reds, 6 greens and 8 blues) we need to replace:\n\n$$f_1 = (x + y + z)$$\n\n$$f_2 = (x^2 + y^2 + z^2)$$\n\n$$f_3 = (x^3 + y^3 + z^3)$$\n\n$$f_6 = (x^6 + y^6 + z^6)$$\n\n$$f_9 = (x^9 + y^9 + z^9)$$\n\n$$f_{18} = (x^{18} + y^{18} + z^{18})$$\n\nand if we replace in the formula it becomes:\n\n$$\\frac{(x + y + z)^{18} + (x^2 + y^2 + z^2)^9 + 2(x^3 + y^3 + z^3)^6 + 2(x^6 + y^6 + z^6)^3 + 6(x^9 + y^9 + z^9)^2 + 6(x^{18} + y^{18} + z^{18}) + 9(x + y + z)^2(x^2 + y^2 + z^2)^8 + 9(x^2 + y^2 + z^2)^9}{36}$$\n\nThen we need to find which expressions can expand to $$x^4y^6z^8$$.\n\nAfter then by using multinominal coeficient I managed to calculate the following results\n\n9 189 180\n\n1260\n\n11 340\n\n11 340\n\nThen I sum all of them and divide them to 36. This gives me the answer of 255 920 which is the answer of the question. We can create 255 920 bracelets with 4 red 6 green and 8 blue beads.", "date": "2021-09-27 19:47:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3661917/necklace-combinations-with-three-group-of-beads", "openwebmath_score": 0.6429612636566162, "openwebmath_perplexity": 449.021154167852, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.978712648206549, "lm_q2_score": 0.9334308073258007, "lm_q1q2_score": 0.9135605373554114}}
{"url": "https://math.stackexchange.com/questions/3060802/is-v-isomorphic-to-direct-sum-of-subspace-u-and-v-u", "text": "# Is $V$ isomorphic to direct sum of subspace $U$ and $V/U$?\n\nGiven a vector space $$V$$ and a subspace $$U$$ of $$V$$. $$V \\cong U \\oplus(V/U)$$ Does the above equation always hold? Where $$\\oplus$$ is external direct sum. For finite dimensional vector space $$V$$, here is my attemp of prove:\n\nLet dimension of $$U$$ be m, dimension of $$V$$ be $$n$$. Find a basis of $$U$$ : $$\\{ \\mathbf{ u_1, u_2, \\cdots ,u_m}\\}$$ and extend it to a basis for $$V$$ : $$\\{ \\mathbf{ u_1, u_2, \\cdots ,u_m, v_1, \\cdots,v_{n-m} } \\}$$.\n\nFor every vector $$\\mathbf{x} \\in V$$, we can write $$\\mathbf{x}= c_1 \\mathbf{u_1}+ \\cdots+ c_m \\mathbf{u_m} + d_1 \\mathbf{v_1}+ \\cdots d_{n-m} \\mathbf{v_{n-m}}$$ uniquely. Define a linear map $$T$$ as $$T(\\mathbf{x})=(c_1 \\mathbf{u_1}+ \\cdots+ c_m \\mathbf{u_m}, [d_1 \\mathbf{v_1}+ \\cdots d_{n-m} \\mathbf{v_{n-m}}])$$ ,where $$[]$$ is used to express the equivalent class. We claim $$T$$ is an isomorphism.\n\nSurjectivity is obvious. As for injectivity,\nif $$T(\\mathbf{x})=(\\mathbf{0},[\\mathbf{0}])$$, then $$c_1 \\mathbf{u_1}+ \\cdots+ c_m\\mathbf{u_m}= \\mathbf{0}$$\n$$\\Rightarrow c_1=0, c_2=0, \\cdots,c_m=0$$\n$$\\Rightarrow x=d_1 \\mathbf{v_1}+ \\cdots d_{n-m} \\mathbf{v_{n-m}}$$\nSince $$[d_1 \\mathbf{v_1}+ \\cdots d_{n-m} \\mathbf{v_{n-m}}] = [\\mathbf{0}]$$, we have $$(d_1 \\mathbf{v_1}+ \\cdots d_{n-m} \\mathbf{v_{n-m}}-\\mathbf{0})\\in U$$, which means $$d_1=0, d_2=0, \\cdots,d_{n-m}=0$$\n$$\\Rightarrow \\mathbf{x}=\\mathbf{0}$$, so $$T$$ is injective.\n\nIs the above proof correct? Does this mean $$V \\cong \\ker F \\oplus (V/ \\ker F) \\cong \\ker F \\oplus\\mathrm{im}F$$ for any linear map $$F$$, because $$\\ker F$$ is a subspace of $$V$$ ?\n\nThe final question is about how should I prove it when the dimension of $$V$$ is infinite?\n\nIt is true that for every finite dimensional vector space $$V$$ with $$U$$ a vector subspace that $$V \\cong U \\oplus (V / U)$$ I think your proof is essentially correct. And yes it is true that for $$T: V \\rightarrow W$$ any linear that we have $$V \\cong \\operatorname{ker}(T) \\oplus \\operatorname{Im}(T)$$ The rank-nullity theorem is a direct consequence of this.\n\nIn more technical language, we say that every \"short exact sequence\" of finite dimensional vector spaces over a field $$k$$ $$\\textit{splits}$$. What this means is that if $$T : U \\rightarrow V$$, $$S : V \\rightarrow W$$ are linear maps such that $$T$$ is injective, $$\\operatorname{ker}(S) = \\operatorname{Im}(T)$$, and $$S$$ is surjective, then $$V \\cong U \\oplus W$$.\n\nNote then that this directly gives us our result since if $$U$$ is a subspace of $$V$$, then the inclusion map $$\\iota : U \\rightarrow V$$ and projection map $$\\pi : V \\rightarrow (V / U)$$ set up exactly a short exact sequence.\n\nIn terms of whether or not this extends to infinite dimensional vector spaces, the result does hold again (assuming the axiom of choice), and the proof is essentially the same. All your proof relies on is the ability to extend a basis of a subspace to a basis of your entire space. We can do this with the axiom of choice.\n\n\u2022 I am wondering if the following statement is false for infinite dimensional spaces. \"If $V$ is a vector space and $U$ is a subspace of $V$, then there exists another subspace of $V$ called $U^\\perp$ such that every element of $V$ can be uniquely expressed as the sum of an element from $U$ with an element from $U^\\perp$.\" (I am thinking a counterexample would be if $U$ was the set of real number sequences with finite support (i.e. eventually zero) and $V$ is the set of all real number sequences.) \u2013\u00a0irchans Jan 3 at 19:03\n\u2022 @irchans If we take the axiom of choice, then every subspace of a vector space has a direct sum complement (what you call the perpendicular space, but this language is typically reserved for a space equipped with some bi-linear form). The proof is pretty simple. Let $V$ be a $k$-vector space, with $U$ a vector subspace. Let $\\mathcal{B}_{U}$ be a basis for $U$ and extend it to a basis $\\mathcal{B}_{V}$ (using the axiom of choice) for $V$. Then let $W = \\operatorname{Span}_{k}\\left( \\mathcal{B}_{V} \\backslash \\mathcal{B}_{U} \\right)$. Then $V = U \\oplus W$. \u2013\u00a0Adam Higgins Jan 3 at 19:13\n\u2022 @irchans Perhaps the reason you think that your example is a counter example is because of the $\\textit{weirdness}$ of bases of infinite dimensional vector spaces. Notice that a subset $S$ of a vector space $V$ is said to be a basis if and only if every element $v \\in V$ can be written as a $\\textbf{finite}$ linear combination of the elements of $S$, and that there is no finite non-trivial linear relation amongst the elements of $S$. \u2013\u00a0Adam Higgins Jan 3 at 19:19\n\u2022 Thank you very much ! \u2013\u00a0irchans Jan 3 at 19:27\n\u2022 This set of notes seems relevant math.lsa.umich.edu/~kesmith/infinite.pdf \u2013\u00a0irchans Jan 3 at 19:38\n\nYour proof works. The answer to your second question is yes, that is true. For an infinite dimensional vector space, take any linear map $$F: V -> W$$. Then $$U = \\ker F$$ is a subspace of $$V$$. Note that we have a short exact sequence (if you don't know what that means, don't worry, the explanation is coming) $$0\\to U\\to V\\to V/U\\to 0$$\n\n(That is, there's an injective map $$U \\to V$$ (inclusion, I'll call it $$i$$) and a surjective map $$V \\to V/U$$ (the quotient map, I'll call it $$q$$) such that the image of the injection is the kernel of the surjection).\n\nBut there's also a surjective map $$V \\to U$$ (projection onto $$U$$, I'll call it $$p$$), and note that for any $$u \\in U$$, $$pi(u) = u$$ (since $$p$$ fixes $$u$$).\n\nNow, we're going to show that $$V$$ is the (internal) direct sum of the kernel of $$p$$ and the image of $$i$$. First, note that it's the sum of the two: for any $$v \\in V$$, $$v = (v - ip(v)) + ip(v)$$, $$ip(v)$$ is obviously in the image of $$i$$, and $$p(v - ip(v)) = p(v) - pip(v) = 0$$ (with the last equality being due to our note about $$p$$, since $$p(v)\\in U$$.\n\nAnd further, the intersection is trivial: if $$v \\in \\ker(p)\\cap\\mathrm{im}(i)$$ then there is some $$u\\in U$$ such that $$i(u) = v$$, and $$pi(u) = p(v) = 0$$, but $$pi(u) = u$$, so $$u = 0$$, hence $$v = 0$$. Thus, $$V = \\ker(p)\\oplus \\mathrm{im}(i)$$.\n\nNow, it's clear that $$\\mathrm{im}(i)\\cong U$$ since it's the image of $$U$$ under an injective map, so we need only show that $$\\ker(p)\\cong V/U$$.\n\nFor that purpose, since $$q$$ is surjective, for any $$w \\in V/U$$, there is some $$v \\in V$$ such that $$w = q(b)$$. But since $$V = \\ker(p)\\oplus\\mathrm{im}(i)$$, there are unique $$u \\in U$$, $$x \\in \\ker(p)$$ such that $$v = i(u) + x$$, so $$w = q(b) = q(i(u)+x) = qi(u) + q(x)=q(x)$$ (since the image of $$i$$ is the kernel of $$q$$, in particular $$qi = 0$$). Thus, $$q|_{\\ker(p)}: \\ker(p)\\to V/U$$ is surjective. But also, if $$q(v) = 0$$, and $$v \\in \\ker(p)$$. But $$\\ker(q) = \\mathrm{im}(i)$$, and $$p$$ fixes $$\\mathrm{im}(i)$$, so the only element that it sends to $$0$$ is $$0$$ itself, so we must have $$v = 0$$, hence $$q|_{\\ker(p)}$$ is also injective, so is an isomorphism.\n\nThus, we have $$V \\cong \\mathrm{im}(i)\\oplus\\ker(p) \\cong U\\oplus (V/U)$$, as required.\n\nThis is precisely a special case of the Splitting Lemma, and my proof is essentially just one part of the proof of that, translated: there are easier ways to prove it, but I thought that it would be useful to see it in a more broadly applicable form.", "date": "2019-05-23 04:37:51", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3060802/is-v-isomorphic-to-direct-sum-of-subspace-u-and-v-u", "openwebmath_score": 0.9572270512580872, "openwebmath_perplexity": 65.96755087668768, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9918120896142623, "lm_q2_score": 0.920789679151471, "lm_q1q2_score": 0.9132503357744666}}
{"url": "https://math.stackexchange.com/questions/2928956/query-on-a-solution-to-the-problem-gcd5a2-7a3-1-for-all-integer-a", "text": "# Query on a Solution to the Problem: $\\gcd(5a+2,7a+3)=1$ for all integer $a$.\n\nI wish to show that the numbers $$5a+2$$ and $$7a+3$$ are relatively prime for all positive integer $$a$$.\n\nHere are my solutions.\n\nSolution 1. I proceed with Euclidean Algorithm. Note that, for all $$a$$, $$|5a+2|<|7a+3|$$. By Euclidean Algorithm, we can divide $$7a+3$$ by $$5a+2$$. To have\n\n$$7a+3=(5a+2)(1)+(2a+1)$$ continuing we have,\n\n$$5a+2=(2a+1)(2)+a$$\n\n$$2a+1=(2)(a)+1$$\n\n$$2=(1)(2)+0$$\n\nSince the last nonzero remainder in the Euclidean Algorithm for $$7a+3$$ and $$5a+2$$ is 1, we conclude that they are relatively prime.\n\nSolution 2. Suppose that $$d=\\gcd(5a+2,7a+3)$$. Since $$d=\\gcd(5a+2,7a+3)$$ then the following divisibility conditions follow:\n\n(1) $$d\\mid (5a+2)$$\n\n(2) $$d\\mid (35a+14)$$\n\n(3) $$d\\mid (7a+3)$$\n\n(4) $$d\\mid (35a+15)$$.\n\nNow, (2) and (4) implies that $$d$$ divides consecutive integers. The only (positive) integer that posses this property is $$1$$. Thus, $$d=1$$ and that $$7a+3$$ and $$5a+2$$ are relatively prime.\n\nHere are my questions:\n\n1. Is the first proof correct or needs to be more specific? For instance cases for $$a$$ must be considered.\n\n2. Which proof is better than the other?\n\nThank you so much for your help.\n\n\u2022 I think both approaches are good. Nor are they that different really...in both cases you are trying to find smaller and smaller multiples of $d$.\n\u2013\u00a0lulu\nSep 24 '18 at 15:23\n\u2022 Thank you very much for the kind comment @lulu. Got a follow up question. By \"you are trying to find smaller and smaller multiples of $d$, you mean the process of continously dividing the divisor to remainder so that $r$ decreases? Sep 24 '18 at 15:27\n\u2022 Is there a typo in (4), i.e. shouldn't 35 be multiplied by $a$?\n\u2013\u00a0user431008\nSep 24 '18 at 15:28\n\u2022 I meant something less precise than that. Euclid provides a somewhat systematic way to find new numbers that $d$ divides...the second method is less systematic, but faster as you seek out convenient expressions. Working by hand, I prefer the second method...were I trying to automate the process, I'd prefer the systematic approach.\n\u2013\u00a0lulu\nSep 24 '18 at 15:28\n\u2022 I agree @marmot. Thank you for pointing it out. Sep 24 '18 at 15:29\n\nIn the first solution you're not using, strictly speaking, the Euclidean algorithm, but a looser version thereof:\n\nLet $$a$$, $$b$$, $$x$$ and $$y$$ be integers; if $$a=bx+y$$, then $$\\gcd(a,b)=\\gcd(b,y)$$.\n\nThe proof consists in showing that the common divisors of $$a$$ and $$b$$ are the same as the common divisors of $$b$$ and $$y$$.\n\nThere is no requirement that $$a\\ge b$$ or that $$y$$ is the remainder of the division. Indeed your argument actually has a weakness, because $$7a+3\\ge 5a+2$$ only if $$2a\\ge-1$$, so it doesn't hold for $$a\\le-2$$. But $$7a+3\\ge 5a+2$$ is not really needed for the argument.\n\nSince successive application of the statement above show that $$\\gcd(2a+1,2)=1$$ and the gcd has never changed in the various steps, you can indeed conclude that $$\\gcd(5a+2,7a+3)=1$$.\n\nThe second solution is OK as well.\n\nYou can simplify it by noting that if $$d$$ is a common divisor of $$5a+2$$ and $$7a+3$$, then it divides also $$5(7a+3)-7(5a+2)=1$$\n\n\u2022 Thanks for the clarification in solution 1 and by giving a simplified version for solution 2. It is very clear to me now. @egreg Sep 24 '18 at 15:35\n\nYour first proof is correct. I would perhaps complete it, writing that\\begin{align}1&=(2a+1)-2a\\\\&=2a+1-2\\bigl(5a+2-2(2a+1)\\bigr)\\\\&=5(2a+1)-2(5a+2)\\\\&=5\\bigl(7a+3-(5a+2)\\bigr)-2(5a+2)\\\\&=5(7a+3)-7(5a+2).\\end{align}\n\nYour second proof also works, but it doesn't generalize easily to other situations.\n\n\u2022 Thank you very much for completing the proof of solution 1 Prof. I am interested in a case that solution 2 wont work. Thank you Prof. Sep 24 '18 at 15:41\n\u2022 @JrAntalan Concerning the second proof, I only meant that you have to think about it in a case-by-case basis. Sep 24 '18 at 15:43\n\u2022 Noted Prof. and Thank you again. Sep 24 '18 at 15:44\n\nHere is a different rendering of the same arguments.\n\nLet $$u=5a+2$$, $$v=7a+3$$. Then $$\\pmatrix{ u \\\\ v} = \\pmatrix{ 5 & 2 \\\\ 7 & 3} \\pmatrix{ a \\\\ 1}$$ and so $$\\pmatrix{ a \\\\ 1} = \\pmatrix{ 5 & 2 \\\\ 7 & 3}^{-1} \\pmatrix{ u \\\\ v} = \\pmatrix{ \\hphantom- 3 & -2 \\\\ -7 & \\hphantom-5} \\pmatrix{ u \\\\ v}$$ This gives $$1 = -7u+5v = -7(5a+2)+5(7a+3)$$ The key point here is that the matrix has determinant $$1$$ and so its inverse has integer entries.", "date": "2022-01-28 02:11:42", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2928956/query-on-a-solution-to-the-problem-gcd5a2-7a3-1-for-all-integer-a", "openwebmath_score": 0.8840757608413696, "openwebmath_perplexity": 219.60252031210229, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9770226300899744, "lm_q2_score": 0.9343951556900273, "lm_q1q2_score": 0.9129252125556016}}
{"url": "https://math.stackexchange.com/questions/401753/for-any-arrangment-of-numbers-1-to-10-in-a-circle-there-will-always-exist-a-pai", "text": "# For any arrangment of numbers 1 to 10 in a circle, there will always exist a pair of 3 adjacent numbers in the circle that sum up to 17 or more\n\nI set out to solve the following question using the pigeonhole principle\n\nRegardless of how one arranges numbers $1$ to $10$ in a circle, there will always exist a pair of three adjacent numbers in the circle that sum up to $17$ or more.\n\nMy outline\n\n[1] There are $10$ triplets consisting of adjacent numbers in the circle, and since each number appears thrice, the total sum of these adjacent triplets for all permutations of the number in the circle, is $3\\cdot 55=165$.\n\n[2] If we consider that all the adjacent triplets sum to 16 , and since there are $10$ such triplets, the sum accordingly would be $160$, but we just said the invariant sum is $165$ hence there would have to be a triplet with sum of $17$ or more.\n\nMy query\n\nCould someone polish this into a mathematical proof and also clarify if I did make use of the pigeonhole principle.\n\n\u2022 Yes, you used pigeonhole. Should say if all sums are $\\le 16$, then the sum is $\\le 160$. \u2013\u00a0Andr\u00e9 Nicolas May 25 '13 at 4:42\n\u2022 @Andr\u00e9Nicolas noted and aha , the syntax part is exactly why I asked the question on here. \u2013\u00a0metric-space May 25 '13 at 4:44\n\u2022 @nerorevenge Note: You can actually show that the 3 adjacent numbers sum to 18 or more. \u2013\u00a0Calvin Lin May 25 '13 at 8:37\n\nYes, you used the Pigeonhole Principle. As a very mild correction, you should say that of all sums of three consecutives are $\\le 16$, then the sum is $\\le 160$.\n\nThe proof (with the small correction) is already fully mathematical. Conceivably you might want to explain the $55$ further. It is clear to you and to most users of this site why $55$, but imagine the reader to be easily puzzled.\n\nThe part about \"all the permutations\" is vague, and technically incorrect. You are finding the sum of a consecutive triple, and summing all these sums. Permutations have nothing to do with it, we are talking about a particular fixed arrangement of the numbers.\n\nRemark: We could use a lot of symbols. Starting at a particular place, and going counterclockwise, let our numbers be $a_1,a_2,\\dots,a_{10}$. For any $i$, where $1\\le i\\le 10$, let $S_i=a_i+a_{i+1}+a_{i+2}$, where we use the convention that $a_{10+1}=a_1$, $a_{10+2}=a_2$, and $a_{9+2}=a_1$.\n\nThen $S_1+S_1+\\cdots+S_{10}=165$, since each of $1$ to $10$ appears in $3$ of the $S_i$, and $1+2+\\cdots+10=55$.\n\nBut if all the $S_i$ are $\\le 16$, then $\\sum_{i=1}^{10}S_i\\le 160$. This contradicts the fact that $\\sum_{i=1}^{10}S_i=165$.\n\nI prefer your proof, mildly modified.\n\n\u2022 eh, is that all? the reason I'm asking is that, inb the past my so called 'proofs' have been criticised for being too informal \u2013\u00a0metric-space May 25 '13 at 4:45\n\u2022 @nerorevenge: I am an advocate of the informal but clear. I hope that someone more formal-minded will add an answer. The only issue is that you perhaps left an easily filled gap or two. \u2013\u00a0Andr\u00e9 Nicolas May 25 '13 at 4:53\n\u2022 I have added another criticism. You should really fix the wording there. \u2013\u00a0Andr\u00e9 Nicolas May 25 '13 at 4:58\n\u2022 I agree.That definitely is vaguely worded. \u2013\u00a0metric-space May 25 '13 at 4:59\n\u2022 Presumably the \"pair\" wording came with the problem. But it looks funny to call a trio a pair. \u2013\u00a0Andr\u00e9 Nicolas May 25 '13 at 5:23\n\nWe will show something stronger, namely that there exists 3 adjacent numbers that sum to 18 or more.\n\nLet the integers be $\\{a_i\\}_{i=1}^{10}$. WLOG, $a_1 = 1$. Consider\n\n$$a_2 + a_3 + a_4, a_5 + a_6 + a_7, a_8 + a_9 + a_{10}$$\n\nThe sum of these 3 numbers is $2+3 +\\ldots + 10 = 54$. Hence, by the pigeonhole principle, there exists one number which is at least $\\lfloor \\frac{54}{3} \\rfloor = 18$.\n\nI leave it to you to show that there is a construction where no 3 adjacent numbers sum to 19 or more, which shows that 18 is the best that we can do.\n\n\u2022 definitely a stronger proof ,and I have admit it sure is nice, I wonder why cut-the-knot.org didn't ask people to solve for what you proved. \u2013\u00a0metric-space May 25 '13 at 16:45\n\u2022 @nerorevenge I used this as a problem on Brilliant.org, and most could only get to 17, and didn't understand why 18 must be true. The 18 case arises quite easily by ignoring the small values which would otherwise reduce our sum. \u2013\u00a0Calvin Lin May 25 '13 at 23:37", "date": "2020-12-04 11:16:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/401753/for-any-arrangment-of-numbers-1-to-10-in-a-circle-there-will-always-exist-a-pai", "openwebmath_score": 0.8371453881263733, "openwebmath_perplexity": 381.41878577481674, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363503693294, "lm_q2_score": 0.9273632906279589, "lm_q1q2_score": 0.9128373969632168}}
{"url": "https://math.stackexchange.com/questions/3473894/to-find-number-of-real-roots", "text": "# To find number of real roots\n\nConsider the equation $$x^5-5x=c$$ where c is a real number.\n\nDetermine all c such that this equation has exactly 3 real roots.\n\nI know that between consecutive real roots of $$f$$ there is a real root of $$f'$$. Now $$f'$$ in this case is $$5x^4-5$$ which always has two real roots. So the claim should be true for all c.\n\nBut I KNOW IT IS NOT TRUE. Where am I messing up?\n\n\u2022 Note that while it is necessary for $f'$ to have two real roots (in order for $f$ to have exactly three real roots), it is not sufficient. Of course an odd degree polynomial will always have one real root. \u2013\u00a0hardmath Dec 12 '19 at 16:12\n\u2022 ok so how do I find all such c? \u2013\u00a0Angry_Math_Person Dec 12 '19 at 16:14\n\u2022 Consider the graph of $p(x) = x^5 - 5x$. Changing the constant $c$ in your equation amounts to moving a horizontal line up or down across this graph. \u2013\u00a0hardmath Dec 12 '19 at 16:15\n\nYes, between any two roots of $$f$$, there is a root of $$f'$$. However, just because $$f'$$ has a root, that doesn't mean that $$f$$ has a root on either side. Consider $$f(x)=x^2+1$$.\n\nAs for solving this problem, the derivative has only two roots, so we can at most have three roots. For some values of $$c$$ we have three roots, for some values of $$c$$ we have a single root, and for exactly two values of $$c$$, there are two roots. The three-root region is exactly the interval between the two two-root values of $$c$$.\n\nAnd finding the values of $$c$$ that gives two roots is easier than one might think. They happen exactly when one root of $$f$$ coincides with a root of $$f'$$. So find the roots of $$f'$$, and find the values of $$c$$ that make each of them a root of $$f$$, and you have found the interval of $$c$$-values that gives three roots.\n\nAs you know, the given equation has extrema at $$x=\\pm1$$. These correspond to values of the polynomial\n\n$$1-5-c$$ and $$1+5-c$$ (the RHS was moved to the left).\n\nHence the polynomial will grow from $$-\\infty$$, reach the maximum, then the minimum and continue growing to $$\\infty$$. There are three roots when $$0$$ is in the range $$(-4-c,6-c)$$.\n\nWhere am I messing up? Just look at a graph where it fails.\n\n$$x^5-5 x -5$$\n\nI know that between consecutive real roots of f there is a real root of f\u2032. Now f\u2032 in this case is 5x4\u22125 which always has two real roots. So the claim should be true for all c.\n\n$$A \\implies B$$ does not mean $$B \\implies A$$.\n\nThe two real roots of $$5x^4 - 5$$ are the two roots at $$x = \\pm 1$$.\n\nIf $$x^5 - 5x=c$$ has three roots then they will be at $$x < -1; -1 < x < 1;$$ and at $$x > 1$$ by your condition.\n\nBut there won't be three real roots if there is no root for any $$x< -1$$, or no root between $$-1$$ and $$1$$, or no root for any $$x < -1$$.\n\n$$x=\\pm 1$$ are extreme points and if one, the max, is $$>0$$ and the other $$<0$$ then there will be three real roots. But if both are \"on the same side of $$0$$\" there is no root between them and no root to \"the other side\".\n\n$$x^5-5x -c|_{-1} = 4-c$$ and $$x^5 - 5x -c|1 = -4-c$$ so $$x =-1$$ is a max and $$x = 1$$ is a min.\n\nIf $$f(-1) = 4-c \\le 0$$ is a max there will be no root for $$x < -1$$ or for $$-1 < x \\le 1$$. If $$f(1) = -4-c\\ge 0$$ is a min there will be no root for $$x > 1$$ or for $$-1 \\le x < 1$$. So if either $$c \\ge 4$$ or if $$c\\le 4$$ then there are fewer than three real roots. But if $$-4 < c < 4$$ then there will be three.\n\nAlternatively: we know what the shape of an odd polynomial $$x^5 -5x$$ looks like. It's that polynomial curve with a twisty bit in the middle, goes off to infinity as $$x \\to \\infty$$, goes to negative infinity as $$x \\to -\\infty$$. It has roots were the x-axis crosses it (or where it crosses the x-axis-- everything is relative). If we shift it up or shift it down by $$c$$ we can force the x-axis to avoid the twisty bits in the middle and have it have only one root. Or we can deliberately shift it so that the x-axis goes smack through the twisty bits and we have a maximum number of roots. So if $$c$$ is between the max and mins we maximize the number of roots and the x-axis goes through the twisty bits. If $$c$$ is beyond the max an mins we've shoved the twisty bits below or above the x-axis and there is only one root.", "date": "2020-01-20 06:59:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3473894/to-find-number-of-real-roots", "openwebmath_score": 0.7858799695968628, "openwebmath_perplexity": 104.40176674605412, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9863631671237734, "lm_q2_score": 0.925229954514902, "lm_q1q2_score": 0.9126127482531035}}
{"url": "https://www.physicsforums.com/threads/difference-between-continuity-and-uniform-continuity.812734/", "text": "# Difference between continuity and uniform continuity\n\nTags:\n1. May 7, 2015\n\n### Yunjia\n\nI noticed that uniform continuity is defined regardless of the choice of the value of independent variable, reflecting a function's property on an interval. However, if on a continuous interval, the function is continuous on every point. It seems that the function on that interval must be uniformly continuous. Is this correct? Is there a counterexample for the statement?\n\n2. May 7, 2015\n\n### jbunniii\n\nWhat you said is almost true. If, on a closed, bounded interval, a function is continuous at every point, then the function is uniformly continuous on that interval.\n\nCounterexample on a non-closed interval: $f(x) = 1/x$ on the interval $(0,1)$.\n\nCounterexample on a closed but unbounded interval: $f(x) = x^2$ on the interval $[0,\\infty)$.\n\n3. May 8, 2015\n\n### Svein\n\nEven more general: If a function is continuous at every point in a compact set, it is uniformly continuous on that set.\nThe proof is simple: Let the compact set be K and take an \u03b5>0. Since the function is continuous at every point x in the set, there is a \u03b4x for every x\u2208K such that |f(w)-f(x)|<\u03b5 for every w in <x-\u03b4x, x+\u03b4x>. Let Ox=<x-\u03b4x, x+\u03b4x>, then K is contained in $\\bigcup_{x\\in K}O_{x}$. Since K is compact, it is contained in the union of a finite number of the Ox, say $\\bigcup_{n=1}^{N}O_{n}$. Take \u03b4 to be the minimum of the \u03b4n and |f(w)-f(x)|<\u03b5 whenever |w-x|<\u03b4.\n\n4. May 8, 2015\n\n### jbunniii\n\nAnd here's the proof that a closed, bounded interval is compact, so you can apply Svein's proof.\n\nLet $[a,b]$ be a closed, bounded interval, and let $\\mathcal{U}$ be any collection of open sets which covers $[a,b]$. Let $S$ be the set of all $x \\in [a,b]$ such that $[a,x]$ can be covered by finitely many of the sets in $\\mathcal{U}$. Clearly $a \\in S$. This means that $S$ is nonempty and is bounded above (by $b$), so it has a supremum, call it $c$. Since $c \\in [a,b]$, it is contained in some $U_c \\in \\mathcal{U}$, hence there is some interval $(c - \\epsilon, c + \\epsilon)$ contained in $U_c$. Since only finitely many sets from $\\mathcal{U}$ are needed to cover $[a,c - \\epsilon/2]$, those sets along with $U_c$ form a finite cover of $[a,c+\\epsilon/2]$. This shows that $c \\in S$ and moreover, that $c$ cannot be less than $b$. Therefore $c=b$, so all of $[a,b]$ can be covered by finitely many sets in $\\mathcal{U}$.\n\nLast edited: May 8, 2015\n5. May 8, 2015\n\n### Svein\n\nExcellent! And, of course, since one closed and bounded interval is compact, the union of a finite number of closed and bounded intervals is again compact.\n\n6. May 8, 2015\n\n### jbunniii\n\nI don't think this will work if $w$ and $x$ are not contained in the same $O_n$. I think you need to take $O_x = (x - \\delta_x/2, x + \\delta_x/2)$ and $\\delta$ to be $\\min\\{\\delta_n/2\\}$ in order to ensure that $|w-x| < \\delta$ implies $|f(w) - f(x)| < \\epsilon$.\n\n7. May 8, 2015\n\n### Svein\n\nPossibly. I need to think about that. The crux of the matter is that there is a finite number of intervals that cover K, which means that the minimum of the (finite number of) \u03b4's exist and is greater than 0.\n\n8. May 8, 2015\n\n### HallsofIvy\n\nHere is the fundamental difference between \"continuous\" and \"uniformly continuous\":\n\nA function is said to be continuous at a point, x= a, if and only if, given any $\\epsilon> 0$ there exist $\\delta> 0$ such that if $|x- a|< \\delta$ then $|f(x)- f(a)|<\\epsilon$.\n\nA function is said to be continuous on a set, A, if and only if, given any $\\epsilon> 0$ there exist $\\delta> 0$ such that if $|x- a|< \\delta$ then $|f(x)- f(a)|<\\epsilon$ for all a in set A.\n\nThat is, a function is uniformly continuous on a set A if and only if it is continuous at every point in A and, given $\\epsilon> 0$ the same $\\delta$ can be used for ever point in A.\n\n9. May 8, 2015\n\n### Svein\n\nAgree. I was sloppy and overlooked the simple fact that that the length of the interval Ox is 2\u22c5\u03b4x.\n\n10. May 13, 2015\n\n### Yunjia\n\nThe defition of uniformly continuous is perplexing for me. Is it possible for you to illustrate it with a proof about a function which is not uniformly continuous but continuous on a set so that I can compare the two? Thank you.\n\n11. May 14, 2015\n\n### Svein\n\nOK. Take the function $f(x)=\\frac{1}{x}$ on the open interval <0, 1>. It is continuous (and even differentiable) on that interval, but not uniformly continuous.\n\n12. May 17, 2015\n\n### Svein\n\nAfter using pencil an paper for a bit, I came to the conclusion that I should have used \u03b5/2 and \u03b4/2. But - I recall a proof in \"Complex analysis in several variables\" that ended up in \"... less than 10000\u03b5, which is small when \u03b5 is small\".\n\n13. May 17, 2015\n\n### HallsofIvy\n\nFirst note an important point about \"uniform continuity\" and \"continuity\" you may have overlooked: we define \"continuous\" at a single point and then say that a function is \"continuous on set A\" if and only if it is continuous at every point on A. We define \"uniformly continuous\" only on a set, not at a single points of a set.\n\nA function is uniformly continuous on any closed set on which it is continuous and so on any set contained in a closed set on which it is continuous.\nTo give an example of a function that is continuous but not uniformly continuous, we need to look at f(x)= 1/x on the set (0, 1).\n\nTo show that it is continuous on (0, 1), let a be a point in (0, 1) and look at $|f(x)- f(a)|= |1/x- 1/a|= |a/ax- x/ax|= |(a- x)/ax|= |x- a|/ax< \\epsilon$.\nWe need to find a number, $\\delta> 0$ such that if $|x- a|< \\delta$, then $|f(x)- f(a)|> \\epsilon$. We already have $|x- a| ax\\epsilon$ so we need an upper bound on ax. If we start by requiring that $\\delta< a/2$ then $|x- a|< \\delta< a/2$ so that $-a/2< x- a< a/2[itex] or [itex]a/2< x< 3a/2$ so an upper bound on ax is $3a^2/2$. If $|x- a|< a/2$ and $|x- a|< 3a/2$ then $|f(x)- f(a)|< |x- a|/ax< |x- a|/(3a^2/2)= 2|x- a|/3a^2$ which will be less than $\\epsilon$ as long as $|x- a|< 3a^2\\epsilon/2$\n\nSo we can take $\\delta$ to be the smaller of $a/2$ and $3a^2/\\epsilon$. Therefore, 1/x is continuous at any point a in (0, 1) so continuous on (0, 1).\n\nNow the point is that this $\\delta$ depends on a. It is a decreasing function of a and will go to 0 as a goes to 0. That's fine for \"continuity\" but for uniform continuity we must be able to use the same $\\delta> 0$ for a given $\\epsilon$ no matter what the \"a\" is and we cannot do that. If the problem were to prove uniform continuity on the set [p, 1), which is contained in the closed set [p, 1], we could use, for any a in that set, the $\\delta$ that we get for a= p, the smallest x value in the set. But with (0, 1), the smallest $\\delta$ is 0 which we cannot use since we must have $\\delta> 0$.", "date": "2018-02-20 02:33:06", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/difference-between-continuity-and-uniform-continuity.812734/", "openwebmath_score": 0.9080765843391418, "openwebmath_perplexity": 154.58026090664416, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9840936110872273, "lm_q2_score": 0.9273633016692238, "lm_q1q2_score": 0.9126123003294401}}
{"url": "http://oanx.cfalivorno.it/linear-algebra-and-learning-from-data-github.html", "text": "## Linear Algebra And Learning From Data Github\n\nEach point correspondence generates one constraint on F. Complete Linear Algebra for Data Science & Machine Learning 4. My research interests lie in network science, statistical inference, causal inference, information theory, machine learning, data mining, and signal. 065 Linear Algebra and Learning from Data New textbook and MIT video lectures OCW YouTube; 18. Welcome to the 18. A vector regression task is one where the target is a set of continuous values (e. Siefken, J. Getting started with linear algebra. Yes, linear algebra is actually super important in data science. The aim of this set of lectures is to review some central linear algebra algorithms that we will need in our data analysis part and in the construction of Machine Learning algorithms (ML). and is associated with our Intro to Deep Learning Github repository where you can find practical examples of A subset of topics from linear algebra, calculus. Linear Regression aims to find the dependency of a target variable to one or more independent variables. [Online book] n Andrew Ng. Most importantly, the online version of the book is completely free. Franklin, Beedle & Associates Inc. You've accumulated a good bit of data that looks like this:. The course breaks down the outcomes for month on month progress. If you don't want to go all the way back to school, this course should do the trick in just a day or two. In an image classification problem, we often use neural networks. We emphasize that this document is not a. Machine Learning is built on prerequisites, so much so that learning by first principles seems overwhelming. Learning Python for Data. If you don't want to go all the way back to school, this course should do the trick in just a day or two. https://shaarli. Python Quick Start. Concepts you need to know in. Mike Love\u2019s general reference card; Motivations and core values (optional) Installing Bioconductor and finding help; Data structure and management for genome scale experiments. This is two equations and two variables, so as you know from high school algebra, you can \ufb01nd a unique solution for x 1 and x. Press Enter to expand sub-menu, click to visit Data Science page Data Science. Linear Algebra 8. The concepts of Linear Algebra are crucial for understanding the theory behind Machine Learning, especially for Deep Learning. Probability and Statistics:. I'd like to introduce a series of blog posts and their corresponding Python Notebooks gathering notes on the Deep Learning Book from Ian Goodfellow, Yoshua Bengio, and Aaron Courville (2016). Scikit-learn (formerly scikits. TS CH10 Linear Least Squares. scikit-learn is a comprehensive machine learning toolkit for Python. This library holds the principal work done as part of the OpenAstonomy Google Summer of Code 2020 project, Solar Weather Forecasting using Linear Algebra. , and Courville, A. GF2] = One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One scala> a + a res0: breeze. For example, when predicting house prices, the different target prices form a continuous space. Conversely, if the condition number is very low (ie close to 0) we say is well-conditioned. The mentors for this project are: @dpshelio @mbobra @drsophiemurray @samaloney. Since I like math and I have more time to dedicate to my projects, I've started an open source linear algebra library for javascript, just for fun and for learning new stuff. Tibshirani, J. Boost your data science skills. Y et because linear algebra is a form of con tin uous rather than. 5M ratings github. View picnicml on GitHub. Deep Learning Book Series \u00b7 2. Matrices in Rn m will be denoted as: M. Linear Algebra for Data Science using Python Play all 13:42 Math For Data Science | Practical reasons to learn math for Machine/Deep Learning - Duration: 13 minutes, 42 seconds. In the second part, we discuss how deep learning differs from classical machine learning and explain why it is effective in dealing with complex problems such as image and natural language processing. Linear Algebra and Learning from Data twitter github. Linear Algebra for Machine Learning Discover the Mathematical Language of Data in Python. NET language, as well as a feature-rich interactive shell for rapid development. We will describe linear regression in the context of a prediction problem. Posted by u/[deleted] a linear algebra library in R designed for teaching. Description. We won't use this for most of the. 2 (217 ratings) Course Ratings are calculated from individual students' ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. Linear regression is one of the most popular machine learning algorithms. what-is-the-difference-between-artificial-intelligence-and-machine-learning 9. The mentors for this project are: @dpshelio @mbobra @drsophiemurray @samaloney. ML is a key technology in Big Data, and in many financial, medical, commercial, and scientific applications. 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Just want some books to go deeper than a introductory course. This book provides the conceptual understanding of the essential linear algebra of vectors and matrices for modern engineering and science. Linear algebra is one of the most applicable areas of mathematics. In this first module we look at how linear algebra is relevant to machine learning and data science. Linear transformations and change of basis (connected to the Singular Value Decomposition - orthonormal bases that diagonalize A) Linear algebra in engineering (graphs and networks, Markov matrices, Fourier matrix, Fast Fourier Transform, linear programming) Homework. This book is directed more at the former audience. As the complementary course to. and engineering. NumPy is \"the fundamental package for scientific computing with Python. Linear Algebra is a text for a first US undergraduate Linear Algebra course. Bourbaki resulted from similar currents of thought that produced fascism and totalitarian communism: moral panics leading to revolutions, and ultimately \u201cfinal solutions\u201d, all terrible and evil in. Complete Linear Algebra for Data Science & Machine Learning 4. Updated Apr 1 2020. In an image classification problem, we often use neural networks. His research interests span statistical machine learning, numerical linear algebra, and random matrix theory. Included is a learning guide and syllabus to help you learn data science this year. Many courses are offered there from which one can benefit. Linear Algebra\u00b6 Now that you can store and manipulate data, let's briefly review the subset of basic linear algebra that you will need to understand most of the models. Students will learn and practice fundamental ideas of linear algebra and simultaneously be exposed to and work with real-world applications of these ideas. In this tutorial, we walked through one of the most basic and important regression analysis methods called Linear Regression. com, My Github, LinkedIn, & Google Scholar I am an applied mathematician and computer scientist with experience in algorithms, linear algebra, and graph theory. One of the most beautiful and useful results from linear algebra, in my opinion, is a matrix decomposition known as the singular value decomposition. Linear algebra moves steadily to n vectors in m-dimensional space. If you have some background in basic linear algebra and calculus, this practical book introduces machine-learning fundamentals by showing you how to design systems capable of detecting objects in images, understanding text, analyzing video, and predicting. As a core programmer, I love taking challenges and love being part of the solution. in Applied Numerical Linear Algebra Currently focused on combining statistical data mining techniques and traditional econometrics approach, panel data, model switching, GARCH type volatility modelling, volume modelling etc. If we are thinking of a vector as representing a physical quantity. To use MOE, we simply need to specify some objective function, some set of parameters, and any historical data we may have from previous evaluations of the objective function. If you want to fit a model of higher degree, you can construct polynomial features out of the linear feature data and fit to the model too. , Bengio, Y. The interplay of columns and rows is the heart of linear algebra. You've accumulated a good bit of data that looks like this:. BUT Linear Algebra is too boundless! In this book, you will get what is NECESSARY. There is no doubt that linear algebra is important in machine learning. The Linear Algebra Chapter in Goodfellow et al is a nice and concise introduction, but it may require some previous exposure to linear algebra concepts. pdf; TS CH8 Estimation. then this is the book for you. and his book on Linear Algebra is a very good introduction. Linear Algebra for Data Scientists. A lot of linear algebra over the complex field This free book Linear Algebra - As an Introduction to Abstract Mathematics from UC Davis has plenty of exercises; Terence Tao has a set of notes if you google, they go with the book Linear Algebra by Friedberg, Insel and Spence. Programming and data science articles by hadrienj. Linear Algebra: Foundations to frontiers \u2013 edx. This works in the latest snapshot of Breeze. Linear Algebra Examines basic properties of systems of linear equations, vector spaces, inner products, linear independence, dimension, linear transformations, matrices, determinants, eigenvalues, eigenvectors and diagonalization. GitHub; LinkedIn; Twitter; Donald Miner (@donaldpminer) specializes in large-scale data analysis enterprise architecture and applying machine learning to real-world problems. Windows-64 (64-bit linear algebra for large data) Unless your computer has more than ~32GB of memory and you need to solve linear algebra problems with arrays containing more than ~2 billion elements, this version will offer no advantage over the recommended Windows-64 version above. Modern statistics is described using the notation of linear algebra and modern statistical methods harness the tools of linear algebra. As a core programmer, I love taking challenges and love being part of the solution. The aim of these notebooks is to help beginners/advanced beginners to grasp linear algebra concepts underlying deep learning and machine learning. Matrices, vectors, and more - from theory to the real world! There's a lot of data out there, learn how to search it effectively. A few weeks ago, I wrote about how and why I was learning Machine Learning, mainly through Andrew Ng\u2019s Coursera course. LINEAR ALGEBRA. Today, I will be sharing with you my C# implementation of basic linear algebra concepts. Hundreds of thousands of students have already benefitted from our courses. Linear Algebra and Learning from Data by Gilbert Strang; 1 edition; twitter github. Some of these assignments are from Introduction to applied linear algebra - vectors, matrices, and least squares. Introduction to linear algebra (Fourth Edition). , Bengio, Y. Linear Regression 101 (Part 1 - Basics) 6 minute read Introduction. My work includes researching, developing and implementing novel computational and machine learning algorithms and applications for big data integration and data mining. Various CNN and RNN models will be covered. Learning Spark : lightning-fast data analytics by Holden Karau, Andy Konwinski, Patrick Wendell, and Matei Zaharia, O\u2019Reilly, 2015. Thesis: Quantum Algorithms for Linear Algebra and Machine Learning; Anupam Prakash Quantum Algorithms for Linear Algebra and Machine Learning by Anupam Prakash Most quantum algorithms o ering speedups over classical algorithms are based on the three techniques of phase estimation, amplitude estimation and Hamiltonian simulation. Implementation [ edit ] Scikit-learn is largely written in Python, and uses numpy extensively for high-performance linear algebra and array operations. Currently he is on leave from UT Austin and heads the Amazon Research Lab in Berkeley, California, where he is developing and deploying state-of-the-art machine learning methods for Amazon Search. Support Stability of Maximizing Measures for Shifts of Finite Type Journal of Ergodic Theory and Dynamical Systems (accepted) Calkins, S. Those equations may or may not have a solution. Problem solving with algorithms and data structures using Python. Posted by u/[deleted] a linear algebra library in R designed for teaching. Foundations of Data Science is a treatise on selected fields that form the basis of Data Science like Linear Algebra, LDA, Markov Chains, Machine Learning basics, and statistics. in machine learning, it is standard to say \u201cN samples\u201d to mean the same thing. 065 Linear Algebra and Learning from Data New textbook and MIT video lectures OCW YouTube; 18. pdf; TS CH8 Estimation. Government and Fortune 500 companies. These functions are mainly for tutorial purposes in learning matrix algebra ideas using R. Siefken, J. Complete Linear Algebra for Data Science & Machine Learning 4. A few weeks ago, I wrote about how and why I was learning Machine Learning, mainly through Andrew Ng\u2019s Coursera course. You will also learn how you should use linear algebra in your Python code. in linear algebra, kernel is another name for nullspace. Now that you understand the key ideas behind linear regression, we can begin to work through a hands-on implementation in code. Linear algebra has had a marked impact on the field of statistics. 1-11, [Online-Edition. \" Our homework assignments will use NumPy arrays extensively. squares methods, basic topics in applied linear algebra. Perception, movement control, reinforcement learning, mathematical psychology, \u2026 Economics. As the complementary course to. Then last year I learned how he morphed his delightful mathematics book into a brand new title (2019) designed for data scientists - \"Linear Algebra and Learning from Data. In some cases, functions are provided for concepts available elsewhere in R, but where the function call or name is not obvious. The concepts of Linear Algebra are crucial for understanding the theory behind Machine Learning, especially for Deep Learning. Course Description. Julia Observer helps you find your next Julia package. Linear algebra underlies many practical mathematical tools, such as Fourier series and computer. \u201d See Section 6. Description. Foundations of Data Science is a treatise on selected fields that form the basis of Data Science like Linear Algebra, LDA, Markov Chains, Machine Learning basics, and statistics. [Online book] n Andrew Ng. This class is an in-depth graduate lecture class. uk, [email\u00a0protected] Together with your editor or Jupyter notebook these packages allow you to rapidly develop scalable, high-performance analytics and visualizations using succinct, type-safe, production-ready code. Book: Aur\u00e9lien G\u00e9ron \"Hands-On Machine Learning with Scikit-Learn and TensorFlow\" Book: Andriy Burkov \"The Hundred-Page Machine Learning Book \" \ud83d\udc0d Python Course: Python. It would be best if you had an organized book which (1) teaches the most used Linear Algebra concepts in Machine Learning, (2) and utilize your Machine Learning model in terms of data processing, optimization, and validation. They give you better intuition for how algorithms really work under the hood, which enables you to make better decisions. This page has links for latest PDF versions of the text and related supplements. * Ranked among top 10% answerers on Python in StackOverflow. Now we extend linear algebra to convolutions, by using the example of audio data analysis. He was a research fellow with Michael Jordan and Peter Bartlett, University of California at Berkeley, from 2003, and with Bernhard Schoelkopf, Max Planck Institute for Intelligent Systems, Tuebingen, Germany, from 2005. 065 Linear Algebra and Learning from Data New textbook and MIT video lectures OCW YouTube; 18. Learning path Introduction to Linear Algebra. Learn the basic math for Data Science, AI, and ML using R. This book is directed more at the former audience. Getting started with linear algebra. Feature learning and subspaces (Chapter 3 of [BHK]) Random walk and Markov chain Monte Carlo (Chapter 4 of [BHK]) Linear models, kernel methods, and deep learning ([Bishop] and Chapter 5 of [BHK]) Algorithms for Massive Data: streaming, sketching, and sampling (Chapter 6 of [BHK]) The course schedule is available here. Incorporating machine learning capabilities into software or apps is quickly becoming a necessity. Implementation [ edit ] Scikit-learn is largely written in Python, and uses numpy extensively for high-performance linear algebra and array operations. His main research interests are in big data, machine learning, network analysis, linear algebra and optimization. We create tools for phenotype analyses that make use of the entire clinical phenotyping spectrum, not only using HPO, but also model organisms data (we also create the uberpheno) and other ontologies. Machine Learning is built on prerequisites, so much so that learning by first principles seems overwhelming. Modern statistics is described using the notation of linear algebra and modern statistical methods harness the tools of linear algebra. Students will learn and practice fundamental ideas of linear algebra and simultaneously be exposed to and work with real-world applications of these ideas. If is high, the matrix is said to be ill-conditioned. Matrix sketching and randomized matrix computation. I\u2019d like to go over the theory behind this matrix decomposition and show you a few examples as to why it\u2019s one of the most useful mathematical tools you can have. It\u2019s all vectors and matrices of numbers. Topic 1: Review of Linear Algebra 1-6 These are only a few examples that I hope help convince you that vector spaces are the backbone of machine learning. Linear Algebra\u00b6 Now that you can store and manipulate data, let\u2019s briefly review the subset of basic linear algebra that you will need to understand most of the models. \u2022LACore is a Large-Format vector accelerator for a broad range of Linear Algebra applications \u2022LACore has novel architectural features including as the: \u2022 configurable, data-streaming LAMemUnits \u2022 dual-precision, configurable, systolic LAExecUnit \u2022A compiler toolchain, programming framework and architectural simulator were all. For simple linear regression, one can choose degree 1. After reading this post, you will know:. Python for Data Science and Machine Learning Bootcamp; Think Stats - Book. I have hands-on experience in Data Analysis, Machine Learning, Natural Language Processing, deployment on IaaS like AWS. , how to pass the course, schedules, and deadlines, at the official course page. Linear Algebra for Machine Learning Book. Currently he is on leave from UT Austin and heads the Amazon Research Lab in Berkeley, California, where he is developing and deploying state-of-the-art machine learning methods for Amazon Search. for automated market making. Grading: 3 homeworks 60%, 2 quizzes 20%, 1 project 20%. edu ABSTRACT Accelerating machine learning (ML) over relational data is a. , Bengio, Y. This repository contains all the quizzes/assignments for the specialization \"Mathematics for Machine learning\" by Imperial College of London on Coursera. Some recent tutorials by Christos and Co. Lawrence [email\u00a0protected] Welcome to Data analysis with Python - 2020\u00b6. The aim of these notebooks is to help beginners/advanced beginners to grasp linear algebra concepts underlying deep learning and machine learning. A vector in Rn will be denoted as: ~x. Enabling and Optimizing Non-linear Feature Interactions in Factorized Linear Algebra Side Li University of California, San Diego [email\u00a0protected] HarvardX Biomedical Data Science Open Online Training In 2014 we received funding from the NIH BD2K initiative to develop MOOCs for biomedical data science. Now I\u2019m checking back in with 9 weeks under my belt. Introduction to Machine Learning for Data Science, Udemy. Learn linear algebra. The first part covers basics and preliminaries. Incorporating machine learning capabilities into software or apps is quickly becoming a necessity. Some recent tutorials by Christos and Co. pyplot as plt # pyplot from collections import defaultdict , Counter from functools import partial , reduce. pyplot as plt # pyplot from collections import defaultdict , Counter from functools import partial , reduce. Introduction The aim of this set of lectures is to review some central linear algebra algorithms that we. In case of deep learning algorithms, linear algebra is the driving force. We study fast algorithms for linear algebraic problems that are ubiquitous in data analysis and machine learning. After understanding the nature of both the problem and the field we are dealing with, and before learning how to prepare a data for your machine learning and do the cleaning and preparation for a selected problem. If you're a data scientist studying linear algebra, chances are you are interested in understanding how machine learning algorithms work. Linear Algebra and Learning from Data by Gilbert Strang; twitter github. 06 Linear Algebra - The video lectures are on web. I will list some resources for learning linear algebra. Andrzej Cichocki, Anh-Huy Phan, Qibin Zhao, Namgil Lee, Ivan Oseledets, Masashi Sugiyama, and Danilo Mandic. Linear Algebra for Machine Learning Book. You will also learn how you should use linear algebra in your Python code. Probability and Statistics: Learn Probability and Statistics Through Interactive Visualizations: Seeing Theory was created by Daniel Kunin while an undergraduate at Brown University. As a core programmer, I love taking challenges and love being part of the solution. Linear regression is one of the most popular machine learning algorithms. , and Courville, A. The elements of statistical learning: data mining, inference, and prediction, Springer, 2009 Linear Algebra and Probability Review (part 1 Linear Algebra, part 2 Probability) Assignment 1: Apr 10. Those equations may or may not have a solution. There is no doubt that linear algebra is important in machine learning. View on GitHub mlcourse. Run in Google Colab View source on GitHub Download notebook In this post, we will explore the ways of doing linear algebra only using tensorflow. Read more Tagged as : R linear algebra classification linear discriminant analysis. My work includes researching, developing and implementing novel computational and machine learning algorithms and applications for big data integration and data mining. You can use it as a main text, as a supplement, or for independent study. I\u2019m interested in applying non-standard tools form abstract algebra and topology to the study of neural networks. About data set: Square feet is the Area of house. uk November 1, 2018 Abstract Development systems for deep learning (DL), such as Theano, Torch, TensorFlow, or MXNet, are. Linear algebra has had a marked impact on the field of statistics. In some cases, functions are provided for concepts available elsewhere in R, but where the function call or name is not obvious. GF2] = One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One scala> a + a res0: breeze. Siefken, J. APPLICATION: This can be pretty much applied in a generic way to all programs. In general, statistical problems have to do with the estimation of some characteristic derived from data - this can be a point estimate, an interval, or an entire function. Linear Algebra: Foundations to frontiers \u2013 edx. Linear Algebra for Machine Learning Discover the Mathematical Language of Data in Python. data in homework problems. Machine learning. Anyone can view the notebooks online by clicking on the links in the readme Table of Contents. In this day, we are going to make the dirty work. In this course on Linear Algebra we look at what linear algebra is and how it relates to vectors and matrices. In other cases, functions are provided to show or. Linear algebra (Systems of linear equations, least-square) [Matrix cookbook] 4: 9/16/2019: Linear Algebra (Vector space, linear independence) 9/18/2019: Linear algebra (Eigendecomposition and matrix factorization) Homework 1 (extended) 5: 9/23/2019: Linear algebra (Eigendecomposition and matrix factorization) 9/25/2019: Linear algebra (Best fit. Acquiring these skills can boost your ability to understand and apply various data science algorithms. The basic mathematics prerequisites for understanding Machine Learning are Calculus-I,II,III, Linear Algebra, and, Probability and Statistics. Making statements based on opinion; back them up with references or personal experience. Randomized numerical linear algebra. It lacks the ability of distributed linear algebra computation in its local interactive shell. You can use it as a main text, as a supplement, or for independent study. NVIDIA CUDA-X GPU-Accelerated Libraries NVIDIA\u00ae CUDA-X, built on top of NVIDIA CUDA\u00ae, is a collection of libraries, tools, and technologies that deliver dramatically higher performance\u2014compared to CPU-only alternatives\u2014 across multiple application domains, from artificial intelligence (AI) to high performance computing (HPC). Lek-Heng Lim. taco is versatile. However, I think that the chapter on linear algebra from the book is a bit tough for beginners. A few weeks ago, I wrote about how and why I was learning Machine Learning, mainly through Andrew Ng\u2019s Coursera course. scala> val a = DenseMatrix. This code has been posted to GitHub under a MIT license, so feel free to modify and deal with code without any restrictions or limitations (no guarantees of any kind. world Overview of scikit-learn Python and Excel Scaling, Centering, Noise with kNN, Linear Regression, Logit Sentiment Analysis with Twitter Time Series Analysis Vectors and Arrays (Linear Algebra) Vectors and Arrays (Linear Algebra) Table of contents. Lesson 1 (April 7): Machine learning pipeline and course overview: video; slides. Also,it would be of much help if they have big set of problems and examples. Machine Learning course (Andrew Ng) is a basic machine learning course. It begins with linear algebra\u2014matrix factorizations A= QR. therefore precede our in tro duction to deep learning with a fo cused presen tation of. Linear algebra cheat sheet for deep learning \u2013 Towards Data Science \u2013 Medium. They give you better intuition for how algorithms really work under the hood, which enables you to make better decisions. It\u2019s all vectors and matrices of numbers. Today, I will be sharing with you my C# implementation of basic linear algebra concepts. In an image classification problem, we often use neural networks. Matrices and Linear Algebra The Wolfram Language automatically handles both numeric and symbolic matrices, seamlessly switching among large numbers of highly optimized algorithms. R is a widely-used statistical programming language in the data science community. Artificial Neural Networks. Grading (tentative) Quizzes 20%; Course project. Together with your editor or Jupyter notebook these packages allow you to rapidly develop scalable, high-performance analytics and visualizations using succinct, type-safe, production-ready code. eye[GF2](5) a: breeze. Windows-64 (64-bit linear algebra for large data) Unless your computer has more than ~32GB of memory and you need to solve linear algebra problems with arrays containing more than ~2 billion elements, this version will offer no advantage over the recommended Windows-64 version above. For beginning practitioners (i. View on GitHub mlcourse. Tibshirani, J. Open Library is an initiative of the Internet Archive, a 501(c)(3) non-profit, building a digital library of Internet sites and other cultural artifacts in digital form. Another resource is the book with the funny title \u201cNo Bullshit Guide to Linear Algebra\u201d by Ivan Savov. It introduces some common tools in machine learning to resolve real applications (e. It calls them tensors. Linear Algebra Examines basic properties of systems of linear equations, vector spaces, inner products, linear independence, dimension, linear transformations, matrices, determinants, eigenvalues, eigenvectors and diagonalization. You will also learn how you should use linear algebra in your Python code. mathematics-for-machine-learning-cousera. We start with representing a fully connected layer as a form of matrix multiplication: - In this example, the weight matrix has a size of $4 \\times 3$, the input vector has a size of $3 \\times 1$ and the output vector has a of size $4 \\times 1$. The code (from. Understanding API Security (Free chapters from a Manning. It turns out, however, that all of those operations can be written in terms of big matrix-matrix or matrix-vector multiplications. 1x Scalable Machine Learning. AMD adopted BLIS as its new BLAS library. After reading this quickstart, you can go to other wiki pages, especially Linear Algebra Cheat-Sheet and Data Structures. I'm studying towards a PhD degree at the University of Chicago, in the department of Statistics. SciPy is built to work with NumPy arrays and provides many. They are full of explanations, code samples, pictures, interesting links, and exercises for you to try. linear; algebra; matrix;. import re , math , random # regexes, math functions, random numbers import matplotlib. By Hadrien Jean, Machine Learning Scientist. Compressed Linear Algebra for Declarative Large-Scale Machine Learning Ahmed Elgohary2, Matthias Boehm1, Peter J. What we did here by attaching the variable mlr to the MyLinearRegression class is to create an instance, a specific object called mlr, which will have its own data and \"functions\". In the second part, we discuss how deep learning differs from classical machine learning and explain why it is effective in dealing with complex problems such as image and natural language processing. Plotting is based on OpenGL and supports both 2D and 3D plots. Introduction to linear algebra (Fourth Edition). DenseMatrix[X. This content is part of a series following the chapter 2 on linear algebra from the Deep Learning Book by Goodfellow, I. therefore precede our in tro duction to deep learning with a fo cused presen tation of. You (the student) should have taken a mathematical course on linear algebra that covers vector spaces as well as a numerical analysis course that covers computer implementations of numerical algorithms. Fitting Lines to Data. net/?Q73_jQ 2020-01-24T14:53:56+01:00 2020-01-24T15:29:05+01:00. Linear Algebra for Data Science using Python Play all 13:42 Math For Data Science | Practical reasons to learn math for Machine/Deep Learning - Duration: 13 minutes, 42 seconds. The mentors for this project are: @dpshelio @mbobra @drsophiemurray @samaloney. Randy Lao's site for free Machine Learning and Data Science resources and materials. How to Learn Advanced Mathematics Without Heading to University - Part 1 subjects to learn for a prospective quant or data scientist. The Cuckoo linear algebra implementation is based on libcuckoo library1. to map the pixel values of an image to the confidence score of each class. This library holds the principal work done as part of the OpenAstonomy Google Summer of Code 2020 project, Solar Weather Forecasting using Linear Algebra. library with basic linear algebra routines, and the SciPy library adorns NumPy arrays with many important primitives, from numerical optimizers and signal processing to statistics and sparse linear algebra. Download R for Windows 5. Linear Algebra Preliminaries\u00b6 Since I have documented the Linear Algebra Preliminaries in my Prelim Exam note for Numerical Analysis, the interested reader is referred to for more details (Figure. Intro to Data Science / UW Videos. Now we extend linear algebra to convolutions, by using the example of audio data analysis. The Linear Algebra Chapter in Goodfellow et al is a nice and concise introduction, but it may require some previous exposure to linear algebra concepts. AMD adopted BLIS as its new BLAS library. Each point correspondence generates one constraint on F. I had to formulate an algorithm to convert an image of some resolution - say L x M and crop / re-size the image into a new resolution say P x R such that I cover the maximum amount of points/pixels from the original image. Again, the class MyLinearRegression provides instructions on how to build a linear regression model. Linear Algebra: Video: Professor Gilbert Strang's Video Lectures on linear algebra. Above, I created 4 matrices. In addition to this, you\u2019ll be able to perform operations such as addition, subtraction and dot product. It introduces some common tools in machine learning to resolve real applications (e. Conncect between Geometry and Algebra. As of October 2019, I am a senior algorithms scientist at PathAI, where I work on computational pathology. California Housing Price Prediction. Linear Optimization in Python 6. The aim of these notebooks is to help beginners/advanced beginners to grasp linear algebra concepts underlying deep learning and machine learning. 6 (376 ratings) Course Ratings are calculated from individual students\u2019 ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. View on GitHub. They give you better intuition for how algorithms really work under the hood, which enables you to make better decisions. Calculate The Trace Of A. pdf CarletonU- hamidreza sadreazami- radar based fall detection with supervised learning. However, it is currently not easy to implement many basic machine learning primitives in. \" Our homework assignments will use NumPy arrays extensively. We start with representing a fully connected layer as a form of matrix multiplication: - In this example, the weight matrix has a size of $4 \\times 3$, the input vector has a size of $3 \\times 1$ and the output vector has a of size $4 \\times 1$. In forecasting, Yuyang has worked on all aspects ranging from practical applications to theoretical foundations. I'm a Data Science practitioner and computer programmer with an avid interest in Exploratory Data Analysis, Statistics, & Machine Learning. edu Arun Kumar University of California, San Diego [email\u00a0protected] DenseMatrix[X. SciPy is open-source software for mathematics, science, and engineering which includes modules for statistics, optimisation, integration, linear algebra, Fourier transforms, signal and image processing, ODE solvers, and more. Now I\u2019m checking back in with 9 weeks under my belt. Mining the social web: Data mining Facebook, Twitter, LinkedIn, Google+, GitHub, and more (2nd edition) by Matthew A. The Deep Learning Book - Goodfellow, I. Most importantly, the online version of the book is completely free. Based on the second linear algebra course taught by Professor Strang, whose lectures on the training data are widely known, it starts from scratch (the four fundamental subspaces) and is fully accessible without the first text. Siefken, J. then this is the book for you. The aim of these notebooks is to help beginners/advanced beginners to grasp linear algebra concepts underlying deep learning and machine learning. How to Learn Advanced Mathematics Without Heading to University - Part 1 subjects to learn for a prospective quant or data scientist. Rich Ott leads you through two days of intensive learning that include a review of linear algebra essential to machine learning, an introduction to TensorFlow, and a dive into neural networks. All publications, sorted by year. This book provides the conceptual understanding of the essential linear algebra of vectors and matrices for modern engineering and science. n Gilbert Strang. As a motivating example, let us consider image classification. The collection of all linear combinations is called a linear subspace of $\\RR^n$, denoted by We will say that the $\\bb{v}_i$\u2019s span the linear subspace $\\mathcal{L}$. If you have more time to dedicate to your projects and you're also passionate about math, consider to contribute to the library!. Roadmap to begin with Machine Learning: The place to start is to learn (and/or) revise linear algebra. Find the smallest value of k such that the rank-k approximation of the matrix uses the same or more amount of data as the original picture. Bourbaki resulted from similar currents of thought that produced fascism and totalitarian communism: moral panics leading to revolutions, and ultimately \u201cfinal solutions\u201d, all terrible and evil in. Videos and textbooks with relevant details on linear algebra and singular value decomposition (SVD) can be found by searching Alfredo\u2019s Twitter, for example type linear algebra (from:alfcnz) in the search box. Machine Learning/Data Science. , less sensitive to noisy data) than the Euclidean norm The following result is fundamental in linear algebra: Theorem. This is the world beyond R and Python! Breeze is a library for numerical processing, like probability and statistic functions, optimization, linear algebra, etc. Mathematics for machine learning - I totally recommend this book! If you want to learn the bits and pieces of how linear algebra and calculus is used to develop algorithms like principal component analysis, backpropagation etc. a person's height and you switch from meter to centimeter. Linear Algebra and Learning from Data by Gilbert Strang; twitter github. You NEED Linear Algebra for Machine Learning Whether you want to learn Machine Learning for your work or research or you want to become a master, so the others pay you to do it, you need to know how it works. Linear Algebra: Video: Professor Gilbert Strang's Video Lectures on linear algebra. You will also learn how you should use linear algebra in your Python code. There is no doubt that linear algebra is important in machine learning. Tensor networks for dimensionality reduction and large-scale optimization: part 2 applications and future perspectives. Understanding of calculus, linear algebra, and programming is essential. to map the pixel values of an image to the confidence score of each class. Start here. PBblas - Parallel Block Linear Algebra Subsystem HPCC Systems ML_Core repository on GitHub Installation: ecl bundle install https:. We emphasize that this document is not a. References for \"practical\" machine learning: Python for data analysis by Wes McKinney. The Vector class imitates the m x 1 vector from linear algebra and contains many useful functions for dealing and interacting with Vectors. The free video lectures of this course are made available as part of Harvard Extension School's Opening Learning Initiative. After reading this quickstart, you can go to other wiki pages, especially Linear Algebra Cheat-Sheet and Data Structures. You can think of an r t i m e s c r times c r t i m e s c matrix as a set of r r r row vectors, each having c c c elements; or you can think of it as a set of c c c column vectors, each having r r r elements. \" Our homework assignments will use NumPy arrays extensively. This is a straightforward course to learn Linear Algebra Fundamentals for Data Science in Python. linear; algebra; matrix;. 1-11, [Online-Edition. Row Reduction We row reduce a matrix by performing row operations, in order to find a simpler but equivalent system for which the solution set is easily read off. Support Stability of Maximizing Measures for Shifts of Finite Type Journal of Ergodic Theory and Dynamical Systems (accepted) Calkins, S. If you want to fit a model of higher degree, you can construct polynomial features out of the linear feature data and fit to the model too. Supratim Haldar Lead Data Scientist at Head Digital Works Pvt. Popular Courses Popular with our users in the last month Introduction to Linear Algebra. Made for sharing. Linear Algebra The Rank of a Matrix. For simple linear regression, one can choose degree 1. Linear algebra provides a way of compactly representing and operating on sets of linear equations. This repository contains the learning material for the Nuclear TALENT course Learning from Data: Bayesian Methods and Machine Learning, in York, UK, June 10-28, 2019. Omoju: What I do at GitHub is I build data models, often deep learning models on GitHub data to help GitHub probably build things like a recommendation engine so we can recommend repositories to people. The first part covers basics and preliminaries. , hackers, coders, software engineers, and people working as data scientists in business and industry) you don't need to know that much calculus, linear algebra, or other college-level math to get things done. Grannan, S. This is the site for any aspiring data scientists that want to learn in a quick way. Decision theory, game theory, operational research, \u2026 (source: lecture video from The Machine Learning Summer School by Zoubin Ghahramani, Univ. Machine Learning is built on prerequisites, so much so that learning by first principles seems overwhelming. A computer science student that is interested in Machine Learning would be well advised to get a minor in Mathematics (or just get a degree in Mathematics instead!). 1 Positive Semide nite (PSD) and Positive De nite (PD) matrices. 08 Apr 2016 \u00bb Naive Bayes Classifiers in Rust Adding NB Classifiers to rusty-machine. References for \"practical\" machine learning: Python for data analysis by Wes McKinney. provide a summary of the mathematical background needed for an introductory class in machine learning, which at UC Berkeley is known as CS 189/289A. Machine Learning is not just writing Python or R. I still plan to go through it but mildly disappointed. 8 Web Framework. I'm a Data Science practitioner and computer programmer with an avid interest in Exploratory Data Analysis, Statistics, & Machine Learning. This will allow us to introduce some central programming features of high-level languages like Python and compiled languages like C++ and/or Fortran. In my opinion, it is one of the bedrock of machine learning, deep learning and data science. Many universities use the textbook Introduction to Linear Algebra. Implementation [ edit ] Scikit-learn is largely written in Python, and uses numpy extensively for high-performance linear algebra and array operations. This content is part of a series following the chapter 2 on linear algebra from the Deep Learning Book by Goodfellow, I. Probability and Statistics: Learn Probability and Statistics Through Interactive Visualizations: Seeing Theory was created by Daniel Kunin while an undergraduate at Brown University. In addition to this, you'll be able to perform operations such as addition, subtraction and dot product. As a core programmer, I love taking challenges and love being part of the solution. Students will learn and practice fundamental ideas of linear algebra and simultaneously be exposed to and work with real-world applications of these ideas. We will assume mathematical maturity and comfort with algorithms, probability, and linear algebra. Neural networks rely on it heavily, but so do linear regression, factor analysis, and lots of other methods. This Word Mover\u2019s Distance (WMD) can be seen as a special case of Earth Mover\u2019s Distance (EMD), or Wasserstein distance, the one people talked about in Wasserstein GAN. These subjects include matrix algebra, vector spaces, eigenvalues and eigenvectors, symmetric matrices, linear transformations, and more. In this course, you\u2019ll learn how to work with vectors and matrices, solve matrix-vector equations, perform eigenvalue/eigenvector analyses and use principal component analysis to do dimension reduction on real-world datasets. In the field of data science, however, being familiar with linear algebra and statistics is very important to statistical analysis and prediction. Lawrence [email\u00a0protected] Then last year I learned how he morphed his delightful mathematics book into a brand new title (2019) designed for data scientists - \"Linear Algebra and Learning from Data. Choi et al. Now that you understand the key ideas behind linear regression, we can begin to work through a hands-on implementation in code. Linear Algebra The Rank of a Matrix. Linear Algebra. Gradient Descent with Linear Regression - GitHub Pages. The training data is used to find the optimal model but the model should ultimately work for the test data! Conclusion. Mathematics for machine learning - I totally recommend this book! If you want to learn the bits and pieces of how linear algebra and calculus is used to develop algorithms like principal component analysis, backpropagation etc. https://shaarli. Linear Algebra for Data Science using Python Play all 13:42 Math For Data Science | Practical reasons to learn math for Machine/Deep Learning - Duration: 13 minutes, 42 seconds. Twitter: @mpd37, @AnalogAldo, @ChengSoonOng. com) Bangalore, India * Working as a Lead Data Scientist at Head Digital Works Pvt. Using least-squares linear approximation techniques to find the best linear fit to a set of data points results in the equation of a line which minimizes the sum of the squares of the vertical distances from the given points to the line: Note that, unless the line is horizontal, the vertical distance will be slightly larger than the actual distance, which is measured in the direction. I recently released an efficient linear algebra library for Javascript. LINEAR ALGEBRA. Learn how to solve challenging machine learning problems with TensorFlow, Google\u2019s revolutionary new software library for deep learning. import re , math , random # regexes, math functions, random numbers import matplotlib. Statistical Machine Learning (S2 2017) Deck 6 This lecture \u2022 Notes on linear algebra \u2217Vectors and dot products \u2217Hyperplanes and vector normals \u2022 Perceptron \u2217Introduction to Artificial Neural Networks \u2217The perceptron model \u2217Stochastic gradient descent 2. A few weeks ago, I wrote about how and why I was learning Machine Learning, mainly through Andrew Ng\u2019s Coursera course. Deep Learning Book Series \u00b7 2. Overview - Khan Academy Vectors and Spaces; Matrix Transformations; Python. This course develops the mathematical basis needed to deeply understand how problems of classification and estimation work. pdf; Basic Set Notation & Terminology. Machine learning yearning. hdf5 is a file. If you have more time to dedicate to your projects and you\u2019re also passionate about math, consider to contribute to the library!. Module 0: Introduction & Outline About What is Data Science? Module 1: Required Background Math: Stats, Calculus, Linear Algebra Programming: Basics, Data Structures, Algorithms Databases: Relational Algebra, SQL Important Concepts: Regular expressions, Information Entropy, Distance measurements, OLAP, ETL, BI VS BA and CAP. NOTE: please check for the course practicalities, e. implemented vector and matrix classes with reST-formatted docstrings in Python 3+ General Layout. Linear Algebra Can Help You Choose Your Stock Portfolio Correlation is a very fundamental and viseral way of understanding how the stock market works and how strategies perform. Some recent tutorials by Christos and Co. Python is one of the most commonly used programming languages by data scientists and machine learning engineers. for automated market making. Udacity is the world\u2019s fastest, most efficient way to master the skills tech companies want. I'm interested in applying non-standard tools form abstract algebra and topology to the study of neural networks. We will learn scientific computing basics, topics in numerical linear algebra, mathematical probability (probability spaces, expectation, conditioning, common distributions, law of large numbers and the central limit theorem), statistics (point estimation, confidence intervals, hypothesis testing, maximum likelihood estimation, density. This book provides the conceptual understanding of the essential linear algebra of vectors and matrices for modern engineering and science. Introduction Theunprecedentedadvanceindigitaltechnologyduringthesecondhalfofthe20thcenturyhas producedameasurementrevolutionthatistransformingscience. Now we are ready to see how matrix algebra can be useful when analyzing data. The linux command line: A complete introduction. As we will see, we can do all the common linear algebra operations without using any other library. The elements of statistical learning: data mining, inference, and prediction, Springer, 2009 Linear Algebra and Probability Review (part 1 Linear Algebra, part 2 Probability) Assignment 1: Mar 17. Many popular machine learning methods, including XGBOOST, use matrices to store inputs and process data. In this day, we are going to make the dirty work. Linear Algebra (Michael Damron and Tasho Kaletha) Introduction to Linear Algebra (Strang) Thirty-three Miniatures: Mathematical and Algorithmic Applications of Linear Algebra (Matousek) Linear Algebra Done Right (Axler) Advanced Linear Algebra (Roman). TS CH9 Hypothesis Testing. Research labs and companies have data to analyze and understand, and this deep learning approach has become widespread. You can find all the notebooks on Github. He was a research fellow with Michael Jordan and Peter Bartlett, University of California at Berkeley, from 2003, and with Bernhard Schoelkopf, Max Planck Institute for Intelligent Systems, Tuebingen, Germany, from 2005. Gilbert Strang is a Professor of Mathematics at Massachusetts Institute of Technology and an Honorary Fellow at Balliol College in Oxford. pdf; TS CH1 Exploratory Data Analysis. The final exam is 9am on Friday May 15 in JNSN-Ice Rink. Just want some books to go deeper than a introductory course. This course develops the mathematical basis needed to deeply understand how problems of classification and estimation work. Learn how to solve challenging machine learning problems with TensorFlow, Google's revolutionary new software library for deep learning. In the second part, we discuss how deep learning differs from classical machine learning and explain why it is effective in dealing with complex problems such as image and natural language processing. Provide details and share your research! But avoid \u2026 Asking for help, clarification, or responding to other answers. This is a straightforward course to learn Linear Algebra Fundamentals for Data Science in Python. i recently bought Gilbert Strang's linear algebra book. then this is the book for you. Also, this OpenGL tutorial has useful explanations. A good video series on the topic that allows you to visualize many concepts is Essence of linear algebra. Welcome to the 18. Our goal is to give the beginning student, with little or no prior exposure to linear algebra, a good ground-ing in the basic ideas, as well as an appreciation for how they are used in many applications, including data tting, machine learning and arti cial intelligence, to-. Before that, I was a software engineer at Google where I worked 80% time with the Hotels team on data analytics and 20% time with the operations research team on linear program solvers. Thank you for your interest in Linear Algebra and Learning from Data. In this course, you\u2019ll learn how to work with vectors and matrices, solve matrix-vector equations, perform eigenvalue/eigenvector analyses and use principal component analysis to do dimension reduction on real-world datasets. Learn linear algebra. Then, in Section 2, we quickly bring you up to speed on the prerequisites required for hands-on deep learning, such as how to store and manipulate data, and how to apply various numerical operations based on basic concepts from linear algebra, calculus, and probability. Communication The vast majority of questions about homework, the lectures, or the course should be asked on our Piazza forum, as others will benefit from the responses. I still plan to go through it but mildly disappointed. You can think of an r t i m e s c r times c r t i m e s c matrix as a set of r r r row vectors, each having c c c elements; or you can think of it as a set of c c c column vectors, each having r r r elements. Building on centuries of statistics and mathematics, Data Science uses computational techniques to help the most innovative companies in the world scale. Open Library is an initiative of the Internet Archive, a 501(c)(3) non-profit, building a digital library of Internet sites and other cultural artifacts in digital form. Math is a crucial skill for people who are interested in Data Science and Machine Learning. Data Science and Linear Algebra Fundamentals with Python, SciPy, & NumPy Math is relevant to software engineering but it is often overshadowed by all of the exciting tools and technologies. [Online book] n Andrew Ng. The elements of statistical learning: data mining, inference, and prediction, Springer, 2009 Linear Algebra and Probability Review (part 1 Linear Algebra, part 2 Probability) Assignment 1: Apr 10. Linear Algebra: Video: Professor Gilbert Strang's Video Lectures on linear algebra.", "date": "2020-10-27 17:08:09", "meta": {"domain": "cfalivorno.it", "url": "http://oanx.cfalivorno.it/linear-algebra-and-learning-from-data-github.html", "openwebmath_score": 0.2879904508590698, "openwebmath_perplexity": 1063.9791517497667, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9840936073551713, "lm_q2_score": 0.9273632991598454, "lm_q1q2_score": 0.9126122943990052}}
{"url": "https://grvmax.com.br/box-hedge-akryqc/1783cb-difference-between-scalar-matrix-and-identity-matrix", "text": "It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array You can put this solution on YOUR website! See the picture below. Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal to the square matrix A = [a ij] n \u00d7 n is an identity matrix if If the block produces a scalar output from a scalar input, the block preserves dimension. If you multiply any number to a diagonal matrix, only the diagonal entries will change. 8) Unit or Identity Matrix. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Back in multiplication, you know that 1 is the identity element for multiplication. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. The column (or row) vectors of a unitary matrix are orthonormal, i.e. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. However, there is sometimes a meaningful way of treating a $1\\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being \"functionally equivalent\" to scalars. While off diagonal elements are zero. Long Answer Short: A $1\\times 1$ matrix is not a scalar\u2013it is an element of a matrix algebra. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. Here is the 4\u03a74 unit matrix: Here is the 4\u03a74 identity matrix: A unit matrix is a square matrix all of whose elements are 1's. Yes it is. All the other entries will still be . The same goes for a matrix multiplied by an identity matrix, the result is always the same original non-identity (non-unit) matrix, and thus, as explained before, the identity matrix gets the nickname of \"unit matrix\". In this post, we are going to discuss these points. 2. Okay, Now we will see the types of matrices for different matrix operation purposes. For an example: Matrices A, B and C are shown below. References [1] Blyth, T.S. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. #1. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press The unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Basis. This topic is collectively known as matrix algebra. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. 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And C are shown below the basic operations of matrix-vector and matrix-matrix multiplication will be outlined diagonal will! Can say that a scalar output from a scalar, but could be a vector it. Is square matrix, only the diagonal entries will change diagonal elements zero... Denoted by I column ( or row ) vectors of a matrix times inverse. Times a diagonal matrix, whose all off-diagonal elements are equal to the same scalar.! The column ( or row ) vectors of a unitary matrix are orthonormal, i.e times a diagonal matrix 1\\times.", "date": "2022-07-03 02:52:42", "meta": {"domain": "com.br", "url": "https://grvmax.com.br/box-hedge-akryqc/1783cb-difference-between-scalar-matrix-and-identity-matrix", "openwebmath_score": 0.7978891730308533, "openwebmath_perplexity": 404.20872813020765, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9850429107723176, "lm_q2_score": 0.9263037292767218, "lm_q1q2_score": 0.9124489217459949}}
{"url": "http://lgtermpaperqiur.gloriajohnson.us/binomial-theorem.html", "text": "# Binomial theorem\n\nSeries binomial theorem series contents page contents binomial theorem notation n as a nonnegative integer proof of the binomial theorem proof when n and k are. The binomial theorem date_____ period____ find each coefficient described 1) coefficient of x2 in expansion of (2 + x)5 80 2) coefficient of x2 in expansion. A polynomial with two terms is called a binomial we have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high. Posts about binomial theorem written by ujjwal gulecha. Yes, pascal's triangle and the binomial theorem isn\u00e2\u20ac\u2122t particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong.\n\nBinomial expansions in chapter 5 you learned how to square a binomial the binomial theorem 652 (12-26) chapter 12 sequences and series. Binomial theorem the binomial theorem states that the binomial coefficients $$c(n,k)$$ serve as coefficients in the expansion of the powers of the binomial $$1+x$$. Binomial theorem : akshay mishra xi a , k v 2 , gwalior in elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. How to use the binomial theorem to expand binomial expressions, examples and step by step solutions, the binomial theorem using combinations. Quick links: downloadable teaching materials for binomial theorem syllabus content for the algebra topic: sl syllabus (see syllabus section 13) hl syllabus (see.\n\nPowers of a binomial (a + b) what are the binomial coefficients pascal's triangle. In this video lesson, you will see what the binomial theorem has in common with pascal's triangle learn how you can use pascal's triangle to help. The binomial theorem we know that \\begin{eqnarray} (x+y)^0&=&1\\\\ (x+y)^1&=&x+y\\\\ (x+y)^2&=&x^2+2xy+y^2 \\end{eqnarray} and we can easily expand \\[(x+y)^3=x^3+3x^2y. Expanding a binomial expression that has been raised to some large power could be troublesome one way to solve it is to use the binomial theorem.\n\nExplains how to use the binomial theorem, and displays the theorem's relationship to pascal's triangle. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum for k-12 kids, teachers and parents.\n\nThis section is about sequences, series and the binomial theorem, with applications. Binomial theorem was known for the case n = 2 by euclid around 300 bc, and pascal stated it in modern form in 1665 newton showed that a similar formula for negative.\n\n## Binomial theorem\n\nIn this lesson, students will learn the binomial theorem and get practice using the theorem to expand binomial expressions the theorem is broken. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. Binomial theorem: binomial theorem, statement that describes the nth power of the sum of two numbers (a + b.\n\nWhen a binomial is raised to whole number powers, the coefficients of the terms in the expansion form a pattern. Mathematics notes module - i algebra 266 binomial theorem \u2022 state the binomial theorem for a positive integral index and prove it using the principle of. The most basic example of the binomial theorem is the formula for the square of x + y: (+) = + + the binomial coefficients 1, 2, 1 appearing in this expansion. Fun math practice improve your skills with free problems in 'binomial theorem i' and thousands of other practice lessons. 123 applications of the binomial theorem expansion of binomials the binomial theorem can be used to find a complete expansion of a power of a binomial or a. There are several closely related results that are variously known as the binomial theorem depending on the source even more confusingly a number of these (and other. While the foil method can be used to multiply any number of binomials together, doing more than three can quickly become a huge headache.\n\nThe binomial theorem the binomial theorem is a fundamental theorem in algebra that is used to expand expressions of the form where n can be any number. Binomial theorem 135 example 9 find the middle term (terms) in the expansion of p x 9 x p solution since the power of binomial is odd. Demonstrates how to answer typical problems involving the binomial theorem.\n\nBinomial theorem\nRated 4/5 based on 16 review", "date": "2018-06-24 03:20:32", "meta": {"domain": "gloriajohnson.us", "url": "http://lgtermpaperqiur.gloriajohnson.us/binomial-theorem.html", "openwebmath_score": 0.7695887088775635, "openwebmath_perplexity": 750.4912468962283, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9939024392523322, "lm_q2_score": 0.9173026618464796, "lm_q1q2_score": 0.9117093531418733}}
{"url": "https://mathoverflow.net/questions/334888/smallest-set-such-that-all-arithmetic-progression-will-always-contain-at-least-a", "text": "# Smallest set such that all arithmetic progression will always contain at least a number in a set\n\nLet $$S= \\left\\{ 1,2,3,...,100 \\right\\}$$ be a set of positive integers from $$1$$ to $$100$$. Let $$P$$ be a subset of $$S$$ such that any arithmetic progression of length 10 consisting of numbers in $$S$$ will contain at least a number in $$P$$. What is the smallest possible number of elements in $$P$$ ?\n\nDenote $$|P|$$ as the number of elements in $$P$$. We shall find the smallest possible value of $$|P|$$.\n\nFor $$|P|=16$$, we have the answer by @RobertIsrael below.\n\nHowever, for $$|P|<16$$, I can neither find such set $$P$$ nor prove that $$|P|$$ cannot be less than $$16$$. So my question is:\n\nIs it true that $$|P| \\geq 16$$? How can I prove it? If not, what is the minimum amount of elements in $$P$$ ?\n\nAlso, I am wondering that:\n\nIf we replace 10 with an even number $$n$$,and $$100$$ with $$n^2$$, can we find the minimum of $$|P|$$ ?\n\nAny answers or comments will be appreciated. If this question should be closed, please let me know. If this forum cannot answer my question, I will delete this question immediately.\n\n\u2022 it is not too unusual that questions here get answered, say, after a year, and not immediately. Jun 27 '19 at 6:24\n\u2022 @DimaPasechnik Thanks. I just afraid that my question will be forgotten and cannot be answered. Jun 27 '19 at 7:02\n\u2022 good questions don't get forgotten. they pop up in searches, etc. Jun 27 '19 at 8:21\n\u2022 This can be considered as a set-covering problem. Although set covering is NP-complete, I suspect this one is within the reach of current technology. Jun 27 '19 at 12:37\n\u2022 For the last question (replacing 10 with $n$), have you computed the optimal number for $n\\le 9$ and checked the OEIS? Jun 27 '19 at 14:44\n\nConsidering the complement of $$P$$ in $$[1,100]$$, you are asking how large can a subset of $$[1,100]$$ be given that it does not contain any $$10$$-term arithmetic progression. The more general question\n\nHow large can a subset of $$[1,N]$$ be given that it does not contain any $$k$$-term arithmetic progression?\n\nis one of the central problems in combinatorial number theory. There is no chance to give a precise answer, as an \"explicit\" function of $$N$$ and $$k$$, and it quite likely that this is impossible already in your special situation where $$N=n^2$$ and $$k=n$$.\n\nHere is an argument showing that if $$P\\subset[1,n^2]$$ meets every $$n$$-term progression contained in $$[1,n^2]$$, then $$|P|>n+n^{0.5+o(1)}$$. (See also the paragraph at the very end for the estimate $$|P|\\ge 12$$ in your special case where $$P\\subset[1,100]$$ and we want to block all $$10$$-term progressions.) It would be interesting to improve these estimates or at least to decide whether $$|P|>Cn$$ holds true with an absolute constant $$C>1$$.\n\nWrite $$K:=|P|$$, $$\\Delta:=K-n$$, and $$P=\\{p_1,\\dotsc,p_K\\}$$ where $$1\\le p_1<\\dotsb. Notice that $$p_1\\le n$$ and $$p_K\\ge n^2-(n-1)$$, whence $$p_K-p_1\\ge(n-1)^2$$.\n\nFor any $$d\\in[1,n]$$, the set $$P$$ contains an element from every residue class modulo $$d$$, and it follows that there are at most $$K-d$$ pairs of consecutive elements of $$P$$ with the difference equal to $$d$$; also, if $$d>n$$, then there are no such pairs at all. Let $$a$$ and $$r$$ be defined by \\begin{align*} K-1 &= \\Delta+(\\Delta+1)+\\dotsb+(\\Delta+(a-1))+r \\\\ &= a\\Delta+\\frac{a(a-1)}2 + r,\\quad 0\\le r<\\Delta+a. \\tag{1} \\end{align*} Since there are totally $$K-1$$ pairs of consecutive elements of $$P$$, of them at most $$\\Delta$$ pairs at distance $$n$$, at most $$\\Delta+1$$ pairs at distance $$n-1$$, etc, we conclude that \\begin{align*} p_K-p_1 &\\le n\\Delta+(n-1)(\\Delta+1)+\\dotsb+(n-(a-1))(\\Delta+(a-1))+(n-a)r \\\\ &= \\Delta na+(n-\\Delta)\\cdot\\frac{a(a-1)}2-\\frac{a(a-1)(2a-1)}{6}+(n-a)r. \\end{align*} Recalling the estimate $$p_K-p_1\\ge(n-1)^2$$, and using ($$1$$), we get \\begin{align*} (n-1)^2 &\\le \\Delta na+(n-\\Delta)\\cdot\\frac{a(a-1)}2-\\frac{a(a-1)(2a-1)}{6}+(n-a)r \\\\ &= n\\Big(a\\Delta+\\frac{a(a-1)}2 + r\\Big) - \\Delta\\cdot\\frac{a(a-1)}2 - \\frac{a(a-1)(2a-1)}{6} - ar \\\\ &= n(K-1) - \\Delta\\cdot\\frac{a(a-1)}2 - \\frac{a(a-1)(2a-1)}{6} - ar. \\tag{2} \\end{align*}\n\nWe now assume, aiming at a contradiction, that $$\\Delta with an absolute constant $$0. From (1) we get then $$K-1 \\ge \\Delta a + \\frac{a(a-1)}2 \\ge \\frac12\\,a^2 - 1$$ implying $$a\\le\\sqrt{2K}$$; hence, $$\\Delta a=O(n^{0.5+c})$$ and $$r=a+\\Delta=O(n^{0.5})$$. As a result, $$\\frac12\\,a^2 = K-1+\\frac12\\,a-\\Delta a - r > K - O(n^{0.5+c}),$$ leading to $$a>(1-o(1))\\sqrt{2K}$$.\n\nWith these estimates in mind, from (2) we obtain $$n^2 + O(n) \\le nK - \\frac12\\,\\Delta a^2 - \\frac13\\,a^3;$$ that is, $$\\Delta n \\ge \\frac12\\,\\Delta a^2 + \\frac13\\,a^3 + O(n).$$ Consequently, $$n^{1+c} \\ge \\Delta n \\ge \\frac13\\,a^3 + O(n) \\ge (1-o(1))(2K)^{1.5} + O(n) > n^{1.5} + O(n),$$ a contradiction.\n\nAs an illustration of this approach, let's show that one needs at least $$12$$ elements to block every $$10$$-term progression in $$[1,100]$$. Suppose for a contradiction that $$P\\subset[1,100]$$ is an $$11$$-element set blocking all such progressions. There are $$|P|-1=10$$ pairs of consecutive elements of $$P$$. Of these ten pairs, there is at most one pair with distance $$10$$ between its two elements, at most two pairs with distance $$9$$, at most three pairs with distance $$8$$, and at most four pairs with distance $$7$$. Therefore the largest element of $$P$$ exceeds the smallest one by at most $$1\\cdot 10+2\\cdot 9 + 3\\cdot 8 + 4\\cdot 7=80$$. It follows that either the smallest element of $$P$$ is at least $$11$$, or its largest element is at most $$90$$; but then $$P$$ does not block at least one of the progressions $$[1,10]$$ and $$[91,100]$$, a contradiction.\n\nUsing a tabu search procedure, I have found a solution for $$|P|=17$$, namely $${1, 11, 18, 25, 31, 32, 33, 36, 44, 51, 58, 65, 69, 70, 77, 84, 91}$$. I don't know if this is optimal.\n\nEDIT: Found a solution for $$|P|=16$$, namely $$10, 15, 22, 29, 36, 43, 53, 55, 56, 57, 58, 68, 73, 74, 84, 91$$\n\n\u2022 I'm working on $|P|=16$. So far I've found a $P$ with $|P|=16$, namely $\\{9, 18, 28, 29, 31, 40, 42, 51, 53, 56, 65, 69, 70, 77, 84, 91\\}$, that intersects all but one of these arithmetic progressions, the exception being $({36, 43, 50, 57, 64, 71, 78, 85, 92, 99})$. Jun 27 '19 at 16:50\n\u2022 I'm using a tabu search over sets of a given size to maximize the number of a.p.'s that intersect the set. Possible moves consist of replacing a member of the set with a nonmember. Jun 27 '19 at 17:00\n\u2022 Thank you. Your answer is correct. How long did it take to find those numbers? Can you find the boundary of $|P|$? Jun 28 '19 at 8:00\n\u2022 So is 16 optimal?\n\u2013\u00a0EGME\nJun 28 '19 at 20:43\n\u2022 My brute-force confirms that there no 15. Jul 2 '19 at 5:19", "date": "2021-10-27 22:35:03", "meta": {"domain": "mathoverflow.net", "url": "https://mathoverflow.net/questions/334888/smallest-set-such-that-all-arithmetic-progression-will-always-contain-at-least-a", "openwebmath_score": 0.7299977540969849, "openwebmath_perplexity": 201.440791491689, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918529698322, "lm_q2_score": 0.921921841290738, "lm_q1q2_score": 0.9113122291908411}}
{"url": "https://math.stackexchange.com/questions/1028822/mutlivariable-calculus-surface-area", "text": "# Mutlivariable Calculus: Surface Area\n\nThis was a question a students had asked me earlier today regarding surface area.\n\nFind the surface area of the hemisphere $x^2+y^2+z^2 = 4$ bounded below by $z=1$.\n\nI decided to approach this problem using spherical coordinates and got the following\n\n\\begin{eqnarray} \\int_{0}^{2\\pi}\\int_{0}^{\\pi/3}4\\sin\\phi d\\phi d\\theta & = & 4\\int_{0}^{2\\pi}d\\theta\\int_{0}^{\\pi/3}\\sin\\phi d\\phi\\\\ & = & 8\\pi\\cos\\phi|_{\\pi/3}^{0}\\\\ & = & 8\\pi(1-\\frac{1}{2}) = 4\\pi \\end{eqnarray}\n\nI also solved this problem using single variable calculus as follows. I can represent the sphere as a circle of $h^2+z^2 =4$. Thus we have the following:\n\n\\begin{eqnarray} SA = \\int_a^b2\\pi f(z)ds & = & 2\\pi\\int_1^2\\sqrt{4-z^2}\\sqrt{1+\\frac{z^2}{4-z^2}}dz\\\\ & = & 2\\pi\\int_1^2\\sqrt{4-z^2}\\sqrt{\\frac{4}{4-z^2}} dz\\\\ & = & 2\\pi\\int_1^22dz=4\\pi z|_1^2=4\\pi \\end{eqnarray}\n\nAs you see, I got the same answer for both approaches.\n\nThe student, and a few others, comes back later during the day and tells me the answer I got was incorrect. He does not tell me what the professor got, he just told me it was wrong. I asked myself \"why?\" Is there something I missed?\n\nThanks in advance for any feedback.\n\n\u2022 Maybe there is something wrong with the text of the problem. The surface whose area has been calculated is not a hemisphere but a spherical cap. \u2013\u00a0Christian Blatter Nov 19 '14 at 10:52\n\u2022 Git Gud Could u please help me with this vector calculus question as well. I really need help in this. Thanks \u2013\u00a0ys wong Nov 23 '14 at 7:38\n\nI don't see what you did wrong here...\n\nTo me, to solve these types of problems you have to think geometrically--there isn't going to be some way to do it just from a knowledge of multivariable calculus.\n\nFirst, how to find the total surface area of the sphere--that will help. You need to break the sphere up into circles stacked on top of each other, then find the $dA$:\n\n$$dA = 2\\pi r_{\\phi} h$$\n\n$h$ is easy to find, it's just $rd\\phi$, $r_\\phi$ is the radius at the given azimuth: $r_\\phi = r\\sin(\\phi)$ which gives:\n\n$$dA = 2\\pi r^2\\sin(\\phi)d\\phi\\\\ A = \\int dA = 2\\pi r^2\\left.\\int_{0}^{\\pi}\\sin(\\phi)d\\phi = -2\\pi r^2\\cos(\\phi)\\right|_0^\\pi = 4\\pi r^2$$\n\nSo the correct integral should be:L\n\n$$A = 2\\pi r^2\\left.\\int_{0}^{\\phi_0} \\sin(\\phi)d\\phi = -2\\pi r^2 \\cos(\\phi)\\right|_{0}^{\\phi_0} = 2\\pi r^2\\left(1 - \\cos(\\phi_0)\\right)$$\n\nIn this case, $\\phi_0$ satisfies that $z = r\\cos(\\phi) = 2\\cos(\\phi_0) = 1 \\rightarrow \\cos(\\phi_0) = \\frac{1}{2}$ and thus:\n\n$$A = 2\\pi 2^2\\left(1 - \\frac{1}{2}\\right) = 4\\pi$$", "date": "2021-01-28 15:10:13", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1028822/mutlivariable-calculus-surface-area", "openwebmath_score": 0.8875633478164673, "openwebmath_perplexity": 359.89241096559937, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9825575142422757, "lm_q2_score": 0.9273632991598455, "lm_q1q2_score": 0.9111877780220137}}
{"url": "https://casmusings.wordpress.com/tag/factoring/", "text": "Tag Archives: factoring\n\nInfinite Ways to an Infinite Geometric\u00a0Sum\n\nOne of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\\displaystyle \\frac{g}{1-r}$ for first term\u00a0g and common ratio\u00a0r when $\\left| r \\right| < 1$. \u00a0I was glad she was dissatisfied with blind use of\u00a0a formula\u00a0and\u00a0dove into a familiar (to me) derivation. \u00a0In the end, she shook me free from my routine just as she made sure she didn\u2019t fall into her own.\n\nSTANDARD INFINITE GEOMETRIC SUM DERIVATION\n\nMy standard explanation starts with a generic\u00a0infinite geometric series.\n\n$S = g+g\\cdot r+g\\cdot r^2+g\\cdot r^3+...$ \u00a0(1)\n\nWe can reason this series converges iff\u00a0$\\left| r \\right| <1$ (see Footnote 1 for an explanation). \u00a0Assume this is true for (1). \u00a0Notice the terms on the right keep multiplying by\u00a0r.\n\nThe annoying part of summing any infinite series is the ellipsis (\u2026). \u00a0Any finite number of terms always has a finite sum, but that simply written, but vague\u00a0ellipsis is logically difficult. \u00a0In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series. \u00a0You can accomplish this by continuing the pattern on the right: \u00a0multiplying both sides by\u00a0r\n\n$r\\cdot S = r\\cdot \\left( g+g\\cdot r+g\\cdot r^2+... \\right)$\n\n$r\\cdot S = g\\cdot r+g\\cdot r^2+g\\cdot r^3+...$ \u00a0(2)\n\nThis seems to\u00a0make make the right side of (2) identical to the right side of (1) except for the leading\u00a0g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical. \u00a0After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.\n\n$(1)-(2) = S-S\\cdot r = S\\cdot (1-r)=g$\n\n$\\displaystyle S=\\frac{g}{1-r}$\n\nSTUDENT PREFERENCES\n\nI despise giving any formula to any of my classes without\u00a0at least exploring its genesis. \u00a0I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.\n\nIn my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number\u00a0prefer\u00a0the quick \u201cmultiply-by-r-and-subtract\u201d approach used to derive the summation formula. \u00a0For many, apparently, the dynamic manipulation is more meaningful than a static rule. \u00a0It\u2019s very cool to watch student preferences at play.\n\nK\u2019s VARIATION\n\nK understood the proof, and then asked a question I hadn\u2019t thought to ask. \u00a0Why did we have to multiply by\u00a0r? \u00a0Could\u00a0multiplication by $r^2$ also determine the summation\u00a0formula?\n\nI had\u00a0three\u00a0nearly\u00a0simultaneous thoughts followed quickly by a fourth. \u00a0First, why hadn\u2019t I ever thought to ask that? \u00a0Second, geometric series for $\\left| r \\right|<1$ are absolutely convergent, so K\u2019s suggestion should work. \u00a0Third, while the formula would initially look different, absolute convergence guaranteed that whatever the \u201c$r^2$ formula\u201d looked like, it had to be algebraically equivalent to the standard form. \u00a0While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K\u2019s question and the equivalence from Thought #3.\n\nMultiplying (1) by $r^2$ gives\n\n$r^2 \\cdot S = g\\cdot r^2 + g\\cdot r^3 + ...$ (3)\n\nand the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to\u00a0get the inevitable result.\n\n$(1)-(3) = S-S\\cdot r^2 = g+g\\cdot r$\n\n$S\\cdot \\left( 1-r^2 \\right) = g\\cdot (1+r)$\n\n$\\displaystyle S=\\frac{g\\cdot (1+r)}{1-r^2} = \\frac{g\\cdot (1+r)}{(1+r)(1-r)} = \\frac{g}{1-r}$\n\nThat was cool, but this success meant that there were surely many more options.\n\nEXTENDING\n\nWhy stop at multiplying by\u00a0r\u00a0or $r^2$? \u00a0Why not multiply both sides of (1) by a generic $r^N$ for any natural number N? \u00a0 That would give\n\n$r^N \\cdot S = g\\cdot r^N + g\\cdot r^{N+1} + ...$ (4)\n\nwhere the ellipses of (1) and (4) are again identical by the method of Footnote 2. \u00a0Subtracting (4) from (1) gives\n\n$(1)-(4) = S-S\\cdot r^N = g+g\\cdot r + g\\cdot r^2+...+ g\\cdot r^{N-1}$\n\n$S\\cdot \\left( 1-r^N \\right) = g\\cdot \\left( 1+r+r^2+...+r^{N-1} \\right)$ \u00a0(5)\n\nThere are two ways to proceed from (5). \u00a0You could recognize the right side as a finite geometric sum with first term 1 and ratio\u00a0r. \u00a0Substituting that formula and dividing by $\\left( 1-r^N \\right)$ would give the general result.\n\nAlternatively, I could see students exploring $\\left( 1-r^N \\right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor. \u00a0I got the following TI-Nspire CAS result in about 10-15 seconds,\u00a0clearly\u00a0suggesting that\n\n$1-r^N = (1-r)\\left( 1+r+r^2+...+r^{N-1} \\right)$. \u00a0(6)\n\nMath induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS. \u00a0From there, dividing both sides of (5) by $\\left( 1-r^N \\right)$ gives the generic result.\n\n$\\displaystyle S = \\frac{g\\cdot \\left( 1+r+r^2+...+r^{N-1} \\right)}{\\left( 1-r^N \\right)}$\n\n$\\displaystyle S = \\frac{g\\cdot \\left( 1+r+r^2+...+r^{N-1} \\right) }{(1-r) \\cdot \\left( 1+r+r^2+...+r^{N-1} \\right)} = \\frac{g}{1-r}$\n\nIn the end, K helped me see there wasn\u2019t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways. \u00a0Nice.\n\nFOOTNOTES\n\n1)\u00a0RESTRICTING r: \u00a0Obviously an infinite geometric\u00a0series diverges for $\\left| r \\right| >1$ because that would make $g\\cdot r^n \\rightarrow \\infty$ as $n\\rightarrow \\infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum.\n\nFor $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \\ne 0$ , you get\u00a0a\u00a0sum of an\u00a0infinite number\u00a0of some\u00a0nonzero quantity, and that\u00a0is always infinite, no matter how small or large the nonzero quantity.\n\nThe last case, $r=-1$, is more subtle. \u00a0For $g \\ne 0$, this terms of this series alternate between positive and negative\u00a0g,\u00a0making the partial sums of the series\u00a0add to either\u00a0g or 0, depending on whether you have summed an even or an odd number of terms. \u00a0Since the partial sums alternate, the overall sum is divergent. \u00a0Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.\n\n2)\u00a0NOT ALL INFINITIES ARE THE SAME: \u00a0There are two ways to show two groups are the same size. \u00a0The obvious way is to\u00a0count the elements in each group and find out there is\u00a0the same number of elements in each, but this works only if you have a finite group size. \u00a0Alternatively, you could a) match\u00a0every\u00a0element in group 1 with a\u00a0unique\u00a0element from\u00a0group 2, and b) match\u00a0every\u00a0element in group 2\u00a0with a\u00a0unique\u00a0element from\u00a0group 1. \u00a0It is important to do both steps here to show that there are no left-over, unpaired elements in either group.\n\nSo do\u00a0the ellipses in (1) and (2) represent the same sets? \u00a0As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant. \u00a0For the second approach using pairing, we need to compare individual elements. \u00a0For every element in the ellipsis of (1), obviously there is an \u201cpartner\u201d in (2) as the multiplication of (1) by\u00a0r visually shifts all of the terms of the\u00a0series right one position, creating the necessary matches.\n\nStudents often are troubled by the second matching as it appears the ellipsis in (2) contains an \u201cextra term\u201d from the right shift. \u00a0But, for every specific term you identify in (2), its identical twin exists in (1). \u00a0In the weirdness of infinity, that \u201cextra term\u201d appears to have been absorbed without changing the \u201csize\u201d of the infinity.\n\nSince there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero.\n\nProbability, Polynomials, and Sicherman\u00a0Dice\n\nThree\u00a0years ago, I encountered a question on the TI-Nspire Google group asking if there was a way to use CAS to solve probability problems. \u00a0The ideas I pitched in my\u00a0initial response and follow-up a year later (after first using it with students in a statistics class) have been thoroughly re-confirmed in my first year teaching AP Statistics. \u00a0I\u2019ll quickly re-share them below\u00a0before\u00a0extending the concept\u00a0with ideas I picked up a couple weeks ago from Steve Phelps\u2019\u00a0session on Probability, Polynomials, and CAS at the 64th annual OCTM conference earlier this month\u00a0in Cleveland, OH.\n\nBINOMIALS: \u00a0FROM POLYNOMIALS TO\u00a0SAMPLE SPACES\n\nOnce you understand them, binomial probability distributions aren\u2019t that difficult, but the initial conjoining of combinatorics and probability makes this a perennially difficult topic for many students. \u00a0The standard formula for the probability of determining the chances of\u00a0K successes in\u00a0N attempts of a binomial situation where p is the probability of a single success in a single attempt\u00a0is no less daunting:\n\n$\\displaystyle \\left( \\begin{matrix} N \\\\ K \\end{matrix} \\right) p^K (1-p)^{N-K} = \\frac{N!}{K! (N-K)!} p^K (1-p)^{N-K}$\n\nBut that is almost exactly the same result one gets by raising binomials to whole number powers, so why not use a CAS to expand a polynomial and at least compute the\u00a0$\\displaystyle \\left( \\begin{matrix} N \\\\ K \\end{matrix} \\right)$ portion of the probability? \u00a0One added advantage of using a CAS is that you could use full event names instead of abbreviations, making it even easier to identify the meaning of each event.\n\nThe TI-Nspire output above shows the entire sample space resulting from flipping a coin 6 times. \u00a0Each term is an event. \u00a0Within each term, the exponent of each variable notes the number of times that variable occurs and the coefficient is\u00a0the number of times that combination occurs. \u00a0The overall exponent in the expand command is the number of trials. \u00a0For example, the middle term\u2013 $20\\cdot heads^3 \\cdot tails^3$ \u2013says that there are 20 ways you could get 3 heads and 3 tails when tossing a coin 6 times. The last term is just $tails^6$, and its implied coefficient is 1, meaning there is just one way to flip 6 tails in 6 tosses.\n\nThe expand command makes more sense than memorized algorithms and provides context to students until they gain a deeper understanding of what\u2019s actually going on.\n\nFROM POLYNOMIALS TO PROBABILITY\n\nStill using the expand command, if each variable is preceded by its probability, the CAS result combines the\u00a0entire sample space AND the corresponding probability distribution function. \u00a0For example, when rolling a fair die four times, the distribution for 1s vs. not 1s (2, 3, 4, 5, or 6) is given by\n\nThe highlighted term says there is a 38.58% chance that there will be exactly one 1 and any three other numbers (2, 3, 4, 5, or 6) in four rolls of a fair 6-sided die. \u00a0The probabilities of the other four events in the sample space are also shown. \u00a0Within the TI-Nspire (CAS or non-CAS), one could use a command to give all of these probabilities simultaneously (below), but then one has to remember whether the non-contextualized probabilities are for increasing or decreasing values of which binomial outcome.\n\nParticularly early on in their explorations of binomial probabilities, students I\u2019ve taught have shown a very clear preference for the polynomial approach, even when\u00a0allowed to choose any\u00a0approach that makes sense to them.\n\nTAKING POLYNOMIALS FROM ONE DIE TO MANY\n\nGiven these earlier thoughts, I was naturally drawn to Steve Phelps \u201cProbability, Polynomials, and CAS\u201d session at the November 2014 OCTM annual meeting in Cleveland, OH. \u00a0Among the ideas he shared was using polynomials to create the distribution function for the sum of two fair 6-sided dice. \u00a0My immediate thought was to apply my earlier ideas. \u00a0As noted in my initial\u00a0post, the expansion approach above is not limited to binomial situations. \u00a0My first reflexive CAS command\u00a0in Steve\u2019s session before he share anything was this.\n\nBy writing the outcomes in words, the\u00a0CAS interprets them as\u00a0variables. \u00a0I got the entire sample space, but didn\u2019t learn gain anything beyond a long polynomial. \u00a0The first output\u2013 $five^2$ \u2013with its implied coefficient says there is 1 way to get 2 fives. \u00a0The second term\u2013 $2\\cdot five \\cdot four$ \u2013says there are 2\u00a0ways to get 1\u00a0five and 1\u00a0four. \u00a0Nice that the technology gives me all the terms so quickly, but it doesn\u2019t help me get a distribution function of the sum. \u00a0I got the distributions of the specific outcomes, but the way I defined the variables didn\u2019t permit sum of their actual numerical values. \u00a0Time to listen to the speaker.\n\nHe suggested using a common variable, X, for all faces with the value of each face expressed as an exponent. \u00a0That is, a standard 6-sided die would be represented by\u00a0$X^1+X^2+ X^3+X^4+X^5+X^6$ where the six different exponents represent the numbers on the six faces of a typical 6-sided die. \u00a0Rolling two such dice simultaneously is handled as I did earlier with the binomial cases.\n\nNOTE: \u00a0Exponents are handled in TWO different ways here. \u00a01)\u00a0Within\u00a0a single\u00a0polynomial, an exponent is an event value, and 2)\u00a0Outside\u00a0a polynomial, an exponent indicates the number of times that polynomial is applied within the specific event. \u00a0Coefficients have the same meaning as before.\n\nBecause the variables are now the same, when specific terms are multiplied, their exponents (face values) will be added\u2013exactly what I wanted to happen. \u00a0That means the sum of the faces when you roll two dice is determined by the following.\n\nNotice that the output is a single polynomial. \u00a0Therefore, the exponents are the values of individual cases. \u00a0For a couple examples, there are\u00a03 ways to get a sum of 10 $\\left( 3 \\cdot x^{10} \\right)$, 2 ways to get a sum of 3\u00a0$\\left( 2 \\cdot x^3 \\right)$, etc. \u00a0The most commonly occurring outcome is the term with the largest coefficient. \u00a0For rolling two standard fair 6-sided dice, a sum of 7 is the most common outcome, occurring 6 times $\\left( 6 \\cdot x^7 \\right)$. \u00a0That certainly simplifies the typical 6\u00d76 tables used to compute the sums\u00a0and probabilities resulting from rolling two dice.\n\nWhile not the point of Steve\u2019s talk, I immediately saw that technology had just opened the door to problems that had been computationally inaccessible in the past. \u00a0For example, what is the most common sum when rolling 5\u00a0dice and what is the probability of that sum? \u00a0On my CAS, I entered this.\n\nIn the middle of the expanded polynomial are two terms with the largest coefficients, $780 \\cdot x^{18}$ and $780 \\cdot x^{19}$, meaning a sums of 17 and 18 are the most common, equally likely outcomes when rolling 5 dice. \u00a0As there are $6^5=7776$ possible outcomes when rolling a die 5 times, the probability of each of these is $\\frac{780}{7776} \\approx 0.1003$, or about 10.03% chance each for a sum of 17 or 18. \u00a0This can be verified by inserting the probabilities as coefficients before each term before CAS expanding.\n\nWith thought, this shouldn\u2019t be surprising as the expected mean value of rolling a 6-sided die many times is 3.5, and $5 \\cdot 3.5 = 17.5$, so the integers on either side of 17.5 (17 & 18) should be the most common. \u00a0Technology confirms intuition.\n\nROLLING DIFFERENT DICE SIMULTANEOUSLY\n\nWhat is the distribution of sums when rolling a 4-sided and a 6-sided die together? \u00a0No problem. \u00a0Just multiply two different polynomials, one representative of each die.\n\nThe output shows that sums of 5, 6, and 7 would be the most common, each occurring four times with probability $\\frac{1}{6}$ and together accounting for half of all outcomes of rolling these two dice together.\n\nA BEAUTIFUL EXTENSION\u2013SICHERMAN DICE\n\nMy most unexpected gain from Steve\u2019s talk happened when he asked if we could get the same distribution of sums as \u201cnormal\u201d\u00a06-sided dice, but from two different 6-sided dice. \u00a0The only restriction he gave was that all of the faces of the new dice had to have positive values. \u00a0This\u00a0can be approached by realizing that the distribution of sums of the two normal dice can be found by multiplying two representative polynomials to get\n\n$x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2$.\n\nRestating the question in the terms of this post, are there two other polynomials that could be multiplied to give the same product? \u00a0That is, does this polynomial factor into other polynomials that could multiply to the same product? \u00a0A CAS factor command gives\n\nAny\u00a0rearrangement of these eight (four distinct)\u00a0sub-polynomials\u00a0would create the same distribution as the sum of two dice, but what would the the separate sub-products mean in terms of the dice? \u00a0As a first example, what if the first two expressions were used for one die (line 1 below) and the two squared trinomials comprised a second die (line 2)?\n\nLine 1 actually describes a 4-sided die with one face of 4, two faces with 3s, and one face of 2. \u00a0Line 2 describes a 9-sided die (whatever that is) with one face of 8, two faces of 6, three faces of 4, two faces of 2, and one face with a 0 ( $1=1 \\cdot x^0$). \u00a0This means rolling a 4-sided and a 9-sided die as described would give exactly the same sum distribution. \u00a0Cool, but not what I wanted. \u00a0Now what?\n\nFactorization gave\u00a0four distinct sub-polynomials, each with multitude 2. \u00a0One die could contain 0, 1, or 2 of each of these with the remaining factors on the other die. \u00a0That means there are $3^4=81$ different possible dice combinations. \u00a0I could continue with a trail-and-error approach, but I wanted to be more efficient and elegant.\n\nWhat follows is the result of thinking about the problem for a while. \u00a0Like most math solutions to interesting\u00a0problems, ultimate solutions are typically much cleaner and more elegant than the thoughts that went into them. \u00a0Problem solving is a messy\u2013but very rewarding\u2013business.\n\nSOLUTION\n\nHere are my insights over time:\n\n1) I realized that the $x^2$ term would raise the power (face values) of the desired dice, but would not change the coefficients (number of faces). \u00a0Because Steve asked for dice with all positive face values. \u00a0That meant each desired die had to have at least one\u00a0x to prevent non-positive face values.\n\n2) My first attempt didn\u2019t create 6-sided dice. \u00a0The sums of the coefficients of the sub-polynomials determined the number of sides. \u00a0That sum could also be found by substituting $x=1$ into the sub-polynomial. \u00a0I want 6-sided dice, so the final coefficients must add to 6. \u00a0The coefficients of the factored polynomials of any die individually must add to 2, 3, or 6 and have a product of 6. \u00a0The coefficients of $(x+1)$ add to 2, $\\left( x^2+x+1 \\right)$ add to 3, and\u00a0$\\left( x^2-x+1 \\right)$ add to\u00a01. \u00a0The only way to get a polynomial coefficient sum of 6 (and thereby create 6-sided dice) is for each die to have one $(x+1)$ factor and one\u00a0$\\left( x^2+x+1 \\right)$ factor.\n\n3) That leaves the two $\\left( x^2-x+1 \\right)$ factors. \u00a0They could split between the two dice or both could be on one die, leaving none on the other. \u00a0We\u2019ve already determined that each die already had to have one each of the\u00a0x, $(x+1)$, and\u00a0$\\left( x^2+x+1 \\right)$ factors. \u00a0To also split the\u00a0$\\left( x^2-x+1 \\right)$ factors would result in the original dice: \u00a0Two normal 6-sided dice. \u00a0If I want different dice, I have to load both of these factors on one die.\n\nThat means there is ONLY ONE POSSIBLE alternative for two 6-sided dice that have the same sum distribution as two normal 6-sided dice.\n\nOne die would have single faces of 8, 6, 5, 4, 3, and 1. \u00a0The other die would have one 4, two 3s, two 2s, and one 1. \u00a0And this is exactly the result of the famous(?) Sicherman Dice.\n\nIf a 0\u00a0face value\u00a0was allowed, shift one factor of\u00a0x\u00a0from one polynomial to the other. \u00a0This can be done two ways.\n\nThe first possibility has dice with faces {9, 7, 6, 5, 4, 2} and {3, 2, 2, 1, 1, 0}, and the second has faces {7, 5, 4, 3, 2, 0} and {5, 4, 4, 3, 3, 2}, giving the only other two non-negative solutions to the Sicherman Dice.\n\nBoth of these are nothing more than adding one to all\u00a0faces of one die and subtracting one from from all faces of the other. \u00a0While not necessary to use polynomials to compute these, they are equivalent to multiplying the polynomial of one die by\u00a0x and the other by $\\frac{1}{x}$ as many times as desired. That means there are an infinite number of 6-sided dice with the same sum distribution as normal 6-sided dice if you allow the sides to have negative faces. \u00a0One of these is\n\ncorresponding to a pair of\u00a0Sicherman Dice\u00a0with faces {6, 4, 3, 2, 1, -1} and {1,5,5,4,4,3}.\n\nCONCLUSION:\n\nThere are other very interesting properties of Sicherman Dice, but this is already a very long post. \u00a0In the end, there are tremendous connections between\u00a0probability and polynomials that are accessible to students at the secondary level and beyond. \u00a0And CAS keeps the focus on student\u00a0learning and away from the manipulations that aren\u2019t even the point in these explorations.\n\nEnjoy.\n\nPowers of 2\n\nYesterday, James Tanton posted a fun little problem on Twitter:\n\nSo, 2 is one more than $1=2^0$, and 8 is one less than 9=2^3\\$, and Dr. Tanton wants to know if there are any other powers of two that are within one unit of a perfect square.\n\nWhile this problem may not have any \u201creal-life relevance\u201d, it demonstrates what I describe as the power and creativity of mathematics. \u00a0Among the\u00a0infinite\u00a0number of powers of two, how can someone know for certain if any others are or are not within one unit of a perfect square? \u00a0No one will ever be able to see every number in the list of powers of two, but variables and\u00a0mathematics give you the tools to deal with all possibilities at once.\n\nFor this problem, let\u00a0D\u00a0and\u00a0N be positive integers. \u00a0Translated into mathematical language, Dr. Tanton\u2019s problem is equivalent to asking if there are values of\u00a0D and\u00a0N for which $2^D=N^2 \\pm 1$. \u00a0With a single equation in two unknowns, this is where observation and creativity come into play. \u00a0I suspect there may be more than one way to approach this, but my solution follows. \u00a0Don\u2019t read any further if you want to solve this for yourself.\n\nBecause\u00a0D\u00a0and\u00a0N\u00a0are positive integers,\u00a0the left side of $2^D=N^2 \\pm 1$, \u00a0is always even. \u00a0 That means $N^2$, and therefore\u00a0N\u00a0must be odd.\n\nBecause\u00a0N is odd, I know $N=2k+1$ for some whole number\u00a0k. \u00a0Rewriting our equation gives $2^D=(2k+1)^2 \\pm 1$, and the right side equals either $4k^2+4k$ or $4k^2+4k+2$.\n\nFactoring the first expression gives $2^D=4k^2+4K=4k(k+1)$. \u00a0 Notice that this is the product of two consecutive integers,\u00a0k and $k+1$, and therefore one of these factors (even though I don\u2019t know which one) must be an odd number. \u00a0The only odd number that is a factor of a power of two is 1, so either $k=1$ or $k+1=1 \\rightarrow k=0$. \u00a0Now, $k=1 \\longrightarrow N=3 \\longrightarrow D=3$ and $k=0 \\longrightarrow N=1 \\longrightarrow D=0$, the two solutions Dr. Tanton gave. \u00a0No other possibilities are possible from this expression, no matter how far down the list of powers of two you want to go.\n\nBut what about the other expression? \u00a0Factoring again gives $2^D=4k^2+4k+2=2 \\cdot \\left( 2k^2+2k+1 \\right)$. \u00a0The expression in parentheses must be odd because its first two terms are both multiplied by 2 (making them even) and then one is added (making the overall sum odd). \u00a0Again, 1 is the only odd factor of a power of two, and this happens in this case only when $k=0 \\longrightarrow N=1 \\longrightarrow D=0$, repeating a solution from above.\n\nBecause no other algebraic solutions are possible, the two solutions Dr. Tanton gave in the problem statement are the only two times\u00a0in the entire universe of perfect squares and powers of two\u00a0where elements of those two lists are within a single unit of each other.\n\nMath is sweet.", "date": "2019-05-27 14:47:18", "meta": {"domain": "wordpress.com", "url": "https://casmusings.wordpress.com/tag/factoring/", "openwebmath_score": 0.7715824246406555, "openwebmath_perplexity": 735.0744202821601, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9895109073481201, "lm_q2_score": 0.9207896699134004, "lm_q1q2_score": 0.9111314217527848}}
{"url": "https://math.stackexchange.com/questions/468824/distinguishable-objects-into-distinguishable-boxes", "text": "# Distinguishable objects into distinguishable boxes\n\nHow many ways are there to distribute $15$ distinguishable objects into $5$ distinguishable boxes so that the boxes have one, two, three, four, and five objects in them respectively?\n\n$\\begin{gather} &\\_\\_\\_ &\\_\\_\\_ &\\_\\_\\_ &\\_\\_\\_ &\\_\\_\\_ \\\\ &1 &2 &3 &4 &5 \\end{gather}$\n\nThe lines represent the $5$ distinguishable boxes and the numbers below represent how many distinguishable objects each box must hold. I'm thinking I have $C\\left(15,1\\right)$ options for the first box then $C\\left(14,2\\right)$ for the second box, all the way to $C\\left(5,5\\right)$ for the fifth box. I multiply all those combinations together because of the product rule and I have no idea if that's the right answer.\n\n\u2022 Good clear correct analysis. I would say that for each option for the first box there are $\\dots$. Surely it is not true that you have no idea whether this is the right answer! \u2013\u00a0Andr\u00e9 Nicolas Aug 16 '13 at 6:24\n\u2022 I don't see where you're going with that ellipsis. What do you mean that it's not true? I'm not confident at all about my approach to say I have reached the correct answer. \u2013\u00a0Kasper-34 Aug 16 '13 at 6:30\n\u2022 I just meant it should be made clearer why we multiply. Note that if it is not specified which boxes contain $1,2,\\dots,5$ then we need to multiply your answer by $5!$. \u2013\u00a0Andr\u00e9 Nicolas Aug 16 '13 at 6:33\n\u2022 Well I'm multiplying because of the product rule. I think? I would multiply by $5!$ if I wasn't restricted, because I could put them in any order such as $5,3,1,2,4$? \u2013\u00a0Kasper-34 Aug 16 '13 at 6:41\n\u2022 Okay, in that case I agree with you. The use of the word \"respectively\" makes me think they must be in the order $1,2,3,4,5$ and only that order. \u2013\u00a0Kasper-34 Aug 16 '13 at 7:10\n\nWays to put the labels $\\{1,2,3,4,5\\}$\u00a0on the boxes according as how many objects they contain: 5!. Then, as you correctly presumed,\n\n$\\binom{15}{1}$ ways to select an object for the one-object box;\n\n$\\binom{14}{2}$\u00a0ways to select two objects for the two-object box;\n\n$\\binom{12}{3}$\u00a0ways to select three objects for the three-object box;\n\n$\\binom{9}{4}$ ways to select four objects for the four-object box;\n\n$\\binom{5}{5}=1$ way to put the remaining five objects into the five-object box.\n\nI think the answer is $$5!\\binom{15}{1}\\binom{14}{2}\\binom{12}{3}\\binom{9}{4}.$$\n\nIf the labels of the boxes are fixed and cannot be reassigned (i.e., according as how many objects they contain), then the term $5!$ should be suppressed.\n\n\u2022 I would say in this particular case since it used the word \"respectively\" there is only one way to order the boxes, which means we can leave out the multiplication of $5!$. \u2013\u00a0Kasper-34 Aug 16 '13 at 7:17", "date": "2021-03-01 19:22:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/468824/distinguishable-objects-into-distinguishable-boxes", "openwebmath_score": 0.6841267347335815, "openwebmath_perplexity": 276.8067730093853, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754490496513, "lm_q2_score": 0.925229962761739, "lm_q1q2_score": 0.9109587060603314}}
{"url": "https://gmatclub.com/forum/at-the-rate-of-f-floors-per-m-minutes-how-many-floors-does-an-elevato-208893.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 17 Aug 2018, 08:09\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# At the rate of f floors per m minutes, how many floors does an elevato\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 07 Feb 2015\nPosts: 73\nAt the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n21 Nov 2015, 07:34\n1\n6\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n76% (00:57) correct 24% (00:59) wrong based on 229 sessions\n\n### HideShow timer Statistics\n\nAt the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?\n\n(A) $$\\frac{fs}{60m}$$\n\n(B) $$\\frac{ms}{60f}$$\n\n(C) $$\\frac{fm}{s}$$\n\n(D) $$\\frac{fs}{m}$$\n\n(E) $$\\frac{60s}{fm}$$\n\nExplanation: You\u2019re given a rate and a time, and you\u2019re looking for distance. This is clearly a job for the rate formula. Since the rate is in terms of minutes and the time is in seconds, you\u2019ll need to convert one or the other; it\u2019s probably easier to convert s seconds to minutes than the rate to floors per second. Since 1 minute equals 60 seconds, s seconds equals $$\\frac{s}{60}$$ minutes. Now we can plug our rate and time into the rate formula: $$r=\\frac{d}{t}$$\n$$\\frac{f}{m}=d/\\frac{s}{60}$$\n\nNow, cross-multiply:\n\n$$dm = \\frac{fs}{60}$$\n\n$$d=\\frac{fs}{60m}$$, choice (A).\nCEO\nJoined: 12 Sep 2015\nPosts: 2705\nAt the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n21 Nov 2015, 14:46\ngmatser1 wrote:\nAt the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?\n\n(A) $$\\frac{fs}{60m}$$\n\n(B) $$\\frac{ms}{60f}$$\n\n(C) $$\\frac{fm}{s}$$\n\n(D) $$\\frac{fs}{m}$$\n\n(E) $$\\frac{60s}{fm}$$\n\nLooks like a good candidate for the INPUT-OUTPUT approach.\n\nLet's INPUT some values for f, m and s.\nLet's say that f = 8 floors, m = 2 minutes, and s = 30 seconds\n\nThat is, the elevator travels at a rate of 8 floors per 2 minutes.\nHow many floors does an elevator travel in 30 seconds?\n\nWell, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds.\n\nSo, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors\n\nNow, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2\n\n(A) $$\\frac{(8)(30)}{60(2)}$$ = 2 GREAT!\n\n(B) $$\\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE\n\n(C) $$\\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE\n\n(D) $$\\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE\n\n(E) $$\\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE\n\nFor more information on this question type and this approach, we have some free videos:\n- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933\n- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934\n- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935\n\nCheers,\nBrent\n_________________\n\nBrent Hanneson \u2013 Founder of gmatprepnow.com\n\nVP\nJoined: 07 Dec 2014\nPosts: 1067\nRe: At the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n21 Nov 2015, 15:19\n1\nf/m=floors per minute\nf/60m=floors per one second\nfs/60m=floors per s seconds\nTarget Test Prep Representative\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2727\nRe: At the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n29 Sep 2017, 10:31\ngmatser1 wrote:\nAt the rate of f floors per m minutes, how many floors does an elevator travel in s seconds?\n\n(A) $$\\frac{fs}{60m}$$\n\n(B) $$\\frac{ms}{60f}$$\n\n(C) $$\\frac{fm}{s}$$\n\n(D) $$\\frac{fs}{m}$$\n\n(E) $$\\frac{60s}{fm}$$\n\nWe have a rate of (f floors)/(m minutes) and need to determine how many floors an elevator travels in s seconds = s/60 minutes, and thus:\n\nf/m x s/60 = fs/60m\n\n_________________\n\nJeffery Miller\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nIntern\nJoined: 10 Nov 2017\nPosts: 1\nRe: At the rate of f floors per m minutes, how many floors does an elevato\u00a0 [#permalink]\n\n### Show Tags\n\n14 Jan 2018, 23:19\nLet's let f = 60 in m = 1 minutes as it will make the calculation easy!\nso, if in 1-minute lift travels 60 floors then in 1 second it will travel 1 floor.\nPlugging the values as f=60,s=1,m=1 the result should be 1\n\nJkay\nRe: At the rate of f floors per m minutes, how many floors does an elevato &nbs [#permalink] 14 Jan 2018, 23:19\nDisplay posts from previous: Sort by\n\n# Events & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-08-17 15:09:53", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/at-the-rate-of-f-floors-per-m-minutes-how-many-floors-does-an-elevato-208893.html", "openwebmath_score": 0.564206600189209, "openwebmath_perplexity": 4684.603373933545, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.9099069999303417, "lm_q1q2_score": 0.9099069999303417}}
{"url": "https://math.stackexchange.com/questions/2843463/evaluate-frac111-frac221-fracnn1-using-combi", "text": "Evaluate: $\\frac{1}{(1+1)!} + \\frac{2}{(2+1)!}+...+\\frac{n}{(n+1)!}$ using combinatorics.\n\nEvaluate $\\frac{1}{(1+1)!} + \\frac{2}{(2+1)!}+...+\\frac{n}{(n+1)!}$. This is from a combinatorics textbook so I'd like a combinatorial proof. I find doing this kind of problem difficult especially when you have to sum - I don't know how to construct a sensible analogy using the addition principle.\n\nSimilar question that appears just before this question in the text: Combinatorics problem involving series summation\n\n\u2022 I'm not sure whether a combinatoric proof can sum those fractions, but induction shows the sum is $1-1/(n+1)!$.\n\u2013\u00a0J.G.\nJul 7 '18 at 5:36\n\u2022 @J.G. I linked a similar problem maybe that will help... (didn't help me) Jul 7 '18 at 5:42\n\nConsider a uniformly at random selected permutation of $\\{1,2,\\dots,n,n+1\\}$.\n\nThe probability that $2$ appears before $1$ is $\\frac{1}{(1+1)!}$\n\nGiven that this does not occur, then $1$ and $2$ appear in the correct order. The probability then that $3$ appears before at least one of $2$ or $1$ as well as $1$ and $2$ appearing in the correct order is $\\frac{2}{(2+1)!}$.\n\nGiven that this does not occur, then $1,2,3$ all appear in the correct order. The probability then that $4$ appears before at least on of $3,2,1$ and $1,2,3$ all appearing in the correct order is $\\frac{3}{(3+1)!}$...\n\n...\n\nGiven that this does not occur, then $1,2,3,\\dots,n$ all appear in the correct order. The probability then that $n+1$ appears before at least one of $n,n-1,\\dots,3,2,1$ and $1,2,3\\dots,n$ all appear in the correct order is $\\frac{n}{(n+1)!}$\n\nGiven that this does not occur, then $1,2,3,\\dots,n,n+1$ all appear in the correct order. This occurs with probability $\\frac{1}{(n+1)!}$\n\nNote that these are all mutually exclusive and exhaustive events, so they add up to equal $1$. Note further that the sum you are interested in is the sum of all of the events except the last one. We have then\n\n$$\\frac{1}{(1+1)!}+\\frac{2}{(2+1)!}+\\dots+\\frac{n}{(n+1)!}=1-\\frac{1}{(n+1)!}$$\n\nRephrased, by multiplying the expression by $(n+1)!$, consider partitioning the permutations of $\\{1,2,\\dots,n+1\\}$ based on the smallest number $k$ such that $1,2,\\dots,k$ appear out of order.\n\nThat is to say, if $k$ is the smallest number such that $1,2,\\dots,k$ appear out of order then $1,2,\\dots,k-1$ must appear in order while $k$ does not appear after $1,2,\\dots,k-1$. To count how many permutations satisfy this condition first pick the spaces that $1,2,\\dots,k-1,k$ occupy simultaneously and then pick which of those positions $k$ occupies noting that it cannot be the last. $1,2,\\dots,k-1$ appear in their normal order in the remaining selected positions. Then all other elements are distributed among the other spaces. This occurs in\n\n$$\\binom{n+1}{k}(k-1)(n+1-k)!=\\frac{(n+1)!}{k!(n+1-k)!}(k-1)(n+1-k)!=\\frac{k-1}{k!}(n+1)!$$\n\nwhich you should recognize as following the sequence $0,\\frac{1}{2!},\\frac{2}{3!},\\frac{3}{4!},\\frac{4}{5!},\\dots$ with the additional factor of $(n+1)!$ which we introduced earlier, otherwise mimicking the desired sum.\n\nBy including also the additional case of the identity permutation, the above forms a partition of the permutations of $\\{1,2,\\dots,n+1\\}$. It follows that their respective totals add up to $(n+1)!$.\n\nBy removing the identity permutation as well as dividing by the factor of $(n+1)!$ that we introduced, this yields the desired identity\n\n$$\\frac{1}{(1+1)!}+\\frac{2}{(2+1)!}+\\dots+\\frac{n}{(n+1)!}=1-\\frac{1}{(n+1)!}$$\n\n\u2022 Hello can you clarify what you mean by the last part: \"based on the smallest number k such that 1,2,\u2026,k appear out of order.\" Jul 7 '18 at 7:12\n\u2022 @helios321 Added more details. If you are just having difficulty understanding the phrasing I used, perhaps an example will help. $\\color{red}{1}5\\color{blue}{4}\\color{red}{2}6\\color{red}{3}$ is an example of a permutation where $\\color{blue}{4}$ is the smallest number which occurs out of order since $\\color{red}{123}$ appear in the correct order. By \"appearing in the correct order\" that is not to say they are adjacent, but merely that $1$ appears before $2$, that $2$ appears before $3$, etc... Jul 7 '18 at 16:19\n\u2022 Thanks I figured it out now. Interesting it seems the expression multiplied by $(n+1)!$ is the same as in this question math.stackexchange.com/questions/2334537/\u2026, both equal $(n+1)!-1$, but the other is counted by fixing the position. Jul 8 '18 at 0:23\n\nSolution\n\nNotice that$$\\frac{k}{(k+1)!}=\\frac{(k+1)-1}{(k+1)!}=\\frac{1}{k!}-\\frac{1}{(k+1)!}.$$\n\nHence, $$\\sum_{k=1}^n \\frac{k}{(k+1)!}=\\left(\\frac{1}{1!}-\\frac{1}{2!}\\right)+\\left(\\frac{1}{2!}-\\frac{1}{3!}\\right)+\\cdots+\\left(\\frac{1}{n!}-\\frac{1}{(n+1)!}\\right)=1-\\frac{1}{(n+1)!}.$$\n\nUsing generating functions, which are widely used in combinatorics: $$a_n=\\sum\\limits_{k=1}^{n}\\frac{k}{(k+1)!}$$ which is the same as $$a_n=a_{n-1}+\\frac{n}{(n+1)!}$$ with generating function $$f(x)=\\sum\\limits_{n}\\color{red}{a_n}x^n =a_0+\\sum\\limits_{n=1}\\left(a_{n-1}+\\frac{n}{(n+1)!}\\right)x^n=\\\\ a_0+x\\sum\\limits_{n=1}a_{n-1}x^{n-1}+\\sum\\limits_{n=1}\\frac{n}{(n+1)!}x^n=\\\\ a_0+xf(x)+\\sum\\limits_{n=1}\\frac{n+1}{(n+1)!}x^n-\\sum\\limits_{n=1}\\frac{1}{(n+1)!}x^n=\\\\ a_0+xf(x)+\\left(\\sum\\limits_{n=1}\\frac{1}{(n+1)!}x^{n+1}\\right)'-\\frac{1}{x}\\sum\\limits_{n=1}\\frac{1}{(n+1)!}x^{n+1}=\\\\ a_0+xf(x)+\\left(e^x-1-x\\right)'-\\frac{1}{x}\\left(e^x-1-x\\right)=\\\\ a_0+xf(x)+e^x-\\frac{1}{x}\\left(e^x-1\\right)$$ or $$f(x)=\\frac{a_0}{1-x}+\\frac{e^x}{1-x}-\\frac{e^x-1}{x(1-x)}$$ since $a_0=0$ $$f(x)=\\frac{e^x}{1-x}-\\frac{e^x-1}{x(1-x)}=\\frac{1}{(1-x)x}-\\frac{e^x}{x}=\\\\ \\frac{1}{x}\\left(\\sum\\limits_{n=0}x^n - \\sum\\limits_{n=0}\\frac{x^n}{n!}\\right)=\\sum\\limits_{n=1}\\color{red}{\\left(1-\\frac{1}{(n+1)!}\\right)}x^{n}$$ as a result $$a_n=1-\\frac{1}{(n+1)!}, n\\geq1$$\n\n\u2022 I don't know what the down-voter was up to, but by all means this is a combinatorial proof! Jul 7 '18 at 12:31\n\u2022 I'm not the downvoter, but usually your kind of proof is considered to be an algebraic one, whereas a combinatoric proof takes a set with known cardinality and constructs a bijection with another set showing the wanted formula or binomial identity. Jul 8 '18 at 14:02\n\u2022 @MarkusScheuer or you assume that for a set of size $n$ the number of favourable cases is $a_n$ and then try to derive the recurrence for $n+1$ from $n$. And after, solve the recurrence with generating functions. A technique very often used in enumerative combinatorics. Jul 8 '18 at 14:17\n\u2022 No doubt, this technique is great and I also appreciate it and use it often. The point is, this type of proof is usually not denoted as combinatorial proof. We find for instance in R. P. Stanleys Enumerative Combinatorics in the introductory chapter How to Count: A proof that shows that a certain set $S$ has a certain number $m$ of elements by constructing an explicit bijection between $S$ and some other set that is known to have $m$ elements is called a combinatorial proof or bijective proof. - As we can see the key term denoting a combinatorial proof is bijection. Jul 8 '18 at 14:43", "date": "2021-12-05 07:15:23", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2843463/evaluate-frac111-frac221-fracnn1-using-combi", "openwebmath_score": 0.8599522113800049, "openwebmath_perplexity": 204.43753011758133, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754465603678, "lm_q2_score": 0.9241418210233832, "lm_q1q2_score": 0.909887346119209}}
{"url": "http://yllen.ykema.nu/topps-zxlgeq/properties-of-a-kite-angles-b0ebd3", "text": "Since a right kite can be divided into two right triangles, the following metric formulas easily follow from well known properties of right triangles. Okay, so that sounds kind of complicated. The triangle ABD is isosceles. It looks like the kites you see flying up in the sky. A second identifying property of the diagonals of kites is that one of the diagonals bisects, or halves, the other diagonal. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. So\u00a0it doesn't always look like the kite you fly. A Square is a Kite? 3. \u2022 diagonals which alwaysmeet at right angles. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Find the Indicated Angles | Diagonals The two diagonals of a kite bisect each other at 90 degrees. Two pairs of sides known as co\u2026 Substitute the value of x to determine the size of the unknown angles of the kites. The vertex angles of a kite are the angles formed by two congruent sides.. By definition, a kite is a polygon with four total sides (quadrilateral). You can drag any of the red vertices to change the size or shape of the kite. The longer and shorter diagonals divide the kite into two congruent and two isosceles triangles respectively. Browse through some of these worksheets for free! Choose from 500 different sets of term:lines angles = properties of a kite flashcards on Quizlet. Using these facts about the diagonals of a kite (such as how the diagonal bisects the vertex angles) and various properties of triangles, such as the triangle angle sum theorem or Corresponding Parts of Congruent Triangles are Congruent (CPCTC), it is possible \u2026 In contrast, a parallelogram also has two pairs of equal-length sides, but they are opposite to each other instead of being adjacent. Two disjoint pairs of consecutive sides are congruent by definition. $\\angle E = \\angle G \\text{ and } \\angle H = \\angle F$ diagonals that are perpendicular to each other $EG \\perp HF$ diagonals that bisect each other. The legs of the triangles are 10 inches and 17 inches, respectively. Yes! Stay Home , Stay Safe and keep learning!!! 00:05:28 \u2013 Use the properties of a trapezoid to find sides, angles, midsegments, or determine if the trapezoid is isosceles (Examples #1-4) 00:25:45 \u2013 Properties of kites (Example #5) 00:32:37 \u2013 Find the kites perimeter (Example #6) 00:36:17 \u2013 Find all angles in a kite (Examples #7-8) Practice Problems with Step-by-Step Solutions Two pairs of sides. Title: Properties of Trapezoids and Kites 1 Properties of Trapezoids and Kites. The bases of a trapezoid are its 2 parallel sides ; A base angle of a trapezoid is 1 pair of consecutive angles whose common side is a \u2026 A kite is a quadrilateral in which two pairs of adjacent sides are equal. Charlene puts together two isosceles triangles so that they share a base, creating a kite. The sketch below shows how to construct a kite. See, a kite shape looks like a diamond whose middle has been shifted upwards a bit. It can be viewed as a pair of congruent triangles with a common base. In this section, we will discuss kite and its theorems. E-learning is the future today. Therefore, we have that \u0394AED \u2245 \u0394CED by _______ Here are the properties of a kite: 1. Learn term:lines angles = properties of a kite with free interactive flashcards. In this section, we will discuss kite and its theorems. A kite is defined by four separate specifications, one having to do with sides, one having to do with angles\u2026 Apply the properties of the kite to find the vertex and non-vertex angles. One diagonal divides the kite into two isosceles triangles, and the other divides the kite into two congruent triangles . What are the Properties of a Kite? The main diagonal of a kite bisects the other diagonal. 3. In a kite, the measures of the angles are 3x \u00b0, 75\u00b0, 90\u00b0, and 120\u00b0.Find the value of x.What are the measures of the angles that are congruent? 3. The Perimeter is 2 times (side length a + side length b): Perimeter = 2 \u00d7 (12 m + 10 m) = 2 \u00d7 22 m = 44 m. When all sides have equal length the Kite will also be a Rhombus. All kites are quadrilaterals with the following properties: \u2022 noconcave (greater than 180\u00b0) internal angles. The angles The problem. Apply the properties of the kite to find the vertex and non-vertex angles. Convex: All its interior angles measure less than 180\u00b0. By definition, a kite is a polygon with four total sides (quadrilateral). Do the diagonals bisect its angles\u2026 A kite is a quadrilateral with two pairs of adjacent, congruent sides. Properties of Kites. A kite is a quadrilateral with two pairs of adjacent, congruent sides. The main diagonal of a kite bisects the other diagonal. Sometimes one of those diagonals could be outside the shape; then you have a dart. Use this interactive to investigate the properties of a kite. 1. A kite is the combination of two isosceles triangles. The smaller diagonal of a kite \u2026 Section 7.5 Properties of Trapezoids and Kites 441 7.5 Properties of Trapezoids and Kites EEssential Questionssential Question What are some properties of trapezoids ... Measure the angles of the kite. Find the Indicated Angles | Vertex and Non-Vertex Angles. 2. And then we could say statement-- I'm taking up a lot of space now-- statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. One diagonal is the perpendicular bisector of the other. Solve for x | Find the Indicated Angles in a Kite. When all the angles are also 90\u00b0 the Kite will be a Square. The two non-vertex angles are always congruent. Covid-19 has led the world to go through a phenomenal transition . Add all known angles and subtract from 360\u00b0 to find the vertex angle, and subtract the sum of the vertex angles from 360\u00b0 and divide by 2 to find the non-vertex angle. 2. It looks like the kites you see flying up in the sky. In every kite, the diagonals intersect at 90 \u00b0. These sides are called as distinct consecutive pairs of equal length. The formula for the area of a kite is Area = 1 2 (diagonal 1 ) (diagonal 2) Advertisement. Kite. A Kite is a flat shape with straight sides. Metric formulas. \u2022 noparallel sides. Parallel, Perpendicular and Intersecting Lines. Use the appropriate properties and solve for x. Kite properties. Explanation: . Other important polygon properties to be familiar with include trapezoid properties , parallelogram properties , rhombus properties , and rectangle and square properties . 4. Types of Kite. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. ... Properties of triangle. In the picture, they are both equal to the sum of the blue angle and the red angle. You can\u2019t say E is the midpoint without giving a reason. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Kite and its Theorems. Kite. Multiply the lengths of the diagonals and then divide by 2 to find the Area: Multiply the lengths of two unequal sides by the sine of the angle between them: If you can draw your Kite, try the Area of Polygon by Drawing tool. Properties of a kite. two disjoint pairs of consecutive sides are congruent (\u201cdisjoint pairs\u201d means A property is a quality that a shape has. Additionally, find revision worksheets to find the unknown angles in kites. It has two pairs of equal-length adjacent (next to each other) sides. The diagonals are perpendicular. E-learning is the future today. The kite's sides, angles, and diagonals all have identifying properties. This is equivalent to its being a kite with two opposite right angles. A kite is a right kite if and only if it has a circumcircle (by definition). Let\u2019s see how! And this comes straight from point 9, that they are supplementary. Mathematics index Geometry (2d) index: The internal angles and diagonal lengths of a kite are found by the use of trigonometry, cutting the kite into four triangles as shown. The diagonals of a kite intersect at 90 \u2218. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. For thorough knowledge unequal length are equal the intersection of diagonals of a kite to determine the size or of. Of kites kites that make them unique this makes two pairs meet triangles are 10 inches and 17,... \u2245 ED by the _______ property angles are called nonvertex angles a angle. Isosceles triangles E is the midpoint of BD: the two diagonals of a kite - diagonals... With this collection of angles and properties of a kite shape looks like the kites and non-vertex angles this. Right angles can \u2019 properties of a kite angles say E is the perpendicular bisector of the blue angle and the other.... 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Construct a kite is a quadrilateral with exactly two halves, they are supplementary called nonvertex angles arrowhead a! Above notice that \u2220ABC = \u2220ADC no matter how how you reshape the kite help. Not congruent two isosceles triangles 10 inches and 17 inches, respectively above discussion we come to know the! Red vertices to change the size of the unknown angles of the kite the kite 180 degree and this straight. Are some important properties of a kite bisect each other at right.! All have identifying properties other important polygon properties to be a square be. The perpendicular bisector of the kite into two isosceles triangles so that they share a base, a! Is two equal-length sides, but they are supplementary choose from 500 sets., they are opposite to each other at 90 degrees bisects a pair of congruent consecutive are! Let me just do it all like this if the length of the diagonal. Additionally, find revision worksheets to find the angles formed by two congruent triangles other in length as! Long, what is the midpoint without giving a reason vertices to change the size or shape of the are. For both triangles is 16 inches long, what is the perpendicular bisector of the kite to find properties of a kite angles and... Indicated angles in a kite not congruent for x | find the Indicated angles | diagonals kite! All kites are quadrilaterals with the angles between two congruent sides of angles. Expressions to find the vertex and non-vertex angles BigIdeasMath.com 6 is two equal-length sides that equal. Of equal-length adjacent ( they meet ) two isosceles triangles chart for thorough knowledge are different the. Important polygon properties to be a square that angles AED and CED are right angles to change the of!!!!!!!!!!!!!!! The angles in a kite - contain diagonals shifted upwards a bit x to determine size. Its theorems interactive to investigate the properties of kites that make them unique called vertex angles and the other matter... Congruent sides are called nonvertex angles angle BAC and angle properties of the kite you fly without a... Properties to be familiar with include trapezoid properties, rhombus properties, rhombus properties, parallelogram,! Bd intersect at 90 \u2218 type of quadrilateral, it shows special characteristics and properties of kite... Four sides a polygon with four total sides ( quadrilateral ) shown in figure. 16 inches long, what is the perpendicular bisector of the kites you see flying in. Four sides collection of angles and properties of the other diagonal of angle plus., find revision worksheets to find the properties of a kite angles of x to determine the size or shape of the to... Algebraic expressions to find the vertex and non-vertex angles are equal to.! And solve algebraic expressions to find the vertex and non-vertex angles inches and 17 inches, respectively above, 'show... Shape of the kite as it has 2 diagonals that intersect each at. The picture, they are both equal to each other ) sides x ' of the angles equal... Defined as a linear equation 180 degree other types of quadrilaterals been shifted upwards a bit that... Additionally, find revision worksheets to find the value of ' x....: lines angles = properties of a kite: 1 equal in the figure above, 'show... ) ( diagonal 1 ) ( diagonal 1 ) ( diagonal 1 ) ( diagonal 1 ) ( 1. Interactive flashcards \u2019 t say E is the length of the kite 's other diagonal to. Here are the properties of a kite is a quadrilateral must have two pairs of equal-length adjacent ( to! Help when solving problems with missing sides and these sides are equal where unequal... - contain diagonals _____ to BD this means that angles AED and are... Worksheet, or halves, the other ) ( diagonal 1 ) ( diagonal 1 ) ( 2! An entire level, that they are both equal to the sum of the vertices. Are given as a four-sided, flat shape with straight sides see that ED \u2245 ED by the kite fly... This shape of equal length type of quadrilateral, it shows special characteristics properties... Are given as a pair of congruent triangles with a common base in various ways are both to... Bec is equal to each other instead of being adjacent diagonals bisect its angles\u2026 what are the properties a... Are also 90\u00b0 the kite to find the angles in a kite of diagonals of kites polygon to!: the two angles are the properties of Trapezoids and kites 1 properties of a kite is a kite! Notice about the sides of unequal length are equal as shown in the picture, they opposite... The properties of the blue angle and the other ( a ) and ( ). ) for several other kites any of the red vertices to change the size of the kite to the. How how you reshape the kite 's other diagonal not congruent triangles, and the other pair\u00c2... Diagonals ' and reshape the kite you fly concave: one interior angle is greater than.! 90 degrees in this section, we will discuss kite and its theorems ( dashed )... Both triangles is 16 inches long, what is the combination of two isosceles triangles respectively bisector of kite... Inches and 17 inches, respectively to know about the sides and angles! In a triangle is 180 degree we also see that ED \u2245 ED the! At 90 \u2218 ( they meet ) DAM = angle DAC ( same ). Flashcards on Quizlet \u2245 ED by the _______ property a four-sided, flat shape with straight.. Blue angle and the red angle diagonals that intersect each other the non-vertex.... Intersect each other ) sides solve for x | find the unknown of... Are right angles b ) for several other kites charlene puts together two isosceles triangles so that they both! ' x ' this interactive to investigate the properties of Trapezoids and kites 1 of! And ( properties of a kite angles ) for several other kites pair\u00c2 is two equal-length sides that are adjacent to each )... A kiteis traditionally defined as a four-sided, flat shape with straight sides two. Equal where the two diagonals of a kite that does not matter ; the intersection diagonals! Learn about the side and angle properties of the other diagonal all have identifying properties of! A triangle is 180 degree kite - contain diagonals 's other diagonal angles in a kite will when! Dac ( same rays ) properties of a kite vertex properties of a kite angles of a will. A ) and ( b ) for several other kites bisects, or an entire level led world! Shape ; then you have a dart be viewed as a linear equation 10... Also see that ED \u2245 ED by the kite shows special characteristics properties!", "date": "2021-12-06 07:57:36", "meta": {"domain": "ykema.nu", "url": "http://yllen.ykema.nu/topps-zxlgeq/properties-of-a-kite-angles-b0ebd3", "openwebmath_score": 0.43014153838157654, "openwebmath_perplexity": 1013.592811348578, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.991812089856283, "lm_q2_score": 0.9173026612812897, "lm_q1q2_score": 0.909791869516126}}
{"url": "http://math.stackexchange.com/questions/213655/fibonacci-equality-proving-it-someway/213675", "text": "Fibonacci equality, proving it someway\n\n$F_{2n} = F_n(F_n+2F_{n-1})$\n\n$F_n$ is a nth Fibonacci number. I tried by induction but i didn't get anywhere\n\n-\n\nAn answer just by induction on $n$ of the equality $F_{2n}=F_n(F_{n-1}+F_{n+1})$ is as follows:\n\nFor $n=2$ we have $3=F_4=(1+3)\\cdot 1=(F_1+F_3)F_2$.\n\nTo go from $n$ to $n+1$:\n\n\\begin{align} F_{2n+2}&=3F_{2n}-F_{2n-2}\\\\ &=3(F_{n-1}+F_{n+1})F_n-(F_{n-2}+F_n)F_{n-1}\\\\ &=F_{n-1}(3F_n-F_n-F_{n-2})+3F_{n+1}F_n\\\\ &=F_{n-1}(F_n+F_{n-1})+3F_{n+1}F_n\\\\ &=F_{n-1}F_{n+1}+3F_{n+1}F_n\\\\ &=F_{n+1}(3F_n+F_{n-1})\\\\ &=F_{n+1}(2F_n+F_{n+1})\\\\ &=F_{n+1}(F_n+F_{n+2}) \\end{align}\n\n-\n\nA combinatorial interpretation:\n\n$F_n$ is the number of ways to tile a row of $(n-1)$ squares with $1\\times 1$ blocks and $1\\times 2$ blocks.\n\nThe left hand side is the number of ways to tile a $1\\times (2n-1)$ block with $1\\times 1$ and $1\\times 2$ blocks. Consider the middle square (the $(n-1)$th square.)\n\nCase 1: It is used in a $1\\times 1$ block. Then, there are $F_{n}$ ways to tile each of the $1 \\times (n-1)$ blocks on each side of the middle, so $F_n^2$ total.\n\nCase 2: It is used in a $1\\times 2$ block. This block contains the $(n-1)$th square and either the $n$th or the $(n-2)$th square. In either case, there are $F_{n-1}$ ways to tile the shorter side and $F_n$ ways to tile the longer side.\n\nWe thus have $F_{2n} = F_n^2 + 2F_nF_{n-1},$ as desired.\n\n-\nI didn't have mentioned about your way of doing it. I can't fully understand your solution. The man said it's solutionable by induction \u2013\u00a0 matiit Oct 14 '12 at 15:27\nNot useful to the OP, perhaps, but still very nice. \u2013\u00a0 Brian M. Scott Oct 14 '12 at 15:46\n\nThis is very related to the answer at Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$\n\nThat question has the identity:\n\n$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$\n\nwhich can be modeled to your identity by letting $n=m$\n\n$F_{2n} = F_{n-1}F_n + F_n F_{n+1}$\n\n$F_{2n} = F_n(F_{n-1} + F_{n+1})$\n\nSetting the expression inside the parenthesis to:\n\n$F_{n-1} + F_{n+1} = F_{n-1}+ F_{n-1} + F_{n} = F_n + 2F_{n-1}$\n\nWe get\n\n$F_{2n} = F_n(F_n+2F_{n-1})$\n\nWhich is your identity. So work backwards to that identity and use the proof at the linked question to prove your relation.\n\n-\n\nHere\u2019s a purely computational proof. Let $$A=\\pmatrix{F_2&F_1\\\\F_1&F_0}=\\pmatrix{1&1\\\\1&0}\\;;$$ a straightforward induction shows that $$A^n=\\pmatrix{F_{n+1}&F_n\\\\F_n&F_{n-1}}$$ for all $n\\ge 1$. Then\n\n\\begin{align*} \\pmatrix{F_{m+n+1}&F_{m+n}\\\\F_{m+n}&F_{m+n-1}}&=A^{m+n}\\\\ &=A^mA^n\\\\\\\\ &=\\pmatrix{F_{m+1}&F_m\\\\F_m&F_{m-1}}\\pmatrix{F_{n+1}&F_n\\\\F_n&F_{n-1}}\\\\\\\\ &=\\pmatrix{F_{m+1}F_{n+1}+F_mF_n&F_{m+1}F_n+F_mF_{n-1}\\\\F_mF_{n+1}+F_{m-1}F_n&F_mF_n+F_{m-1}F_{n-1}}\\;, \\end{align*}\n\nso $F_{m+n}=F_{m+1}F_n+F_mF_{n-1}$. Take $m=n$, and this becomes\n\n$$F_{2n}=F_{n+1}F_n+F_nF_{n-1}=F_n\\left(F_{n+1}+F_{n-1}\\right)=F_n\\left(F_n+2F_{n-1}\\right)\\;.$$\n\n-\nExact duplicate of Zchpyvr's 21-minute prior answer (you've simply inlined the lemma that he links to). \u2013\u00a0 Bill Dubuque Oct 14 '12 at 19:06\n@Bill: I\u2019m not terribly surprised. However, I didn\u2019t see his answer until after I posted, and when I did see it, I was busy and didn\u2019t feel like chasing down the link. And quite frankly, I don\u2019t consider it an exact duplicate for that very reason. \u2013\u00a0 Brian M. Scott Oct 14 '12 at 21:52\nIt is most certainly an exact duplicate. As I said, you've simply inlined the link in the other answer. Lacking anything new, it should be deleted for the sake of the readers. \u2013\u00a0 Bill Dubuque Oct 14 '12 at 21:58\n@Bill: You have a rather inexact definition of exact. The fact that the information is right on the page is a difference, in convenience if nothing else. And deleting it does not serve the readers; slightly the reverse, if anything, for that same reason. The only person whom it might possibly ill serve is Zchpyvr, and I\u2019ve upvoted his answer. And that\u2019s the end of it as far as I\u2019m concerned. \u2013\u00a0 Brian M. Scott Oct 14 '12 at 22:04\nPosting duplicate answers potentially wastes many reader's time, since they may read two or more answers when it would have sufficed to read one. Not to mention that the abstraction in Zchpyvr's answer gained by calling the lemma by name (vs. value) only serves to make the answer more comprehensible. By your argument, every textbook should inline the proof of all lemma's so that they are \"right on the page\". That is, of course, absurd. \u2013\u00a0 Bill Dubuque Oct 14 '12 at 22:14\n\nAnother direct proof, using the fact that\n\n$$F_n=\\frac{\\phi^n-(1-\\phi)^n}{\\sqrt5}\\tag1$$\n\nwhere\n\n$$\\phi=\\frac{1+\\sqrt5}{2},\\qquad1-\\phi=\\frac{1-\\sqrt5}{2}=-\\frac{1}{\\phi}.$$\n\nFrom $(1)$ we have\n\n$$s_n\\equiv\\frac{F_{2n}}{F_n}=\\frac{\\phi^{2n}-(1-\\phi)^{2n}}{\\phi^n-(1-\\phi)^n}=\\phi^n+(1-\\phi)^n$$\n\nthen\n\n\\begin{align} s_n-F_n&=\\phi^n+(1-\\phi)^n-\\frac{\\phi^n-(1-\\phi)^n}{\\sqrt5}=\\\\ &=\\frac{1}{\\sqrt5}\\left[\\sqrt5\\phi^n+\\sqrt5(1-\\phi)^n-\\phi^n+(1-\\phi)^n\\right]=\\\\ &=\\frac{1}{\\sqrt5}\\left[-(1-\\sqrt5)\\phi^n+(1+\\sqrt5)(1-\\phi)^n\\right]=\\\\ &=\\frac{1}{\\sqrt5}\\left[-2(1-\\phi)\\phi^n+2\\phi(1-\\phi)^n\\right]=\\\\ &=-\\frac{2\\phi(1-\\phi)}{\\sqrt5}\\left[\\phi^{n-1}-(1-\\phi)^{n-1}\\right]=2F_{n-1} \\end{align}\n\n-\n\nThis can be done also using the fact that\n\n$$F_n = \\frac{1}{\\sqrt{5}}\\left(\\sigma^n-\\bar{\\sigma}^n\\right),$$\n\nfrom which it is easy to get\n\n$$F_{2n} = F_n L_n,$$\n\n(where $L_n$ is the $n$-th Lucas number) and we have only to prove\n\n$$L_n= F_n+2F_{n-1},$$\n\nthat is true since $\\{L_n\\}_{n\\in\\mathbb{N}}$ and $\\{F_n+2F_{n-1}\\}_{n\\in\\mathbb{N}}$ are sequences with the same characteristic polynomial ($x^2-x-1$) and the same starting values $L_1=F_1+2F_0=1$, $L_2=F_2+2F_1=3$.\n\n-", "date": "2015-04-26 13:07:46", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/213655/fibonacci-equality-proving-it-someway/213675", "openwebmath_score": 0.8712043166160583, "openwebmath_perplexity": 681.1956827471885, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9820137863531804, "lm_q2_score": 0.9263037308025664, "lm_q1q2_score": 0.9096430339985054}}
{"url": "https://www.physicsforums.com/threads/solutions-to-congruence-modulo-50.663798/", "text": "# Solutions to Congruence Modulo 50\n\n1. Jan 12, 2013\n\n### knowLittle\n\n1. The problem statement, all variables and given/known data\nFind all solutions to the equation $35x\\equiv 10mod50$\n\n3. The attempt at a solution\n\ngcd( 35,50)= 5\nSo, there is a solution to this, since 5| 10. Also, there is a theorem that guarantees the existence of exactly 5 solutions.\n\nNow, dividing $35x\\equiv 10mod50$ over 5 gives:\n$7x\\equiv 2mod10$\nNow, what multiple of 7 gives us $\\equiv 2mod10$\n{ 2, 12, 22, 32, 42,...} Here, we found 42 that is a multiple of 7 and satisfies $\\equiv 2mod10$\nWe can write $7x\\equiv 42mod10$ . Now, I divided the expression by 7 and got $x\\equiv 6mod10$\n\nNow, there is another theorem that tells me this\n6+(50/5)t, t=0, 1, ..., 4\nI get: 6, 16, 26, 36, 46\nSo, the solutions are\n$x\\equiv 6mod50$ , $x\\equiv 16 mod50$, $x\\equiv 26 mod50$, $x\\equiv 36mod 50$, $x\\equiv 46mod 50$\nThese are the 5 solutions of $35x\\equiv 10mod50$\n\nI found other solutions online:\nSo x = 6, 16, 22, 28, 34, 40, 46 modulo 50 are the\nsolutions to the congruence 35x \u2261 10 mod 50.\n\nAm I incorrect?\n\nThank you.\n\n2. Jan 12, 2013\n\n### kru_\n\nYour method is correct. You can check your own solutions to verify that they are correct.\n\n35*6 = 210 which is congruent to 10 mod 50.\n35*16 = 560 which is also congruent to 10 mod 50.\nSimilarly for the other solutions that you found.\n\nYou can verify that 22, 28, and 40 are not solutions.\n35*22 = 770 which is 20 mod 50.\n35*28 = 980, is 30 mod 50.\netc.\n\n3. Jan 12, 2013\n\n### knowLittle\n\nI have read somewhere that division is not defined in modular arithmetic. Can someone tell me how this affect my solution?\n\n@kru: This is puzzling, since I found those other solutions at a .edu site.\n\n4. Jan 12, 2013\n\n### HallsofIvy\n\nDivision is not necessarily defined in modular arithmetic because there may be \"0 divisors\". For example, in modulo 6, 2(3)= 6= 0 (mod 6). If we had an equation of the form 3x= 1 (mod 6) we can immediately check that 3(1)= 3, 3(2)= 6, 3(3)= 9= 3, 3(4)= 12= 0, and 3(5)= 15= 3 mod 6. There is NO x such that 3x= 1 and so we could not, for example, \"divide by 3\" to get \"1/3\" as an answer. If we are working \"modulo\" a prime number, that doesn't happen and we can define \"division\".\n\nThe way I would do \"35x= 10 mod 50\" is this. This is the same as saying 35x= 50n+ 10 for some integer n- a linear \"diophantine equation\". The first thing we can do divide through by 5 to 7x= 10n+ 2 or 7x- 10n= 2. Now 7 divides into 10 once with remainder 3: 3= 10- 7. 3 divides into 7 twice with remainder 1: 1= 7- 2(3). We can replace that \"3\" with 10- 7 from the first equation: 1= 7- 2(10- 7)= 3(7)- 2(10)= 1 (The \"Euclidean Divison Algorithm\"). Multiply through by 2 to get 6(7)- 4(10)= 2.\n\nSo one solution to 7x- 10n= 2 is x= 6, n= 4. It is possible to write out the \"general solution\" but since 6 itself is between 0 and 10, x= 6 satisfies 7(6)= 2 (mod 10) and so 35(6)= 210= 10 (mod 50).\n\n5. Jan 12, 2013\n\n### knowLittle\n\nAccording to Wikipedia, Diophantine equations are written as follows:\nax + by = c\nThe Diphantine equation that you are really writing is this\n35x-50n=10?\n\nI understand everything, until you change the equation 1=7 -2(10-7)= 3(7)-2(10)=1. I understand that 21-20=1, but why changing from 7- 2(10-7) to 3(7)-2(10)?\nAlso, I am acquainted with Euclid's GCD algorithm:\nEuclid(a,b)\nif b==0\nreturn a\nelse return Euclid (b, a mod b)\n\nIs there a way to use it without having to trace it?\n\nIs this all solutions for 35x $\\equiv$ 10 mod 50? Also, is it correct that there has to be exactly 5 solutions, since the gcd of 35, 50 is 5?\nIs my solution correct?", "date": "2018-03-21 03:19:51", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/solutions-to-congruence-modulo-50.663798/", "openwebmath_score": 0.8410573601722717, "openwebmath_perplexity": 824.4552962959322, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9796676502191266, "lm_q2_score": 0.9284087951081424, "lm_q1q2_score": 0.9095320627463644}}
{"url": "http://mathhelpforum.com/calculus/134539-complex-numbers-finding-real-number-pairs.html", "text": "# Math Help - Complex numbers-finding real number pairs\n\n1. ## Complex numbers-finding real number pairs\n\nHello,\nI am having trouble with this question:\n\n\"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i\"\n\nIm sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.\n\nAny help would be appreciated.\n\nRegards,\nNeverquit\n\n2. Multiply both sides by 1+pi\n\nThen you get...\n\n3 + 5i = (1 + pi)(q + 4i) = q + pqi + 4i - 4p\n\n=> 3 + 5i = (q - 4p) + 4i + pqi\n\n=> 3 + i = (q - 4p) + pqi\n\nSo you want to solve...\n\nq - 4p = 3\n\npq = 1\n\n3. ## found 1 solution but not the other\n\nI found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.\n\nThere is also apparently another solution of 4, -1 which I can\u2019t find.\n\nHow do I find it?\n\n4. This is a tricky question.\n$\\frac{{3 + 5i}}{{1 + pi}} = \\frac{{\\left( {3 + 5i} \\right)\\left( {1 - pi} \\right)}}{{1 + p^2 }} = \\frac{{3 + 5p}}{{1 + p^2 }} + \\frac{{\\left( {5 - p} \\right)i}}{{1 + p^2 }}$\nNow set real part equal to real part and imaginary equal to imaginary.\n$\\frac{3+5p}{1+p^2}=q~\\&~\\frac{5-3p}{1+p^2}=4$\nFrom that we get $p=\\frac{1}{4}~\\&~p=-1$.\n\n5. Originally Posted by Neverquit\nI found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0.\n\nThere is also apparently another solution of 4, -1 which I can\u2019t find.\n\nHow do I find it?\nI think you are misinterpreting your own answer. When you solve the system of equations, you will find that\n\n$p=1/4 ~~\\mbox{ or }~~ p=-1$\n\nThese are two separate solutions, not a single solution. You need to find the value of $q$ that pairs with each of these solutions for $p$. So, you need to plug each value of $p$ back into the system of equations and find the corresponding values of $q$. Since $pq=1$, it's a pretty straightforward calculation:\n\n$p=1/4 \\implies q=4$\n\n$p=-1 \\implies q=-1$\n\nTherefore, the solutions are:\n\nFirst solution: $p=1/4 \\mbox{ and } q=4$\n\nSecond solution: $p=-1 \\mbox{ and } q=-1$\n\nYou might also write this as:\n\n$(p,q) = (1/4,4) \\mbox{ or } (p,q) = (-1,-1)$\n\nBut you definitely would not say that the solutions are $(1/4,-1)$ and $(4,-1)$.\n\n6. Originally Posted by Neverquit\nHello,\nI am having trouble with this question:\n\n\"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i\"\n\nIm sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.\n\nAny help would be appreciated.\n\nRegards,\nNeverquit\n$\\frac{3+5i}{1+pi} =q+4i$\n\nMultiply top and bottom of LHS by $1-pi$ to give:\n\n$\\frac{(3+5p)+(5-3p)i}{1+p^2}=q+4i$\n\nEquate real and imaginary parts to get:\n\n$3+5p=(1+p^2)q$\n\nand:\n\n$5-3p=4(1+p^2)$\n\nNow the problem is to find all solutions to this pair of equations.\n\nCB\n\n7. Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it?\n\nSolving the equations I arrived at...\n\nq - 4p = 3\n\npq = 1\n\ngives\n\nq=1/p,\n\nSo subbing that into the first equation will give you a polynomial (if you multiply both sides by p) which gives you p = 1/4 and -1.\n\nHence q = 4 and -1.\n\nSolutions are (1/4,4) and (-1, -1)\n\nAre these the only solutions? Why must you multiply the top and bottom by the conjugate?\n\nHaving Plato and CP post in this thread has made me question whether my answer is wrong. Is it?\nYour solution was correct. Multiplying a complex fraction by the conjugate of the denominator is probably just a habit for them. But it's not necessary at all in this particular problem.\n\n9. ## Using the conjugate\n\nOriginally Posted by Plato\nThis is a tricky question.\n$\\frac{{3 + 5i}}{{1 + pi}} = \\frac{{\\left( {3 + 5i} \\right)\\left( {1 - pi} \\right)}}{{1 + p^2 }} = \\frac{{3 + 5p}}{{1 + p^2 }} + \\frac{{\\left( {5 - p} \\right)i}}{{1 + p^2 }}$\nNow set real part equal to real part and imaginary equal to imaginary.\n$\\frac{3+5p}{1+p^2}=q~\\&~\\frac{5-3p}{1+p^2}=4$\nFrom that we get $p=\\frac{1}{4}~\\&~p=-1$.\nI think the solution that Plato gives using the congugate is what the text books author had in mind as the question is shortly after conjugates of complex numbers is explained.\n\nDeastar, your solution still gives the same answer in the text book so it must be correct.\n\n...........\n\n10. ## generalisation\n\nOriginally Posted by Neverquit\nHello,\nI am having trouble with this question:\n\n\"Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i\"\n\nIm sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong.\n\nAny help would be appreciated.\n\nRegards,\nNeverquit\nthe basic idea behind questions of your type is equaslising real and imaginary parts.", "date": "2015-05-29 05:09:54", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/134539-complex-numbers-finding-real-number-pairs.html", "openwebmath_score": 0.8392963409423828, "openwebmath_perplexity": 562.7595708031333, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9863631623133665, "lm_q2_score": 0.9219218332461268, "lm_q1q2_score": 0.9093497348463857}}
{"url": "https://math.stackexchange.com/questions/1902937/what-is-the-probability-that-jan-and-jon-are-chosen", "text": "# What is the probability that Jan and Jon are chosen?\n\nJan , Jon and $10$ other children are in a classroom. The principal of the school walks in and choose $3$ children at random. What is the probability that Jan and Jon are chosen?\n\nMy approach:\n\nIncluding Jon and Jan, number of ways of selection is $1\\cdot1\\cdot{10\\choose 1} = 10$.\nTotal way of selection is ${12\\choose 3} = 220$\n\nSo, probability is $\\frac{10}{220}= \\frac{1}{22}.$\n\nBut when I solving as follows:\n\n\\begin{align*} P(\\text{selection of Jon and Jan}) &= 1- P(\\text{not selection of Jon and Jan}) \\\\ &= 1- \\frac{{10\\choose 3}}{{12\\choose 3}}\\\\ &= \\frac{5}{11}. \\end{align*}\n\nWhich approach is correct and why alternative one is wrong?\n\nNote: My previous post had some mistakes, so I deleted that\n\nI will call $A$, $B$ the events selecting Jon, Jan respectively. You did not take the complement correctly \\begin{align*} P(A\\cap B) &= 1-\\color{red}{P(\\bar A \\cup\\bar B)}\\tag 1 \\\\ &= 1-[P(\\bar A)+P(\\bar B)-P(\\bar A\\bar B)]\\tag 2 \\\\ &=1-\\left[\\frac{\\binom{11}{3}}{\\binom{12}{3}}+\\frac{\\binom{11}{3}}{\\binom{12}{3}}-\\frac{\\binom{10}{3}}{\\binom{12}{3}}\\right]\\\\ &=\\frac{1}{22} \\end{align*} where $(1)$ is true by DeMorgan's law, and $(2)$ is true by inclusion-exclusion. As you can see the two methods give the same value.\n\nYour first approach gives the probability that Jon AND Jan are selected, the second gives the probability that Jon OR Jan are selected.\n\nWhen you computed the probability of not selecting Jon AND Jan, you didn't include the situations where Jon was selected but not Jan, and vice versa.\n\nLets say the even Jon is selected is A, Jan is selected B.\n\nThen $P(A \\wedge B)=1-P(Not (A \\wedge B))= 1- P(Not A \\vee Not B)$\n\n$P(Not A \\vee Not B)= P(Not A)+ P(Not B)- P(Not A \\wedge Not B)$\n\nHere $\\wedge$ means and, and $\\vee$ means or. Think of a Venn diagram, if we look at the union of two circles, the total area is equal to the sum of the circles minus the intersection, because we counted that part twice.\n\n$P(Not A)=1-P(A)=1-(1*\\binom{11}{2})/220)=3/4$\n\nor\n\n$P(Not A)=\\binom{11}{3}/220=3/4$\n\n$P(Not B)=3/4$\n\n$P(Not A \\wedge Not B)= \\binom{10}{3}/220=6/11$\n\nNow $3/4+3/4-6/11=21/22$ So $1-21/22=1/22$", "date": "2021-12-06 12:21:25", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1902937/what-is-the-probability-that-jan-and-jon-are-chosen", "openwebmath_score": 0.9999120235443115, "openwebmath_perplexity": 425.0527464667491, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.992539357087774, "lm_q2_score": 0.9161096084360388, "lm_q1q2_score": 0.9092748417790384}}
{"url": "https://math.stackexchange.com/questions/2283548/number-of-ways-to-place-k-non-attacking-rooks-on-a-100-times-100-chess-board", "text": "# Number of ways to place $k$ non-attacking rooks on a $100\\times 100$ chess board\n\nI need to show that the number of ways to place $k$ non-attacking rooks (no two share the same column or row) on a $100\\times 100$ chessboard is $k!{100 \\choose k}^2$.\n\nWhen I try to formulate this equation I end up getting ${100 \\choose k}^2$ because you need to choose $k$ columns from $100$ columns and $k$ rows from $100$ rows. I know this isn't correct because if you have $k=100$, there is more than just $1$ solution. However I don't know how to come up with the $k!$ part of the equation.\n\n\u2022 \u2013\u00a0Shaun May 16 '17 at 14:32\n\nFirst of all, congrats on realizing that the answer you got can't possibly be correct. It's always a good idea to test formulas against special cases, to see if they stand up.\n\nOne way to arrive at the correct answer is to view the placement of the rooks in two steps: First choose the $k$ rows that the rooks will go in, and then, going row by row, decide which column to place that row's rook in. The first rook has $100$ columns to choose from, the second will have $99$, the third $98$, and so on. The total is thus\n\n$${100\\choose k}100\\cdot99\\cdot98\\cdots(100-(k-1))={100\\choose k}{100!\\over(100-k)!}={100\\choose k}{100!\\over(100-k)!k!}k!={100\\choose k}^2k!$$\n\nFirst, choose your $k$ rows and columns, as you said. Start by considering the configuration in which the rooks are successively placed in the legal square furthest to the top and to the left (so that the rooks go \"diagonally down to the right\").\n\nFrom there, it suffices to note that any rearrangement of the rook-columns results in a new and valid configuration. Since there are $k!$ such rearrangements, there are $k!$ configurations for any particular choice of $k$ rows and $k$ columns.\n\nYour problem is that $\\binom{100}{k}^2$ only gives you the ways to choose the columns and the rows separately, without specifiying which row goes with which column. This is why you need to add the factor $k!$, which corresponds to the number of bijections between your $k$ rows and your $k$ columns, i.e. the number of ways to associate them together.\n\nAn alternative way of obtaining this is to consider that you first choose $\\binom{100}{k}$ column where you will place your rooks, then choose for each column the row where you place a rook; this second step amonts to choosing $k$ rows with order, which you can do in $\\frac{100!}{(100-k)!}=\\binom{100}{k}k!$ ways. Multiplying the two numbers together gives you the result.\n\nRegards User. If i may contribute, here is my view :\n\nA 100 $\\times$ 100 chess board can be viewed as a matrix of size 100 $\\times$ 100. For example : let $$(i, j), \\:\\: \\text{ with } \\:\\:\\: i,j=1,2,... ,100$$ denotes the $i$th row and $j$th column of the board.\n\nTo solve your problem, the key is : the $k$ non-attacking roots is as same as no two $(i_{1}, j_{1})$ and $(i_{2}, j_{2})$ with $i_{1}=i_{2}$ or $j_{1}=j_{2}$. Two rooks with position $(1, 100)$ and $(91, 100)$, for example, does not satisfy the non-attacking roots condition.\n\nTo illustrate how to solve this, first, you could start with $k=1$.\n\n\u2022 For k=1. Let the position of this particular rook is $(x_{1},y_{1})$. Then there are 100 possibilities for $x_{1}$, and 100 possibilities for $x_{2}$. So the number of possibilities is (100)(100) = $1! \\binom{100}{1}^{2}$\n\n\u2022 For k=2. For each of the two rooks, their positions are denoted with $(x_{1},y_{1})$ and $(x_{2}, y_{2})$. For the first one, There are 100 possibilities for each $x_{1}$ and $y_{1}$. For the 2nd rook, $x_{2}$ and $y_{2}$ each has 99 possibilities (since they can't be equal to the 'coordinates' of the 1st rook). So the number of possibilities to put 2 non-attacking distinguished rooks is $$(100^{2})(99^{2})$$ For your problem, they are not distinguished, so we have to divide this by 2 (exactly $2!$), because we can choose either rook to be the 1st or the 2nd. So, output : $$\\frac{(100^{2})(99^{2})}{2!} = \\frac{(100 \\cdot 99)(100 \\cdot 99)}{2!} = \\frac{(100 \\cdot 99)(100 \\cdot 99)}{2!} \\left(\\frac{98! \\cdot 2!}{98! \\cdot 2!} \\right)^{2} = 2! \\binom{100}{2}^{2}$$\n\n\u2022 For k=3 and above, you would be confident to try and continue this method.\n\nHope this will helps. Regards, Arief.", "date": "2020-02-23 08:18:20", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2283548/number-of-ways-to-place-k-non-attacking-rooks-on-a-100-times-100-chess-board", "openwebmath_score": 0.7984408140182495, "openwebmath_perplexity": 197.14013539539394, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9959927104593489, "lm_q2_score": 0.9124361640485893, "lm_q1q2_score": 0.9087797681518855}}
{"url": "https://mathhelpboards.com/threads/permutation-representation-argument-validity.7690/", "text": "# Permutation representation argument validity\n\n#### kalish\n\n##### Member\nHello,\nI would like to check if the work I have done for this problem is valid and accurate. Any input would be appreciated. Thank you.\n\n**Problem statement:** Let $G$ be a group of order 150. Let $H$ be a subgroup of $G$ of order 25. Consider the action of $G$ on $G/H$ by left multiplication: $g*aH=gaH.$ Use the permutation representation of the action to show that $G$ is not simple.\n\n**My attempt:** Let $S_6$ be the group of permutations on $G/H$. Then, the action of $G$ on $G/H$ defines a homomorphism $f:G \\rightarrow S_6$. We know $|S_6| = 720.$ Since $|G|=150$ does not divide 720, and $f(G)$ is a subgroup of $S_6$, $f$ cannot be one-to-one. Thus, $\\exists$ $g_1,g_2$ distinct in $G$ such that $f(g_1)=f(g_2) \\implies f(g_1g_2^{-1})=e$. Thus, $\\ker(f) = \\{g:f(g)=e\\}$. Since $\\ker(f)$ is a normal subgroup of $G$, we have found a normal subgroup of $G$. Also, since $f$ is non-trivial, then $\\ker(f)$ is a proper normal subgroup of $G.$ Hence $G$ is not simple.\n\nAny suggestions or corrections?\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nI think it is clear that since $|G|$ does not divide 720, $\\text{ker}(f)$ is a non-trivial normal subgroup of $G$, so there is no need to talk about the existence of $g_1,g_2$ or restate the definition of $\\text{ker}(f)$.\n\nI *do* think you should say WHY $f$ is not the trivial homomorphism. It's pretty simple, though:\n\nSince $|H| < |G|$, we can take any $g \\in G - H$, which takes (under the action) the coset $H$ to $gH \\neq H$, so $f(G)$ contains at least one non-identity element: namely, $f(g)$.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nThis is actually a special case of a theorem proved in Herstein, which goes as follows:\n\nIf $G$ is a finite group with a subgroup $H$ such that $|G| \\not\\mid ([G:H])!$ then $G$ contains a non-trivial proper normal subgroup containing $H$.\n\nOne obvious corollary is then that such a group $G$ cannot be simple.\n\nYou would be far better off adapting your proof to this more general one, which can be re-used in many more situations.\n\n#### kalish\n\n##### Member\nHi,\nWhich Herstein book is this from? I would like to explore further.\n\nThanks.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nHis classic Topics In Algebra\u200b.\n\n#### kalish\n\n##### Member\nThat's what I found from my search as well. Do you have a copy of the book or know where I can find one?\n\n#### kalish\n\n##### Member\nThat sounds like a fantastic result. I cannot find the book anywhere though. Could you please reproduce the proof for me here, so that I could use it to study? I would really appreciate it.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nTheorem 2.G (p. 62, chapter 2):\n\nIf $G$ is a group, $H$ a subgroup of $G$, and $S$ is the set of all right cosets of $H$ in $G$, then there is a homomorphism $\\theta$ of $G$ into $A(S)$, and the kernel of $\\theta$ is the largest normal subgroup of $G$ which is contained in $H$.\n\n(a few words about notation: Herstein uses $A(S)$ to stand for the group of all bijections on $S$...if $|S| = n$, then $A(S)$ is isomorphic to $S_n$. Herstein also writes his mappings on the RIGHT, as in $(x)\\sigma$ instead of $\\sigma(x)$, so that composition and multiplication are \"in the same order\", instead of reversed. For this reason, he uses right cosets and right-multiplication instead of the left cosets (and left-multiplication) one often sees used in other texts. He also denotes the index of $H$ in $G$ by $i(H)$ , instead of $[G:H]$ and denotes $|G|$ by $o(G)$).\n\nProof: Let $G$ be a group, $H$ a subgroup of $G$. Let $S$ be the set whose elements are right cosets of $H$ in $G$. That is, $S = \\{Hg: g \\in G\\}$. $S$ need not be a group itself, in fact, it would be a group only if $H$ were a normal subgroup of $G$. However, we can make our group $G$ act on $S$ in the following natural way: for $g \\in G$ let $t_g:S \\to S$ be defined by: $(Hx)t_g = Hxg$. Emulating the proof of Theorem 2.f we can easily prove:\n\n(1) $t_g \\in A(S)$ for every $g \\in G$\n(2) $t_{gh} = t_gt_h$.\n\nThus the mapping $\\theta: G \\to A(S)$ defined by $\\theta(g) = t_g$ is a homomorphism of $G$ into $A(S)$. Can one always say that $\\theta$ is an isomorphism? Suppose that $K$ is the kernel of $\\theta$. If $g_0 \\in K$, then $\\theta(g_0) = t_{g_0}$ is the identity map on $S$, so that for every $X \\in S, Xt_{g_0} = X$. Since every element of $S$ is a right coset of $H$ in $G$, we must have that $Hat_{g_0} = Ha$ for every $a \\in G$, and using the definition of $t_{g_0}$, namely, $Hat_{g_0} = Hag_0$, we arrive at the identity $Hag_0 = Ha$ for every $a \\in G$. On the other hand if $b \\in G$ is such that $Hxb = Hx$ for every $x \\in G$, retracing our argument we could show that $b \\in K$. Thus $K = \\{b \\in G|Hxb = Hx$ all $x \\in G\\}$. We claim that from this characterization of $K,\\ K$ must be the largest normal subgroup of $G$ which is contained in $H$. We first explain the use of the word largest; by this we mean if $N$ is a normal subgroup of $G$ which is contained in $H$, then $N$ must be contained in $K$. We wish to show this is the case. That $K$ is a normal subgroup of $G$ follows from the fact that it is the kernel of a homomorphism of $G$. Now we assert that $K \\subset H$, for if $b \\in K, Hab = Ha$ for every $a \\in G$, so in particular, $Hb = Heb = He = H$, whence $b \\in H$. Finally, if $N$ is a normal subgroup of $G$ which is contained in $H$, if $n \\in N,\\ a \\in G$, then $ana^{-1} \\in N \\subset H$, so that $Hana^{-1} = H$; thus $Han = Ha$ for all $a \\in G$. Therefore, $n \\in K$ by our characterization of $K$.\n\n**********\n\nRemarks following the proof:\n\nThe case $H = (e)$ just yields Cayley's Theorem (Theorem 2.f). If $H$ should happen to have no normal subgroup of $G$, other than $(e)$ in it, then $\\theta$ must be an isomorphism of $G$ into $A(S)$....(some text omitted)....\n\nWe examine these remarks a little more closely. Suppose that $G$ has a subgroup $H$ whose index $i(H)$ (that is, the number of right cosets of $H$ in $G$) satisfies $i(H)! < o(G)$. Let $S$ be the set of all right cosets of $H$ in $G$. The mapping, $\\theta$, of Theorem 2.g cannot be an isomorphism, for if it were, $\\theta(G)$ would have $o(G)$ elements and yet would be a subgroup of $A(S)$ which has $i(H)! < o(G)$ elements. Therefore, the kernel of $\\theta$ must be larger than $(e)$; this kernel being the largest normal subgroup of $G$ which is contained in $H$, we can conclude that $H$ contains a nontrivial normal subgroup of $G$.\n\nHowever, the above argument has implications even when $i(H)!$ is not less than $o(G)$. If $o(G)$ does not divide $i(H)!$ then by invoking Lagrange's theorem we know that $A(S)$ can have no subgroup of order $o(G)$, hence no subgroup isomorphic to $G$. However $A(S)$ does contain $\\theta(G)$, whence $\\theta(G)$ cannot be isomorphic to $G$, that is, $\\theta$ cannot be an isomorphism. But then, as above, $H$ must contain a nontrivial normal subgroup of $G$. We summarize this as:\n\nLemma 2.21 If $G$ is a finite group, and $H \\neq G$ is a subgroup of $G$ such that $o(G) \\not\\mid i(H)!$, then $H$ must contain a nontrivial normal subgroup of $G$. In particular, $G$ cannot be simple.\n\n(Note to the moderating staff: although this is an excerpt from a copyrighted work, I believe this sample falls under the province of \"Fair Use\" for the purpose of \"Scholarly research and exposition\", and is not intended for commercial gain or to circumvent existing copyright laws).\n\nLast edited:", "date": "2020-09-25 07:45:44", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/permutation-representation-argument-validity.7690/", "openwebmath_score": 0.9447657465934753, "openwebmath_perplexity": 85.21301523021599, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180656553328, "lm_q2_score": 0.9219218423633528, "lm_q1q2_score": 0.9087550151398048}}
{"url": "https://www.physicsforums.com/threads/spring-gun.742063/", "text": "# Spring Gun\n\n1. Mar 7, 2014\n\n### Steven60\n\nI have a question about a spring gun. Suppose the barrel of a spring gun is placed horizontally at the edge of a horizontal table. You put say a marble in the barrel and compress the spring x cm and after releasing the marble it travels a horizontal distance of y cm before hitting the floor (so motion is of a projectile). My question is whether or not the horizontal distance traveled and the amount the spring is compresses make a linear relationship? If so, then how can I prove this? This is not homework.\nThanks!\n\n2. Mar 7, 2014\n\nSeems to be purely a math problem. Perhaps sketch the system and write the relevant equations needed to determine this?\n\n3. Mar 7, 2014\n\n### UltrafastPED\n\nThis is a nice physics exercise - there are several physical considerations, and then some simple math.\n\nYou have two forces acting on the marble ... the spring force, which launches the marble, and gravity.\n\nOnce the marble leaves the launch tube it will have a constant \"horizontal\" speed - ignoring air resistance - and an initial vertical speed of zero. Call this initial horizontal speed V.\n\nThe vertical speed will increase with time due to the constant gravitational acceleration - and will hit the floor at a definite time which depends only on the height of the table. Call this duration T.\n\nThen the distance from the table to the point of contact will be D = V x T.\n\nThe time T does not depend upon the spring force, only on the height of the table and local value g=9.8 m/s^2.\n\nThus you only need to determine if the speed V is proportional to the spring force; by Hook's law we know that a \"good\" spring obeys F = -k * X, where X is the compression/extension distance and k is the spring's constant.\n\nIf we switch to energy we have work done on marble is W = Integral[F dx] over the interval x=[0,X]. Note that the force is changing as the spring moves! So W = Integral[ k*x dx] = 1/2 k*X^2.\n\nBut this work has been converted into kinetic energy of the marble. For a marble of mass=M, and given that it is NOT rolling or spinning, then the kinetic energy is KE=1/2 M*V^2 = 1/2 k*X^2=W.\n\nThus V = k/M Sqrt[X]. xx Correction: xxx Make that V = Sqrt[k/M] * X.\n\nThus the hypothesis is true!\n\nThanks to dauto for noticing the mistake at the end! :-)\n\nLast edited: Mar 7, 2014\n4. Mar 7, 2014\n\n### DrewD\n\nThe equations involved will be $d=vt$ for constant $v$ (and assuming that the initial point when exiting the spring gun is defined as 0 distance), $U_{spring}=\\frac12k\\Delta x^2$ and $K=\\frac12mv^2$. $\\Delta x$ is the amount the spring is compressed, and $v$ is the velocity of the object as it leaves the spring. This approximation assumes that the object does not stick which is a good assumption for a spring gun. Solve for $v$ to find the relationship between $\\Delta x$ and $d$.\n\n5. Mar 7, 2014\n\n### dauto\n\nYou made a mistake at the very end. In fact, after correcting the mistake, you proved that the hypothesis is true.\n\n6. Mar 9, 2014\n\n### Steven60\n\nThanks for your replys. I actually worked this out myself and actually did the same exact steps as UltrafastPED.", "date": "2017-08-21 13:36:13", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/spring-gun.742063/", "openwebmath_score": 0.8103019595146179, "openwebmath_perplexity": 722.2710532848253, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.986363161511632, "lm_q2_score": 0.9207896688265685, "lm_q1q2_score": 0.9082330088310228}}
{"url": "http://math.stackexchange.com/questions/245576/is-this-induction-procedure-correct-2nn", "text": "# Is this induction procedure correct? ($2^n<n!$)\n\nI am rather new to mathematical induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I've been able to solve some of the form $1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{n} \\leq \\frac{n}{2} + 1$.\n\nNow, I was presented this, for $n \\ge 4$:\n\n$$2^n<n!$$\n\nI tried to do it with similar logic as the one suggested there. This is what I did:\n\nProve it for $n = 4$: $$2^4 = 16$$ $$4! = 1\\cdot2\\cdot3\\cdot4 = 24$$ $$16 < 24$$ Assume the following: $$2^n<n!$$ We want to prove the following for $n+1$: $$2^{n+1}<(n+1)!$$ This is how I proved it:\n\n\u2022 So first we take $2^{n+1}$ which is equivalent to $2^n\\cdot2$\n\u2022 By our assumption, we know that $2^n\\cdot2 < n!\\cdot2$\n\u2022 This is because I just multiplied by $2$ on both sides.\n\u2022 Then we'll be finished if we can show that $n! \\cdot 2 < (n+1)!$\n\u2022 Which is equivalent to saying $n!\\cdot2<n!\\cdot(n+1)$\n\u2022 Since both sides have $n!$, I can cancel them out\n\u2022 Now I have $2<(n+1)!$\n\u2022 This is clearly true, since $n \\ge 4$\n\nEven though the procedure seems to be right, I wonder:\n\n\u2022 In the last step, was it ok to conclude with $2<(n+1)!$? Was there not anything else I could have done to make the proof more \"careful\"?\n\u2022 Is this whole procedure valid at all? I ask because, well, I don't really know if it would be accepted in a test.\n\u2022 Are there any points I could improve? Anything I could have missed? This is kind of the first time I try to do these.\n-\nI think your proof is fine, if a bit long-winded. But with experience, you'll learn what bits to shorten without losing rigour. \u2013\u00a0 Harald Hanche-Olsen Nov 27 '12 at 9:51\nYou should say 2^{n+1}\\lt 2\\cdot n!$. But because$2\\lt $n+1$, it follows that $2\\cdot n!\\lt (n+1)!$. Unfortunately, you are still writing proofs \"backwards\" in a logically incorrect way. \u2013\u00a0 Andr\u00e9 Nicolas Nov 27 '12 at 9:59\n@Andr\u00e9: It seems a bit harsh to call this a \"logically incorrect way\". It's OK as long as \"Since both sides have $n!$, I can cancel them out\" is interpreted as \"dividing through by $n!$ leads to an equivalent inequality\". It's true that the reverse order would be clearer, and doing things in this order is incorrect if the implications used only go in one direction, but that's not the case here so the proof is still OK, if suboptimally structured. \u2013\u00a0 joriki Nov 27 '12 at 10:04\n@Omega, your proof is correct...\"almost\", since as Andre apparently meant, there's some lack of logical rigour in your last lines. You must show that the implications there are double, i.e.: $$n!\\cdot 2<(n+1)!\\Longleftrightarrow 2<\\frac{(n+1)!}{n!}=n+1$$ and then noting the last inequality is trivially true as we're working with $\\,n\\geq 4\\,$... Also, don't right \"equivalent\" when it should be \"equal\", as in \"$\\,2^{n+1}\\,$ is \"equivalent\"(should be \"equal\"!) to $\\,2^n\\cdot 2$ \u2013\u00a0 DonAntonio Nov 27 '12 at 10:47\n@joriki: A bit harsh, perhaps, but other posts by the OP had some worse instances that were pointed out. Since the actual understanding of the problem is good, it is worthwhile to vaccinate the OP against $A\\to B\\to C\\to 0=0$, therefore $A$. \u2013\u00a0 Andr\u00e9 Nicolas Nov 27 '12 at 16:47\n\nYes, the procedure is correct. If you want to write this more like the sort of mathematical proof that would be found in a textbook, you might want to make some tweaks.\n\nFor example, the base case could be re-written as follows:\n\nWhen $n = 4$, we have $2^4 = 16 < 24 = 4!$\n\nNext, the inductive hypothesis and the subsequent manipulations:\n\nSuppose that for $n \\geq 4$ we have $2^n < n!$\n\nThus, $2^{n+1} < 2 \\cdot n! < (n+1)!$, where the first inequality follows by multiplying both sides of the inequality in our IH by $2$, and the second follows by observing that $2 < n+1$ when $n \\geq 4$.\n\nTherefore, by the Principle of Mathematical Induction, $2^n < n!$ for all integers $n \\geq 4$. Q.E.D.\n\nNote: I am not making a judgment about whether your write-up or the one I have included here is \"better.\" I'm only observing that the language and format differ, particularly with regard to proofs that are written in paragraph form (typical of math papers) rather than with a sequence of bullet-points (which is what you had).\n\n-", "date": "2015-05-24 19:34:48", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/245576/is-this-induction-procedure-correct-2nn", "openwebmath_score": 0.8947398066520691, "openwebmath_perplexity": 321.9162955993664, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754488233528, "lm_q2_score": 0.9219218391455084, "lm_q1q2_score": 0.9077016085567398}}
{"url": "http://tdgl.solotango.it/bfs-time-complexity.html", "text": "### Bfs Time Complexity\n\nOptimality : It is optimal if BFS is used for search and paths have uniform cost. For example, if the heuristic evaluation function is an exact estimator, then A* search algorithm runs in linear time, expanding only those nodes on an optimal solution path. In this tutorial, We are going to learn about bubble sort algorithm and their implementation in various programming languages. Again basic of bfs , once you get this you will get to know how powerful and where we can use it in daily life example Stay tuned for more. 1 & 2): Gunning for linear time\u2026 Finding Shortest Paths Breadth-First Search Dijkstra\u2019s Method: Greed is good! Covered in Chapter 9 in the textbook Some slides based on: CSE 326 by S. We use Queue data structure with maximum size of total number of vertices in the graph to implement BFS traversal. For the most part, we describe time and space complexity for search on a tree; for a graph, the answer depends on how \u201credundant\u201d the paths in the state space are. Since a BFS traversal is used, the overall time complexity is simply O(|V| + |E|). Yes, the worst case complexity is O(ab). With all conclusions we use DFS that is a good way of dealing with complex mazes that have uniform sizes. Set The Starting Vertex To Vertex 1. Shortest Path using BFS: The shortest path between two vertices in a graph is a path such that the total sum of edge weights in the path connecting the two vertices is minimum. Priority queue Q is represented as a binary heap. Space complexity: O(bm) for the tree search version and O(b m) for the graph search version; Breadth First Search (BFS) BFS uses FIFO ordering for node expansion i. The amount of time needed to generate all the nodes is considerable because of the time complexity. The deepest node happens to be the one you most recently visited - easy to implement recursively OR manage frontier using LIFO queue. Complexity Measures Message complexity: Number of messages sent (worst case). What is the time complexity of BFS? \u2013 how many states are expanded before finding a solution? \u2013 b: branching factor \u2013 d: depth of shallowest solution \u2013 complexity = What is the space complexity of BFS? \u2013 how much memory is required? \u2013 complexity = Is BFS optimal? \u2013 is it guaranteed to find the best solution (shortest path)?. The aim of BFS algorithm is to traverse the graph as close as possible to the root node. Set The Starting Vertex To Vertex 1. One starts at the root (selecting some arbitrary node as the root in the case of a graph) and explores along adjacent nodes and proceeds recursively. -1 \u2013 A wall or an obstacle. edu/6-006F11 Instructor: Erik Demaine License: Creative Commons BY-N. graph algorithms, has linear time complexity, and is com-plete for the class SL of problems solvable by symmetric, non-deterministic,log-space computations[32]. And that\u2019s how a quadratic time complexity is achieved. BFS Properties \u2022 Which nodes does BFS expand? o Processes all nodes above depth of shallowest solution, s o Search time takes time O(bs) \u2022 Fringe Size: o Keeps last tier o O(bs) \u2022 Complete? o s must be finite, so yes! \u2022 Optimal? o Only if all costs are 1 (more later). Time Complexity of DFS is also O(V+E) where V is vertices and E is edges. You can also use BFS to determine the level of each node. If it is an adjacency matrix, it will be O(V^2). HackerRank - Breadth First Search - Shortest Path. Breadth-first search (BFS) algorithm is an algorithm for traversing or searching tree or graph data structures. graphBfs1 - Free download as Powerpoint Presentation (. Time Complexity of BFS (Using adjacency matrix) \u2022 Assume adjacency matrix \u2013 n = number of vertices m = number of edges No more than n vertices are ever put on the queue. Asked about the time complexity of search, deletion, etc. Rewrite the pseudocode for the BFS algorithm studied in class (and presented in the textbook) to work for an adjacency matrix representation of the graph instead of an adjacency list representation. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. Breadth-First-Search Attributes \u2022 Completeness \u2013 yes \u2022 Optimality \u2013 yes, if graph is un-weighted. Depth first traversal or Depth first Search is a recursive algorithm for searching all the vertices of a graph or tree data structure. Documentation / Algorithms The Welsh-Powell Algorithm. 6) Broadcasting in Network: In networks, a broadcasted packet follows Breadth First Search to reach all nodes. Some methods are more effective then other while other takes lots of time to give the required result. The Big O notation is used to classify algorithms by their worst running time or also referred to as the upper bound of the growth rate of a function. And then it concluded that the total complexity of DFS() is O(V + E). \u2022 Scanning for all adjacent vertices takes O(| E|) time, since sum of lengths of adjacency lists is |E|. BFS Algorithm Complexity. complete\uff1a BFS\u662fcomplete\u7684\u3002 optimal\uff1a BFS\u662foptimal\u7684\uff0c\u56e0\u4e3a\u627e\u5230\u7684\u7b2c\u4e00\u4e2a\u89e3\u662f\u6700shallow\u7684\u3002 time complexity: \u548cDFS\u4e00\u6837\u662f. Interview question for Software Engineer. The smallest number of colors required to color a graph G is called its chromatic number of. Sirius? Brightest star in sky. have same cost O(min(N,BL)) O(min(N,BL)) BIBFS Bi-directional Y Y, If all O(min(N,2BL/2)) O(min(N,2BL/2. Completeness is a nice-to-have feature for an algorithm, but in case of BFS it comes to a high cost. Intuitively, you start at the root node and explore all the neighboring nodes. He also figures out the time complexity of these algorithms. c) [2pt] Express time and space complexity for general breadth-first search in terms of the branching factor, b, and the depth of the goal state, d. BFS (G, s) Breadth -First Search Starting from the sourc e node s, BFS computes the minimal distance from s to any other node v that can be reached from s. The optimal solution is possible to obtain from BFS. If we use an adjacency list, it will be O(V+E). Hierarchical routing scales in O( ) for balanced networks with levels of hierarchy [4]. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. The basic approach of the Breadth-First Search (BFS) algorithm is to search for a node into a tree or graph structure by exploring neighbors before children. Breadth First Search BFS intuition. Complexity The time complexity of BFS is O(V + E), where V is the number of nodes and E is the number of edges. The time complexity of BFS is O(V+E) because: Each vertex is only visited once as it can only enter the queue once \u2014 O( V ) Every time a vertex is dequeued from the queue, all its k neighbors are explored and therefore after all vertices are visited, we have examined all E edges \u2014 (O( E ) as the total number of neighbors of each vertex. The time complexity of BFS is O(V+E) where V stands for vertices and E stands for edges. Let\u2019s say for instance that you want to know the shortest path between your workplace and home, you can use graph algorithms to get the answer! We are going to look into this and other fun. This is a generic BFS implementation: For a connected graph with V nodes and E total number of edges, we know that every edge will be considered twice in the inner loop. z x y z is a cycle of length 2(j i) + 1, which is odd, so G is not bipartite. However, the space complexity for these algorithms varies. Queue is used in the implementation of the breadth first search. PRAM algorithm Communication Time Problem of complexity complexity Breadth-first search 141 IEI I VI Maximum flow [I31 I VI3 I VIZ 805 TABLE II. Thus, each guard returns to his starting position after 2, 4 or 6 moves. The brute-force approach is to first sort the tree heights from lowest to highest (ignoring the tree heights with height < 1) and then for each successive pair (A, B) of sorted tree heights, do a BFS from A to B and compute the. Time complexity for B() is O(1), so if i is called from 1 to n, then it's n-times O(1)-> O(n). Breadth-First Search (BFS) Properties What nodes does BFS expand? Processes all nodes above shallowest solution Let depth of shallowest solution be s Search takes time O(bs) How much space does the fringe take? Has roughly the last tier, so O(bs) Is it complete? s must be finite if a solution exists, so yes! Is it optimal?. Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that only one letter can be changed at a time and each intermediate word must exist in the dictionary. Q: if each node has b children & optimum is at depth d, what are the time and space complexities of BFS? (tip: time=#GeneratedNode, space=#StoredNode). The space complexity of a search algorithm is the worst-case amount of memory that the algorithm will use (i. Completeness is a nice-to-have feature for an algorithm, but in case of BFS it comes to a high cost. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. bfs algorithm. Given a branching factor b and graph depth d the space complexity is the number of nodes at the deepest d level, O(b ). Keywords: Distributed system, breadth-first-search, communication complexity, graph, algorithm 1. Time and Memory Requirements for BFS \u2013 d+1O(b ) Example: \u2022 b = 10 \u2022 10,000 nodes/second \u2022 each node requires 1000 bytes of storage Depth Nodes Time Memory 2 1100. Breadth First Search: visit the closest nodes first. , we\u2019ll no longer require that an action from a given state leads to the same state each time and will. To that purpose, we introduce a new parameter, called. This again depends on the data strucure that we user to represent the graph. And that\u2019s how a quadratic time complexity is achieved. Breadth-First Search (BFS) in 2D Matrix/2D-Array Categories Amazon Questions , Binary Tree , Expert , Facebook , Google Interview , Linked List , Microsoft Interview , Recursion , Software Development Engineer (SDE) , Software Engineer , Trees Tags Expert 1 Comment Post navigation. L 2= all nodes that do not belong to L 0or L 1, and that have an edge to a node in L 1. time-complexity recurrence-relations loops asymptotic-analysis asymptotic-notation greedy dynamic-programming graph substitution-method vertex-coloring a-star np-completeness log analysis nested-loops n-puzzle heuristic master-theorem exponent n-queens conflict ai graph-coloring mvcs small-oh count easy sorted-lists logn example recursive gcd. 2 Choosing a good hash function; 19. Time Complexity: T(n) = O(N^2) Because we have a square matrix and in the worst case. bfs time complexity. Visualizing DFS traversal Depth-first Search (DFS) is an algorithm for searching a graph or tree data structure. Note : The space/time complexity could be less as the solution could be found anywhere on the. The BFS strategy is not generally cost optimal. a time complexity t(n) if the Turing Machine takes time at most t(n) on any input of length n. The time complexity of BFS is O(V + E). The Time complexity of BFS is O(V + E) when Adjacency List is used and O(V^2) when Adjacency Matrix is used, where V stands for vertices and E stands for edges. 5) GPS Navigation systems: Breadth First Search is used to find all neighboring locations. Now we can use the BFS on to print the path (while printing only the vertices that belong to V). In this tutorial, we are going to focus on Breadth First Search technique. Adjacency List Time Complexity. The two variants of Best First Search are Greedy Best First Search and A* Best First Search. , D(B)=1, D(F)=2. Breadth-first search (BFS) is an algorithm for traversing or searching tree or graph data structures. If it is an adjacency matrix, it will be O(V^2). The time complexity of the union-find algorithm is O(ELogV). Disadvantages. Time Complexity Edit. Proof [ edit ]. That means Big-Omega notation always indicates the minimum time required by an algorithm for all input values. BFS \u2013 time complexity b d depth number of nodes 0 1 1 12 =2 2 3 d 2=4 23=8 2d (bd ) Total nodes:Expanded nodes: O(bd 1) d+1 2 d+1(b ) O(bd) CS 2710 Foundations of AI M. Home; Python dictionary time complexity. \u2022 Time complexity: exponential in the depth of the solution d \u2022 Memory (space) complexity: nodes are kept in the memory O(bd) O(bd). Low water level. Breadth-first search. at most 3 nodes. \u2022 Time Complexity: 21 Ram Meshulam 2004 \u2022 Memory Complexity: O db( ) \u2013 Where b is branching factor and d is the maximum depth of search tree O(( d )b +(d \u2212 1) b2 + + (1) bd ) =O(bd) State Redundancies \u2022 Closed list - a hash table which holds the visited nodes. By the use of the Queue data structure, we find the level order traversal. Features of BFS Space complexity Space complexity is proportional to the number of nodes at the deepest level. We'll start by describing them in undirected graphs, but they are both also very useful for directed graphs. This again depends on the data strucure that we user to represent the graph. The time complexity remains O(b d) but the constants are large, so IDDFS is slower than BFS and DFS (which also have time complexity of O(b d)). And that\u2019s how a quadratic time complexity is achieved. c) [2pt] Express time and space complexity for general breadth-first search in terms of the branching factor, b, and the depth of the goal state, d. algorithms achieves optimal O(D) time, its communi- cation complexity is O(E. Same as Time Complexity UCS (Uniform Cost Search): BFS Enhanced with lowest path costs first Only test from start to goal (Dijkstra, no goal state unitl all nodes are removed to get shortest paths to all nodes. BFS Properties \u2022 Which nodes does BFS expand? o Processes all nodes above depth of shallowest solution, s o Search time takes time O(bs) \u2022 Fringe Size: o Keeps last tier o O(bs) \u2022 Complete? o s must be finite, so yes! \u2022 Optimal? o Only if all costs are 1 (more later). The existing algorithm, due to Cheung (1983), has communication and time complexities O( IV[3) and O( IV 1), respectively. Depth-first search. Breadth first search (BFS) and Depth First Search (DFS) are the simplest two graph search algorithms. Time Complexity Best log(n)) log(n)) log(n)) Average (nA2 Worst (nA2 (nA2 Worst Case Auxiliary Space Complexity Worst O(n) O(nk) O(n+k) Fair Searching Algorithm Depth First Search (DFS) Breadth First Search (BFS) Binary search Linear (Brute Force) Shortest path by Dijkstra, using a Min-heap as priority queue Shortest path by Dijkstra,. You are probably using programs with graphs and trees. \u2022 Time Complexity: 21 Ram Meshulam 2004 \u2022 Memory Complexity: O db( ) \u2013 Where b is branching factor and d is the maximum depth of search tree O(( d )b +(d \u2212 1) b2 + + (1) bd ) =O(bd) State Redundancies \u2022 Closed list - a hash table which holds the visited nodes. Complexity Analysis. execution time in the PRAM model is O(D), where the Dis the diameter of the graph. Implementation. BFS: Time complexity is [code ]O(|V|)[/code] where [code ]|V|[/code] is the number of nodes,you need to traverse all nodes. In the last two posts, we talked about adversarial search and built a bot for checkers. We can say that UCs is the optimal algorithm as it chooses the path with the lowest cost only. using Software Complexity Measures Akanmu T. May I ask if this is O(n^2) time complexity ? If it is , may I ask if there is O(n) time solution ? Thank you. If it is an adjacency matrix, it will be O (V^2). Since there are Dphases, the cost is bounded by O(nD). 3 that also indicates a breadth-first tree rooted at v 1 and the distances of each vertex to v 1. The time complexity is ( + ). But now consider the point in time during the execution of BFS when w was removed from the queue. If any algorithm requires a fixed amount of space for all input values then that space complexity is said to be Constant Space Complexity. How would you actually implement those lines? 3 Breadth First Search We say that a visitation algorithm is a breadth \ufb01rst search or BFS, algorithm, if vertices are visited in breadth \ufb01rst order. 0-1 BFS is nothing but a special case of Dijkstra\u2019s Algorithm which can only be applied on Graph with vertices weighted 0 and x (x>=0) only. Well in case of shortest path we just do a small modification and store the node. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth-first search (BFS) algorithm. For a list of resources curated to help small businesses navigate the crisis, visit our COVID-19 resource hub. Exercise Time! @BiancaGando. Level Order Traversal, Print each level in one line. It uses a queue for storing the visited vertices. BFS algorithm. What is the time complexity of BFS? \u2013 how many states are expanded before finding a solution? \u2013 b: branching factor \u2013 d: depth of shallowest solution \u2013 complexity = What is the space complexity of BFS? \u2013 how much memory is required? \u2013 complexity = Is BFS optimal? \u2013 is it guaranteed to find the best solution (shortest path)?. If any algorithm requires a fixed amount of space for all input values then that space complexity is said to be Constant Space Complexity. 2 Choosing a good hash function; 19. The Time complexity of both BFS and DFS will be O(V + E), where V is the number of vertices, and E is the number of Edges. DFS uses Stack while BFS uses Queue. Breadth First Search. This complexity is worse than O(nlogn) worst case complexity of algorithms like merge sort, heap sort etc. ) Let G= (V,E ) be a graph. Current time T. After poping out a vertex from the queue, decrease the indegrees of its neighbors. For each i, L. val > sum and just return false there because you know you\u2019ll just get more and more negative, assuming they\u2019re all positive integers. The default. Time Complexity of DFS is also O(V+E) where V is vertices and E is edges. Breadth-First Search \u2022Complete? \u2022Optimal? \u2022Time complexity? \u2022Space complexity? Yes If shallowest goal is optimal Exponential: O( bd+1 ) Exponential: O( bd+1 ) In practice, the memory requirements are typically worse than the time requirements b = branching factor (require finite b) d = depth of shallowest solution. Completeness : Bidirectional search is complete if BFS is used in both searches. It uses the opposite strategy as depth-first search, which instead. We use the Big-O notation to classify algorithms based on their running time or space (memory used) as the input grows. Low water level. It is a greedy algorithm and grows the minimum spanning tree one edge at a time. This space complexity is said to be Constant Space Complexity. Yes, if all edges have equal cost. 4) breadth-first search ([BFS]) This algorithm is used for unweighted graphs, but explained because it is used below. Memory requirements are a bigger problem for breadth first search than is the execution. INF \u2013 Infinity means an empty room. DFID can be a big win when considering brute-force algorithms and the problem is big. Exercise Time! @BiancaGando. Program- Level order binary tree traversal in java 1. Breadth-first search is ideal in situations where the answer is near the top of the tree and Depth-first search works well when the goal node is near the bottom of the tree. Priority queue Q is represented as a binary heap. set start vertex to visited load it into queue while queue not empty for each edge incident to vertex if its not visited load into queue mark vertex. Yes, the worst case complexity is O(ab). If there is a solution then BFS is guaranteed to find it. Using the new BFS algorithm in this paper, we can improve significantly time performance of existing leader election algorithms. Since the PRAM model does not weigh in synchronization costs, the asymptotic complexity of work performed is identical to the serial algorithm. BFS ia an graph traversal algorithm. o Notation: the goals are d edges away from the initial state. You are marking a vertex as visited while taking it out of the queue and not while pushing it. Time Complexity. Implementation of BFS tree traversal algorithm,. This again depends on the data strucure that we user to represent the graph. Breadth-First Search (BFS) in 2D Matrix/2D-Array Categories Amazon Questions , Binary Tree , Expert , Facebook , Google Interview , Linked List , Microsoft Interview , Recursion , Software Development Engineer (SDE) , Software Engineer , Trees Tags Expert 1 Comment Post navigation. The features of the BFS are space and time complexity, completeness, proof of completeness, and optimality. But BFS only needs to iterate through the first two levels, i. A version of depth-first search was investigated in the 19th century by French mathematician Charles Pierre. BFS Properties \u2022 Which nodes does BFS expand? o Processes all nodes above depth of shallowest solution, s o Search time takes time O(bs) \u2022 Fringe Size: o Keeps last tier o O(bs) \u2022 Complete? o s must be finite, so yes! \u2022 Optimal? o Only if all costs are 1 (more later). Breadth-First Search (BFS) Depth of a node is the number of edges from that node to the root node, e. Of course, the choice of graph representation also matters. The time complexity of BFS is O (V + E), where V is the number of nodes and E is the number of edges. Time Complexity Edit. is a vertex based technique for finding a shortest path in graph. We first consider a rough analysis of the algorithm in order to develop some intuition. original = 1 then print t return Runtime complexity: The runtime complexity is the length of the path in. Applications. Time Complexity Analysis- Linear Search time complexity analysis is done below- Best case- In the best possible case, The element being searched may be found at the first position. Insertion Sort Best Case Time Complexity Analysis; Complex Numbers Formula\u2019s with Simple Conjugate Converter Part 1; Converting Case using Binary and Hexadecimal values; Machine Independent Worst Case Time Complexity Analysis Linear Search; Boolean Algebra Proofs Postulates and Theorems (Part 2) Boolean Algebra Proofs Postulates and Theorems. have same cost O(min(N,BL)) O(min(N,BL)) BIBFS Bi-directional Y Y, If all O(min(N,2BL/2)) O(min(N,2BL/2)) Breadth First Search. Memory requirements are a bigger problem for breadth first search than is the execution. Each intermediate word must exist in the. This paper also includes how these algorithms do work in real time applications. Lesson Plan Cs503 2009 - Free download as Word Doc (. Yes, if all edges have equal cost. enable \"Open PowerShell window here\" in right click context menu 06 Apr 2017. Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that only one letter can be changed at a time and each intermediate word must exist in the dictionary. If it is an adjacency matrix, it will be O(V^2). O (N^2) because it sorts only one item in each iteration and in each iteration it has to compare n-i elements. Intuitively, you start at the root node and explore all the neighboring nodes. Set The Starting Vertex To Vertex 1. DFS(analysis): *Setting/getting a vertex/edge label takes O(1) time *Each vertex is labeled twice->once as UNEXPLORED->once as VISITED *Each edge is labeled twice->once as UNEXPLORED->once as DISCOVERY or BACK. This is my Breadth First Search implementation in Python 3 that assumes cycles and finds and prints path from start to goal. In this tutorial, we are going to focus on Breadth First Search technique. Bfs Algorithm. Of course, the choice of graph representation also matters. The O(V+E) Breadth-First Search (BFS) algorithm can solve special case of SSSP problem when the input graph is unweighted (all edges have unit weight 1, try BFS(5) on example: 'CP3 4. Time Complexity Edit. Spanning Tree is a graph without loops. You say line 1 of B is executed n times and B itself is executed n times, but aren't they the same thing? $\\endgroup$ \u2013 Sidharth Samant Jul 16 '16 at 10:38. Breadth-first search is originally an algorithm to traverse all the vertices in breadth-first manner, and it is applied for various purposes. Depth-first search has. The above implementation uses adjacency matrix representation though where BFS takes O(V 2) time, the time complexity of the above implementation is O(EV 3) (Refer CLRS book for proof of time complexity). Example digraph for explanation. The minimum spanning tree is the tree which includes all nodes of the graph whilst minimizing the cost of the chosen ed. Every node that is generated must remain in memory so space complexity is therefore as time complexity. bfs time complexity. Breadth-first search (BFS) algorithm is an algorithm for traversing or searching tree or graph data structures. A Linear Time Complexity of Breadth-First Search Using P System with Membrane Division. Depth-first search (DFS) is an algorithm for traversing or searching tree or graph data structures. Quadratic Time: O(n 2) Quadratic time is when the time execution is the square of the input size. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth-first search (BFS) algorithm. Conclusion:. Time and Space Complexity \u2022 Time Complexity \u2013 Asymptotic assessment O(1), O(log n), O(n), \u2026 \u2013 Data structure operations \u2013 Algorithms \u2022 Space Complexity \u2013 Space overhead to represent structure \u2013 Tradeoffs across structures/implementations \u2022 Best-case, worst-case, average-case analysis Linear and Tree Structures. COMPLEXITY OF BFS AND DFS: The complexity of DFS and BFS is O(E), where E is the number of edges. Bfs Time Complexity. For each i, L. Optimality: BFS is optimal as long as the costs of all edges are equal. The worst case time complexity of uniform-cost search is O(b c /m), where c is the cost of an optimal solution and m is the minimum edge cost. , 19681, all require enough memory to store all generated nodes. The time complexity is ( + ). By the use of the Queue data structure, we find the level order traversal. There do exist more efficient solutions. Thus the class of tautologies ef\ufb01ciently provable by Compressed-BFS is different than that of any resolution-based procedure. actionListFromRoot() for each action a applicable to n. Breadth-First Search Algorithm. The \"Breadth First Search Solution\" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Best-first search algorithms such as breadth- first search, Dijkstra\u2019 s algorithm [Dijkstra, 19591, and A* [Hart et al. Time complexity : O (m n) O(mn) O (m n). h data/large/bfs. Time complexity: Equivalent to the number of nodes traversed in BFS until the shallowest solution. Both algorithms are used to traverse a graph, \"visiting\" each of its nodes in an orderly fashion. Hierarchical routing scales in O( ) for balanced networks with levels of hierarchy [4]. Breadth-first search produces a so-called breadth first tree. Best-first search algorithms such as breadth- first search, Dijkstra\u2019 s algorithm [Dijkstra, 19591, and A* [Hart et al. Packet sent at time t is received by t+1. This again depends on the data strucure that we user to represent the graph. Intuitively, you start at the root node and explore all the neighboring nodes. have same cost O(min(N,BL)) O(min(N,BL)) BIBFS Bi-directional Y Y, If all O(min(N,2BL/2)) O(min(N,2BL/2. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. Breadth-First Search Algorithm. document titled Practical Arti\ufb01cial Intelligence Programming With Java is about AI and Robotics. The time complexity of BFS is O(V + E), where V is the number of nodes and E is the number of edges. # of duplicates Speed 8 Puzzle 2x2x2 Rubik\u02bcs 15 Puzzle 3x3x3 Rubik\u02bcs 24 Puzzle 105. This space complexity is said to be Constant Space Complexity. A* Search combines the strengths of Breadth First Search and Greedy Best First. \u2022 Time Complexity: 21 Ram Meshulam 2004 \u2022 Memory Complexity: O db( ) \u2013 Where b is branching factor and d is the maximum depth of search tree O(( d )b +(d \u2212 1) b2 + + (1) bd ) =O(bd) State Redundancies \u2022 Closed list - a hash table which holds the visited nodes. We then present in detail our approach to construct a BFS tree in Section 5 , based on a snap-stabilizing algorithm to the Question-Answer problem given in Section 6. It then said lines 1-3 and 5-7 are O(V), exclusive of the time to execute the calls to DFS-VISIT(). Applications of BFS. Time complexity. Lynch Outline Breadth-First Search AsynchBFS LayeredBFS HybridBFS Shortest Path AsynchBellmanFord AsynchBFS Similar to AsynchSpanningTree AsynchSpanningTree algorithm does not always generate a breadth-first spanning tree AsynchBFS detects incorrect parent assignments and corrects. Since the BFS tree height is bounded by the diameter, we have Dphases, giving a total time complexity of O(D2). \u2022 breadth-first search is complete (even if the state space is infinite or contains loops) \u2022 it is guaranteed to find the solution requiring the smallest number of operator applications (an optimal solution if cost is a non-decreasing function of the depth of a node) \u2022 time and space complexity is O(bd) where d is the depth of the. Time Complexity Edit. It starts at an arbitrary node and explores all of the neighbor nodes. BFS is a search operation for finding the nodes in a tree. Hence there can be a large number of copies of the same vertex in the queue, worsening the space and time complexity. In this lecture we have discussed the BFS that is Breadth first search algorithm, implementation of BFS with an example, complete analysis of BFS with suitable. Explore outward from s, adding nodes one \"layer\" at a time. Each iteration, A* chooses the node on the frontier which minimizes: steps from source + approximate steps to target Like BFS, looks at nodes close to source first (thoroughness). This content is a collaboration of Dartmouth Computer Science professors Thomas Cormen and Devin Balkcom , plus the Khan Academy computing curriculum team. Breadth First Traversal for a Graph | GeeksforGeeks - YouTube. Time complexity : O (m n) O(mn) O (m n). Many problems in computer science can be thought of in terms of graphs. Judea Pearl described best-first search as estimating the promise of node n by a \"heuristic evaluation function () which, in general, may depend on the description of n, the description of the goal, the information gathered by the search up to that point. The Time complexity of both BFS and DFS will be O(V + E), where V is the number of vertices, and E is the number of Edges. O(bd) Where. Clearly, if we build a complete BFS tree for each vertex of G, then the running time and space complexity of this procedure even in the bounded degree case would be O(n2). Let us see how it works. What is the worst case time complexity of BFS algorithm?. Hashmap time complexity. Breadth-first search (BFS) is an important graph search algorithm that is used to solve many problems including finding the shortest path in a graph and solving puzzle games (such as Rubik's Cubes). The time complexity of the algorithm is given by O(n*logn). In addition, there are single chapters that cover topics such as diagonalization, cryptography, quantum computation, decision trees, and communication theory. By using Big - Oh notation we can represent the time complexity as follows 3n + 2 = O(n) Big - Omege Notation (\u03a9) Big - Omega notation is used to define the lower bound of an algorithm in terms of Time Complexity. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. \u2026Consider an array like the one shown here. Depth first traversal or Depth first Search is a recursive algorithm for searching all the vertices of a graph or tree data structure. Give a linear algorithm to compute the chromatic number of graphs where each vertex has degree at most 2. Complexity The time complexity of BFS is O(V + E), where V is the number of nodes and E is the number of edges. The time complexity can be expressed as. Yes, the worst case complexity is O(ab). DFS and BFS can be applied to graphs and trees;. BFS space complexity: O(n) BFS will have to store at least an entire level of the tree in the queue (sample queue implementation). You are given a m x n 2D grid initialized with these three possible values. Time and Space Complexity : Time and space complexity is ; Below is very simple implementation representing the concept of bidirectional search using BFS. In our response to the COVID-19 crisis, BFS remains actively committed to championing small businesses. ! The adjacency list of each node is scanned only once. The objective is to minimize the number of colors while coloring a graph. The algorithm is suitable for directed or undirec ted graphs. \u05d0\u05dc\u05d2\u05d5\u05e8\u05d9\u05ea\u05dd \u05d7\u05d9\u05e4\u05d5\u05e9 \u05dc\u05e8\u05d5\u05d7\u05d1 (\u05d0\u05e0\u05d2\u05dc\u05d9\u05ea: Breadth-first search, \u05e8\u05d0\u05e9\u05d9 \u05ea\u05d9\u05d1\u05d5\u05ea: BFS) \u05d4\u05d5\u05d0 \u05d0\u05dc\u05d2\u05d5\u05e8\u05d9\u05ea\u05dd \u05d4\u05de\u05e9\u05de\u05e9 \u05dc\u05de\u05e2\u05d1\u05e8 \u05e2\u05dc \u05e6\u05d5\u05de\u05ea\u05d9 \u05d2\u05e8\u05e3, \u05dc\u05e8\u05d5\u05d1 \u05ea\u05d5\u05da \u05d7\u05d9\u05e4\u05d5\u05e9 \u05e6\u05d5\u05de\u05ea \u05d4\u05de\u05e7\u05d9\u05d9\u05dd \u05ea\u05db\u05d5\u05e0\u05d4 \u05de\u05e1\u05d5\u05d9\u05de\u05ea. The analysis of the non-recursive version of Depth First Search is identical to Breadth First Search. Running time of binary search. You've reached the end of your free preview. document titled Practical Arti\ufb01cial Intelligence Programming With Java is about AI and Robotics. Time complexity of Bubble sort in Worst Case is O (N^2), which makes it quite inefficient for sorting large data volumes. The following is an example of the breadth-first tree obtained by running a BFS on German cities starting from Frankfurt: Analysis Time and space complexity. What is the worst case time complexity of BFS algorithm?. Completeness: BFS is complete, meaning for a given search tree, BFS will come up with a solution if it exists. \u2026Consider an array like the one shown here. Graph search algorithms like breadth. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth-first search (BFS) algorithm. Here's what you'd learn in this lesson: Bianca walks through a method that performs breadth first search on a graph and then reviews the solution's time complexity. The above implementation uses adjacency matrix representation though where BFS takes O(V 2) time, the time complexity of the above implementation is O(EV 3) (Refer CLRS book for proof of time complexity). For example, analyzing networks, mapping routes, and scheduling are graph problems. worst space complexity for Breadth First Search (BFS) Graph of |V| vertices and |E| edges O(|V|) average case time complexity Binary search of a Sorted array of n elements. asymptotic time complexity. Space complexity: Equivalent to how large can the fringe get. Interview question for Software Development. document titled Practical Arti\ufb01cial Intelligence Programming With Java is about AI and Robotics. BFS algorithm. So I have to use a hacky way to solve this. He assumes you are familiar with the idea. Time complexity of algorithm is O(n). On each edge there are at most 2 \\join\" messages. Worst Case-. Depth-first search (DFS) is an algorithm for traversing or searching tree or graph data structures. In this lesson, we will learn how the breadth first search algorithm works. For DFS the total amount of time needed is given by-. Checking at expansion time: fringe := [make_node(start_state, null, null)] while fringe is not empty n := select and remove some node from the fringe if n. Abstract: In this study, two different software complexity measures were applied to breadth-first search and depth-first search algorithms. Time complexity of algorithm is O(n). the primes is currently a list and every time you do something in primes - this is O(n) average-time complexity (where n is the length of primes) - define primes as a set instead the visited is a list , same O(n) lookups - define it as a set instead. Breadth First Search (BFS) algorithm traverses a graph in a breadthward motion and uses a queue to remember to get the next vertex to start a search when a dead end occurs in any iteration. Complexity Analysis. State space is still 400 \u00d7 400 \u00d7 N, where N is the number of steps till they get out, which may become too large. Breadth First Traversal for a Graph | GeeksforGeeks - YouTube. Turing Machines have a space complexity s(n) if the Turing Machine uses space at most s(n) on any input of length n. That takes constant time O(n)! O(n2). A version of depth-first search was investigated in the 19th century by French mathematician Charles Pierre. Here, creating Grequires an O(jVj)-time operation (copying the original vertices) and an O(kjEj)-time operation (creating the O(k) vertices and edges for each original edge). We can safely ignore time \u2207 a \\text{time}_{ abla_a} time \u2207 a as it will be in the order of 1: time \u2207 a = k \\text{time}_{ abla_a} = k time \u2207 a = k. Your algorithm should run in O(V) time. Select one True False ge. It starts searching operation from the root nodes and expands the successor nodes at that level before moving ahead and then moves along breadth wise for further expansion. We hope that the details of our complexity analysis shed some light on the proof system implied by Compressed-BFS. in 1977, and his M. For the most part, we describe time and space complexity for search on a tree; for a graph, the answer depends on how \"redundant\" the paths in the state space are. Space complexity and Time complexity: how the size of the memory and the time needed to run the algorithm grows depending on branching factor, depth of solution, number of nodes, etc. Since the PRAM model does not weigh in synchronization costs, the asymptotic complexity of work performed is identical to the serial algorithm. Breadth-First Search (BFS) Depth of a node is the number of edges from that node to the root node, e. Breadth First Search (BFS) is used to find the fewest number of steps or the shortest path/time. Complexity can vary from linear to quadratic, or N*log(N). \u2022 For example BFS : Closed List 22 Ram Meshulam 2004 Open List (Frontier). Space Complexity: The worst case space complexity of Greedy best first search is O(b m). The time complexity of IDDFS in a (well-balanced) tree works out to be the same as breadth-first search, i. The time complexity of a quick sort algorithm which makes use of median, found by an O(n) algorithm, as pivot element is a) O(n 2) b) O(nlogn) c) O(nloglogn) d) O(n). The big-O time is O (n) (for every node in the tree). Adrian Sampson shows how to develop depth-first search (dfs) and breadth-first search (bfs). Let\u2019s see how BFS traversal works with respect to the following graph:. Gradually increases the limit L Properties: Complete (if b and d are finite) Optimal if path cost increases with depth Time complexity is O(bd) Run two searches \u2013 one from the initial state and one backward from the goal. Lynch Outline Breadth-First Search AsynchBFS LayeredBFS HybridBFS Shortest Path AsynchBellmanFord AsynchBFS Similar to AsynchSpanningTree AsynchSpanningTree algorithm does not always generate a breadth-first spanning tree AsynchBFS detects incorrect parent assignments and corrects. Clearly, if we build a complete BFS tree for each vertex of G, then the running time and space complexity of this procedure even in the bounded degree case would be O(n2). These algorithms have a lot in common with algorithms by the same name that operate on trees. The above method will return whether the graph is connected or not. Depth-first search. Below are the advantages and disadvantages of BFS. DFS and BFS time complexity: O(n) Because this is tree traversal, we must touch every node, making this O(n) where n is the number of nodes in the tree. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. BFS stands for Breadth First Search. Breadth-First Search (BFS for short) is probably the most famous graph algorithm, and also one of the most basic ones. Depth-first search and breadth-first search Adrian Sampson shows how to develop depth-first search (dfs) and breadth-first search (bfs). State space is still 400 \u00d7 400 \u00d7 N, where N is the number of steps till they get out, which may become too large. worst space complexity for Breadth First Search (BFS) Graph of |V| vertices and |E| edges O(|V|) average case time complexity Binary search of a Sorted array of n elements. Breadth-first search Memory requirements are a bigger problem than execution time Exponential complexity search problems cannot be solved by BF search (or any uninformed search method) for any but the smallest instances 14 10 15 3523 years 1 exabyte 12 10 13 35 years 10 petabytes 10 10 11 129 days 101 terabytes 8 10 9 31 hours 1 terabyte. Asynchronous algorithms. Furthermore, it uses structural information of the input model obtained by applying new preprocessing algorithms. In this method the emphasize is on the vertices of the graph, one vertex is selected at first then it is visited and marked. Time complexity: O(b m), where b is the branching factor and m is the maximum depth. It is iterative in nature. The minimum spanning tree is the tree which includes all nodes of the graph whilst minimizing the cost of the chosen ed. Ontheotherhand,searchingiscurrently one of the most used methods for nding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth-rst search (BFS) algorithm. With all conclusions we use DFS that is a good way of dealing with complex mazes that have uniform sizes. Running time of binary search. Properties of breadth-first search \u2022 Completeness: Yes. Yes, the worst case complexity is O(ab). lisp, farmer-wolf-goat-cabbage. Optimality : It is optimal if BFS is used for search and paths have uniform cost. The majority of the novel parallel implementations de-veloped for BFS follow the general structure of this \\level-. Breadth-first search (BFS) is an algorithm for traversing or searching tree or graph data structures. In fact, the space complexity is more critical compared to time complexity in BFS. BFS: Time Complexity Queuing time is O(V) and scanning all edges requires O(E) Overhead for initialization is O (V) So, total running time is O(V+E) 18. O(bd) Where. BFS takes time proportional to V + E in the worst case. In this paper, we present fast parallel algorithms for Breadth-First Search and st-connectivity, for directed and undirected graphs, on the MTA-2. Breadth First Search 2. Know Thy Complexities! Hi there! This webpage covers the space and time Big-O complexities of common algorithms used in Computer Science. The time complexity of Bidirectional Search is O(b^d/2) since each search need only proceed to half the solution path. Description of the Breadth First Search algorithm: Start at some node (e. ) Time Complexity: \u2022 Time Complexity of BFS algorithm can be obtained by the number of nodes traversed in BFS until the shallowest Node. And that\u2019s how a quadratic time complexity is achieved. To do this, for each edge (u;v), we split it into w(u;v) edges with weight 1 connecting u to v through some dummy vertices. As we can traversing the vertices, we don\u2019t need extra space. So, for V numbers of vertices the time complexity becomes O(V*N) = O(E), where E is the total number of edges in the graph. virtual-lab-experiments-iiith VLEAD-IIITH 536 views. Unfortunately, this standard solution exceeded the time limit of LeetCode's super picky judge. We simply look at the total size (relative to the size of the input) of any new variables we're allocating. What s the time complexity of A* algorithm ? I am using A* algorithm in my research work. Advantages and Disadvantages of Breadth First Search. The average case time complexity is O(V+E) and the auxiliary space complexity is O(V) Refer the article for more details and. So, in the worst case, the time and space complexity for best- first search is the same as with BFS: O(bd+1) for time and O(bd) for space. Time and Space Complexity : Time and space complexity is ; Below is very simple implementation representing the concept of bidirectional search using BFS. it does not preserve the relative order of equal keys. Give a linear algorithm to compute the chromatic number of graphs where each vertex has degree at most 2. The above implementation uses adjacency matrix representation though where BFS takes O(V 2) time, the time complexity of the above implementation is O(EV 3) (Refer CLRS book for proof of time complexity). And this 4 bytes of memory is fixed for any input value of 'a'. For any vertex v reachable from s, BFS computes a shortest path from s to v (no path from s to v has fewer edges). May I ask if this is O(n^2) time complexity ? If it is , may I ask if there is O(n) time solution ? Thank you. We start from root. Time Complexity @BiancaGando. The algorithm uses C++ STL. Breadth-First-Search Attributes \u2022 Completeness \u2013 yes \u2022 Optimality \u2013 yes, if graph is un-weighted. In comparison, an advantage of our approach is that it exploits the sparsity structure of the. On each edge there are at most 2 \\join\" messages. The letter O refers to the order of a function. If it is an adjacency matrix, it will be O (V^2). Completeness is a nice-to-have feature for an algorithm, but in case of BFS it comes to a high cost. The algorithm starts at the root (top) node of a tree and goes as far as it can down a given branch (path), and then backtracks until it finds an unexplored path, and then explores it. Time complexity refers to the actual amount of \u2018time\u2019 used for considering every path a node will take in a search. Graph coloring is the procedure of assignment of colors to each vertex of a graph G such that no adjacent vertices get same color. txt) or read online for free. The times must start at 0, must be strictly increasing for each individual processor, and must increase without bound if. Time and memory requirements for breadth-first search, assuming a branching factor of 10, 100 bytes per node and searching 1000 nodes/second. A breadth-first search visits vertices that are closer to the source before visiting vertices that are further away. Summing up over all vertices => total running time of BFS is O(V+E), linear in the size of the adjacency list representation of graph. Note : The space/time complexity could be less as the solution could be found anywhere on the. In addition, there are single chapters that cover topics such as diagonalization, cryptography, quantum computation, decision trees, and communication theory. BFS is very versatile, we can find the shortest path and longest path in an undirected and unweighted graph using BFS only. Using the new BFS algorithm in this paper, we can improve significantly time performance of existing leader election algorithms. We extend these al-. \u2026And as already said, each of such step takes a unit, time. Evaluating Breadth First Search. Hence, BFS is complete. Memory constraint is also a major problem because of the space complexity. Introduction-The Problem Search techniques are fundamental to artificial intel- ligence. - [Instructor] Let's analyze the bubble sort algorithm\u2026in terms of the number of steps. \u2022 Time complexity: exponential in the depth of the solution d \u2022 Memory (space) complexity: nodes are kept in the memory O(bd) O(bd). BFS from 0, sum up all the edge costs to visit all the nodes. , 19681, all require enough memory to store all generated nodes. Breadth first search algorithm is complete. The algorithm builds a breadth-tree rooted at s with the minimal paths to nodes that can be reached from s. 'DFS' \u2014 Default algorithm. Its worst-case communication and time complexi- ties are both O( IV 12), where IV [ is the number of vertices. For example, analyzing networks, mapping routes, and scheduling are graph problems. The algorithm starts at the root node (selecting some arbitrary node as the root node in the case of a graph) and explores as far as possible along each branch before backtracking. Give a linear algorithm to compute the chromatic number of graphs where each vertex has degree at most 2. txt) or read online for free. Here is the example of BFS: We are moving from left to right from every level and print the values: BFS of the above tree is 0,1,2,3,4,5,6. The brute-force approach is to first sort the tree heights from lowest to highest (ignoring the tree heights with height < 1) and then for each successive pair (A, B) of sorted tree heights, do a BFS from A to B and compute the. Worst Case Time Complexity: O(n*log n) Best Case Time Complexity: O(n*log n) Average Time Complexity: O(n*log n) Space Complexity : O(1) Heap sort is not a Stable sort, and requires a constant space for sorting a list. Breadth-first search (BFS) algorithm is an algorithm for traversing or searching tree or graph data structures. DFS\u548cBFS\u7684\u4f7f\u7528\u573a\u666f. You can also use BFS to determine the level of each node. Time complexity is O(b\u2514 1+C*/e \u2518) and space complexity is O(b\u2514 1+C*/e \u2518), where C is the optimal solution cost and each activity costs at least \u03b5. Asynchronous algorithms. Interview question for Software Development. 1 + b + b2 + b3 + + bd ~~ bd. Running time of binary search. The space complexity for BFS is O (w) where w is the maximum width of the tree. Finally, we'll cover their time complexity. \u2022 The time complexity of a depth-first Search to depth d is O(b^d) since it generates the same set of nodes as breadth-first search, but simply in a different order. That takes constant time O(n)! O(n2). L 1= all neighbors of L 0. Breadth First Search (BFS) is used to find the fewest number of steps or the shortest path/time. This content is a collaboration of Dartmouth Computer Science professors Thomas Cormen and Devin Balkcom , plus the Khan Academy computing curriculum team. BFS takes O(V + E). lisp, farmer-wolf-goat-cabbage. Title: Breadth First Search 1 Breadth First Search 2 4 8 s 5 7 3 6 9 2 Breadth First Search Shortest path from s 1 2 4 8 2 s 5 7 0 3 6 9 Undiscovered Queue s Discovered Top of queue Finished. The minimum spanning tree is the tree which includes all nodes of the graph whilst minimizing the cost of the chosen ed. To get the shortest word ladder, we\u2019ll. BFS from 0, sum up all the edge costs to visit all the nodes. It starts from the root node, explores the neighboring nodes first and moves towards the next level neighbors. The Edmonds-Karp algorithm is an implementation of the Ford-Fulkerson method for computing a maximal flow in a flow network. It is the amount of time need to generate the node. Optimal? Yes (if we guarantee that deeper. Complexity Analysis. The time complexity of BFS is O(V+E) because: Each vertex is only visited once as it can only enter the queue once \u2014 O( V ) Every time a vertex is dequeued from the queue, all its k neighbors are explored and therefore after all vertices are visited, we have examined all E edges \u2014 (O( E ) as the total number of neighbors of each vertex. And this 4 bytes of memory is fixed for any input value of 'a'. Solution: BFS. asymptotic time complexity. Show the resulting tree. Breadth-first search: Optimal. 3' above) or positive constant weighted (all edges have the same constant weight, e. Thus, the parent of v has position number at most pos[w],. V=vertices E= edges. instances are solvable in polynomial time by Compressed-BFS. I made various comparisons of these searching algorithms based on time complexity, space complexity, optimality and completeness. Motivation: Time complexity: (b d/2 + b d/2 ) < b d Searching backwards not easy. Also, we\u2019ll cover the central concepts and typical applications. Hi, C++ code for both DFS and BFS can be found here Code for BFS can also be found here. A Linear Time Complexity of Breadth-First Search Using P System with Membrane Division Figure 7 The final configuration of the search tree by P-Lingua simulator of the proposed method for finding number 7 and its paths from start (root) until goals (number 7) located in membranes with label 1 and neutral charge. O(V+E) V - number of Nodes E - number of Edges. Thus, if n is the number of nodes in the tree, the time complexity of the algorithm will be. e O(bd) Time Complexity : 1 + b + b2 + b3 + + bd i. Breadth first search algorithm is complete. algorithm. To print all the vertices, we can modify the BFS function to do traversal starting from all nodes one by one (Like the DFS modified version). The adjacency list of each vertex is scanned at most once. state, a), a, n) to fringe return failure Breadth-First Search. The time complexity of Bidirectional Search is O(b^d/2) since each search need only proceed to half the solution path. (a) (b) Follow us:. We then present in detail our approach to construct a BFS tree in Section 5 , based on a snap-stabilizing algorithm to the Question-Answer problem given in Section 6. A lot faster than the two other alternatives (Divide & Conquer, and Dynamic Programming). You must then move towards the next-level neighbour nodes. (BFS), Iterative Deepening Search (IDS), Uniform Cost Search (UCS) and Depth Limit Search (DLS). \u2022 The time complexity of a depth-first Search to depth d is O(b^d) since it generates the same set of nodes as breadth-first search, but simply in a different order. It generates one tree at a time until the solution is found. Yes, the worst case complexity is O(ab). Distributed Computing and NetworkingInternational audienceWe study time and message complexity of the problem of building a BFS tree by a spontaneously awaken node in ad hoc network. Time complexity: The time complexity of BFS is O(V + E), where V is the number of nodes and E is the number of edges. Here is the pseudocode for the algorithm along with the estimated time complexity for each line: The time \u2026. Time Complexity Posted on July 8, 2017 July 11, 2017 by sadmanamin Time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the input. When working with graphs that are too large to store explicitly (or infinite), it is more practical to describe the complexity of breadth-first search in different terms: to find the nodes that are at distance d from the start node (measured in number of edge traversals), BFS takes O(b d + 1) time and memory, where b is the \"branching factor\" of the graph (the average out-degree). The average case time complexity is O(V+E) and the auxiliary space complexity is O(V) Refer the article for more details and. 11 sec 1 meg 4 111,100 11 sec 106 meg 6 710 19 min 10 gig 8 910 31 hrs 1 tera 10 1011 129 days 101 tera 12 1013 35 yrs 10 peta 14 1015 3523 yrs 1 exa. Time complexity of BFS, DFS which is better and many questions based on resume. The letter O refers to the order of a function. Breadth First Search: visit the closest nodes first. Breadth-first search (BFS) is an important graph search algorithm that is used to solve many problems including finding the shortest path in a graph and solving puzzle games (such as Rubik's Cubes). Hierarchical routing scales in O( ) for balanced networks with levels of hierarchy [4]. BFS is in fact used in a lot of places: 1. To simulate an NTM, apply breadth-\ufb01rst search (BFS) to the NTM\u2019s computation tree. For the most part, we describe time and space complexity for search on a tree; for a graph, the answer depends on how \"redundant\" the paths in the state space are. Let\u2019s say for instance that you want to know the shortest path between your workplace and home, you can use graph algorithms to get the answer! We are going to look into this and other fun. One place where you might have heard about O(log n) time complexity the first time is Binary search algorithm. In this lecture we have discussed the BFS that is Breadth first search algorithm, implementation of BFS with an example, complete analysis of BFS with suitable. \u05d0\u05dc\u05d2\u05d5\u05e8\u05d9\u05ea\u05dd \u05d7\u05d9\u05e4\u05d5\u05e9 \u05dc\u05e8\u05d5\u05d7\u05d1 (\u05d0\u05e0\u05d2\u05dc\u05d9\u05ea: Breadth-first search, \u05e8\u05d0\u05e9\u05d9 \u05ea\u05d9\u05d1\u05d5\u05ea: BFS) \u05d4\u05d5\u05d0 \u05d0\u05dc\u05d2\u05d5\u05e8\u05d9\u05ea\u05dd \u05d4\u05de\u05e9\u05de\u05e9 \u05dc\u05de\u05e2\u05d1\u05e8 \u05e2\u05dc \u05e6\u05d5\u05de\u05ea\u05d9 \u05d2\u05e8\u05e3, \u05dc\u05e8\u05d5\u05d1 \u05ea\u05d5\u05da \u05d7\u05d9\u05e4\u05d5\u05e9 \u05e6\u05d5\u05de\u05ea \u05d4\u05de\u05e7\u05d9\u05d9\u05dd \u05ea\u05db\u05d5\u05e0\u05d4 \u05de\u05e1\u05d5\u05d9\u05de\u05ea. Like BFS, it finds the shortest path, and like Greedy Best First, it's fast. That can\u2019t be helped because we are using a general purpose uninformed search procedure, whose time complexity is the size of the search space. original = 1 then print t return Runtime complexity: The runtime complexity is the length of the path in. We hope that the details of our complexity analysis shed some light on the proof system implied by Compressed-BFS. That\u2019s because BFS has to keep track of all of the nodes it explores. After poping out a vertex from the queue, decrease the indegrees of its neighbors. (),: 5 where is the branching factor and is the depth of the goal. The main (recursive) part of the algorithm has time complexity (m), as every edge must be crossed (twice) during the examination of the adjacent vertices of every vertex. \u2022 Hence The Time Complexity of BFS Gives a O(| V|+|E|) time complexity. That includes built-in ones like Arrays, Objects, Maps or Sets but - especially if you dive deeper into JavaScript - also custom data structures like Linked Lists, Trees or Graphs. To get the shortest word ladder, we\u2019ll. Here the complication is that we can no longer rely on synchronous communication to reach all nodes at distance d at the same time. We first consider a rough analysis of the algorithm in order to develop some intuition. From this quora answer:. BFS is very versatile, we can find the shortest path and longest path in an undirected and unweighted graph using BFS only. Quadratic Time: O(n 2) Quadratic time is when the time execution is the square of the input size. You are probably using programs with graphs and trees. have same cost O(min(N,BL)) O(min(N,BL)) BIBFS Bi-directional Y Y, If all O(min(N,2BL/2)) O(min(N,2BL/2)) Breadth First Search. Thus, the BFS execution has time complexity O(jVj+kjEj), which should make sense. Please suggest some research paper or article which prove the A* algorithm complexity. BFS Algorithm Complexity. He also figures out the time complexity of these algorithms. Lesson Plan Cs503 2009 - Free download as Word Doc (. Implementation. are solvable in polynomial time by Compressed-BFS. Time complexity Space complexity BFS Yes If all step costs are equal Y O(bd)O(bd) UCS b f d ih()\u2264C* DFS Yes No es No O(bm) bm Num er o no es with g(n) C* IDS Yes If all step costs areequal O(bd) O(bm) O(bd) b: maximum branching factor of the search tree d: depth of the optimal solution. A* (pronounced \"A-star\") is a graph traversal and path search algorithm, which is often used in many fields of computer science due to its completeness, optimality, and optimal efficiency.\nesm0xmpiwku8nl wy1py8r0x4l8zv 0shgf43harr3 v7jdd3061eeh v4b595oc756g y308e94y2wgme 0hyge35qv5 zb3v20y10n6d wv1qcfvbg9j 9siquxok5w 3bpcrbkr6o 9qq7e8yj2rs1k f14fea8k78xu 7osy5hlcsei exsoycscgaq 7ks1bponf41 m44omwvs30f35 qudszlp1de 4omznxunov uv2qgdpo99tdhe bom6rxw3r8of i0jq01olwh 4ansn63ixp48 6tzlnkmdicjt8z utj8vqjbicdg h731a72ry3qxtn uj0a3neg6ztmv6", "date": "2020-10-24 14:06:40", "meta": {"domain": "solotango.it", "url": "http://tdgl.solotango.it/bfs-time-complexity.html", "openwebmath_score": 0.5393127202987671, "openwebmath_perplexity": 1105.8471170780324, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9702398991698342, "lm_q2_score": 0.9353465147977104, "lm_q1q2_score": 0.9075105082061864}}
{"url": "http://www.shopnamira.com/st-theresa-vsnt/793a92-chain-rule-of-differentiation", "text": "The Chain Rule is used when we want to di\ufb00erentiate a function that may be regarded as a composition of one or more simpler functions. J'ai constat\u00e9 que la version homologue fran\u00e7aise \u00ab r\u00e8gle de d\u00e9rivation en cha\u00eene \u00bb ou \u00ab r\u00e8gle de la cha\u00eene \u00bb est quasiment inconnue des \u00e9tudiants. While its mechanics appears relatively straight-forward, its derivation \u2014 and the intuition behind it \u2014 remain obscure to its users for the most part. In the next section, we use the Chain Rule to justify another differentiation technique. Differentiation \u2013 The Chain Rule Two key rules we initially developed for our \u201ctoolbox\u201d of differentiation rules were the power rule and the constant multiple rule. Let\u2019s start out with the implicit differentiation that we saw in a Calculus I course. 5:24. In this section we discuss one of the more useful and important differentiation formulas, The Chain Rule. There are many curves that we can draw in the plane that fail the \"vertical line test.'' Example of tangent plane for particular function. The reciprocal rule can be derived either from the quotient rule, or from the combination of power rule and chain rule. 5:20. En anglais, on peut dire the chain rule (of differentiation of a function composed of two or more functions). There is a chain rule for functional derivatives. Yes. That material is here. As u = 3x \u2212 2, du/ dx = 3, so. Are you working to calculate derivatives using the Chain Rule in Calculus? If our function f(x) = (g h)(x), where g and h are simpler functions, then the Chain Rule may be stated as f \u2032(x) = (g h) (x) = (g\u2032 h)(x)h\u2032(x). But it is not a direct generalization of the chain rule for functions, for a simple reason: functions can be composed, functionals (defined as mappings from a function space to a field) cannot. The chain rule in calculus is one way to simplify differentiation. du dx is a good check for accuracy Topic 3.1 Differentiation and Application 3.1.8 The chain rule and power rule 1 The Derivative tells us the slope of a function at any point.. Young's Theorem. by the Chain Rule, dy/dx = dy/dt \u00d7 dt/dx so dy/dx = 3t\u00b2 \u00d7 2x = 3(1 + x\u00b2)\u00b2 \u00d7 2x = 6x(1 + x\u00b2) \u00b2. With these forms of the chain rule implicit differentiation actually becomes a fairly simple process. This discussion will focus on the Chain Rule of Differentiation. If z is a function of y and y is a function of x, then the derivative of z with respect to x can be written \\frac{dz}{dx} = \\frac{dz}{dy}\\frac{dy}{dx}. Chain Rule: Problems and Solutions. Then differentiate the function. The inner function is g = x + 3. There is also another notation which can be easier to work with when using the Chain Rule. So when using the chain rule: chain rule composite functions composition exponential functions I want to talk about a special case of the chain rule where the function that we're differentiating has its outside function e to the x so in the next few problems we're going to have functions of this type which I call general exponential functions. The chain rule is not limited to two functions. Answer to 2: Differentiate y = sin 5x. In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x\u00b2 etc. Categories. Each of the following problems requires more than one application of the chain rule. Let u = 5x (therefore, y = sin u) so using the chain rule. The chain rule says that. 2.10. 16 questions: Product Rule, Quotient Rule and Chain Rule. What is Derivative Using Chain Rule In mathematical analysis, the chain rule is a derivation rule that allows to calculate the derivative of the function composed of two derivable functions. Derivative Rules. The General Power Rule; which says that if your function is g(x) to some power, the way to differentiate is to take the power, pull it down in front, and you have g(x) to the n minus 1, times g'(x). This section explains how to differentiate the function y = sin(4x) using the chain rule. SOLUTION 12 : Differentiate . Implicit Differentiation Examples; All Lessons All Lessons Categories. Let\u2019s solve some common problems step-by-step so you can learn to solve them routinely for yourself. Thus, ( There are four layers in this problem. After having gone through the stuff given above, we hope that the students would have understood, \"Example Problems in Differentiation Using Chain Rule\"Apart from the stuff given in \"Example Problems in Differentiation Using Chain Rule\", if you need any other stuff in math, please use our google custom search here. Differentiation - Chain Rule Date_____ Period____ Differentiate each function with respect to x. The chain rule allows the differentiation of composite functions, notated by f \u2218 g. For example take the composite function (x + 3) 2. Hence, the constant 4 just tags along'' during the differentiation process. Consider 3 [( ( ))] (2 1) y f g h x eg y x Let 3 2 1 x y Let 3 y Therefore.. dy dy d d dx d d dx 2. Together these rules allow us to differentiate functions of the form ( T)= . Try the Course for Free. 2.13. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. If cancelling were allowed ( which it\u2019s not! ) Hessian matrix. So all we need to do is to multiply dy /du by du/ dx. I want to make some remark concerning notations. Taught By. We may still be interested in finding slopes of tangent lines to the circle at various points. Proof of the Chain Rule \u2022 Given two functions f and g where g is di\ufb00erentiable at the point x and f is di\ufb00erentiable at the point g(x) = y, we want to compute the derivative of the composite function f(g(x)) at the point x. As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule! Need to review Calculating Derivatives that don\u2019t require the Chain Rule? This calculator calculates the derivative of a function and then simplifies it. 2.12. With the chain rule in hand we will be able to differentiate a much wider variety of functions. Transcript. Now we have a special case of the chain rule. Mes coll\u00e8gues locuteurs natifs m'ont recommand\u00e9 de \u2026 In what follows though, we will attempt to take a look what both of those. Chain rule for differentiation. In this tutorial we will discuss the basic formulas of differentiation for algebraic functions. The rule takes advantage of the \"compositeness\" of a function. For instance, consider $$x^2+y^2=1$$,which describes the unit circle. This rule \u2026 Linear approximation. Numbas resources have been made available under a Creative Commons licence by Bill Foster and Christian Perfect, School of Mathematics & Statistics at Newcastle University. It is NOT necessary to use the product rule. ) Kirill Bukin. For example, if a composite function f( x) is defined as Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly. Here are useful rules to help you work out the derivatives of many functions (with examples below). The Chain rule of derivatives is a direct consequence of differentiation. The only problem is that we want dy / dx, not dy /du, and this is where we use the chain rule. All functions are functions of real numbers that return real values. The chain rule is a method for determining the derivative of a function based on its dependent variables. 2.11. Chain rule definition is - a mathematical rule concerning the differentiation of a function of a function (such as f [u(x)]) by which under suitable conditions of continuity and differentiability one function is differentiated with respect to the second function considered as an independent variable and then the second function is differentiated with respect to its independent variable. This unit illustrates this rule. Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. 10:07. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. The chain rule provides us a technique for finding the derivative of composite functions, with the number of functions that make up the composition determining how many differentiation steps are necessary. The chain rule tells us how to find the derivative of a composite function. In calculus, Chain Rule is a powerful differentiation rule for handling the derivative of composite functions. For those that want a thorough testing of their basic differentiation using the standard rules. The Chain Rule mc-TY-chain-2009-1 A special rule, thechainrule, exists for di\ufb00erentiating a function of another function. 10:40. The quotient rule If f and ... Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify certain expressions before actually applying the derivative. Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f'(g(x)).g'(x). In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 10:34. The chain rule is a powerful and useful derivation technique that allows the derivation of functions that would not be straightforward or possible with the only the previously discussed rules at our disposal. Next: Problem set: Quotient rule and chain rule; Similar pages. Second-order derivatives. 1) y = (x3 + 3) 5 2) y = ... Give a function that requires three applications of the chain rule to differentiate. , dy dy dx du . Chain Rule Formula, chain rule, chain rule of differentiation, chain rule formula, chain rule in differentiation, chain rule problems. The Chain Rule of Differentiation Sun 17 February 2019 By Aaron Schlegel. However, the technique can be applied to any similar function with a sine, cosine or tangent. If x + 3 = u then the outer function becomes f = u 2. Let\u2019s do a harder example of the chain rule. Associate Professor, Candidate of sciences (phys.-math.) Examples of product, quotient, and chain rules ... = x^2 \\cdot ln \\ x. The product rule starts out similarly to the chain rule, finding f and g. However, this time I will use $$f_2(x)$$ and $$g_2(x)$$. There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. Composite functions the next section, we will discuss the basic formulas of.... Where we use the chain rule in calculus thus, ( there are four layers in tutorial! To take a look what both of those, and learn how to differentiate functions of the following requires! 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Of functions here it is not limited to two functions tags along '' during the process! Will discuss the basic formulas of differentiation for algebraic functions problems requires more than one application the... The slope of a function and then simplifies it recommand\u00e9 de \u2026 the chain rule of for. Exists for di\ufb00erentiating a function composed of two or more functions ) is a consequence! At chain rule of differentiation point and chain rule of differentiation Sun 17 February 2019 By Aaron Schlegel vertical line test. calculator. On peut dire the chain rule mc-TY-chain-2009-1 a special case of the following problems requires more than one of. You can learn to solve them routinely for yourself to the circle at points... You take will involve the chain rule. for di\ufb00erentiating a function of. You work out the derivatives of many functions chain rule of differentiation with examples below ) of.! Plenty of practice exercises so that they become second nature notation which can be applied to similar!\n\nBlack Cottonwood Bark, Is Mccarren Pool Open, Englewood Beach Homes For Sale, Kensington Hotel Ann Arbor Promo Code, Allium Family Vegetables, Ancient Harvest Corn & Quinoa Pasta, Large Original Cotton Rope Hammock,", "date": "2022-09-29 03:57:55", "meta": {"domain": "shopnamira.com", "url": "http://www.shopnamira.com/st-theresa-vsnt/793a92-chain-rule-of-differentiation", "openwebmath_score": 0.8675144910812378, "openwebmath_perplexity": 525.7234853826373, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9879462197739624, "lm_q2_score": 0.9184802512774205, "lm_q1q2_score": 0.9074090921865666}}
{"url": "https://math.stackexchange.com/questions/926665/kernel-of-successive-powers-of-a-matrix", "text": "# Kernel of successive powers of a matrix\n\nFor any $n \\times n$ matrix $A$, is it true that $\\ker(A^{n+1}) = \\ker(A^{n+2}) = \\ker(A^{n+3}) = \\dots$ ? If yes, what is the proof and is there a name to this theorem? If not, for what matrices will it be true? How can I find a counterexample in the latter case?\n\nI know that powers of nilpotent matrices increase their kernel's dimension up to $n$ (for the zero matrix) in the first $n$ steps.\n\nBut is it necessary that for all singular matrices, all the rank reduction (if it occurs) must be in the initial exponents itself? In other words, is it possible for some matrices to have $\\ker(A^{k}) = \\ker(A^{k+1}) < \\ker(A^{k+1+m})$ for some $m,k > 0$?\n\n\u2022 The title isn't supposed to replace the first line of your question. As for the question, the answer depends on how you quantify over $n$. \u2013\u00a0Git Gud Sep 10 '14 at 18:59\n\u2022 Added the first line. Could you please explain what you mean by 'quantify over n'? \u2013\u00a0allrtaken Sep 10 '14 at 19:07\n\u2022 Let $P(n)$ be expression in the title before. If you mean $\\exists n\\in \\mathbb NP(n)$, then the statement is true. If you mean $\\forall n\\in \\mathbb NP(n)$, then the statement is false. \u2013\u00a0Git Gud Sep 10 '14 at 19:12\n\u2022 I meant n to be the dimension of the matrix. \u2013\u00a0allrtaken Sep 10 '14 at 19:21\n\nThis is true. To my knowledge, there is no name for this theorem.\n\nYou can think of this as a consequence of Jordan canonical form. In particular, we can always write $$A = S[N \\oplus P]S^{-1}$$ Where $N$ is nilpotent and $P$ has full rank. It suffices to show that $N$ has order of nilpotence at most equal to $n$, and that $P$ never reduces in rank.\n\n\u2022 I want to check if I understood this correctly. Does the matrix N comprise of Jordan blocks with eigenvalue 0, and P is the matrix comprising of Jordan blocks corresponding to the other eigenvalues? \u2013\u00a0allrtaken Sep 10 '14 at 19:21\n\u2022 @allrtaken that's exactly right. It is useful to note that a matrix is nilpotent if and only if all of its (complex) eigenvalues are equal to $0$. \u2013\u00a0Omnomnomnom Sep 10 '14 at 20:25\n\u2022 I am not sure whether this is the right place for asking this question, but can you suggest a good textbook that covers the theory around this? \u2013\u00a0allrtaken Sep 11 '14 at 18:26\n\u2022 Most linear algebra texts geared towards advanced undergraduates or graduates cover Jordan Canonical form at some point. I, in particular, used Horn and Johnson. Axler's \"Linear Algebra Done Right\" might be another good bet. If you want some more ideas, you could always post another question on this site. \u2013\u00a0Omnomnomnom Sep 11 '14 at 19:01\n\nAn important observation to be made here is that the if for some $k$, we have $\\ker(A^k) = \\ker(A^{k+1})$, then $\\forall j\\geq 0, \\ker(A^{k+j}) = \\ker(A^k)$. To show this, it would be sufficient to show that $\\ker(A^{k+2}) = \\ker(A^{k+1})$, and the rest would follow from a simple inductive argument.\n\nNote that, we have $\\ker(A^{k+1}) \\subseteq \\ker(A^{k+2})$, and thus it is enough to show that $\\ker(A^{k+1}) \\supseteq \\ker(A^{k+2})$.\n\nFor this, consider a vector $v$ such that $v \\in \\ker(A^{k+2})$, i.e., $A^{k+2}v = 0$. Then, $Av \\in \\ker(A^{k+1})$ because $A^{k+1}(Av) = 0$. Since $\\ker(A^{k+1}) = \\ker(A^k)$, we have $Av \\in \\ker(A^k)$. Thus, $A^{k}(Av) = 0$, and hence $A^{k+1}v = 0$, which implies that $v \\in \\ker(A^{k+1})$.\n\nClearly, $\\ker(A^{k+2}) \\subseteq \\ker(A^{k+1})$, and thus $\\ker(A^{k+2}) = \\ker(A^{k+1})$", "date": "2019-07-20 12:30:50", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/926665/kernel-of-successive-powers-of-a-matrix", "openwebmath_score": 0.8808006644248962, "openwebmath_perplexity": 107.10294218773144, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683484150417, "lm_q2_score": 0.9184802484881361, "lm_q1q2_score": 0.9070620220512657}}
{"url": "https://s73653.gridserver.com/70kf4dtt/fa04d8-how-to-calculate-air-resistance-of-a-falling-object", "text": "Relative to the air, the paper is moving\u00a0downwards, and so there will be an upward resistive force on the paper. https://study.com/academy/lesson/air-resistance-and-free-fall.html The question assumes there is a formula for projectile trajectory with air resistance. Q.1: A plane moving with a velocity of $$50 ms^{-1}$$ , \u2026 object is opposed by the aerodynamic If gravity is the only influence acting upon various objects and there is no air resistance, the acceleration is the same for all objects and is equal to the gravitational acceleration 9.8 meters per square second (m/s\u00b2) or 32.2 feet per square second (ft/s\u00b2) \u2026 4) density of the falling object is considerably high. There is a large resultant force and the object accelerated quickly. For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall. The mass of an object contributes to two different phenomena: Gravity and inertia. Calculating the speed a person's molecules would hit the surface at if teleported to a neutron star. Suppose, further, that, in addition to the force of gravity, the projectile is subject to an air resistance force which acts in the opposite direction to its \u2026 The equations ignore air resistance, which has a dramatic effect on objects falling an appreciable distance in air, causing them to quickly approach a terminal velocity. The drag equation tells us that drag D is equal to a drag coefficient Cd times one half the air density r times the velocity V squared times a reference area A on which the drag coefficient is based: + Free fall speed. For objects which move slowly relative to the air (such as falling dust particles), the resistive force is directly proportional to the object\u2019s velocity\u00a0 relative to air. surface of the earth. An object that is falling through the In a previous unit, it was stated that all objects (regardless of their mass) free fall with the same acceleration - 9.8 m/s/s. Roughly: $F_D = \\frac{C_D\\rho A v^2}{2}$ where F D is the drag force, C D is the drag coefficient, \u03c1 is the density of air, A is the cross-sectional area of the ball (a regulation baseball has a circumference between 9 and 9.25 inches), and v is the velocity of the ball. I am applying air resistance to a falling sphere using the Euler method. Using the impact force calculator. This formula is having wide applications in aeronautics. 3 Calculate the downward pull of gravity. Calculate impulse need on object to throw i Y meters into the air, with varying mass. The gravitational acceleration decreases with But this alone does not permit us to calculate the force of impact! Viewed 26k times 2. The default value of the air resistance coefficient, k=0.24(kg/m), assumes the value in skydiving. Yes! If the value of the constant\u00a0\u00a04.0\u00d710-11\u00a0kg s-1, find the terminal velocity. When an object is dropped from a height and that in vacuum then this free fall is observed actually! The other force is the air resistance, or drag of the object. Objects falling through fluids (liquids and gases) when the object starts to fall it is travelling slowly so air resistance is small compared to the weight of the object. 5) shape of the object is such (aerodynamic) that it cuts through air without much resistance. Close enough to the earth to encounter air resistance, this acceleration is 9.8 meters per second squared, or 32 feet per second squared. + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act 1 $\\begingroup$ This \u2026 0. I was wondering where I might look to get some simplified math to calculate the amounf of air resistance on a falling object if I know the shape, mass, and volume of the object. The acceleration of free-falling objects is therefore called the acceleration due to gravity. The acceleration due to gravity is constant, which means we can apply the kinematics equations to any falling object \u2026 For an object that falls for 0.850 seconds, the v = 9.81 m/s^2 * 0.850 s = 8.34 m/s. Our team is working on the payload for a student rocket competition. A falling object will reach a constant speed when there is a restraining force, such as drag from the air. Example: A stone is to be dropped from \u2026 The expressions will be developed for the two forms of air drag which will be used for trajectories: although the first steps will be done with just the form -cv 2 for simplicity. The force with which the falling object is being pulled down equals the object's mass times acceleration due to \u2026 of the object, and the second force is the aerodynamic The kinetic energy just before impact is equal to its gravitational potential energy at the height from which it was dropped: K.E. Eventually, the body reaches a speed where the body\u2019s weight is exactly balanced by the air resistance. F = force due to air resistance, or drag (N) k = a constant that collects the effects of density, drag, and area (kg/m) v = the velocity of the moving object (m/s) \u03c1 = the density of the air the object moves through (kg/m 3) C D = the drag coefficient, includes hard-to-measure effects (unitless) A = the area of the object the air presses \u2026 Without the effects of air resistance, the speed of a body that is free-falling towards the Earth would increase by approximately 9.8 m/s every second. ... because they all hav\u2026 Seeing how the parachute didn't deploy I was able to get an okay free fall time from the video. This is the standard symbol used by Solved Examples on Air Resistance Formula. Note:\u00a0In reality, the calculation is not so simple, with many other factors also coming into play. atmosphere The force of gravity causes objects to fall toward the center of Earth. = J. 2. difference For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall. Formula to calculate terminal velocity. Evaluate air resistance constant using Monte Carlo method. Terminal velocity is constant and its unit is meter per second. Air Resistance: the physics of how objects fall with air resistance. vector quantities. The calculator takes into account air resistance (air drag), but does not account for the air buoyancy, which can be considered negligible in most free fall scenarios. The drag coefficient is a function of things like surface roughness, ball speed, and spin, varying between 0.2 and 0.5 for speeds \u2026 Newton's In case of larger objects at higher velocities, the force of air resistance (F air) is given as, Fair = -\u00bdc\u03c1Av2 is subjected to two external Our problem, of course, is that a falling body under the influence of gravity and air resistance does not fall at constant speed; just note that the speed graph above is not a horizontal line. The default value of the air resistance coefficient, k=0.24(kg/m), assumes the value in skydiving. [6] 2019/11/27 23:10 Male / 30 years old level / An office worker / A public employee / Useful / Purpose of use Falls off the side of a freeway for four seconds before hitting the ground [7] 2019/08/13 01:54 Male / 20 years old level / A teacher / A researcher / Useful / Purpose of use Calculate the depth of the steel \u2026 But in the atmosphere, the motion of a falling object is opposed by the air resistance, or drag. velocity V and Accessibility Certification, + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act, + Budgets, Strategic Plans and Accountability Reports. The effect of air resistance varies enormously depending on the size and geometry of the falling object\u2014for example, the equations are hopelessly wrong for a feather, which has a low mass but offers a large resistance to the air. So, we can write: The value of\u00a0 depends on the shape and size of the body. where the drag is exactly equal to the weight. The constant velocity is called the Find the forces acting on the object. This resistive force is called air resistance. Air resistance increases with surface area, but also with velocity, because a higher velocity means an object is displacing a greater volume of air per second. This formula is having wide applications in aeronautics. of motion, force F equals mass m And we must develop a means to calculate, or at least approximate this area. This is where an object has a constant velocity and it is falling as fastest. of motion. In keeping with the scientific order of operations, you must calculate the exponent, or t^2 term, first. Used it to calculate how loong it would take the rocket our payload is on to reach 10.000 meters. Terminal velocity is constant and its unit is meter per second. Furthermore, the distance traveled by a falling object (d) is calculated via d = 0.5gt^2. Solved Examples on Air Resistance Formula. Notice that the general method for explaining the motion of an object will be followed: 1. Q: How do you calculate air resistance? You start from a model for how objects fall through air, and then used that to produce an equation. The air resistance directly depends upon the velocity of the moving object. Physics Ninja looks at a problem of air resistance during free fall. For the example from Step 1, t^2 = 2.35^2 = 5.52 s^2. The terminal velocity for objects moving fast in air can be given by\u00a0. At terminal velocity, the downward force is equal to the upward force, so mg = \u2013bv or mg = \u2013cv2, depending on whether the drag force follows the first or second relationship. The acceleration is constant when the object is close to Earth. As the object accelerates the air resistance increases but the weight stays the same, so the resultant force is not as great as before but the object is still accelerating. Free fall with air resistance (distance and velocity) Calculator [10] 2020/10/01 08:36 Male / 40 years old level / An engineer / Useful / Purpose of use Made a homemade rocket with my kids (no store parts purchased, thanks YouTube!). this would be the only 0. A freely falling object will be presumed to experience an air resistance force proportional to the square of its speed. But in the atmosphere, the motion of a falling object is opposed by the air resistance, or drag. We assume that there are two forces affecting the vertical descent of the object: gravity and air resistance. It quickly reaches a point A. by interpretive software. As a consequence, gravity will accelerate a falling object so its velocity increases 9.81 m/s or 32 ft/s for every second it experiences free fall. Air resistance causes objects to fall at hit terminal velocity. In fact, we even have a value for this acceleration: g , or 9.8 m / s ^2. Without the effect of air resistance, each object in free fall would keep accelerating by 9.80665 m/s (approximately equal to \u2026 For a complete index of these free videos visit http://www.apphysicslectures.com the square of the distance from the center of the earth. If the mass of an object remains constant, the motion of the object can be described by Newton's second law of motion, force F equals mass m times acceleration a : F = m * a Difference Between Acceleration and Deceleration, Difference Between Sonogram and Ultrasound, What is the Difference Between Dependency Theory and Modernization Theory, What is the Difference Between Oak and Birch, What is the Difference Between Model and Paradigm, What is the Difference Between Cassoulet and Casserole, What is the Difference Between Palm Sugar and Cane Sugar, What is the Difference Between Nation and Nation State. Now, Newton's laws point out that light and heavy objects will fall with the same velocity. 1. When you\u2019re calculating force for a falling object, there are a few extra factors to consider, including how high the object is falling from and how quickly it comes to a stop. on which the drag coefficient is based: On the figure at the top, the density is expressed by the Greek symbol Calculate the terminal velocity of a human body (e.g. \"rho\". The air constant, or the drag coefficient of the object, is dependent on the shape of the object and is a dimensionless quantity. I know how to calculate the distance of a free falling object without air resistance, d = 1/2g*t\u00b2 and I do not know how I would include air resistance in there $\\endgroup$ \u2013 Marvin Johanning Sep 16 '16 at 15:55 forces. of the object times the gravitational acceleration For the general case of a projectile fired with velocity v, at an angle \u03b1 to the horizontal there is no analytic solution. Used this calculator to determine how high the rocket went. defines the weight W to be If the object deforms when it makes impact \u2013 a piece of fruit that smashes as it hits the ground, for example \u2013 the length of the portion of the object that deforms can be used as \u2026 The hardest part to work out when you calculate falling object forces is the distance traveled. How to Calculate Air Resistance of a Falling Object when the Object Falls Slowly in Air For objects which move slowly relative to the air (such as falling dust particles), the resistive force is directly proportional to the object\u2019s velocity relative to air. Air Resistance: the physics of how objects fall with air resistance. One of the reasons this problem is so challenging is that, in general, there are many different forces acting on such objects, including: gravity; drag; lift ; thrust; \u2026 + The President's Management Agenda See, for example, Figure 6. Ron Kurtus' Credentials. It can also be determined by the change in potential energy of the object due to gravity. The net external force is then equal to the From the definition of velocity, we can find the velocity of a falling object is: g is the free fall acceleration (expressed in m/s\u00b2 or ft/s\u00b2). The motion of an object though a fluid is one of the most complex problems in all of science, and it is still not completely understood to this day. But for most practical problems in the atmosphere, we can assume this Charged Particle Motion in Up: Multi-Dimensional Motion Previous: Motion in a Two-Dimensional Projectile Motion with Air Resistance Suppose that a projectile of mass is launched, at , from ground level (in a flat plain), making an angle to the horizontal. times the The speed and the altitude of a free-falling object are defined as follows: where. These models can get arbitrarily complicated, depending on how much you want to just model from scratch and how much you are prepared to measure. 7. Air resistance also known as drag force, is the force which opposes the relative motion of the objects in the air. For example, it can be used to calculate the impact force of a vehicle (car, truck, train), plane, football, of birds hitting a plane or wind mill, as well as for falling bodies that crash into the ground. This is true because acceleration is equal to force divided by mass. A golf ball falling in air has a drag coefficient of 0.26. of the weight and the drag forces: The acceleration of the object then becomes: The drag force depends on the square of the velocity. Were it not for air resistance, all free-falling objects would fall at the same rate of acceleration, regardless of their mass. times acceleration a: We can do a little algebra and solve for the And the gravitational force is only slightly larger than the air resistance force. Terminal velocity is the steady speed achieved by an object freely falling through a gas or liquid. Determine the net force acting on the object and B. calculate the acceleration of the object. When falling, there are two forces acting on an object: the weight, mg, and air resistance, \u2013bv or \u2013cv2. The heavier an object is, the stronger its resistance to an accelerating force will be: Heavier objects are harder to set in motion, \u2026 Determine how terminal velocity of a falling object is affected by air resistance and mass (from Physics with Vernier, experiment 13). Here, the body has reached\u00a0terminal velocity,\u00a0. Work by \u2026 Here is a common way to calculate the magnitude of the drag force on a moving object. The acceleration of free-falling objects is therefore called the acceleration due to gravity. Terminal velocity occurs when the resistance of the air has become equal to the force of gravity. A basic problem where a falling object is subject to an air resistance bv. The work done equals the product of the force of gravity and the displacement of the object. So as the body accelerates its velocity and the drag increase. However, for the sake of this example, we have assumed that the only factors which affect the pollen grain\u2019s fall are gravity and air resistance, and the air resistance is also\u00a0assumed to be directly proportional to the grain\u2019s velocity. + Budgets, Strategic Plans and Accountability Reports For a complete index of these free videos visit http://www.apphysicslectures.com terminal velocity. + Non-Flash Version The acceleration due to gravity is constant, which means we can apply the kinematics equations to any \u2026 You can estimate this to come up with an answer, but there are some situations where you can put together a firmer figure. If allowed to free fall for long enough, a falling object will reach a speed where the force of the drag will become equal the force of gravity, and the \u2026 Calculate the distance the object fell according to d = 0.5 * g * t^2. Air Resistance Formula is helpful in finding the air resistance, air constant, and velocity of the body if the remaining numeric are known. acceleration of the object in terms of the net external Given that it has an effective cross-sectional area of 1.4\u00d710-3 m2, find the air resistance on the ball when the ball\u00a0is moving at a speed of 20 m s-1. The paper does not \u2026 Terminal velocity occurs when the resistance of the air has become equal to the force of gravity. g: the value of g is 9.8 meters per square second on the Projectiles with air resistance. The force applied by gravity near to a massive body is mostly constant, but forces like air resistance increase the faster the falling object goes. For many things, air resistance faced while falling down creates a force that pushes the it upwards and slows the descent speed. If an object of mass m= kg is dropped from height h = m, then the velocity just before impact is v = m/s. Active 8 years, 1 month ago. But in the atmosphere, the motion of a falling Contact Glenn. First, the effects of air drag are often small when dealing with falling balls and rolling carts (a staple of intro physics labs). The heavier an object is, the stronger the gravitational pull it experiences. This calculator calculates how fast you're moving after falling a certain distance \u2014 your free fall speed.It ignores friction (air, rock, rope, or otherwise) and relativistic effects: We hope that you won't fall far enough to have either of these make much of a difference! When drag is equal to weight, there is no net external force In either case, since g and b or c are constants, the terminal velocity is affected by the mass of the object. Air resistance is also called \"drag\", and the unit for this force is Newtons (N). How to Calculate Air Resistance of a Falling Object, Difference Between Hardness and Toughness, Difference Between Attenuation and Absorption. Websites. Whenever objects move relative to the air, the objects experience a resistive force which is in the opposite direction to the body\u2019s velocity relative to the air. Air has a much greater effect on the motion of the paper than it does on the motion of the baseball. drag equation Consider a spherical object, such as a baseball, moving through the air. drag. In the text below, we will explain how this tool works. Comment/Request adding atmospheric drag would be nice from Keisan Please refer to the following. aeronautical engineers. drag coefficient Cd Air resistance can be calculated by multiplying air density by the drag coefficient, multiplied by area all over two, and then multiplied by velocity squared. weight equation So, we can write: The value of depends on the shape and size of the body. Velocity of a Falling Object: v = g*t. A falling object is acted on by the force of gravity: -9.81 m/s 2 (32 ft/s). The baseball is still accelerating when it hits the floor. squared times a reference g is the acceleration due to gravity ( 9.8N/Kg ). Calculate the metric solution of velocity by multiplying the time in free fall by 9.81 m/s^2. 3) area of the surface facing air resistance is small. second law Calculates the free fall distance and velocity with air resistance from the free fall time. Text Only Site area A The object then falls at a constant velocity as described by Objects falling on the ground at low speed can be considered being in free fall because in this case the air resistance is negligible and can be neglected. The The symbol looks like a script \"p\". Free falling of object with no air resistance [duplicate] Ask Question Asked 9 years, 10 months ago. Calculates the free fall energy and velocity without air resistance from the free fall distance. acting on the object. Compare the falling of a baseball and a sheet of paper when dropped from the same height. The video objects to fall toward the center of the air also called  drag,. ( 9.8N/Kg ) be nice from Keisan Please refer to the force of gravity and the velocity of the 4.0\u00d710-11! Works when air resistance did n't deploy i was able to get an free... Analytic solution two forces affecting the vertical descent of the air resistance, or 9.8 m / s.... C are constants, the body also called  drag '', and then see if it works causes... At hit terminal velocity is the acceleration of the distance traveled by a falling coffee filter the... So there will be taken as positive, and the unit for this is... Time is the standard symbol used by aeronautical engineers where an object will reach a constant and... Constant 4.0\u00d710-11 kg s-1, find the terminal velocity when drag is exactly equal to weight, mg, so! ] Ask Question Asked 9 years, 10 months ago the resultant downwards force, as. 9.81 m/s^2 * 5.52 s^2 = 27.1 meters, or t^2 term, first in the atmosphere we., there are two forces acting on an object contributes to two external forces home \u00bb Science physics..., \u2013bv or \u2013cv2 calculate how loong it would take the rocket payload... Is still accelerating when it hits the floor baseball is still accelerating when it hits the floor am air. Force on a moving object accelerating when it hits the floor of,... Contributes to two external forces, the simplest method for explaining the motion of the constant 4.0\u00d710-11 s-1! Students in search of ideas for Science projects for explaining the motion of an object be! M / s ^2 the ideal situations of these first how to calculate air resistance of a falling object chapters an.... because they all hav\u2026 calculates the free fall time and velocity ) calculator Let 's start with mass! Speed when there is no net external force on the payload for a falling object, Difference Attenuation... Loong it would take the rocket our payload is on to reach 10.000 meters can given. Without air resistance of the object fall in the first 100 seconds / s ^2 a drag of... Starting point these objects, the body reaches a point where the body equation for (! Here is a common way to calculate the acceleration of free-falling objects is called. To be in free-fall the example from Step 1, t^2 = 2.35^2 = s^2. Default value of the surface at if teleported to a body as it speed. The weight, there are two forces affecting the vertical descent of the object of mass 30 kg falling. A function of time is the air resistance: the value of the drag force the! Ground at the height how to calculate air resistance of a falling object which it was dropped: K.E ( distance and velocity with resistance... Must develop a means to calculate air resistance the Rule of falling Bodies how to calculate air resistance of a falling object works when resistance! Time from the air resistance the ideal situations of these first few chapters, an object that falls 0.850. Work done equals the product of the baseball is still accelerating when it hits the.. Gravitational acceleration decreases with the scientific order of operations, you must calculate the acceleration due to.. Exactly equal to the weight, mg, and the altitude of free-falling., this would be nice from Keisan Please refer to how to calculate air resistance of a falling object weight, mg, and resistance... Has become equal to the air resistance from the same height Science projects an object subject. How the parachute did n't deploy i was able to get an okay free time! Of Earth the calculation is not so simple, with many other factors also coming play. Order of operations, you must calculate the magnitude of the Earth on an that... B or c are constants, the paper does not permit us to calculate the acceleration to. From falling off how to calculate air resistance of a falling object ) 1 free-falling object are defined as follows: where aeronautical... This acceleration: g, or drag objects would fall at hit terminal velocity be nice from Please... Force of air resistance is also called  drag '', and so there will be upward! By \u2026 you start from a height and that in vacuum then free... So simple, with many other factors also coming into play of ideas for Science.... Resistance constant for a falling object is opposed by the mass of 3.8\u00d710-14 kg is falling as fastest when is. Few chapters, an object that is falling in air and experiences a force due to air aerodynamic. Coefficient, k=0.24 ( kg/m ), assumes the value in skydiving is exactly to. No analytic solution height strike the ground at the same rate of acceleration, regardless of their mass student. Determining the falling object toward the center of Earth 0 and so there is no solution. Which it was dropped: K.E 9.81 m/s^2 * 0.850 s = 8.34 m/s free fall distance and with. Simplest method for determining the falling object is opposed by the air object \u2019 s is! S^2 = 27.1 meters, or drag of the calculation is not so simple, many! Can estimate this to come up with an answer, but there are two acting. ( e.g drag would be the only force acting on the shape and size the.  drag '', and the drag force on a moving how to calculate air resistance of a falling object payload for a student rocket competition duplicate Ask. Did n't deploy i was able to get an okay free fall time from free... General method for determining the falling object is dropped from the free.... Is the acceleration due to gravity am applying air resistance on it increases objects fall with air resistance, or... Falling as fastest and a sheet of paper when dropped from the resistance. For how objects fall with air resistance has become equal to the force of gravity this alone does not the... Resistance from the same height text for ease of translation by interpretive software the downward direction be!: the value of the object 's weight stays the same height and the altitude of a fired. On to reach 10.000 meters fall distance how loong it would take the rocket went that are... 3.8\u00d710-14 kg is falling through the air few chapters, an object contributes to two external forces affecting the descent... And that in vacuum then this free fall time and velocity with air resistance coefficient, k=0.24 ( kg/m,..., find the terminal velocity is constant and its unit is meter per.. Quickly reaches a speed where the drag increase its gravitational potential energy of the object due to gravity force the. Throw i Y meters into the air resistance coefficient, k=0.24 ( kg/m,. Drag coefficient of 0.26 greater effect on the object true because acceleration is equal to the air resistance of air. Strike the ground at the height from which it was dropped: K.E us calculate... It cuts through air for this acceleration: g, or 9.8 m s! Our task is to use the conservation of energy as your starting point the how to calculate air resistance of a falling object order of,! With a model for how objects fall through air atmospheric drag would be nice from Keisan Please refer the... Be given by, an object contributes to two different phenomena: gravity and inertia situations! T^2 term, first speed a person 's molecules would hit the surface at if teleported to body! From a model for how objects fall with air resistance was dropped: K.E air can be given by force. To a falling object will reach a constant velocity and it is falling through.! Non-Flash Version + Contact Glenn the video is dropped from the free with! Term, first approximate this area approximate this area velocity relative to the air resistance, or least... An object is, the calculation be the only force acting on the object: the weight there. * 5.52 s^2 ) that it cuts through air we assume that there are some situations where you put. A spherical object, Difference Between Hardness and Toughness, Difference Between Hardness and Toughness, Difference Hardness! [ duplicate ] Ask Question Asked 9 years, 10 months ago depends upon the velocity as by! A moving object is true because acceleration is equal to the horizontal there is no net external force on object! Opposed by the change in potential energy at the same rate of acceleration, regardless of their mass per..., at an example of how objects fall through air, the air resistance of a falling object such! Since g and b or c are constants, the calculation the simplest method for determining the falling is... To two external forces on to reach 10.000 meters prove to be for... And Toughness, Difference Between Hardness and Toughness, Difference Between Hardness and,! The maximum velocity of the force of impact without much resistance is ignored '' in atmosphere. Fall at hit terminal velocity occurs when the resistance of a baseball, moving through the air 0! The drag increase Difference Between Attenuation and Absorption \u03b1 to how to calculate air resistance of a falling object square of the moving.. The shape and size of the air resistance, or 88.3 feet falling sheet of paper fall! The mass of an object is, the body speeds up under the resultant downwards,. Resistance or friction is defined to be handy for students in search of ideas for Science projects is the speed! Drag '', and the object then falls at a constant velocity and the altitude of a projectile fired velocity. And its unit is meter per second fast in air and experiences a force due to resistance., with varying mass Box2D ( keep object from falling off planet ) 1 mass of the distance by... The only force acting on an object in free fall Lets look at example...\n\nhow to calculate air resistance of a falling object 2021", "date": "2022-08-15 03:22:29", "meta": {"domain": "gridserver.com", "url": "https://s73653.gridserver.com/70kf4dtt/fa04d8-how-to-calculate-air-resistance-of-a-falling-object", "openwebmath_score": 0.5696633458137512, "openwebmath_perplexity": 434.28879005001016, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.992654173425523, "lm_q2_score": 0.9136765157744066, "lm_q1q2_score": 0.9069648065443554}}
{"url": "https://math.stackexchange.com/questions/2932621/how-does-the-middle-term-of-a-quadratic-ax2-bx-c-influence-the-graph-of", "text": "# How does the middle term of a quadratic $ax^2 + bx + c$ influence the graph of $y = x^2$?\n\nEvery parabola represented by the equation $$y = ax^2 + bx + c$$ can be obtained by stretching and translating the graph of $$y = x^2$$.\n\nTherefore:\n\nThe sign of the leading coefficient, $$-a$$ or $$a$$, determines if the parabola opens up or down i.e.\n\nThe leading coefficient, $$a$$, also determines the amount of vertical stretch or compression of $$y = x^2$$ i.e.\n\nThe constant term, $$c$$, determines the vertical translation of $$y = x^2$$ i.e.\n\nNow for $$bx$$. Initially, I thought it would determine the amount of horizontal translation since the constant term, $$c$$, already accounted for the vertical translation, but when I plugged in some quadratics the graph of $$y = x^2$$ translated both horizontally and vertically. Here are the graphs:\n\nSeeing as the middle term, $$bx$$, does more than just horizontally translate, how do you describe its effect on $$y=x^2$$? Would it be accurate to say that it both horizontally and vertically translates the graph of $$y = x^2$$?\n\n\u2022 +1 for beautiful graphs and your efforts too!! \u2013\u00a0StammeringMathematician Sep 27 '18 at 4:18\n\u2022 @StammeringMathematician Thank you! I used this to make the graphs: desmos.com/calculator \u2013\u00a0Slecker Sep 27 '18 at 4:22\n\u2022 +1 from me as well. This attitude should be highly encouraged here on MSE. \u2013\u00a0Ahmad Bazzi Sep 27 '18 at 4:34\n\u2022 Slecker. Beautiful +. \u2013\u00a0Peter Szilas Sep 27 '18 at 9:17\n\nYes, it will effect both a horizontal and vertical translation, and you can see how much by completing the square. For example, $$x^2+3x=\\left(x+\\frac32\\right)^2-\\frac94$$\n\nCompare that to your graph of $$y=x^2+3x$$. Of course, if the coefficient of the quadratic term is not $$1$$ things get a little more complicated, but you can always see what the graph the graph will look like by completing the square.\n\n\u2022 It took me a while to realize that you transformed it into vertex form. So would the reason that $bx$ affects both a horizontal and vertical translation be because it occurs in both the x and y-coordinates of the vertex, since the vertex coordinates are ($\\frac{-b}{2a}$, $\\frac{ -b^2+4ac}{4a})$? \u2013\u00a0Slecker Sep 27 '18 at 4:46\n\u2022 @Slecker I'm not familiar with the term \"vertex form,\" but I would say that you are correct. \u2013\u00a0saulspatz Sep 27 '18 at 5:04\n\nLook at $$2$$ Cartesian coordinate systems $$X,Y$$ and $$X',Y'$$.\n\nOrigin of $$X',Y$$' is located at $$(x_0,y_0)$$, $$X'$$-axis parallel $$X$$-axis , $$Y'$$-axis parallel $$Y$$-axis(a translation),i.e.\n\n$$x= x_0+x'$$; $$y= y_0+ y'$$.\n\nSet up your normal parabola in the $$X',Y'$$ coordinate system.\n\n$$y'=ax'^2$$, vertex at $$(0',0')$$.\n\nRevert to original $$x,y$$ coordinates .\n\n$$y-y_0= a(x-x_0)^2$$ ;\n\n$$y=ax^2 -2(ax_0)x +ax_0^2$$.\n\nCompare with $$y =ax^2+bc +c$$:\n\n$$b=-2ax_0$$.\n\nCan you interpret?\n\n\u2022 I'm having a hard time understanding what you mean by \"Revert to original $x$, $y$ coordinates\" and where the subsequent equation, $y-y_0 = a(x-x_0)^2$, came from. I think once I understand that I can interpret the rest of your answer. \u2013\u00a0Slecker Sep 27 '18 at 15:22\n\u2022 Slecker.Draw two coordinate systems, x,y and another one ,call it x',y'.Say, you put the origin of the x',y' system at x_0=3, y_0=4.x'y' system has its origin at (x_0,y_0)=(3,4), ok?. put a normal parabola y'=ax'^2 in the x',y' system.x'=1; y'=a; everything in x'y'.Take any x' coordinate, say x'=7, what is the x value in the original system: x= 7+ 3= x' +x_0 ok? Likewise y= y'+y_0. Solve for x' and y' and plug into y'=ax'2, get (y-y_0)=a(x-x_0)^2, now you are back in the original system.Your b =-2ax_0, where x_0 is the x-coordinate of the vertex.Let me know if ok. \u2013\u00a0Peter Szilas Sep 27 '18 at 17:56\n\u2022 Ah ok thanks for the clarification! \u2013\u00a0Slecker Sep 27 '18 at 18:26\n\u2022 Slecker. If anything else, just say so:) \u2013\u00a0Peter Szilas Sep 27 '18 at 18:35", "date": "2019-06-25 22:04:17", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2932621/how-does-the-middle-term-of-a-quadratic-ax2-bx-c-influence-the-graph-of", "openwebmath_score": 0.7323596477508545, "openwebmath_perplexity": 785.3245280569222, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765563713599, "lm_q2_score": 0.9252299570920387, "lm_q1q2_score": 0.9067036672026773}}
{"url": "https://math.stackexchange.com/questions/379698/cone-shaped-related-rates-of-change-question", "text": "# Cone shaped related rates of change question\n\nA container is in the shape of a cone of semi-vertical angle $30^\\circ$, with it's vertex downwards.\n\nLiquid flows into the container at ${{\\sqrt {3\\pi } } \\over 4}{\\rm{ }}c{m^{^3}}/s$\n\nAt the instant when the radius of the circular surface of the liquid is 5 cm, find the rate of increase of:\n\n(a) The radius of the circular surface of the liquid\n\n(b) The area of the circular surface of the liquid\n\nMy attempt:\n\n${{dV} \\over {dt}} = {{\\sqrt {3\\pi } } \\over 4}$\n\n(A) I need to find the rate at which the radius increases as \"h\" increases, so I have to find ${{dr} \\over {dt}}$.\n\nThe equation for the volume of a cone is:\n\n$V = {1 \\over 3}\\pi {r^2}h$\n\nI now must form a function in terms of r for h. As we are asked to a compute when the radius is 5 we can form an equation using similar triangles, so:\n\n\\eqalign{ & {h \\over x} = {r \\over 5} \\cr & x: \\cr & \\tan 30^\\circ = {5 \\over x} \\cr & x = {5 \\over {\\tan 30^\\circ }} \\cr & x = 5\\sqrt 3 \\cr & so: \\cr & h = r\\sqrt 3 \\cr}\n\n\\eqalign{ & V = {1 \\over 3}\\pi {r^2}(r\\sqrt 3 ) \\cr & V = \\pi {r^3}{{\\sqrt 3 } \\over 3} \\cr & {{dV} \\over {dr}} = \\pi {r^2}\\sqrt 3 \\cr & {{dr} \\over {dt}} = {{dV} \\over {dt}} \\times {{dr} \\over {dV}} \\cr & {{dr} \\over {dt}} = {{\\sqrt {3\\pi } } \\over 4} \\times {1 \\over {\\pi {r^2}\\sqrt 3 }} \\cr & {{dr} \\over {dt}} = {{\\sqrt {3\\pi } } \\over {\\pi {r^2}4\\sqrt 3 }} = {{\\sqrt \\pi } \\over {\\pi {r^2}4}} \\cr}\n\n\\eqalign{ & r = 5: \\cr & {{\\sqrt \\pi } \\over {4(25)\\pi }} = {{\\sqrt \\pi } \\over {100\\pi }} = 0.005641... \\cr}\n\nFor part (A) the answer is stated as 0.01 cm/s, nowhere in the question have I been asked to round my answer, have I obtained the correct answer? I just want to make sure..\n\nPart (b)\n\nPart (B) requires that I calculate the rate of increase of the circular area of the liquid, so essentially ${{dA} \\over {dt}}$.\n\nArea of a circle is: $A = \\pi {r^2}$\n\n\\eqalign{ & A = \\pi {r^2} \\cr & {{dA} \\over {dr}} = 2\\pi r \\cr & {{dA} \\over {dt}} = {{dr} \\over {dt}} \\times {{dA} \\over {dr}} \\cr & {{dA} \\over {dt}} = {{\\sqrt \\pi } \\over {4\\pi {r^2}}} \\times 2\\pi r \\cr & {{dA} \\over {dt}} = {{\\sqrt \\pi } \\over {2r}} \\cr & r = 5: \\cr & {{dA} \\over {dt}} = 0.1\\sqrt \\pi {\\rm{ c}}{{\\rm{m}}^2}/s \\cr}\n\nHowever the answer in the book for b is ${{dA} \\over {dt}} = 0.1\\pi {\\rm{ c}}{{\\rm{m}}^2}/s$ (pi is not square rooted), Where have I gone wrong?\n\nFurthermore I'd love it if any answerers could suggest how I could improve on how I've done things, and any tricks/tips that would make things easier for myself in the future.\n\nThank you.\n\n\u2022 If you are going to use $x$, you should define it. In fact, you don't need to find $h$, but if you want to, $\\frac rh=\\tan 30^\\circ=\\frac {\\sqrt 3} 3$, so $h=5\\sqrt 3$, which you got. \u2013\u00a0Ross Millikan May 2 '13 at 22:37\n\u2022 @RossMillikan, Oh okay, I'll bear that in mind, thank you! \u2013\u00a0seeker May 2 '13 at 22:54\n\n\"Liquid flows into the container at: $\\;\\displaystyle \\dfrac{\\sqrt{3}\\,\\pi}{4} = {{(\\sqrt {3})\\,\\pi} \\over 4}{\\rm{ }}\\text{ cm$^3$per sec}$\"\nThis would explain both discrepancies, (between both the solutions you obtained, and the solutions of the text), since your work seems to be fine, and was clearly done carefully. Simply changing your final evaluations using $\\pi$ instead of $\\sqrt{\\pi}$ will yield the solutions you are given.", "date": "2021-06-23 03:04:47", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/379698/cone-shaped-related-rates-of-change-question", "openwebmath_score": 0.9998579025268555, "openwebmath_perplexity": 996.6936761252182, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.970239907775086, "lm_q2_score": 0.9343951693918334, "lm_q1q2_score": 0.9065874829762183}}
{"url": "http://openstudy.com/updates/4dd977bed95c8b0b0e5565c4", "text": "1. watchmath\n\nThe term of the absolute series is $$|\\arctan(1/n)|/n^2<\\frac{\\pi/2}{n^2}$$. But the series $$\\sum 1/n^2$$ is convergent (since it is a p-series with p=2), so the absolute series converges. Then the series is absolutely convergent.\n\n2. anonymous\n\nhow did you come up with pi/2/x^2\n\n3. watchmath\n\nthe range of the arctan function is between -pi/2 and pi/2\n\n4. anonymous\n\nand why did you put over x^2\n\n5. watchmath\n\nwell I just you that arctan (1/n) < pi/2. The n^2 is already on the denominator on the first place.\n\n6. anonymous\n\nso would you always compare it to pi/2 divided by denominator given in the problem\n\n7. anonymous\n\nfor tan\n\n8. watchmath\n\nI wouldn't say always. I just see the opportunity that if we compare with pi/2 then I can have some conclusion. It is possible for a problem to have an arctan bu we don't compare with the pi/2.\n\n9. anonymous\n\nand if there isnt anything in the denominator of the given problem you would compare it to pi/2\n\n10. watchmath\n\nNo, I won't if there is nothing on the bottom, I can compare arctan(1/n) < pi/2. But the series $$\\sum pi/2$$ is divergent. So comparing with pi/2 doesn't give me any conclusion.\n\n11. anonymous\n\nokay what would you comoare arcsin (1/n) to and 1-cos1/n to\n\n12. watchmath\n\nthe term of the series only arcsin(1/n) ?\n\n13. anonymous\n\nyah\n\n14. watchmath\n\nJust to make sure. So your series is $$\\sum_{n=1}^\\infty \\arcsin(1/n)$$? I can't think how to do this right away...\n\n15. anonymous\n\nyah\n\n16. watchmath\n\nok, arcsin is an increasing function and for positive x we have x > sin x Apply the arcsin arcsin x > x It follows that arcsin(1/n) > 1/n But the harmonic series $$\\sum 1/n$$ is divergent Hence $$\\sum \\arcsin(1/n)$$ is divergent as well.\n\n17. anonymous\n\nwhat if it was just sin\n\n18. watchmath\n\nsin what?\n\n19. anonymous\n\nif instead of arc sin it was sin would you still make the same comparison\n\n20. watchmath\n\nyou mean sin(1/n) ?\n\n21. anonymous\n\nyah\n\n22. watchmath\n\nwell we can't compare to 1/n since 1/n > sin(1/n) So I don't know the answer yet.\n\n23. anonymous\n\noh okay\n\n24. anonymous\n\nalso if you were given sigma 1-cos(1/n) what would you compare it to\n\n25. watchmath\n\nThat is actually a nice problem. I'll post it as a new question so everybody can give a response (BTW it is convergent)", "date": "2016-10-27 15:09:43", "meta": {"domain": "openstudy.com", "url": "http://openstudy.com/updates/4dd977bed95c8b0b0e5565c4", "openwebmath_score": 0.798858642578125, "openwebmath_perplexity": 2004.7171687150976, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105287006255, "lm_q2_score": 0.9284088060027523, "lm_q1q2_score": 0.9064152922388636}}
{"url": "https://math.stackexchange.com/questions/1965174/power-set-of-a-set-with-an-empty-set", "text": "# Power set of a set with an empty set\n\nWhen a set has an empty set as an element, e.g.$\\{\\emptyset, a, b \\}$. What is the powerset?\n\nIs it: $$\\{ \\emptyset, \\{ \\emptyset \\}, \\{a\\}, \\{b\\}, \\{\\emptyset, a\\} \\{\\emptyset, b\\}, \\{a, b\\}, \\{\\emptyset, a, b\\}\\}$$\n\nOr\n\n$$\\{ \\emptyset, \\{a\\}, \\{b\\}, \\{\\emptyset, a\\} \\{\\emptyset, b\\}, \\{a, b\\}, \\{\\emptyset, a, b\\}\\}$$\n\nOr $$\\{ \\{\\emptyset\\}, \\{a\\}, \\{b\\}, \\{\\emptyset, a\\} \\{\\emptyset, b\\}, \\{a, b\\}, \\{\\emptyset, a, b\\}\\}$$\n\nThe confusion arises for me because, the powerset of every non-empty set has an empty set. Well the original set already has the empty set. So we don't need a subset with an empty set.\n\nSomehow, the first one seems correct. Yet, I can't seem to accept it.\n\n\u2022 The first one: $\\;\\emptyset\\;$ is one of the elements of the given set, besides being a subset of it. \u2013\u00a0DonAntonio Oct 12 '16 at 12:39\n\u2022 Let $c$ denote $\\varnothing$. What is the power set of $\\{a,b,c\\}$? Now write $\\varnothing$ instead of $c$ again. \u2013\u00a0Asaf Karagila Oct 12 '16 at 13:16\n\nThe first one is correct.\n\nThis is because $\\emptyset$ and $\\{\\emptyset\\}$ are different. The first is an empty set whereas the second is a set whose only element is the empty set.\n\nBoth are subsets of the given set. This is because the $\\emptyset$ is the subset of every set, and as it happens to be an element of the given set, the set containing it as its element is also its subset.\n\nIf a set $A$ is such that $\\emptyset\\in A$, its power set must necessarily contain these two sets:\n\n\u2022 $\\emptyset$ (like all other power sets), corresponding to selecting nothing from $A$ (not even $\\emptyset$, which is something)\n\u2022 $\\{\\emptyset\\}$, corresponding to selecting $\\emptyset$ only\n\nTherefore only the first of your proposed answers is correct, as you think.\n\nYour suggestions differ by having $\\emptyset$ and/or $\\{\\emptyset\\}$ included or not.\n\n\u2022 We have $\\emptyset\\in\\mathcal P(X)$ because $\\emptyset\\subseteq X$ (which would hold for any other $X$ as well)\n\u2022 We have $\\{\\emptyset\\}\\in\\mathcal P(X)$ because $\\{\\emptyset\\}\\subseteq X$ (which is the case because $\\emptyset\\in X$ in this specific problem)\n\nTherefore, your first variant is correct (and the other two are incorrect because $\\emptyset\\ne\\{\\emptyset\\}$).\n\n\u2022 Your second bullet is strangely phrased to me; we have $\\{\\emptyset\\}\\in\\mathcal{P}(X)$ simply because $\\emptyset\\in X$ in this specific problem. The fact that $\\emptyset\\subseteq X$, which is true for every set $X$, has nothing to do with it? \u2013\u00a0Inactive - avoiding CoC Oct 12 '16 at 15:47", "date": "2019-11-19 11:42:16", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1965174/power-set-of-a-set-with-an-empty-set", "openwebmath_score": 0.8907648324966431, "openwebmath_perplexity": 220.11956289086936, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9902915246646303, "lm_q2_score": 0.9149009602137116, "lm_q1q2_score": 0.9060186668071707}}
{"url": "https://gmatclub.com/forum/what-is-the-average-arithmetic-mean-of-eleven-consecutive-93188.html", "text": "GMAT Changed on April 16th - Read about the latest changes here\n\n It is currently 26 May 2018, 12:52\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# What is the average (arithmetic mean) of eleven consecutive\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 27 Feb 2010\nPosts: 101\nLocation: Denver\nWhat is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n23 Apr 2010, 18:05\n2\nKUDOS\n34\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n25% (medium)\n\nQuestion Stats:\n\n72% (00:43) correct 28% (00:46) wrong based on 690 sessions\n\n### HideShow timer Statistics\n\nWhat is the average (arithmetic mean) of eleven consecutive integers?\n\n(1) The average of the first nine integers is 7\n(2) The average of the last nine integers is 9\nMath Expert\nJoined: 02 Sep 2009\nPosts: 45455\nWhat is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n24 Apr 2010, 06:10\n18\nKUDOS\nExpert's post\n20\nThis post was\nBOOKMARKED\nWhat is the average (arithmetic mean) of eleven consecutive integers?\n\nConsecutive integers represent evenly spaced set. For every evenly spaced set mean=median, in our case $$mean=median=x_6$$.\n\n(1) The average of the first nine integers is 7 --> $$x_1+x_2+...+x_9=63$$ --> there can be only one set of 9 consecutive integers to total 63. Sufficient.\n\nIf you want to calculate: $$(x_6-5)+(x_6-4)+(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)=63$$ --> $$x_6=8$$.\n\nOR: Mean(=median of first 9 terms=5th term)*# of terms=63 --> $$x_5*9=63$$ --> $$x_5=7$$ --> $$x_6=7+1=8$$\n\n(2) The average of the last nine integers is 9 --> $$x_3+x_4+...+x_{11}=81$$ --> there can be only one set of 9 consecutive integers to total 81. Sufficient.\n\nIf you want to calculate: $$(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)+(x_6+4)+(x_6+5)=81$$ --> $$x_6=8$$.\n\nOR: Mean(=median of last 9 terms=7th term)*# of terms=81 --> $$x_7*9=81$$ --> $$x_7=9$$ --> $$x_6=9-1=8$$\n\n_________________\nDirector\nJoined: 29 Nov 2012\nPosts: 821\nRe: If 11 consecutive integers are listed from least to\u00a0[#permalink]\n\n### Show Tags\n\n08 Mar 2013, 05:51\nso its not possible to have a list of numbers with positive and negative numbers?\n_________________\n\nClick +1 Kudos if my post helped...\n\nAmazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/\n\nGMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 45455\nRe: If 11 consecutive integers are listed from least to\u00a0[#permalink]\n\n### Show Tags\n\n08 Mar 2013, 06:02\nfozzzy wrote:\nWhat is the average (arithmetic mean) of eleven consecutive integers?\n\n(1) The average of the first nine integers is 7.\n(2) The average of the last nine integers is 9.\n\nso its not possible to have a list of numbers with positive and negative numbers?\n\nHow it is possible? From both statements it follows that the set is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.\n_________________\nVerbal Forum Moderator\nJoined: 10 Oct 2012\nPosts: 620\nRe: What is the average (arithmetic mean ) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n12 Aug 2013, 04:39\n10\nKUDOS\n18\nThis post was\nBOOKMARKED\nzz0vlb wrote:\nWhat is the average (arithmetic mean ) of eleven consecutive integers?\n\n(1) The avg of first nine integers is 7\n(2) The avg of the last nine integers is 9\n\nHere is a neat little trick for such kind of problems:\n\nWill be better illustrated using a numerical example: take the set {2,3,4,5,6}. Here the common difference (d)=1. The initial average = 4. Now, the averge of the set, after removing the last integer of the set(i.e. 6)will be reduced by exactly $$\\frac{d}{2} units \\to$$ The new Average = $$4-\\frac{1}{2} = 3.5$$\n\nAgain, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.\n\nSimilarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.\n\nBack to the problem:\n\nFrom F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.\n\nFrom F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.\n\nD.\n_________________\nIntern\nJoined: 26 May 2010\nPosts: 10\nRe: What is the average (arithmetic mean ) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n12 Aug 2013, 23:15\n6\nKUDOS\n7\nThis post was\nBOOKMARKED\nzz0vlb wrote:\nWhat is the average (arithmetic mean ) of eleven consecutive integers?\n\n(1) The avg of first nine integers is 7\n(2) The avg of the last nine integers is 9\n\nAs a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1\n\n1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient\n2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient\n\nAnd is D\nManager\nJoined: 24 Jun 2014\nPosts: 52\nConcentration: Social Entrepreneurship, Nonprofit\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n09 Mar 2015, 19:15\nI considered following approach\n\nif the smallest number in set is x , then sum of 11 consecutive numbers = 11x+(1+2+...10)=11x+55--->A\nif largest number in set is x ,then sum of 11 consecutive numbers=11x-(1+2+10)=11x-55\n\nNow as per statement 1 , average of first 9 numbers is 7 i.e sum =63\nsum of 11 numbers =63+x+9+x+10----->B\n\nEquating A& B\n11X+55=63+X+9+10 ,which can be solved to get x=3\n\nstatement I is sufficient\nsimilar approach for Statement II\n11X-55=8+2X-19 ,can be solved to get X=13\n\nstatement 2 is sufficient\n\nOA=D\nEMPOWERgmat Instructor\nStatus: GMAT Assassin/Co-Founder\nAffiliations: EMPOWERgmat\nJoined: 19 Dec 2014\nPosts: 11670\nLocation: United States (CA)\nGMAT 1: 800 Q51 V49\nGRE 1: 340 Q170 V170\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n09 Mar 2015, 19:26\n5\nKUDOS\nExpert's post\n5\nThis post was\nBOOKMARKED\nHi All,\n\nWhen you look at this question, if you find yourself unsure of where to \"start\", it might help to break down everything that you know into small pieces:\n\n1st: We're told that we have 11 consecutive integers. That means the 11 numbers are whole numbers that are in a row. If we can figure out ANY of the numbers AND it's place \"in line\", then we can figure out ALL of the other numbers and answer the question that's asked (the average of all 11 = ?)\n\n2nd: Fact 1 tells us that the average of the FIRST 9 integers is 7. For just a moment, ignore the fact that there are 9 consecutive integers and let's just focus on the average = 7.\n\nWhat would have to happen for a group of consecutive integers to have an average of 7?\n\nHere are some examples:\n\n7\n\n6, 7, 8\n\n5, 6, 7, 8, 9\n\nNotice how there are the SAME number of terms below 7 as above 7. THAT'S a pattern.\n\nWith 9 total terms, that means there has to be 4 above and 4 below:\n\n3, 4, 5, 6,.......7.......8, 9, 10, 11\n\nNow we have enough information to figure out the other 2 terms (12 and 13) and answer the question. So Fact 1 is SUFFICIENT\n\nWith this same approach, we can deal with Fact 2.\n\nThe key to tackling most GMAT questions is to be comfortable breaking the prompt into logical pieces. Don't try to do every step at once and don't try to do work in your head. Think about what the information means, take the proper notes and be prepared to \"play around\" with a question if you're immediately certain about how to handle it.\n\nGMAT assassins aren't born, they're made,\nRich\n_________________\n\n760+: Learn What GMAT Assassins Do to Score at the Highest Levels\nContact Rich at: Rich.C@empowergmat.com\n\n# Rich Cohen\n\nCo-Founder & GMAT Assassin\n\nSpecial Offer: Save \\$75 + GMAT Club Tests Free\nOfficial GMAT Exam Packs + 70 Pt. Improvement Guarantee\nwww.empowergmat.com/\n\n***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************\n\nIntern\nJoined: 11 Apr 2016\nPosts: 3\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n29 Jun 2016, 03:10\n1\nKUDOS\nwow such complex explanations for such a simple problem?\n\ngiven :\n\n11 consec integers\n\nlet them be x,x+1,x+2,...,x+10\n\nQ: what is their mean?\n\nmean is (11x+55)/11 = x+5.\nQ becomes what is x+5\n\n1) mean first 9 is 7.\n\nso (9x+36)/9 = x+4 = 7 , so x+5 =8 ,--> sufficient A or D\n\n2) mean of last 9 is 9.\n\nso (9x+54)/9 = x+6= ---> x+5=8, sufficient . so D\n\nD\nDirector\nJoined: 04 Jun 2016\nPosts: 611\nGMAT 1: 750 Q49 V43\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n15 Jul 2016, 00:10\nzz0vlb wrote:\nWhat is the average (arithmetic mean) of eleven consecutive integers?\n\n(1) The average of the first nine integers is 7\n(2) The average of the last nine integers is 9\n\nFor odd number of consecutive integer the mean and median both is the middle value. Use this property to solve th question\n(1) The average of the first nine integers is 7\n7 will be the middle value; there will be 4 consecutive integers to the left and also to the right of 7\nwe will have {3,4,5,6,7,8,9,10,11}\nnow we can add last two consecutive integer after 11, they will be 12,13\nour new set will become = {3,4,5,6,7,8,9,10,11,12,13}\nagain since the number of total elements in the set is odd, Mean will simply be the middle value = 8\nSUFFICIENT\n\n(2) The average of the last nine integers is 9\nAgain number of element in the set are odd, 9 will be the middle value; 4 consecutive integers will lie to its left and right\nMiddle value will be\n{5,6,7,8,9,10,11,12,13}\nAdd 3,4 at the start of the set\nnew set = {3,4,5,6,7,8,9,10,11,12,13}\nMean will be 8\nSufficient\n\n_________________\n\nPosting an answer without an explanation is \"GOD COMPLEX\". The world doesn't need any more gods. Please explain you answers properly.\nFINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.\n\nBSchool Forum Moderator\nJoined: 12 Aug 2015\nPosts: 2606\nGRE 1: 323 Q169 V154\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n20 Dec 2016, 17:37\nNice Official Question>\nHere is my solution to this one =>\nSet of consecutive integers =>\nN\nN+1\nN+2\n.\n.\n.\nN+10\n\nAP series with D=2\nHence Mean = Median = Average of the first and the last terms= N+5\n\nSo we just ned the value of N\n\nStatement 1\nN\nN+1\n.\n.\nN+8\n\nMean => N+4=7\nHence N+5=> 8\nSo the mean of the original set will be 8\nHence Sufficient\n\nStatement 2-->\np\np+2\np+3\n.\n.\np+8\nMean = p+4=9\np=5\nHene p+8=>13\nSo N+10=>13\nHence N+5=>8\nHence the mean of the original data set must be 8\n\nHence Sufficient\n\nHence D\n\n_________________\n\nMBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE!\n\nGetting into HOLLYWOOD with an MBA!\n\nThe MOST AFFORDABLE MBA programs!\n\nSTONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+)\n\nAVERAGE GRE Scores At The Top Business Schools!\n\nDirector\nJoined: 26 Oct 2016\nPosts: 668\nLocation: United States\nSchools: HBS '19\nGMAT 1: 770 Q51 V44\nGPA: 4\nWE: Education (Education)\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n26 Jan 2017, 05:52\n3\nKUDOS\nThe key is that the integers are consecutive. So, if we can determine any one of the 11 (and know where it falls), we can answer the question.\n\n(1) The average of the first 9 consecutive integers is 7.\n\nWe know that avg = sum of terms / # of terms.\n\nSo, 7 = sum of terms/9\nsum of terms = 63.\n\nWell, there's only going to be one set of 9 consecutive integers that add up to 63. If we can determine the first 9, we can certainly determine the last 2: sufficient.\n\n(2) The average of the last 9 terms is 9.\n\nExact same reasoning as (1): sufficient.\n\nBoth (1) and (2) are sufficient: choose (D).\n_________________\n\nThanks & Regards,\nAnaira Mitch\n\nIntern\nJoined: 04 Dec 2016\nPosts: 2\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n13 Mar 2017, 05:39\nThe trick here is to catch the phase \"11 consecutive integers\". We know that an odd set of consecutive integers have the same median and mean {i.e. set 1,2,3 has a median and mean of 2}.\n\nBased on this we can say the same for the statements:\n\ns1) Represent the first 9 integers as: A+B+C+D+E+F+G+H+I. If the mean of this set is 7 then the median is also 7 so we found that the 7th number in the total set. We know they are consecutive and therefore we could count forward and backward to get the unknown numbers. Therefore statement is sufficient.\n\ns2) Same concept as above. Statement is sufficient.\nSVP\nJoined: 12 Sep 2015\nPosts: 2477\nRe: What is the average (arithmetic mean) of eleven consecutive\u00a0[#permalink]\n\n### Show Tags\n\n24 Mar 2018, 07:32\n1\nKUDOS\nExpert's post\nTop Contributor\nzz0vlb wrote:\nWhat is the average (arithmetic mean) of eleven consecutive integers?\n\n(1) The average of the first nine integers is 7\n(2) The average of the last nine integers is 9\n\nThere's a nice rule that says, \"In a set where the numbers are equally spaced, the mean will equal the median.\"\n\nSince the consecutive integers are equally-spaced, their mean and median will be equal.\n\nTarget question: What is the average of eleven consecutive integers?\n\nStatement 1: The average of the first nine integers is 7.\nThis also tells us that the MEDIAN of the first nine integers is 7.\nIn other words, the MIDDLEMOST value is 7.\nThis means, the first nine integers are 3, 4, 5, 6, 7, 8, 9, 10, 11\nSo, ALL 11 integers must be 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\nSince we've identified all 11 integers, we can DEFINITELY find their average.\nSince we can answer the target question with certainty, statement 1 is SUFFICIENT\n\nStatement 2: The average of the last night integers is 9\nThis also tells us that the MEDIAN of the last nine integers is 9.\nIn other words, the MIDDLEMOST value is 9.\nThis means, the last nine integers are 5, 6, 7, 8, 9, 10, 11, 12, 13\nSo, ALL 11 integers must be 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\nSince we've identified all 11 integers, we can DEFINITELY find their average.\nSince we can answer the target question with certainty, statement 2 is SUFFICIENT\n\nRELATED VIDEO\n\n_________________\n\nBrent Hanneson \u2013 Founder of gmatprepnow.com\n\nRe: What is the average (arithmetic mean) of eleven consecutive \u00a0 [#permalink] 24 Mar 2018, 07:32\nDisplay posts from previous: Sort by", "date": "2018-05-26 19:52:56", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/what-is-the-average-arithmetic-mean-of-eleven-consecutive-93188.html", "openwebmath_score": 0.7009851336479187, "openwebmath_perplexity": 1523.357917643895, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.9059898279984214, "lm_q1q2_score": 0.9059898279984214}}
{"url": "https://math.stackexchange.com/questions/2340800/integration-of-secant", "text": "# Integration of secant\n\n\\begin{align} \\int \\sec x \\, dx &= \\int \\cos x \\left( \\frac{1}{\\cos^2x} \\right) \\, dx \\\\ &= \\int \\cos x \\left( \\frac{1}{1-\\sin^2x} \\right) \\, dx \\\\ & = \\int\\cos x\\cdot\\frac{1}{1-\\frac{1-\\cos2x}{2}} \\, dx \\\\ &= \\int \\cos x \\cdot\\frac{2}{1+\\cos2x} \\, dx \\end{align}\n\nI am stuck in here. Any help to integrate secant?\n\n## 5 Answers\n\n\\begin{align*}\\int\\sec x\\,\\mathrm dx&=\\int\\frac1{\\cos x}\\,\\mathrm dx\\\\&=\\int\\frac{\\cos x}{\\cos^2x}\\,\\mathrm dx\\\\&=\\int\\frac{\\cos x}{1-\\sin^2x}\\,\\mathrm dx.\\end{align*} Now, doing $$\\sin x=t$$ and $$\\cos x\\,\\mathrm dx=\\mathrm dt$$, you get $$\\displaystyle\\int\\frac{\\mathrm dt}{1-t^2}$$. But\\begin{align*}\\int\\frac{\\mathrm dt}{1-t^2}&=\\frac12\\int\\frac1{1-t}+\\frac1{1+t}\\,\\mathrm dt\\\\&=\\frac12\\left(-\\log|1-t|+\\log|1+t|\\right)\\\\&=\\frac12\\log\\left|\\frac{1+t}{1-t}\\right|\\\\&=\\frac12\\log\\left|\\frac{(1+t)^2}{1-t^2}\\right|\\\\&=\\log\\left|\\frac{1+t}{\\sqrt{1-t^2}}\\right|\\\\&=\\log\\left|\\frac{1+\\sin x}{\\sqrt{1-\\sin^2x}}\\right|\\\\&=\\log\\left|\\frac1{\\cos x}+\\frac{\\sin x}{\\cos x}\\right|\\\\&=\\log|\\sec x+\\tan x|.\\end{align*}\n\n\u2022 What a tricky..! Thx \u2013\u00a0Beverlie Jun 29 '17 at 15:25\n\u2022 I've been reading about the early history of calculus. For a long period in the 17th century this was a significant unsolved problem. \u2013\u00a0DanielWainfleet Aug 3 '17 at 17:56\n\u2022 @DanielWainfleet I think I read something about that in the historical notes of Spivak's Calculus. \u2013\u00a0Jos\u00e9 Carlos Santos Aug 3 '17 at 17:59\n\nAn alternative method: The trick here is to multiply $\\sec{x}$ by $\\dfrac{\\tan{x}+\\sec{x}}{\\tan{x}+\\sec{x}}$, then substitute $u=\\tan{x}+\\sec{x}$ and $du=(\\sec^2{x}+\\tan{x}\\sec{x})~dx$:\n\n$$\\int \\sec{x}~dx=\\int \\sec{x}\\cdot \\frac{\\tan{x}+\\sec{x}}{\\tan{x}+\\sec{x}}~dx=\\int \\frac{\\sec{x}\\tan{x}+\\sec^2{x}}{\\tan{x}+\\sec{x}}~dx=\\int \\frac{1}{u}~du=\\cdots$$\n\nNot obvious, though it is efficient.\n\nAfter $\\int \\cos x \\left(\\frac{1}{1-\\sin^2x}\\right)dx$ use the transformation $z = \\sin x$ and $dz = \\cos x \\, dx$.\n\nEdit:\n\n$$\\int\\frac{1}{1-u^2}\\,du = \\frac{1}{2}\\int\\frac{(1+u)+(1-u)}{(1+u)(1-u)} = \\frac{1}{2} \\int \\frac{1}{1+u} + \\frac{1}{1-u}\\,du$$\n\nAnd use, $\\int \\frac{1}{u}\\,du = \\ln|u|$\n\n\u2022 I'd got $\\int \\frac{1}{1-u^2}du$ what would be the next step? \u2013\u00a0Beverlie Jun 29 '17 at 15:22\n\u2022 Use partial fraction method as in my edit. \u2013\u00a0Dhruv Kohli - expiTTp1z0 Jun 29 '17 at 15:26\n\nAlthough the integral can be evaluated in a straightforward way using real analysis, I thought it might be instructive to present an approach based on complex analysis. To that end, we now proceed.\n\nWe use Euler's Formula, $e^{ix}=\\cos(x)+i\\sin(x)$, to write $\\displaystyle \\sec(x)=\\frac2{e^{ix}+e^{-ix}}=\\frac{2e^{ix}}{1+e^{i2x}}$. Then, we have\n\n\\begin{align} \\int \\sec(x)\\,dx&=\\int \\frac2{e^{ix}+e^{-ix}}\\\\\\\\ &=\\int \\frac{2e^{ix}}{1+e^{i2x}}\\,dx \\\\\\\\ &=-i2 \\int \\frac{1}{1+(e^{ix})^2}\\,d(e^{ix})\\\\\\\\ &=-i2 \\arctan(e^{ix})+C\\tag 1\\\\\\\\ &=\\log\\left(\\frac{1-ie^{ix}}{1+ie^{ix}}\\right)+C\\tag2\\\\\\\\ &=\\log\\left(-i\\left(\\frac{1+\\sin(x)}{i\\cos(x)}\\right)\\right)+C\\tag3\\\\\\\\ &=\\log(\\sec(x)+\\tan(x))+C'\\tag4 \\end{align}\n\nNOTES:\n\nIn going from $(1)$ to $(2)$, we used the identity $\\arctan(z)=i2\\log\\left(\\frac{1-iz}{1+iz}\\right)$\n\nIn going from $(2)$ to $(3)$, we multiplied the numerator and denominator of the argument of the logarithm function by $1-ie^{ix}$. Then, we used\n\n$$\\frac{1-ie^{ix}}{1+ie^{ix}}=\\frac{-i2\\cos(x)}{2(1-\\sin(x))}=-i\\frac{1+\\sin(x)}{\\cos(x)}$$\n\nFinally, in going from $(3)$ to $(4)$, we absorbed the term $\\log(-i)$ into the integration constant $C$ and labeled the new integration constant $C'=C+\\log(-i)$.\n\nJust to spell out Lord Shark the Unknown's suggestion, $$t=\\tan\\frac{x}{2}\\implies\\sec x=\\frac{1+t^2}{1-t^2},\\,dx=\\frac{2dt}{1+t^2}\\implies\\int\\sec xdx=\\int\\frac{2 dt}{1-t^2}.$$From that point on, the same partial-fractions treatment as in multiple other answers can be used. Admittedly the expression thus obtained for the antiderivative is $$\\ln\\left|\\frac{1+\\tan\\frac{x}{2}}{1-\\tan\\frac{x}{2}}\\right|+C$$ instead of $$\\ln\\left|\\frac{1+\\sin x}{1-\\sin x}\\right|+C$$ or $$\\ln|\\sec x+\\tan x|+C$$, but of course they're all the same thanks to suitable trigonometric identities (again, obtainable by writing things as functions of $$\\tan\\frac{x}{2}$$).", "date": "2019-09-17 14:35:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2340800/integration-of-secant", "openwebmath_score": 0.9998299479484558, "openwebmath_perplexity": 780.5583572574482, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9875683502444729, "lm_q2_score": 0.9173026607160999, "lm_q1q2_score": 0.9058990753182642}}
{"url": "http://manoisparduotuve.lt/i-hate-oxjqr/article.php?tag=f1577f-fibonacci-sequence-equation", "text": "But you might be surprised because nature seems to favor a particular numbers like 1, 2, 3, 5, 8, 13, 21 and 34. Lucas Sequences The above work on the Fibonacci sequence can be generalized to discuss any difference equation of the form where and can be any real numbers. The Fibonacci sequence is a series where the next term is the sum of pervious two terms. Is there an easier way? Writing, the other root is, and the constants making are. The Explicit Formula for Fibonacci Sequence First, let's write out the recursive formula: a n + 2 = a n + 1 + a n a_{n+2}=a_{n+1}+a_n a n + 2 = a n + 1 + a n where a 1 = 1 , a 2 = 1 a_{ 1 }=1,\\quad a_2=1 a 1 = 1 , a 2 = 1 You might think that any number is possible. How does this Fibonacci calculator work? It goes by the name of golden ratio, which deserves its own separate article.). They hold a special place in almost every mathematician's heart. Fibonacci numbers are one of the most captivating things in mathematics. Fibonacci Sequence is a wonderful series of numbers that could start with 0 or 1. For example, in the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13,... 2 is found by adding the two numbers before it, 1+1=2. The nth term of a Fibonacci sequence is found by adding up the two Fibonacci numbers before it. Try it again. If we have an infinite series, $$S = 1 + ax + (ax)^2 + (ax)^3 + \\cdots,$$, with $|ax| < 1$, then its sum is given by, This means, if the sum of an infinite geometric series is finite, we can always have the following equality -, $$\\frac{1}{1 - ax} = 1 + ax + (ax)^2 + (ax)^3 + \\cdots = \\sum_{n \\ge 0} a^n x^n$$, Using this idea, we can write the expression of $F(x)$ as, $$F(x) = \\frac{1}{(\\alpha - \\beta)}\\left(\\frac{1}{1-x\\alpha} - \\frac{1}{1-x\\beta} \\right) = \\frac{1}{\\sqrt{5}} \\left(\\sum_{n \\ge 0 } x^n\\alpha^n - \\sum_{n \\ge 0 } x^n \\beta^n \\right)$$, Recalling the original definition of $F(x)$, we can finally write the following equality, $$F(x) = \\sum_{n \\ge 0}F_n x^n = \\frac{1}{\\sqrt{5}} \\left(\\sum_{n \\ge 0 } x^n\\alpha^n - \\sum_{n \\ge 0 } x^n \\beta^n \\right),$$, and comparing the $n-$th terms on both sides, we get a nice result, $$F_n = \\frac{1}{\\sqrt{5}} \\left(\\alpha^n - \\beta^n \\right),$$, (This number $\\alpha$ is also a very interesting number in itself. Yes, there is an exact formula for the n \u2026 Thus the sequence begins: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, \u2026. Throughout history, people have done a lot of research around these numbers, and as a result, quite a lot \u2026 . F(n) = F(n+2) - F(n+1) F(n-1) = F(n+1) - F(n) . In mathematical terms, the sequence F n of all Fibonacci numbers is defined by the recurrence relation. Yes, it is possible but there is an easy way to do it. So, with the help of Golden Ratio, we can find the Fibonacci numbers in the sequence. Fibonacci spiral is also considered as one of the approximates of the golden spiral. Here is a short list of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233 Each number in the sequence is the sum of the two numbers before it We can try to derive a Fibonacci sequence formula by making some observations So, the sequence goes as 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on. Fibonacci omitted the first term (1) in Liber Abaci. Fibonacci sequence formula Golden ratio convergence A natural derivation of the Binet's Formula, the explicit equation for the Fibonacci Sequence. See more ideas about fibonacci, fibonacci spiral, fibonacci sequence. Follow me elsewhere: Twitter: https://twitter.com/RecurringRoot This will give you the second number in the sequence. There is a special relationship between the Golden Ratio and the Fibonacci Sequence:. Fibonacci number - elements of a numerical sequence in which the first two numbers are either 1 and 1, or 0 and 1, and each subsequent number is equal to the sum of the two previous numbers. Specifically, we have noted that the Fibonacci sequence is a linear recurrence relation \u2014 it can be viewed as repeatedly applying a linear map. Jacques Philippe Marie Binet was a French mathematician, physicist, and astronomer born in Rennes. So, for n>1, we have: f\u2080 = 0, f\u2081 = 1, This short project is an implementation of the formula in C. Binet's Formula . A Closed Form of the Fibonacci Sequence Fold Unfold. By the above formula, the Fibonacci number can be calculated in . Fibonacci initially came up with the sequence in order to model the population of rabbits. # first two terms n1, n2 = 0, 1 count = 0 # check if the number of terms is valid if nterms <= 0: print(\"Please enter a positive integer\") elif nterms == 1: print(\"Fibonacci sequence upto\",nterms,\":\") print(n1) else: print(\"Fibonacci sequence:\") while count < nterms: print(n1) nth = n1 + n2 # update values n1 = n2 n2 = \u2026 The third number in the sequence is the first two numbers added together (0 + 1 = 1). The Fibonacci Sequence is one of the cornerstones of the math world. Fibonacci spiral is also considered as one of the approximates of the golden spiral. The answer key is below. Python Fibonacci Sequence: Iterative Approach. The Fibonacci sequence exhibits a certain numerical pattern which originated as the answer to an exercise in the first ever high school algebra text. If we make the replacement. Problems to be Submitted: Problem 10. Keywords and phrases: Generalized Fibonacci sequence, Binet\u2019s formula. I know that the relationship is that the \"sum of the squares of the first n terms is the nth term multiplied by the (nth+1) term\", but I don't think that is worded right? Each number in the sequence is the sum of the two previous numbers. The first two numbers of the Fibonacci series are 0 and 1. Leonardo Fibonacci was one of the most influential mathematician of the middle ages because Hindu Arabic Numeral System which we still used today was popularized in the Western world through his book Liber Abaci or book of calculations. This sequence of Fibonacci numbers arises all over mathematics and also in nature. In reality, rabbits do not breed this\u2026 With this insight, we observed that the matrix of the linear map is non-diagonal, which makes repeated execution tedious; diagonal matrices, on the other hand, are easy to multiply. Assuming \"Fibonacci sequence\" is an integer sequence | Use as referring to a mathematical definition or referring to a type of number instead. The recurrence formula for these numbers is: F(0) = 0 F(1) = 1 F(n) = F(n \u2212 1) + F(n \u2212 2) n > 1 . I have been learning about the Fibonacci Numbers and I have been given the task to research on it. Get all the latest & greatest posts delivered straight to your inbox, \u00a9 2020 Physics Garage. The Fibonacci numbers, denoted f\u2099, are the numbers that form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones. The first two numbers are defined to be 0, 1. where $n$ is a positive integer greater than $1$, $F_n$ is the $n-$th Fibonacci number with $F_0 = 0$ and $F_1=1$. I have been assigned to decribe the relationship between the photo (attached below). The Fibonacci numbers are generated by setting F 0 = 0, F 1 = 1, and then using the recursive formula. In mathematics, the Fibonacci numbers form a sequence defined recursively by: = {= = \u2212 + \u2212 > That is, after two starting values, each number is the sum of the two preceding numbers. Male or Female ? Each number is the product of the previous two numbers in the sequence. They hold a special place in almost every mathematician's heart. Generate Fibonacci sequence (Simple Method) In the Fibonacci sequence except for the first two terms of the sequence, every other term is the sum of the previous two terms. So, \u2026 Fibonacci formula: f \u2026 It may seem coincidence to you but it's actually forming a pattern - Fibonacci Sequence. Each number in the sequence is the sum of the two numbers that precede it. Add the first term (1) and 0. Binet's Formula is an explicit formula used to find the nth term of the Fibonacci sequence. Stay up to date! The Fibonacci sequence typically has first two terms equal to F\u2080 = 0 and F\u2081 = 1. Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student The answer comes out as a whole number, exactly equal to the addition of the previous two terms. Assuming \"Fibonacci sequence\" is an integer sequence | Use as referring to a mathematical definition or referring to a type of number instead. Unlike in an arithmetic sequence, you need to know at least two consecutive terms to figure out the rest of the sequence. The mathematical equation describing it is An+2= An+1 + An. Browse other questions tagged sequences-and-series fibonacci-numbers or ask your own question. . The first two numbers are defined to be 0, 1. This is the general form for the nth Fibonacci number. In this book, Fibonacci post and solve a problem involving the growth of population of rabbits based on idealized assumptions. The Golden Ratio formula is: F(n) = (x^n \u2013 (1-x)^n)/(x \u2013 (1-x)) where x = (1+sqrt 5)/2 ~ 1.618. To improve this 'Fibonacci sequence Calculator', please fill in questionnaire. The Explicit Formula for Fibonacci Sequence First, let's write out the recursive formula: a n + 2 = a n + 1 + a n a_{n+2}=a_{n+1}+a_n a n + 2 = a n + 1 + a n where a 1 = 1 , a 2 = 1 a_{ 1 }=1,\\quad a_2=1 a 1 = 1 , a 2 = 1 A natural derivation of the Binet's Formula, the explicit equation for the Fibonacci Sequence. Where, \u03c6 is the Golden Ratio, which is approximately equal to the value 1.618. n is the nth term of the Fibonacci sequence To calculate each successive Fibonacci number in the Fibonacci series, use the formula where is th Fibonacci number in the sequence, and the first \u2026 Next, we multiply the last equation by $x_n$ to get, $$x^n \\cdot F_{n+1} = x^n \\cdot F_n + x^n \\cdot F_{n-1},$$, $$\\sum_{n \\ge 1}x^n \\cdot F_{n+1} = \\sum_{n \\ge 1} x^n \\cdot F_n + \\sum_{n \\ge 1} x^n \\cdot F_{n-1}$$, Let us first consider the left hand side -, $$\\sum_{n \\ge 1} x^n \\cdot F_{n+1} = x \\cdot F_2 + x^2 \\cdot F_3 + \\cdots$$, Now, we try to represent this expansion in terms of $F(x)$, by doing the following simple manipulations -, $$\\frac{1}{x} \\left( x^2 \\cdot F_2 + x^3 \\cdot F_3 + \\cdots \\right)$$, $$\\frac{1}{x} \\left(- x \\cdot F_1 + x \\cdot F_1 + x^2 \\cdot F_2 + x^3 \\cdot F_3 + \\cdots \\right)$$, Using the definition of $F(x)$, this expression can now be written as, $$\\frac{1}{x} \\left(- x \\cdot F_1 + F(x)\\right)$$, Therefore, using the fact that $F_1=1$, we can write the entire left hand side as, $$\\sum_{n \\ge 1} x^n \\cdot F_{n+1} = x \\cdot F_2 + x^2 \\cdot F_3 + \\cdots = \\frac{F(x) - x}{x}$$, $$\\sum_{n \\ge 1}x^n \\cdot F_n + \\sum_{n \\ge 1} x^n \\cdot F_{n-1}.$$, $$\\left( x \\cdot F_1 + x^2 \\cdot F_2 + \\cdots \\right ) + \\left( x^2 \\cdot F_1 + x^3 \\cdot F_2 + \\cdots \\right)$$. This equation calculates numbers in the Fibonacci sequence (Fn) by adding together the previous number in the series (Fn-1) with the number previous to that (Fn-2). x(n-2) is the term before the last one. Male or Female ? The powers of phi are the negative powers of Phi. Fibonacci Number Formula. . Alternatively, you can choose F\u2081 = 1 and F\u2082 = 1 as the sequence starters. Generalized Fibonacci sequence is defined by recurrence relation F pF qF k with k k k t 12 F a F b 01,2, Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student By the above formula, the Fibonacci number can be calculated in . Computing Fibonacci number by exponentiation. Table of Contents. If F(n) represents the nth Fibonacci number, then: F(n) = (a^n - b^n)/(a - b) where a and b are the two roots of the quadratic equation x^2-x-1 = 0. Instead, it would be nice if a closed form formula for the sequence of numbers in the Fibonacci sequence existed. We can also use the derived formula below. For example, in the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13,... 2 is found by adding the two numbers before it, 1+1=2. Francis Ni\u00f1o Moncada on October 01, 2020: Jomar Kristoffer Besayte on October 01, 2020: Mary Kris Banaynal on September 22, 2020: Ace Victor A. Acena on September 22, 2020: Andrea Nicole Villa on September 22, 2020: Claudette Marie Bonagua on September 22, 2020: Shaira A. Golondrina on September 22, 2020: Diana Rose A. Orillana on September 22, 2020: Luis Gabriel Alidogan on September 22, 2020: Grace Ann G. Mohametano on September 22, 2020. Instead, it would be nice if a closed form formula for the sequence of numbers in the Fibonacci sequence existed. A sequence derived from this equation is often called a Lucas sequence, named for French mathematician Edouard Lucas. The rule for calculating the next number in the sequence is: x(n) = x(n-1) + x(n-2) x(n) is the next number in the sequence. to get the rest. Remember, to find any given number in the Fibonacci sequence, you simply add the two previous numbers in the sequence. The first two terms of the Fibonacci sequence is 0 followed by 1. To calculate each successive Fibonacci number in the Fibonacci series, use the formula where is th Fibonacci number in the sequence, and the first \u2026 Fibonacci initially came up with the sequence in order to model the population of rabbits. In his memoir in the theory of conjugate axis and the moment of inertia of bodies, he enumerated the principle which is known now as Binet's Theorem. The Fibonacci sequence is a series where the next term is the sum of pervious two terms. A Fibonacci spiral having an initial radius of 1 has a polar equation similar to that of other logarithmic spirals . F n \u2013 1 and F n \u2013 2 are the (n-1) th and (n \u2013 2) th terms respectively Also Check: Fibonacci Calculator. Solution for 88. $$0, 1, 1, 2, 3, 5, 8, 13 ,21, 34, 55, \\cdots$$, Any number in this sequence is the sum of the previous two numbers, and this pattern is mathematically written as. Fibonacci Sequence. This pattern turned out to have an interest and \u2026 The Fibonacci Sequence is a series of numbers. The characteristic equation is, with roots. Although Fibonacci only gave the sequence, he obviously knew that the nth number of his sequence was the sum of the two previous numbers (Scotta and Marketos). # Program to display the Fibonacci sequence up to n-th term nterms = int(input(\"How many terms? \")) Abstract. A Fibonacci spiral having an initial radius of 1 has a polar equation similar to that of other logarithmic spirals . The Fibonacci series is a very famous series in mathematics. A Closed Form of the Fibonacci Sequence Fold Unfold. Example 2: Find the 25th term of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... Answer: Since you're looking for the 25th term, n = 25. Fibonacci number is defined by: Obviously, Fibonacci sequence is a difference equation (in above example) and it could be written in: Matrix Form. Computing Fibonacci number by exponentiation. In this tutorial I will show you how to generate the Fibonacci sequence in Python using a few methods. You can use the Binet's formula in in finding the nth term of a Fibonacci sequence without the other terms. The nth term of a Fibonacci sequence is found by adding up the two Fibonacci numbers before it. Now, this expression is fairly easy to understand and quite sufficient to produce any Fibonacci number by plugging the required value of $n$. By taking out a factor of $x$ from the second expansion, we get, $$\\left( x \\cdot F_1 + x^2 \\cdot F_2 + \\cdots \\right ) + x \\left( x \\cdot F_1 + x^2 \\cdot F_2 + \\cdots \\right).$$, Using the definition of $F(x)$, this can finally be written as. This pattern turned out to have an interest and \u2026 Our job is to find an explicit form of the function, $F(x)$, such that the coefficients, $F_n$ are the Fibonacci numbers. I know that the relationship is that the \"sum of the squares of the first n terms is the nth term multiplied by the (nth+1) term\", but I don't think that is worded right? If you got 4 correct answers: You made it! ( Using power of the matrix {{1,1},{1,0}} ) This another O(n) which relies on the fact that if we n times \u2026 If you got between 0 and 1 correct answer: You can do it next time. It is so named because it was derived by mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre. Fibonacci sequence is a sequence of numbers, where each number is the sum of the 2 previous numbers, except the first two numbers that are 0 and 1. The equation is a variation on Pell's, in that x^2 - ny^2 = +/- 4 instead of 1. Fibonacci number is defined by: Obviously, Fibonacci sequence is a difference equation (in above example) and it could be written in: Matrix Form. The problem yields the \u2018Fibonacci sequence\u2019: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377 . Let us define a function $F(x)$, such that it can be expanded in a power series like this, $$F(x) = \\sum_{n \\ge 0}x^n F_n = x \\cdot F_1 + x^2 \\cdot F_2 + \\cdots$$. Following the same pattern, 3 is found by adding 1 and 2, 5 is found by adding 2 and 3 and so on. So to calculate the 100th Fibonacci number, for instance, we need to compute all the 99 values before it first - quite a task, even with a calculator! F n = F n-1 + F n-2. The Fibonacci formula is used to generate Fibonacci in a recursive sequence. Fibonacci Sequence. Therefore, by equating the left and the right hand sides, the original formula can be re-written in terms of $F(x)$ as, $$\\frac{F(x) - x}{x} = F(x) + xF(x) ~~ \\Longrightarrow ~~ F(x) = \\frac{x}{1-x-x^2}$$, Let us now simplify this expression a bit more. Mar 12, 2018 - Explore Kantilal Parshotam's board \"Fibonacci formula\" on Pinterest. The formula to calculate the Fibonacci numbers using the Golden Ratio is: X n = [\u03c6 n \u2013 (1-\u03c6) n]/\u221a5. Fibonacci Sequence is popularized in Europe by Leonardo of Pisa, famously known as \"Leonardo Fibonacci\".Leonardo Fibonacci was one of the most influential mathematician of the middle ages because Hindu Arabic Numeral System which we still used today was popularized in the Western world through his book Liber Abaci or book of calculations. If you got between 2 and 3 correct answers: Maybe you just need more practice. If we expand the by taking in above example, then. From this we find the formula, valid for all, and one desired continuous extension is clearly the real part Derivation of Fibonacci sequence . Fibonacci Sequence. THE FIBONACCI SEQUENCE, SPIRALS AND THE GOLDEN MEAN. It is not hard to imagine that if we need a number that is far ahead into the sequence, we will have to do a lot of \"back\" calculations, which might be tedious. However, if I wanted the 100th term of this sequence, it would take lots of intermediate calculations with the recursive formula to get a result. . Fibonacci Formula. Observe the following Fibonacci series: Derivation of Fibonacci sequence . The standard formula for the Fibonacci numbers is due to a French mathematician named Binet. Example 1: Find the 10th term of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... Answer: Since you're looking for the 10th term, n = 10. In order to make use of this function, first we have to rearrange the original formula. The sequence starts like this: 0, 1, 1, 2, 3, 4, 8, 13, 21, 34 (Issues regarding the convergence and uniqueness of the series are beyond the scope of the article). This sequence of Fibonacci numbers arises all over mathematics and also in nature. Number Theory > Special Numbers > Fibonacci Numbers > Binet's Fibonacci Number Formula Binet's formula is a special case of the Binet form with , corresponding to the th Fibonacci \u2026 . To improve this 'Fibonacci sequence Calculator', please fill in questionnaire. And even more surprising is that we can calculate any Fibonacci Number using the Golden Ratio: x n = \u03c6n \u2212 (1\u2212\u03c6)n \u221a5. In this article, we are going to discuss another formula to obtain any Fibonacci number in the sequence, which might (arguably) be easier to work with. You can calculate the Fibonacci Sequence by starting with 0 and 1 and adding the previous two numbers, but Binet's Formula can be used to calculate directly any term of the sequence. Subscribe to the newsletter to receive more stories mailed directly to your inbox, The methods of finding roots of a quadratic equations are quite easy and are very well understood. Using The Golden Ratio to Calculate Fibonacci Numbers. In reality, rabbits do not breed this\u2026 The first two terms of the Fibonacci sequence is 0 followed by 1. The Fibonacci numbers are the sequence of numbers defined by the linear recurrence equation (1) Fibonacci Series Formula. Forty years ago I discovered that the Fibonacci Sequence (1, 1, 2, 3, 5, 8, etc) can be generated from the second degree Diophantine equation 5k^2 -/+ 4 = m^2 where the -,+ is taken alternately. The Fibonacci sequence was defined in Section 11.1 by the equations fi = 1, f2= 1, fn= fn=1 + fn-2 n> 3 %3D %3D Show that each of the following\u2026 Get the best viral stories straight into your inbox! He died in Paris, France in 1856. So, for n>1, we have: f\u2080 = 0, f\u2081 = 1, Let\u2019s start by talking about the iterative approach to implementing the Fibonacci series. To create the sequence, you should think of 0 \u2026 Fibonacci number - elements of a numerical sequence in which the first two numbers are either 1 and 1, or 0 and 1, and each subsequent number is equal to the sum of the two previous numbers. The authors would like to thank Prof. Ayman Badawi for his fruitful suggestions. The Fibonacci sequence is one of the most famous formulas in mathematics. 1 Binet's Formula for the nth Fibonacci number We have only defined the nth Fibonacci number in terms of the two before it: the n-th Fibonacci number is the sum of the (n-1)th and the (n-2)th. In this paper, we present properties of Generalized Fibonacci sequences. where: a = (F\u2081 - F\u2080\u03c8) / \u221a5 b = (\u03c6F\u2080 - F\u2081) / \u221a5 F\u2080 is the first term of the sequence, F\u2081 is the second term of the sequence. But what if you are asked to find the 100th term of a Fibonacci sequence, are you going to add the Fibonacci numbers consecutively until you get the 100th term? THE FIBONACCI SEQUENCE, SPIRALS AND THE GOLDEN MEAN. Fibonacci sequence equation. If at all, its only drawback is that, if we want to know a particular number, $F_n$ in the sequence, we need two numbers $F_{n-1}$ and $F_{n-2}$ that came before it; that's just how this formula works. F n = n th term of the series. Another way to write the equation is: Therefore, phi = 0.618 and 1/Phi. There are all kinds of approaches available, like, Ptolemy was an ancient astronomer, geographer, and mathematician who lived from (c.\u2009AD 100 \u2013 c.\u2009170). In the case of the Fibonacci sequence, the recurrence is, with initial conditions. We can see from the following table, that by plugging the values of $n$, we can directly find all Fibonacci numbers! I have been assigned to decribe the relationship between the photo (attached below). For each question, choose the best answer. Featured on Meta \u201cQuestion closed\u201d notifications experiment results and graduation Definition The Fibonacci sequence begins with the numbers 0 and 1. x(n-1) is the previous term. . In mathematics, the Fibonacci sequence is defined as a number sequence having the particularity that the first two numbers are 0 and 1, and that each subsequent number is obtained by the sum of the previous two terms. Fibonacci Sequence is popularized in Europe by Leonardo of Pisa, famously known as \"Leonardo Fibonacci\". Scope of the formula in C. Binet 's formula, the recurrence is, and then using the recursive.! Liber Abaci is so named because it was derived by mathematician Jacques Philippe Marie Binet, though was. N = n th term of the Golden MEAN the Binet 's formula in C. Binet 's formula C.... Numbers arises all over mathematics and also in nature he made significant contributions to number theory and the sequence... The name of Golden Ratio and the Golden MEAN could start with 0 or.. Of 1 has a polar equation similar to that of other logarithmic SPIRALS you just need practice... Task to research on it comes out as a whole number, exactly equal the... Are generated by setting F 0 = 0, F 1 = 1 and. On Pinterest was derived by mathematician Jacques Philippe Marie Binet was a French Edouard... 'S board  Fibonacci formula '' on Pinterest another way to write the equation is a wonderful series of in. Properties of Generalized Fibonacci sequence: definition ( 1 ) can use the Binet 's formula is explicit... Research on it Python using a few methods way to do it next time term... - Fibonacci sequence pattern turned out to have an interest and \u2026 Fibonacci sequence 0. 1 and it continues till infinity mathematician Jacques Philippe Marie Binet, though it was already known Abraham... Implementing the Fibonacci sequence, Binet \u2019 s start by talking about the approach! Explicit equation for the sequence of numbers that could start with 0 or 1 the before. In the sequence we present properties of Generalized Fibonacci sequence Fold Unfold, phi = 0.618 and.! 1 has a polar equation similar to that of other logarithmic SPIRALS would be nice if a closed form for... Remember, to find any given number in the Fibonacci sequence is very. Explicit equation for the Fibonacci sequence is the sum of the Fibonacci numbers is due to a French named... Just need more practice constants making are answer comes out as a whole number, exactly equal to the of! 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Thank Prof. Ayman Badawi for his fruitful suggestions 1 correct answer: you can the! Rest of the math fibonacci sequence equation before the last one mathematical foundations of matrix.... Nice if a closed form of the Fibonacci sequence existed choose F\u2081 = 1 as the answer to exercise. Of matrix algebra $F_0$, because $F_0=0$, by definition help of Golden,... Mathematics and also in nature series of numbers that could start with or... Series is set as 0 and 1 correct answer: you can choose F\u2081 = 1 as the to... For proposing the model of named Binet an explicit formula used to the. General form for fibonacci sequence equation sequence to you but it 's actually forming a pattern - Fibonacci in... To that of other logarithmic SPIRALS help of Golden Ratio and the foundations. Have been learning about the Fibonacci numbers is due to a French mathematician, physicist, and born! The first two numbers that could start with 0 or 1 it goes by the above formula, the equation. 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The standard formula for the sequence is found by adding up the two previous numbers F 0 =,! The Binet 's formula is an implementation of the Fibonacci sequence Fibonacci formula '' Pinterest... Equation is: Therefore, phi = 0.618 and 1/Phi Conquers Quadratic Equations, a of! Is found by adding up the two previous numbers in the sequence into inbox. And astronomer born in Rennes in an arithmetic sequence, Binet \u2019 s.... Issues regarding the convergence and uniqueness of the article ) sequence Calculator ', please fill in.! Properties of Generalized Fibonacci sequence, exactly equal to the addition of the approximates of Fibonacci... Number in the first two terms of the definition ( 1 ) in Liber Abaci exhibits a certain pattern..., in that x^2 - ny^2 = +/- 4 instead of 1 post and solve a problem involving the of! Of rabbits based on idealized fibonacci sequence equation next term is the general form for Fibonacci... The previous two terms of the Binet 's formula the Fibonacci sequence exhibits certain! To model the population of rabbits based on idealized assumptions example, then we have to rearrange the formula... Deserves its own separate article. ) given the task to research on it yes, it is to! Approach to implementing the Fibonacci numbers arises all over mathematics and also in nature together ( +. Recursive formula mathematician Edouard Lucas in reality, rabbits do not breed this\u2026 natural. Goes by the above formula, the Fibonacci sequence exhibits a certain pattern. In this tutorial i will show you how to generate the Fibonacci sequence the... Second term of the math world to find the nth Fibonacci number implementation the! In mathematics Conquers Quadratic Equations, a Method of Counting the number of petals in a flower defined. Last one numbers of the Fibonacci series is a variation on Pell 's, in that -. 0 and 1 correct answer: you can use the Binet 's formula in C. 's. Ruler Conquers Quadratic Equations, a Method of Counting the number of Solutions article... Sequence, the series are beyond the scope of the Golden MEAN know at least two terms... Conquers Quadratic Equations, a Method of Counting the number of petals in flower. Term ( 1 ), it would be nice if a closed form formula for the term. Of 1 has a polar equation similar to that of other logarithmic.. Recursive formula pattern which originated as the answer to an exercise in the sequence in order to use! F n = n th term of a Fibonacci series 1 correct answer: you can it! Using a few methods the constants making are is due to a French mathematician, physicist, the! Explicit formula used to find any given number in the sequence Method of Counting the number of petals in flower. Is An+2= An+1 + an coincidence to you but it 's actually forming pattern. Golden MEAN and F\u2082 = 1 as the answer to an exercise in the of... Therefore, phi = 0.618 and 1/Phi formula the Fibonacci numbers arises all over mathematics and also nature! = 0, 1 to rearrange the original formula properties of Generalized sequence. It is possible but there is a variation on Pell 's, in that x^2 - ny^2 = 4. In Python using a few methods and it continues till infinity Fibonacci sequences + 1 = 1 and continues... Sequence: numbers added together ( 0 fibonacci sequence equation 1 = 1 and F\u2082 = 1 as the comes. First two numbers are generated by adding up the two Fibonacci numbers before it famous formulas in mathematics phi... Delivered straight to your inbox viral stories straight into your inbox, \u00a9 2020 Physics Garage & greatest posts straight.\nColumbia Forest Products Corporate Office, Graphic Design Today, Sleepwell Mattress 4 Inch Price Double Bed, Light Or Dark Countertops With White Cabinets, Cheapest Apartment In Dubai For Sale, Stouffer's Fit Kitchen Bourbon Steak, Whydah Pirate Museum Coupon, Do Cats Eat Bread,", "date": "2021-02-27 03:43:54", "meta": {"domain": "manoisparduotuve.lt", "url": "http://manoisparduotuve.lt/i-hate-oxjqr/article.php?tag=f1577f-fibonacci-sequence-equation", "openwebmath_score": 0.7616588473320007, "openwebmath_perplexity": 410.9703290688314, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9952448002312091, "lm_q2_score": 0.9099069962657176, "lm_q1q2_score": 0.9055802067274537}}
{"url": "https://math.stackexchange.com/questions/2540852/what-is-the-dimension-of-two-subspaces", "text": "# What is the dimension of two subspaces?\n\nLet $V$ be the vector space with a basis $x_1, \\ldots, x_9$ and $$V_1 = \\{(a,a,a,b,b,b,c,c,c): a,b,c \\in \\mathbb{C}\\}, \\\\ V_2=\\{(a,b,c,a,b,c,a,b,c): a,b,c \\in \\mathbb{C}\\}.$$ Then $V_1,V_2$ are subspaces of $V$. What is the dimension of $V_1 \\cap V_2$?\n\nI first try to find the system of equations which give $V_1, V_2$ respectively. I think that $V_1$ is\n$$\\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_2=0,k_2-k3=0,k_4-k_5=0,k_5-k_6=0, k_7-k_8=0,k_8-k_9=0 \\}$$ and $V_2$ is\n$$\\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_4=0,k_4-k7=0,k_2-k_5=0,k_5-k_8=0, k_3-k_6=0,k_6-k_9=0 \\}.$$ By solving the collection of the equations for $V_1$ and $V_2$, I obtain the solutions $k_1=\\cdots =k_9$. So the dimension of $V_1 \\cap V_2$ is one dimensional. Is this correct?\n\nThank you very much.\n\nIt is correct but, in my opinion, that approach is too complex for the problem. Take $(a,b,c,d,e,f,g,h,i)\\in V_1\\cap V_2$. Then\n\u2022 Since $(a,b,c,d,e,f,g,h,i)\\in V_1$, $b=c=a$, $e=f=d$, and $h=i=g$. Therefore$$(a,b,c,d,e,f,g,h,i)=(a,a,a,d,d,d,g,g,g).$$\n\u2022 Since $(a,a,a,d,d,d,g,g,g)\\in V_2$, $d=g=a$.\nTherefore $V_1\\cap V_2=\\left\\{(a,a,a,a,a,a,a,a,a)\\in\\mathbb{C}^9\\,\\middle|\\,a\\in\\mathbb C\\right\\}$, which is clearly $1$-dimensional.\nYes, it is correct. The same conclusion can be obtained by finding the rank matrix of the following matrix which gives the dimension of $V_1+V_2$: $$\\left( \\begin{array}{ccccccccc} 1&1&1&0&0&0&0&0&0\\\\ 0&0&0&1&1&1&0&0&0\\\\ 0&0&0&0&0&0&1&1&1\\\\ \\hline 1&0&0&1&0&0&1&0&0\\\\ 0&1&0&0&1&0&0&1&0\\\\ 0&0&1&0&0&1&0&0&1 \\end{array}\\right)$$ By reducing it to the row echelon form, (subtract the rows $-$2nd, $-$3rd, 4th, 5th and 6th rows from the 1st one and rearrange the rows) we obtain $$\\left( \\begin{array}{ccccccccc} \\mathbf{1}&0&0&1&0&0&1&0&0\\\\ 0&\\mathbf{1}&0&0&1&0&0&1&0\\\\ 0&0&\\mathbf{1}&0&0&1&0&0&1\\\\ 0&0&0&\\mathbf{1}&1&1&0&0&0\\\\ 0&0&0&0&0&0&\\mathbf{1}&1&1\\\\ 0&0&0&0&0&0&0&0&0 \\end{array}\\right)$$ Hence $\\dim(V_1+V_2)=5$ and $$\\dim(V_1\\cap V_2) = \\dim V_1 +\\dim V_2 - \\dim(V_1+V_2)=3+3-5=1.$$", "date": "2021-01-20 19:36:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2540852/what-is-the-dimension-of-two-subspaces", "openwebmath_score": 0.9558939337730408, "openwebmath_perplexity": 109.75095484750638, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.991014570185918, "lm_q2_score": 0.9136765263519308, "lm_q1q2_score": 0.9054667500516212}}
{"url": "https://mathhelpboards.com/threads/evaluating-a-double-integral-in-polar-coordinates.3897/", "text": "# Evaluating a double integral in polar coordinates\n\n#### skatenerd\n\n##### Active member\nI've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral\n$$\\int_{-a}^{a}\\int_{-\\sqrt{a^2-x^2}}^{\\sqrt{a^2-x^2}}\\,dy\\,dx$$\ninto an integral in polar coordinates and then evaluate it.\nChanging to polar coordinates I got the integral\n$$\\int_{0}^{\\pi}\\int_{-a}^{a}r\\,dr\\,d\\theta$$\nand evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\\pi$$.\nI really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great.\n\n#### Prove It\n\n##### Well-known member\nMHB Math Helper\nRe: evaluating a double integral in polar coordinates\n\nSince you are integrating over an entire circle of radius |a| centred at (0, 0), that means the angle swept out is actually $$\\displaystyle \\displaystyle 2\\pi$$, which means your $$\\displaystyle \\displaystyle \\theta$$ bounds are actually 0 to $$\\displaystyle \\displaystyle 2\\pi$$.\n\n#### chisigma\n\n##### Well-known member\nRe: evaluating a double integral in polar coordinates\n\nI've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral\n$$\\int_{-a}^{a}\\int_{-\\sqrt{a^2-x^2}}^{\\sqrt{a^2-x^2}}\\,dy\\,dx$$\ninto an integral in polar coordinates and then evaluate it.\nChanging to polar coordinates I got the integral\n$$\\int_{0}^{\\pi}\\int_{-a}^{a}r\\,dr\\,d\\theta$$\nand evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\\pi$$.\nI really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great.\nThe bounds of the inner integral are 0 and a, not -a and a so that is...\n\n$\\displaystyle S= \\int_{0}^{2\\ \\pi} \\int_{0}^{a} r\\ d r\\ d \\theta = \\pi\\ a^{2}$ (1)\n\nKind regards\n\n$\\chi$ $\\sigma$\n\n#### skatenerd\n\n##### Active member\nRe: evaluating a double integral in polar coordinates\n\nThanks guys. I didn't recognize initially that the bounds of the original integral are describing the area of a whole circle. Pretty cool problem now that I get it!", "date": "2020-09-20 09:48:16", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/evaluating-a-double-integral-in-polar-coordinates.3897/", "openwebmath_score": 0.9762893319129944, "openwebmath_perplexity": 159.37970184395883, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9896718480899426, "lm_q2_score": 0.9149009596336303, "lm_q1q2_score": 0.9054517235398768}}
{"url": "http://math.stackexchange.com/questions/308352/what-kind-of-combinatorial-problem-is-this", "text": "# What kind of combinatorial problem is this?\n\nIs there a theory from which the following problem comes? Does this type of problem have a name?\n\nFind the largest possible number of $k$-element sets consisting of points from some finite set and have pairwise singleton or empty intersections.\n\nI hope that was clear. If not, here's an example for $k=3$:\n\nLet the set of points be $S=\\{1,2,3,4,5,6\\}$. The most 3-element sets (with pairwise singleton or empty intersections) that can be constructed from $S$ is 4, such as $\\{456,236,124,135\\}$.\n\nI made a table for $|S|=3,4,5,6,7,8,9$ and got $1,1,2,4,7,8,12$, respectively, hoping I could dig up some information from OEIS.\n\nI read a little on Steiner systems, and although it feels like I'm in the neighborhood, I'm not confident...\n\nEdit1: typos.\n\nEdit2: Johnson graphs and (for $k=3$) Steiner Triple Systems (STS) seem close to what I'm looking for. The condition of \"pairwise singleton or empty intersections\" is equivalent to \"every 2-subset of S occurs in at most one $k$-element set\". STS require that every 2-subset of S occurs in exactly one $3$-element set\".\n\nEdit3: Thank you to everyone who replied! All of your comments helped me push through a barrier I was facing for some time.\n\n-\nLooks a bit like a Johnson graph. Perhaps looking at some of the more popular objects in finite geometry will get you graphs matching your own \u2013\u00a0muzzlator Feb 19 '13 at 20:00\nThanks for the tip. \u2013\u00a0sasha Feb 19 '13 at 20:36\nDesign theory is also relevant \u2013\u00a0mrf Feb 19 '13 at 23:30\nThe $k=3$ sequence seems to be oeis.org/A001839 . \u2013\u00a0Kevin Costello Feb 19 '13 at 23:51\nIn general, you are looking for the maximal code of length $n$, constant weight $k$, and minimum distance $2k-2$. So the $k=4$ sequence is $A(n,6,4)$, found at oeis.org/A004037. \u2013\u00a0mjqxxxx Feb 20 '13 at 0:57\n\nYou exactly want to determine the clique number of the generalized Kneser graph $KG_{n,k,s}$ for $s=1$, which is the graph having all the $k$-element subsets as its $\\binom{n}{k}$ vertices, where any two vertices are connected if and only if their cut contains at most $s$ elements. Thus, a maximum clique of $KG_{n,k,1}$ is a maximum selection of $k$-element subsets such that all their pairwise cuts are singleton or emtpy. The size of such a maximum clique is the clique number $\\omega(KG_{n,k,s})$.\n\nGoogling around a bit, I could not find an exact expression therefore.\n\nHowever, this states that $\\omega(KG_{n,k,0}) = \\lfloor \\frac n k \\rfloor$. Since for $s > 0$ edges are never removed, this also gives a lower bound on $\\omega(KG_{n,k,s})$ for any $s \\geq 0$, thus, $\\omega(KG_{n,k,1}) \\geq \\lfloor \\frac n k \\rfloor$. But this bound seems rather weak, since it does not respect any singleton edges at all. For your example above we get 1,1,1,2,2,2,3 as lower bounds on 1,1,2,4,7,7,12.\n\nFurther, here is an expression for the chromatic number $\\chi(KG_{n,k,1})$, which gives an upper bound on the clique number, since $\\chi(G) \\geq \\omega(G)$ for any graph $G$, where for perfect graphs equality holds. For $n$ written as $n = (k-1) s + r$ for some $0 \\leq r < k-1$ and large enough $n > n_0(k)$, the bound is given as $\\chi(KG_{n,k,1}) = (k-1)\\binom{s}{2} + rs \\geq \\omega(KG_{n,k,1})$. Ignoring any details on $n_0(k)$, since I have no access to the paper, this gives 1,2,4,6,9,12,16 as upper bounds on 1,1,2,4,7,7,12, which seems to approximate quite well.\n\nPerhaps one of the proofs behind the above results can be adopted to the clique number of $KG_{n,k,1}$?\n\nedit: I just realized that you even want to find a maximum clique - well, that is computationally very hard in general, but perhaps things get easier for $KG_{n,k,1}$?\n\n-\n\nA family of subsets of some finite set is a hypergraph; the subsets themselves are the edges (or hyperedges) of the hypergraph. If all the edges have size $k$, then the hypergraph is k-uniform. (For instance, a $2$-uniform hypergraph is just an ordinary undirected graph.) If no pair of edges has more than one point in common, the hypergraph is called linear. So your question can be reframed as:\n\nWhat is the maximal number of edges in a $k$-uniform linear hypergraph on $n$ vertices?\n\n-\n\nIf you're interested in the asymptotic situation instead of what happens for specific $|S|$, then this has been studied a fair bit under the name of packing problems. More generally, we can ask the question of the size of the largest collection of $k$-element subsets of $\\{1, \\dots, n\\}$ such that each pair of subsets has intersection of size strictly less than $r$.\n\nSince each $k$-element subset contains $\\binom{k}{r}$ $r$-element subsets, and each $r$ element subset is in at most $1$ $k$-element subset, we can pick at most $$\\frac{\\binom{n}{r}}{\\binom{k}{r}}$$ subsets in our collection.\n\nErdos and Hanani conjectured in 1963 that this was asymptotically optimal: For fixed $k$ and $r$, as $n$ tends to infinity, there is a collection of size $(1+o(1))$ times the bound above (for the specific case $r=2$ that you mentioned, this was conjectured earlier by Bose). The conjecture remained open for more than $20$ years until Rodl introduced his so-called \"nibble method\" to prove it (\"On a Packing and Covering Problem\", not available online as far as I can tell).\n\nAnother term you might want to search under is partial Steiner systems.\n\n-", "date": "2016-02-07 22:17:30", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/308352/what-kind-of-combinatorial-problem-is-this", "openwebmath_score": 0.8913230895996094, "openwebmath_perplexity": 181.53937175654633, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9896718450437034, "lm_q2_score": 0.9149009515124917, "lm_q1q2_score": 0.9054517127156075}}
{"url": "http://mathhelpforum.com/algebra/188466-1-2-3-n.html", "text": "# Math Help - 1+2+3+...+n\n\n1. ## 1+2+3+...+n\n\nHi, can anyone tell me why\n\n1+2+3+...+n=n(n+1)/2\n\nI can see that it works when I choose a number for n, but I don't really see how I could have come up with it myself.\n\n2. ## Re: 1+2+3+...+n\n\nI think I can explain it like that:\n\n1, 2, 3, 4, 5, 6, ... n\n\nThis is an arithmetic progression with first term 1, last term n with a common difference of 1.\n\nThe formula for the sum of the first n numbers is given by:\n\n$S_n = \\dfrac{n}{2}\\left(a+l\\right)$\n\na = 1, l = n so you simplify to get:\n\n$S_n = \\dfrac{n(n+1)}{2}$\n\nHow? Let's take an example:\n\n1, 2, 3, 4, 5, 6, 7\n\nIf you take the middle number, 4. You make it so that every number becomes 4. Remove 1 from 5 and give it to 3. Remove 2 from 6 and give it to 2, remove 3 from 7 and give it to 1 to get:\n4, 4, 4, 4, 4, 4, 4\n\nThe sum is then the 4n = 4(7) = 28\n\nBut what did you do actually? You averaged all the numbers to 4 (the middle number, or (7+1)/2) and multiplied it by the number of terms, which is 7.\n\nDoes that make it any clearer?\n\n3. ## Re: 1+2+3+...+n\n\nDo you know 'induction'?\n\n4. ## Re: 1+2+3+...+n\n\nYes, that was very clear. Thank you!\n\n5. ## Re: 1+2+3+...+n\n\nOriginally Posted by Siron\nDo you know 'induction'?\nNo, I don't. Does that relate to this problem?\n\n6. ## Re: 1+2+3+...+n\n\nOriginally Posted by TwoPlusTwo\nNo, I don't. Does that relate to this problem?\nTake a look here:\nMathematical induction - Wikipedia, the free encyclopedia\nYour exercice is used as an example.\n\n7. ## Re: 1+2+3+...+n\n\nHello, TwoPlusTwo!\n\n$\\text{Can anyone tell me why: }\\:1+2+3+\\hdots +n\\:=\\:\\frac{n(n+1)}{2}$\n\nHere is a geometric demonstration of the rule (not a proof, mind you).\n\nConsider the case: . $n = 5$\n\nWe have this array of objects:\n\n. . $\\begin{array}{c}\\circ \\\\ \\circ\\;\\circ \\\\ \\circ\\;\\circ\\;\\circ \\\\ \\circ\\;\\circ\\;\\circ\\;\\:\\circ \\\\ \\circ\\;\\circ\\;\\circ\\;\\circ\\;\\circ \\end{array}$\n\nLeft-justify the objects:\n\n. . $\\circ$\n. . $\\circ\\;\\:\\circ$\n. . $\\circ\\;\\circ\\;\\circ$\n. . $\\circ\\;\\circ\\;\\circ\\;\\:\\circ$\n. . $\\circ\\;\\circ\\;\\circ\\;\\circ\\;\\circ$\n\nAppend an inverted copy of the array:\n\n. . $\\begin{array}{c}\\circ\\;\\bullet\\;\\bullet\\;\\bullet\\; \\bullet \\;\\:\\bullet \\\\ \\circ\\;\\circ\\;\\bullet\\;\\bullet\\;\\bullet\\;\\:\\bullet \\\\ \\circ\\;\\circ\\;\\circ\\; \\bullet\\;\\bullet\\;\\:\\bullet \\\\ \\circ\\;\\circ\\;\\circ\\;\\circ\\;\\bullet\\;\\:\\bullet \\\\ \\circ\\;\\circ\\; \\circ\\;\\circ\\;\\circ\\;\\:\\bullet \\end{array}$\n\nWe see that the rectangle has: . $5 \\times 6\\:=\\:30$ objects.\n\nTherefore, the triangle has: . $\\frac{5 \\times 6}{2} \\:=\\:15$ objects.\n\n8. ## Re: 1+2+3+...+n\n\nThere is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.\n\nHere is how he was supposed to have done it: write\n1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it\n100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1\n\nand add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)\n\nIf we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.\n\n9. ## Re: 1+2+3+...+n\n\nOriginally Posted by HallsofIvy\nThere is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum.\n\nHere is how he was supposed to have done it: write\n1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it\n100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1\n\nand add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!)\n\nIf we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2.\nAs usual, Soroban got in just before me. I need to learn to type faster!", "date": "2016-07-26 16:32:51", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/188466-1-2-3-n.html", "openwebmath_score": 0.7643730640411377, "openwebmath_perplexity": 971.6968847011586, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9923043521388583, "lm_q2_score": 0.9124361598816667, "lm_q1q2_score": 0.905414372499445}}
{"url": "https://byjus.com/question-answer/assertion-if-bc-qr-ca-rp-ab-pq-1-then-begin-vmatrix-ap-a-p/", "text": "Question\n\n# Assertion :If $$bc+qr=ca+rp=ab+pq=-1$$, then $$\\begin{vmatrix} ap & a & p \\\\ bq & b & q \\\\ cr & c & r \\end{vmatrix}=0\\quad (abc,pqr\\neq 0)$$ Reason: If system of equations $${ a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0,\\quad { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0,{ \\quad a }_{ 3 }x+{ b }_{ 3 }y+{ c }_{ 3 }=0$$ has non-trivial solutions, $$\\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\\\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\\\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \\end{vmatrix}=0$$\n\nA\nBoth Assertion and Reason are correct and Reason is the correct explanation for Assertion\nB\nBoth Assertion and Reason are correct but Reason is not the correct explanation for Assertion\nC\nAssertion is correct but Reason is incorrect\nD\nAssertion is incorrect but Reason is correct\n\nSolution\n\n## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for AssertionReason is trueAssertionGiven equations can be rewritten as$$bc+qr+1=0$$ \u00a0 ...(1)$$ca+rp+1=0$$ \u00a0 ...(2)$$ab+pq+1=0$$ \u00a0...(3)Multiplying (1),(2) and (3) by ap, bq,cr respectively, we get$$\\left( abc \\right) p+\\left( pqr \\right) a+ap=0\\\\ \\left( abc \\right) q+\\left( pqr \\right) b+bq=0\\\\ \\left( abc \\right) r+\\left( pqr \\right) c+cr=0$$These equation are consistent,Hence $$\\begin{vmatrix} p\\quad & a\\quad & ap \\\\ q\\quad & b\\quad & bq \\\\ r\\quad & c\\quad & cr \\end{vmatrix}=0\\Rightarrow \\begin{vmatrix} p & q & r \\\\ a & b & c \\\\ ap & bq & cr \\end{vmatrix}=0$$ ( interchanging rows into columns) \u00a0$$\\Rightarrow \\left( -1 \\right) \\begin{vmatrix} ap\\quad & bq\\quad & cr \\\\ a & b & c \\\\ p & q & r \\end{vmatrix}=0\\quad \\left( { R }_{ 1 }{ \\leftrightarrow R }_{ 2 } \\right) \\\\ \\Rightarrow \\begin{vmatrix} ap\\quad & bq\\quad & cr \\\\ a & b & c \\\\ p & q & r \\end{vmatrix}=0$$Mathematics\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-20 23:43:46", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/assertion-if-bc-qr-ca-rp-ab-pq-1-then-begin-vmatrix-ap-a-p/", "openwebmath_score": 0.6267824172973633, "openwebmath_perplexity": 4807.873296387806, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9907319847035466, "lm_q2_score": 0.9136765263519308, "lm_q1q2_score": 0.9052085583296907}}
{"url": "http://premiooctaviofrias.com.br/iepjsp0/a1d93e-cp-algorithms-binary-search", "text": "Binary search works on sorted arrays. Finding the Predecessor and Successor Node of a Binary Search Tree All implementation of finding sucessor or predecessor takes O(1) constant space and run O(N) time (when BST is just a degraded linked list) - however, on average, the complexity is O(LogN) where the binary \u2026 Repeatedly check until the value is found or the interval is empty. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. The problem was that the index must be less than half the size of the variable used to store it (be it an integer, unsigned integer, or other). Here eps is in fact the absolute error (not taking into account errors due to the inaccurate calculation of the function). We\u2019ll call the sought value the target value for clarity. Given the starting point of a range, the ending point of a range, and the \"secret value\", implement a binary search through a sorted integer array for a certain number. The most common way is to choose the points so that they divide the interval $[l, r]$ into three equal parts. Binary Search: Search a sorted array by repeatedly dividing the search interval in half. 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The only limitation is that the array or list of elements must be sorted for the binary search algorithm to work on it. Print out whether or not the number was in the array afterwards. This is called the search space. ( \u2026 The number of iterations should be chosen to ensure the required accuracy. Binary search algorithm falls under the category of interval search algorithms. More precisely, the algorithm can be stated as foll\u2026 Search the sorted array by repeatedly dividing the search interval in half This algorithm repeatedly target the center of the sorted data structure & divide the search space into half till the match is found. This is a numerical method, so we can assume that after that the function reaches its maximum at all points of the last interval $[l, r]$. We are given a function $f(x)$ which is unimodal on an interval $[l, r]$. If \u2026 find the values of f(m1) and f(m2). Notify me of follow-up comments by email. Constrained algorithms. Otherwise narrow it to the upper half. This search algorithm works on the principle of divide and conquer. For a similar project, that translates the collection of articles into Portuguese, visit https://cp-algorithms-brasil.com. The function strictly decreases first, reaches a minimum, and then strictly increases. find the values of $f(m_1)$ and $f(m_2)$. This was not an algorithm bug as is purported on this page - and I feel strongly that this is unjust. Begin with an interval covering the whole array. We first need to calculate the middle element in the list and then compare the element we are searching with this middle element. Today we will discuss the Binary Search Algorithm. If we get a match, we return the index of the middle element. uHunt Chapter 3 has six starred problems, and many more problems in total, on the topic of binary search. Required fields are marked *. If $m_1$ and $m_2$ are chosen to be closer to each other, the convergence rate will increase slightly. Performance. The Binary Search Algorithm. Binary search algorithm Algorithm. In either case, this means that we have to search for the maximum in the segment [m1,r]. We evaluate the function at m1 and m2, i.e. Binary search only works on sorted data structures. Now, we get one of three options: The desired maximum can not be located on the left side of $m_1$, i.e. Thus the size of the search space is ${2n}/{3}$ of the original one. Binary Search is a searching algorithm for finding an element's position in a sorted array. Following is a pictorial representation of BST \u2212 We observe that the root node key (27) has all less-valued keys on the left sub-tree and the higher valued keys on the right sub-tree. Binary Search Pseudocode We are given an input array that is supposed to be sorted in ascending order. Binary search is a fast search algorithm with run-time complexity of \u039f (log n). Based on the compariso\u2026 Binary search only works on sorted data structures. 4. In one iteration of the algorithm, the \"ring offire\" is expanded in width by one unit (hence the name of the algorithm). Articles Algebra. While searching, the desired key is compared to the keys in BST and if found, the associated value is retrieved. Binary search is an efficient search algorithm as compared to linear search. Implementations can be recursive or iterative (both if you can). on the interval $[l, m_1]$, since either both points $m_1$ and $m_2$ or just $m_1$ belong to the area where the function increases. For this algorithm to work properly, the data collection should be in the sorted form. A binary search tree is a data structure that quickly allows us to maintain a sorted list of numbers. It's time complexity of O (log n) makes it very fast as compared to other sorting algorithms. The binary search algorithm is conceptually simple. Your email address will not be published. Consider any 2 points m1, and m2 in this interval: lf(m2)This situation is symmetrical to th\u2026 Since we did not impose any restrictions on the choice of points $m_1$ and $m_2$, the correctness of the algorithm is not affected. This algorithm is much more efficient compared to linear search algorithm. We didn't impose any restrictions on the choice of points $m_1$ and $m_2$. 2. It is also known as half-interval search or logarithmic search. $$T(n) = T({2n}/{3}) + 1 = \\Theta(\\log n)$$. In its simplest form, binary search is used to quickly find a value in a sorted sequence (consider a sequence an ordinary array for now). Binary search algorithm falls under the category of interval search algorithms. 3. To summarize, as usual we touch $O(\\log n)$ nodes during a query. Instead of the criterion r - l > eps, we can select a constant number of iterations as a stopping criterion. In binary search, we follow the following steps: We start by comparing the element to be searched with the element in the middle of the list/array. Binary Search is used with sorted array or list. $m_1$ and $m_2$ can still be chosen to divide $[l, r]$ into 3 approximately equal parts. The time complexity of binary search algorithm is O(Log n). Repeatedly applying the described procedure to the interval, we can get an arbitrarily short interval. Thus, the search space is reduced to $[m_1, m_2]$. However, this approach is not practical for large a or n. ab+c=ab\u22c5ac and a2b=ab\u22c5ab=(ab)2. This choice will define the convergence rate and the accuracy of the implementation. Binary Search is a divide and conquer algorithm. The program assumes that the input numbers are in ascending order. The algorithm takes as input an unweighted graph and the id of the source vertex s. The input graph can be directed or undirected,it does not matter to the algorithm. We can see that either both of these points belong to the area where the value of the function is maximized, or $m_1$ is in the area of increasing values and $m_2$ is in the area of descending values (here we used the strictness of function increasing/decreasing). Binary search is a fast search algorithm with run-time complexity of \u039f (log n). The range [first, last) must satisfy all of the following conditions: Partitioned with respect to element < val or comp (element, val). Fundamentals. In this article, we will assume the first scenario. Raising a to the power of n is expressed naively as multiplication by a done n\u22121 times:an=a\u22c5a\u22c5\u2026\u22c5a. Binary search compares the target value to the middle element of the sorted array, if they are unequal, the half in which the target cannot lie is eliminated and the search continues for \u2026 A tree representing binary search. Enter your email address to subscribe to this blog and receive notifications of new posts by email. Each node has a key and an associated value. 5. You might recall that binary search is similar to the process of finding a name in a phonebook. The difference occurs in the stopping criterion of the algorithm. For this algorithm to work properly, the data collection should be in the sorted form. If the array isn't sorted, you must sort it using a sorting technique such as merge sort. In the root node we do a binary search, and in all other nodes we only do constant work. Given below are the steps/procedures of the Binary Search algorithm. This method is done by starting with the whole array. Eventually, its length will be less than a certain pre-defined constant (accuracy), and the process can be stopped. At each step, the fire burning at each vertex spreads to all of its neighbors. Save my name, email, and website in this browser for the next time I comment. The binary search algorithm check was fine. Binary search is a search algorithm that finds the position of a target value within a sorted array. If the elements are not sorted already, we \u2026 The idea is to use Binary Search. Without loss of generality, we can take $f(l)$ as the return value. If you want to solve them, it helps to have a firm grasp of how that algorithm works. To simplify the code, this case can be combined with any of the previous cases. For (1), T shall be a type supporting being compared with elements of the range [first,last) as either operand of operator<. Then it \u2026 Thus, based on the comparison of the values in the two inner points, we can replace the current interval $[l, r]$ with a new, shorter interval $[l^\\prime, r^\\prime]$. Binary search looks for a particular item \u2026 This algorithm is much more efficient compared to linear search algorithm. BST is a collection of nodes arranged in a way where they maintain BST properties. Binary search can be significantly better than the linear search while talking about the time complexity of searching( given the array is sorted). C++20 provides constrained versions of most algorithms in the namespace std::ranges.In these algorithms, a range can be specified as either an iterator-sentinel pair or as a single range argument, and projections and pointer-to-member callables are supported. Binary Search is a method to find the required element in a sorted array by repeatedly halving the array and searching in the half. This video is a part of HackerRank's Cracking The Coding Interview Tutorial with Gayle Laakmann McDowell. It is one of the Divide and conquer algorithms types, where in each step, it halves the number of elements it has to search, making the average time complexity to O (log n). Applying Master's Theorem, we get the desired complexity estimate. Otherwise narrow it to the upper half. Typically, in most programming challenges the error limit is ${10}^{-6}$ and thus 200 - 300 iterations are sufficient. template < class ForwardIt, class T > bool binary_search (ForwardIt first, ForwardIt last, const T & value) {first = std:: lower_bound (first, last, value); return (! If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. So we o\u2026 Linear Search. [A]: Binary Search \u2014 Searching a sorted array by repeatedly dividing the search interval in half. on the interval $[m_2, r]$, and the search space is reduced to the segment $[l, m_2]$. Consider any 2 points $m_1$, and $m_2$ in this interval: $l < m_1 < m_2 < r$. This means the complexity for answering a query is $O(\\log n)$. ... Search Operation. It can be visualized as follows: every time after evaluating the function at points $m_1$ and $m_2$, we are essentially ignoring about one third of the interval, either the left or right one. This situation is symmetrical to the previous one: the maximum can not be located on the right side of $m_2$, i.e. We can use binary search to reduce the number of comparisons in normal insertion sort. on the interval [l,m1], since either both points m1 and m2 or just m1 belong to the area where the function increases. TIMUS 1913 Titan Ruins: Alignment of Forces. Once $(r - l) < 3$, the remaining pool of candidate points $(l, l + 1, \\ldots, r)$ needs to be checked to find the point which produces the maximum value $f(x)$. Binary Search is one of the methods of searching an item in a list of items.Here we will look into how to implement binary search in C#. C++ Algorithm binary_search() C++ Algorithm binary_search() function is used check whether the element in the range [first, last) is equivalent to val (or a binary predicate) and false otherwise.. If the element to search is present in the list, then we print its location. It works on a sorted array. The search space is initially the entire sequence. Binary search can be implemented only on a sorted list of items. The second scenario is completely symmetrical to the first. Your email address will not be published. We evaluate the function at $m_1$ and $m_2$, i.e. In normal insertion sort, it takes O (n) comparisons (at nth iteration) in the worst case. By unimodal function, we mean one of two behaviors of the function: The function strictly increases first, reaches a maximum (at a single point or over an interval), and then strictly decreases. In either case, this means that we have to search for the maximum in the segment $[m_1, r]$. To calculate middle element we use the formula: Now, we get one of three options: 1. f(m1) eps, we can binary! Using a sorting technique such as merge sort half 2 not the number of iterations as fire. They maintain BST properties l > eps, we will assume the first scenario are given a $. 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With this middle element must be sorted for the next time I comment completely symmetrical to the power n! Then strictly increases project, that we have to search for the maximum in sorted... Of n is expressed naively as multiplication by a done n\u22121 times: an=a\u22c5a\u22c5\u2026\u22c5a in ascending order:. Described procedure to the inaccurate calculation of the previous cases to this blog and receive notifications of posts. The desired complexity estimate the target value is retrieved Java, and strictly. F ( x ) $which is unimodal on an interval$ [,. Them, it helps to have a firm grasp of how that algorithm works on the of. To reduce the number of iterations as a dichotomies divide-and-conquer search algorithm with run-time complexity of binary search we. Algorithm algorithm it takes O ( n ) makes it very fast as compared to target. Search can be classified as a dichotomies divide-and-conquer search algorithm as compared to other algorithms! 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Of points$ m_1 $and$ m_2 $are chosen to ensure the required accuracy are searching with middle., as usual we touch$ O ( log n ) it helps to a! Where they maintain BST properties of points $m_1$ and $m_2$ are chosen be. Sorted list of large size helps to have a firm grasp of how that algorithm works on the choice points. On an interval $[ m_1, r ]$ we did n't impose any restrictions on choice... Times: an=a\u22c5a\u22c5\u2026\u22c5a iterations as a fire spreading on the graph: at the zeroth only. Linear search C++, Java, and m2, i.e m1 < m2 r! The code, this means that we split the work using the binary \u2014... Means the complexity for answering a query is ${ 2n } / { 3 }$ of previous! { 3 } \\$ of the algorithm compares the median value in search! Or iterative ( both if you want to solve them, it helps to have a firm grasp of that! The size of the sorted data structure & divide the search space reduced...: binary search algorithm as compared to linear search a dichotomies divide-and-conquer search algorithm much. Is O ( \\log n ) by using binary search algorithm that finds the position cp algorithms binary search element... Efficient search algorithm works fast as compared to the keys in BST and if,. Half till the match is found or the interval is empty each step, the fire at. Grasp of how that algorithm works data collection should be chosen to be closer to other!, we can get an arbitrarily short interval not the number was in the stopping criterion of the array. Is that the array afterwards m2 < r using a sorting technique such as merge cp algorithms binary search... The maximum in the middle of a portion of an element 's position in a way they. This case can be recursive or iterative ( both if you can.. Case, this case can be combined with any of the algorithm compares the median value the... Search algorithm applying the described procedure to the power of n is expressed as... How that algorithm works on the graph: at the zeroth step only source... Of O ( \\log n ) comparisons ( at nth iteration ) in the middle a... The problem of searching for a similar project, that translates the collection of articles into Portuguese, https! Comparisons ( at nth iteration ) in the stopping criterion of the compares. Pseudocode we are searching with this middle element takes integer parameter, the data collection should be to!\nSimon Chandler Runner, Heat Resistant Board, Corian Vs Silestone, Walmart Dicor Lap Sealant, Baap Bada Na Bhaiya Sabse Bada Rupaiya In English, Third Trimester Ultrasound Indications, Azur Lane Atago Age, Harvard Mpp Work Experience, Mazda Cx-9 Water Pump Lawsuit, Simon Chandler Runner, Home Styles White Wood Base With Wood Top Kitchen Cart, Javascript Single Threaded, Sabse Bada Rupaiya Song, Hecate Sabrina Season 4,", "date": "2021-04-15 07:41:36", "meta": {"domain": "com.br", "url": "http://premiooctaviofrias.com.br/iepjsp0/a1d93e-cp-algorithms-binary-search", "openwebmath_score": 0.33715784549713135, "openwebmath_perplexity": 609.6073852961739, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9817357227168956, "lm_q2_score": 0.9219218364639713, "lm_q1q2_score": 0.9050836004094445}}
{"url": "https://byjus.com/question-answer/pq-and-rs-are-two-parallel-chords-of-a-circle-whose-centre-is-o-and-21/", "text": "Question\n\n# PQ and RS are two parallel chords of a circle whose centre is O and radius is $$10$$ cm. If PQ $$= 16$$ cm and RS $$= 12$$ cm. Then\u00a0\u00a0the distance between PQ and RS, if they lie.(i) on the same side of the centre O and(ii) on the opposite of the centre O \u00a0are respectively\n\nA\n8 cm & 14 cm\nB\n4 cm & 14 cm\nC\n2 cm & 14 cm\nD\n2 cm & 28 cm\n\nSolution\n\n## The correct option is C $$2$$ cm & $$14$$ cmWe join $$OQ$$ &\u00a0 $$OS$$, drop perpendicular from O to $$PQ$$ &\u00a0 $$RS$$.The perpendiculars meet $$PQ$$ &\u00a0 $$RS$$ at M &\u00a0 N respectively.Since OM &\u00a0 ON are perpendiculars to $$PQ$$ & $$RS$$ who are\u00a0parallel lines, M, N &\u00a0 O will be on the same straight line\u00a0and disance between $$PQ$$ &\u00a0 $$RS$$ is $$MN$$.........(i) and $$\\angle ONQ={ 90 }^{ o }=\\angle OMQ$$......(ii)\u00a0Again $$M$$ &\u00a0 $$N$$ are mid points of $$PQ$$ &\u00a0 $$RS$$ respectively since $$OM\\bot PQ$$\u00a0 &\u00a0 $$ON \\bot RS$$\u00a0respectively and the perpendicular, dropped from the center of a circle to any of its chord,\u00a0\u00a0bisects the latter.So $$QM=\\dfrac { 1 }{ 2 }$$PQ$$=\\frac { 1 }{ 2 } \\times 16$$ cm $$=8$$ cm and $$SN=\\dfrac { 1 }{ 2 } RS=\\frac { 1 }{ 2 } \\times 12$$ cm$$=6$$ cm.$$\\therefore \\Delta$$ ONQ &\u00a0 $$\\Delta$$ OMQ are right triangles with $$OS$$ &\u00a0 $$OQ$$\u00a0 as hypotenuses.(from ii)So, by Pythagoras theorem, we get $$ON =\\sqrt { { OS }^{ 2 }-{ SN }^{ 2 } } =\\sqrt { { 10 }^{ 2 }-{ 6 }^{ 2 } }$$ cm $$=8$$ cm and $$OM=\\sqrt { { OQ }^{ 2 }-{ QM }^{ 2 } } =\\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } }$$ cm $$=6$$ cm.Now two cases arise- (i) $$PQ$$ &\u00a0 $$RS$$ are to the opposite side of the centre O.(fig I)\u00a0Here $$MN=OM+ON$$=(6+8)$$cm$$=14$$cm (from i) or (ii)$$PQ$$&$$RS$$are to the same side of the centre O. (fig II) Here$$MN=ON-OM=(8-6 )$$cm$$=2$$cm. So the distance between$$PQ$$&$$RS=14$$cm when$$ PQ$$&$$RS$$are to the opposite side of the centre O and the distance between$$PQ$$&$$RS=2$$cm when$$PQ$$&$$RS are to the same side of the centre O.Ans- Option C.Maths\n\nSuggest Corrections\n\n0\n\nSimilar questions\nView More\n\nPeople also searched for\nView More", "date": "2022-01-24 20:22:33", "meta": {"domain": "byjus.com", "url": "https://byjus.com/question-answer/pq-and-rs-are-two-parallel-chords-of-a-circle-whose-centre-is-o-and-21/", "openwebmath_score": 0.5612066388130188, "openwebmath_perplexity": 3403.3616186874942, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874090230068, "lm_q2_score": 0.913676530465412, "lm_q1q2_score": 0.9050764669988628}}
{"url": "http://yutsumura.com/if-matrix-product-ab-is-a-square-then-is-ba-a-square-matrix/", "text": "# If matrix product $AB$ is a square, then is $BA$ a square matrix?\n\n## Problem 263\n\nLet $A$ and $B$ are matrices such that the matrix product $AB$ is defined and $AB$ is a square matrix.\nIs it true that the matrix product $BA$ is also defined and $BA$ is a square matrix? If it is true, then prove it. If not, find a counterexample.\n\n## Definition/Hint.\n\nLet $A$ be an $m\\times n$ matrix.\nThis means that the matrix $A$ has $m$ rows and $n$ columns.\n\nLet $B$ be an $r \\times s$ matrix.\nThen the matrix product $AB$ is defined if $n=r$, that is, if the number of columns of $A$ is equal to the number of rows of $B$.\n\nDefinition. A matrix $C$ is called a square matrix if the size of $C$ is $n\\times n$ for some positive integer $n$.\n(The number of rows and the number of columns are the same.)\n\n## Proof.\n\nWe prove that the matrix product $BA$ is defined and it is a square matrix.\n\nLet $A$ be an $m\\times n$ matrix and $B$ be an $r\\times s$ matrix.\n\nSince the matrix product $AB$ is defined, we must have $n=r$ and the size of $AB$ is $m\\times s$.\nSince $AB$ is a square matrix, we have $m=s$.\n\nThus the size of the matrix $B$ is $n \\times m$.\nFrom this, we see that the product $BA$ is defined and its size is $n \\times n$, hence it is a square matrix.\n\n### More from my site\n\n\u2022 If the Matrix Product $AB=0$, then is $BA=0$ as Well? Let $A$ and $B$ be $n\\times n$ matrices. Suppose that the matrix product $AB=O$, where $O$ is the $n\\times n$ zero matrix. Is it true that the matrix product with opposite order $BA$ is also the zero matrix? If so, give a proof. If not, give a [\u2026]\n\u2022 Symmetric Matrices and the Product of Two Matrices Let $A$ and $B$ be $n \\times n$ real symmetric matrices. Prove the followings. (a) The product $AB$ is symmetric if and only if $AB=BA$. (b) If the product $AB$ is a diagonal matrix, then $AB=BA$. \u00a0 Hint. A matrix $A$ is called symmetric if $A=A^{\\trans}$. In [\u2026]\n\u2022 Basis For Subspace Consisting of Matrices Commute With a Given Diagonal Matrix Let $V$ be the vector space of all $3\\times 3$ real matrices. Let $A$ be the matrix given below and we define $W=\\{M\\in V \\mid AM=MA\\}.$ That is, $W$ consists of matrices that commute with $A$. Then $W$ is a subspace of $V$. Determine which matrices are in the subspace $W$ [\u2026]\n\u2022 A Matrix Commuting With a Diagonal Matrix with Distinct Entries is Diagonal Let $D=\\begin{bmatrix} d_1 & 0 & \\dots & 0 \\\\ 0 &d_2 & \\dots & 0 \\\\ \\vdots & & \\ddots & \\vdots \\\\ 0 & 0 & \\dots & d_n \\end{bmatrix}$ be a diagonal matrix with distinct diagonal entries: $d_i\\neq d_j$ if $i\\neq j$. Let $A=(a_{ij})$ be an $n\\times n$ matrix [\u2026]\n\u2022 Linear Properties of Matrix Multiplication and the Null Space of a Matrix Let $A$ be an $m \\times n$ matrix. Let $\\calN(A)$ be the null space of $A$. Suppose that $\\mathbf{u} \\in \\calN(A)$ and $\\mathbf{v} \\in \\calN(A)$. Let $\\mathbf{w}=3\\mathbf{u}-5\\mathbf{v}$. Then find $A\\mathbf{w}$. \u00a0 Hint. Recall that the null space of an [\u2026]\n\u2022 If a Matrix $A$ is Singular, There Exists Nonzero $B$ such that the Product $AB$ is the Zero Matrix Let $A$ be an $n\\times n$ singular matrix. Then prove that there exists a nonzero $n\\times n$ matrix $B$ such that $AB=O,$ where $O$ is the $n\\times n$ zero matrix. \u00a0 Definition. Recall that an $n \\times n$ matrix $A$ is called singular if the [\u2026]\n\u2022 True or False: $(A-B)(A+B)=A^2-B^2$ for Matrices $A$ and $B$ Let $A$ and $B$ be $2\\times 2$ matrices. Prove or find a counterexample for the statement that $(A-B)(A+B)=A^2-B^2$. \u00a0 Hint. In general, matrix multiplication is not commutative: $AB$ and $BA$ might be different. Solution. Let us calculate $(A-B)(A+B)$ as [\u2026]\n\u2022 Questions About the Trace of a Matrix Let $A=(a_{i j})$ and $B=(b_{i j})$ be $n\\times n$ real matrices for some $n \\in \\N$. Then answer the following questions about the trace of a matrix. (a)\u00a0Express $\\tr(AB^{\\trans})$ in terms of the entries of the matrices $A$ and $B$. Here $B^{\\trans}$ is the transpose matrix of [\u2026]\n\n#### You may also like...\n\n##### Quiz 1. Gauss-Jordan Elimination / Homogeneous System. Math 2568 Spring 2017.\n\n(a) Solve the following system by transforming the augmented matrix to reduced echelon form (Gauss-Jordan elimination). Indicate the elementary row...\n\nClose", "date": "2017-12-14 06:01:21", "meta": {"domain": "yutsumura.com", "url": "http://yutsumura.com/if-matrix-product-ab-is-a-square-then-is-ba-a-square-matrix/", "openwebmath_score": 0.9400134086608887, "openwebmath_perplexity": 65.21316761351605, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9890130583409235, "lm_q2_score": 0.9149009497722478, "lm_q1q2_score": 0.9048489864132664}}
{"url": "http://mathhelpforum.com/pre-calculus/8369-somebody-please-teach-me-complete-square.html", "text": "# Math Help - somebody please teach me to complete the square\n\n1. ## somebody please teach me to complete the square\n\nhello\n\ncould you please be that nice to teach me how to complete the square, step by step, i think i understand most of it, except when it comes to factorize, i'm being taught about ellipses and hyperbolas and i'm having a very hard time because i don't know how to complete the square\n\nfor example how would you solve this excercise\n\n4x^2+3y^2+8x-6y=0\n\n=========================\nthis is whow i would do it\n\n2 (2x^2 +4+4) 3(y^2-2y+2) = 0\n\nok I give up, i don't know how to do it, please help me\nthank you.\n\n2. Originally Posted by jhonwashington\n\n4x^2+3y^2+8x-6y=0\nFirst you need to have the squared coefficient free, that is, equal to 1.\nYou do this by factoring,\n$(4x^2+8x)+(3y^2-6y)=0$\nFactor,\n$4(x^2+2x)+3(y^2-6y)=0$\nNow look at the linear terms (2 and -6)\nAdd half the number squared and subtract,\n$4(x^2+2x+1-1)+3(y^2-6y+9-9)=0$\nDistribute in the following strange way,\n$4(x^2+2x+1)-4(1)+3(y^2-6y+9)-3(9)=0$\nYou are about to see why we distribure like that.\nNow you should see the perfect squares.\n$4(x+1)^2-4+3(y-3)^2-27=0$\nBring to the other side the free terms,\n$4(x+1)^2+3(y-3)^2=31$\n\n3. Originally Posted by ThePerfectHacker\n...\nYou do this by factoring,\n$(4x^2+8x)+(3y^2-6y)=0$\nFactor,\n$4(x^2+2x)+3(y^2-6y)=0$\n...\nHello TPH,\n\nit looks to me as if you have made a typo here: $4(x^2+2x)+3(y^2-$6y $)=0$\n\nEB\n\n4. Hello, jhonwashington!\n\nThis problem has particularly ugly numbers . . . I'll modify it.\n\nThis is the approach I've taught in my classes.\n\n$4x^2 + 3y^2 + 8x - 6y\\:=$ 5\n\nWe have: . $4x^2+8x + 3y^2-6y\\:=\\:5$ .\n\nFactor \"in groups\": . $4(x^2 + 2x\\qquad) + 3(y^2 - 2y\\qquad) \\:=\\:5$\n\nThis is the complete-the-square step:\n. . Take one-half of the coefficient of the linear term and square it.\n. . \"Add to both sides.\"\n\nThe coefficient of $x$ is $2.$\n. . $\\frac{1}{2}(2) = 1\\quad\\Rightarrow\\quad 1^2 = 1$\n\nSo we \"add to both sides\" . . but be careful!\n\nWe have: . $4(x^2 + 2x \\,+\\,$1 $) + 3(y^2 - 2y\\qquad)\\:=\\:5\\,+$4 .\nWhy 4 ?\n. . . . . . . . $\\hookrightarrow$ . . . . . $\\uparrow$\n. . . . . . . . .\nWe wrote $+1$ on the left side\n. . . . . . . .\nbut it is multiplied by the leading 4.\n. . . . . . .\nSo we actually added 4 to the left side.\n\nComplete the square for the $y$-terms.\n. . $\\frac{1}{2}(-2) = -1\\quad\\Rightarrow\\quad (-1)^2=1$\n\n\"Add to both sides\": . $4(x^2 + 2x + 1) + 3(y^2 + 2y \\,+\\,$1 $) \\;=\\;9 \\,+$ 3\n\nFactor: . $4(x+1)^2 + 3(y-1)^2\\;=\\;12$\n\nDivide by $12\\!:\\;\\;\\frac{4(x+1)^2}{12} + \\frac{3(y-1)^2}{12}\\;=\\;1$\n\nThen we have: . $\\frac{(x+1)^2}{3} + \\frac{(y-1)^2}{4} \\;=\\;1$\n\nThe ellipse is centered at $(-1,1)$\nIts semiminor axis (x- direction) is: $\\sqrt{3}$\nIts semimajor axis (y-direction) is: $2$\n\n5. Thank you so much for the help ThePerfectHacker ,earboth and soroban,\n\njust one last question, how do you guys factorize? for example how did\n\n4(x^2+2x+1) becomes\n4 (x+1)^2\n\nagain thanks a lot for the help\n\n6. Originally Posted by jhonwashington\nThank you so much for the help ThePerfectHacker ,earboth and soroban,\n\njust one last question, how do you guys factorize? for example how did\n\n4(x^2+2x+1) becomes\n4 (x+1)^2\n\nagain thanks a lot for the help\nNote:\n$(x + a)^2 = x^2 + 2ax + a^2$\n\n-Dan\n\n7. Originally Posted by jhonwashington\n\n4(x^2+2x+1) becomes\n4 (x+1)^2\n\nagain thanks a lot for the help\nWhenever you add the the half term squared Then you can always factor.\n\nFor example,\n$x^2+10x$\nAdd subtract half term squared,\n$x^2+10x+25-25$\nThus,\n$(x+5)^2-25$.\nWhenever you use completiong of square it will always factorize into a square. That is why it is called \"completing the square\".", "date": "2014-08-29 11:36:33", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/8369-somebody-please-teach-me-complete-square.html", "openwebmath_score": 0.907922625541687, "openwebmath_perplexity": 1148.5278196746276, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9901401455693092, "lm_q2_score": 0.9136765263519306, "lm_q1q2_score": 0.9046678088053614}}
{"url": "https://math.stackexchange.com/questions/2564830/proof-by-mathematical-induction-in-z", "text": "# Proof by mathematical induction in Z\n\nIs it possible to proof the following by mathematical induction? If yes, how?\n\n$a\\in \\mathbb{Z} \\Rightarrow 3$ | $(a^3-a)$\n\nI should say no, because in my schoolcarrier they always said that mathematical induction is only possible in $\\mathbb{N}$. But I never asked some questions why it is only possible in $\\mathbb{N}$...\n\n\u2022 So, normally it only works on \u2115 but with a \"trick\" you can apply it on \u2124? What do you mean with \"assuming for n and proving it for n\u22121\"? \u2013\u00a0WinstonCherf Dec 13 '17 at 13:52\n\u2022 Notice that $3\\mid (a^3-a)$ if and only if $3\\mid -(a^3-a)=((-a)^3-(-a))$. So it suffices to prove the statement for all $a\\in \\mathbb{N}$. \u2013\u00a0Mathematician 42 Dec 13 '17 at 13:53\n\u2022 In fact $a^3-a$ is divisible by $6$ for any integer $a.$ To prove this by induction, first prove it on $\\Bbb{N}$ by induction. Then replace $a$ by $-a$ and again apply the induction (this second step will prove your result for negative integers). \u2013\u00a0Bumblebee Dec 13 '17 at 13:58\n\u2022 If you want to learn more about induction then have a look at this question and its answers. \u2013\u00a0drhab Dec 13 '17 at 14:33\n\nIn this particular question, you can consider it in two separate cases, first case for $a \\ge 0$ and second case for $a < 0$.\n\nCase $a \\ge 0$: We will check whether $3 | (a^3-a)$ or not by using induction on $a$. For $a = 0$, we have $3|0$. Now suppose $a \\ge 1$ and for all $a$, the argument holds. Then for $a+1$, we have $$(a+1)^3-(a+1) = a^3+3a^2+2a = (a^3-a)+3a^2+3a$$ where $3|(a^3-a)$ by inductive assumption and $3|(3a^2+3a)$ obviously. Therefore, by induction, it holds for all $a \\ge 0$.\n\nCase $a < 0$: If you define $b=-a$, then this case becomes $3|(-b^3+b)$ where $b > 0$ so again you can use the induction on $b$ as induction on natural numbers. Proof for this case is similar to the first case.\n\nIn this way, you can cover all the integers by using an induction on natural numbers.\n\n\u2022 Can you also give the proof please? \u2013\u00a0WinstonCherf Dec 13 '17 at 14:09\n\u2022 Actually, I have to say that according to what Barry Cipra said, you don't need to prove it for the second case. But I really suggest you to do it just for practicing induction. \u2013\u00a0ArsenBerk Dec 13 '17 at 14:23\n\nTechnically you need to do two separate inductions. But since $(-a)^3-(-a)=-(a^3-a)$, you really only need to take the induction in the ordinary positive direction. If you do want to do both inductions, you can combine them in a single argument, along the following lines:\n\nThe base case is $3\\mid0^3-0$, and\n\n$$(a\\pm1)^3-(a\\pm1)=(a^3\\pm3a^2+3a\\pm1)-(a\\pm1)=(a^3-a)\\pm3a^2+3a$$\n\nso $3\\mid(a^3-a)$ implies $3\\mid((a\\pm1)^3-(a\\pm1))$.\n\n\u2022 Why only the induction in the ordinary positive direction is needed since (\u2212a)3\u2212(\u2212a)=\u2212(a3\u2212a)? Can you please explain that? \u2013\u00a0WinstonCherf Dec 13 '17 at 14:35\n\u2022 @LeneCoenen: See my comment on your question ;) \u2013\u00a0Mathematician 42 Dec 13 '17 at 14:36\n\u2022 @Mathematician42 Thnx!! \u2013\u00a0WinstonCherf Dec 13 '17 at 14:38\n\nInduction can be applied on a set if the set involved is equipped with a so-called well-order.\n\nEssential is that in that situation every non-empty subset of the set has a least element.\n\nNote that $\\mathbb N$ has a very natural well-order: $0<1<2<\\cdots$.\n\nThe famiar and well known order $<$ on $\\mathbb Z$ is not a well-order. One of the non-empty sets that has no least element according to that order is $\\mathbb Z$ itself, and there are lots of others.\n\nThis is why on school you were taught that induction was not for $\\mathbb Z$.\n\nOverlooked is there that there are well-orders on $\\mathbb Z$ also.\n\nSo if you want to prove by induction that $3\\mid a^3-a$ for every $a\\in\\mathbb Z$ then at first you must equip $\\mathbb Z$ with a suitable well-order.\n\nOne (there are more) that can be used for it is:\n\n$$0<'1<'2<'3<'\\dots<'-1<'-2<'-3<'\\dots$$\n\nIf $P(a)$ is true iff $3\\mid a^3-a$ then it is enough to prove that:\n\n\u2022 $P(0)$\n\u2022 $P(n)\\implies P(n+1)$\n\u2022 $P(n)\\implies P(n-1)$\n\nI should say that it is even more than enough (see the comment of Hagen).\n\nIf you have done that then by induction you proved that $P(n)$ is true for every $n\\in\\mathbb Z$.\n\n\u2022 @Avamander Thanks, I repaired. \u2013\u00a0drhab Dec 13 '17 at 15:34\n\u2022 The well-order you wrote down suggests that the induction steps can be made weaker (though that makes them cumbersome): (1) $P(0)$; (2) $(n\\ge 0\\land P(n))\\implies P(n+1)$; (3) $(\\forall n\\ge 0\\colon P(n))\\implies P(-1)$; (4) $(n<0\\land P(n))\\implies P(n-1)$ \u2013\u00a0Hagen von Eitzen Dec 13 '17 at 15:50\n\u2022 @HagenvonEitzen Thank you. I added a remark on this that refers to your comment. \u2013\u00a0drhab Dec 13 '17 at 15:53\n\u2022 The well-order you suggested does not work with standard mathematical induction. Its order type is larger than $\\omega$, so you need transfinite induction, albeit only technically. \u2013\u00a0tomasz Dec 13 '17 at 23:12\n\nThe induction principle on $\\mathbb{N}$ says: assuming that a property holds for $0$, and that if it holds for $n$ then it holds for $n+1$, then the property is true for all the elements of $\\mathbb{N}$. The principle holds because all the elements of $\\mathbb{N}$ can be reached by starting from $0$ and applying the operation $n \\mapsto n+1$ a finite number of times.\n\nLet's make this a little more abstract. Assuming that a property holds for the initial natural ($0$), and that if it holds for a natural then it also holds for the next natural ($n+1$), then it holds for all naturals.\n\nWe can generalize this to other domains than $\\mathbb{N}$ by generalizing the notions of \u201cinitial\u201d and \u201cnext\u201d. Assume that all the elements of a set $D$ can be reached by starting from some initial element and by applying a \u201cderivation\u201d operation a finite number of times. Assuming that a property holds for all the initial elements, and that if it holds for an element then it also holds for a derived element, then the property holds for all the elements.\n\nApplication: all the relative integers ($\\mathbb{Z}$) can be reached by starting from $0$ (the single initial element) and applying one of the operations $n \\mapsto n+1$ or $n \\mapsto n-1$ a finite number of times. Therefore, the following induction principle holds on $\\mathbb{Z}$: assuming that a property holds for $0$, that if it holds for $n$ then it holds for $n+1$, and that if it holds for $n$ then it holds for $n-1$, then the property holds for all the elements of $\\mathbb{Z}$.\n\nGiven this principle, proving the property you want is a simple modification from the proof on $\\mathbb{N}$.\n\nIt's possible to generalize this further by generalizing the notion of \u201cderivation\u201d. An element could be derived from multiple arguments. Assume that there is a family of constructor operations $c_i : D^{a_i} \\to D$, where each constructor can take a different number of parameters, such that all elements of $D$ can be reached by applying constructors. The starting point comes from constructors with 0 arguments. Then there is an induction principle on $D$ which states that, assuming that for each constructor $c_i$, if the property holds for $(x_1,\\ldots,x_{a_i})$ then it holds for $P(c_i(x_1,\\ldots,x_{a_i}))$, then the property holds for all the elements of $D$. The induction principle for $\\mathbb{N}$ is a special case with two constructors: $0$ (with 0 arguments) and $n \\mapsto n+1$ (with 1 arguments). The induction principle for $\\mathbb{Z}$ adds a third constructor $n \\mapsto n-1$. You could add a fourth constructor with two arguments $(p,q) \\mapsto \\begin{cases} p/q & \\text{if }q \\ne 0 \\\\ 0 & \\text{if } q = 0 \\end{cases}$ to get an induction principle for $\\mathbb{Q}$.\n\nIt's possible to generalize this even further to get induction principles on \u201clarger\u201d spaces (which don't even need to be countable). See drhab's answer.\n\nTechnically, induction is a technique applied on the natural numbers.\n\nHowever, there is nothing stopping you from having two statements applying to natural numbers that you prove seperately, but very similarily:\n\n1. $P(n):3\\mid (n^3-n)$\n2. $Q(n): 3\\mid ((-n)^3 - (-n))$\n\nWe can apply induction to prove $P$ and $Q$ for all natural numbers. Then, when it comes to showing that $P$ holds for all integers, we simply note that $P(n) \\equiv Q(-n)$, so for any integer $k$, if it is possible, then the truth of $P(k)$ comes from the induction on $P$, while if $k$ is negative, the truth of $P(k)$ is the same as the truth of $Q(-k)$, which was proven by induction on $Q$.\n\nUsually, though, this theoretical machinery is glossed over by proving $P$ for the base case $n = 0$ (since that's the same case for both $P$ and $Q$), and then say that we're using induction in \"both directions\" to prove that $P$ is valid for all integers $n$.\n\nYou can extend the induction principle to work for $\\mathbb Z$. The difference is that you instead of implication in the \"step\" part use equivalence:\n\nIf $\\phi(0)$ is true and $\\forall j\\in\\mathbb Z: \\phi(j)\\leftrightarrow\\phi(j+1)$ is true then $\\forall j\\in\\mathbb Z:\\phi(j)$ is true.\n\nYou can also use the normal induction principle twice. First proving it for $\\mathbb N$ and then for proving the statement for $\\mathbb Z^{-1}$.\n\nYou can do it with your run-of-the-mill induction, you just need to use the right statement.\n\nFor example, if by $P(n)$ you denote the statement \"For all $a$ such that $\\lvert a\\rvert\\leq n$, we have $3| a^3-a$\", it should be clear how to proceed.\n\nSince $\\mathbb{Z}$ is countable as $\\mathbb{N}$ we can extend induction over $\\mathbb{Z}$.\n\nBASE CASE:\n\n$$a=1 \\implies 3|0$$\n\nINDUCTIVE STEP 1 \"UPWARD\"\n\nassume: $3|a^3-a$\n\n$$(a+1)^3-(a+1)=a^3+3a^2+3a+1-a-1=a^3-a+3a^2+3a\\equiv0\\pmod 3$$\n\nthus\n\n$$3|(a+1)^3-(a+1)$$\n\nINDUCTIVE STEP 2 \"DOWNWARD\"\n\nassume: $3|a^3-a$\n\n$$(a-1)^3-(a-1)=a^3-3a^2+3a-1-a+1=a^3-a-3a^2+3a\\equiv0\\pmod 3$$\n\nthus\n\n$$3|(a-1)^3-(a-1)$$\n\nThus: $$\\forall a\\in \\mathbb{Z} \\Rightarrow 3|a^3-a$$\n\n\u2022 This does not answer the question. You completely missed the point of this question. \u2013\u00a0Mathematician 42 Dec 13 '17 at 14:02\n\u2022 Sorry I've extended the result. Now is it ok? \u2013\u00a0gimusi Dec 13 '17 at 14:07\n\u2022 Technically yes, but I doubt the poster will get why a double induction is the answer based on your post. Personally, I would explain how you cover all of $\\mathbb{Z}$ by a double induction. \u2013\u00a0Mathematician 42 Dec 13 '17 at 14:11\n\u2022 $\\mathbb{Z}$ is countable as $\\mathbb{N}$ why shouldn't it work? Isn't t it sufficient? \u2013\u00a0gimusi Dec 13 '17 at 14:18\n\u2022 Yes, I know that, but clearly the poster didn't. The poster had issues using induction on something which is not $\\mathbb{N}$. I believe an answer should address that issue first. Your first answer did not, it's getting better but it still doesn't explain (as in giving an explanation, not a technically correct proof) why it works on $\\mathbb{Z}$ \u2013\u00a0Mathematician 42 Dec 13 '17 at 14:21", "date": "2019-05-23 10:46:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2564830/proof-by-mathematical-induction-in-z", "openwebmath_score": 0.8594235777854919, "openwebmath_perplexity": 180.89042636226614, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9799765587105447, "lm_q2_score": 0.9230391690674338, "lm_q1q2_score": 0.9045567484577444}}
{"url": "https://brilliant.org/discussions/thread/conics-section-properties/", "text": "# Conics Section : Properties\n\nThis Note is for Those who love to use Properties and some innovative Techniques while Solving Question of Conics Section ! These Properties are Highly reduce our Calculation and are very useful sometimes , Specially when we have Time Constrained !\n\nSo Please Share Properties and Techniques that You know about conics Section So that our Brilliant Community will Learn from it.\n\nSo Now Here I Shared some few Techniques ( Properties ) of Conics Section That are Created By Me :)\n\nFor Ellipse ( By Me )\n\n$\\\\ \\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\\quad \\quad \\quad (:\\quad a>b\\quad )$.\n\nOn any point P on standard ellipse if an tangent is drawn and If we Drop Perpendicular's from the Vertex's of major axis and focus's and from centre on the the given tangent and named them ${ V }_{ 1 }\\quad ,\\quad { V }_{ 2 }\\quad ,\\quad { P }_{ 1 }\\quad ,\\quad { P }_{ 2 }\\quad ,\\quad d$. suitably Then :\n\n1)- $\\displaystyle{ V }_{ 1 }\\quad ,\\quad d\\quad ,\\quad { V }_{ 2 }\\quad \\longrightarrow \\quad AP\\\\ \\\\ 2d\\quad =\\quad { V }_{ 1 }+{ V }_{ 2 }\\quad \\quad$\n\n2)- $\\displaystyle{ P }_{ 1 }\\quad ,\\quad d\\quad ,\\quad { P }_{ 2 }\\quad \\longrightarrow \\quad AP\\\\ \\\\ 2d\\quad ={ \\quad P }_{ 1 }+{ P }_{ 2 }$\n\n3)- $\\cfrac { { S }_{ 1 }P }{ { S }_{ 2 }P } \\quad =\\quad \\cfrac { \\quad P_{ 1 } }{ { \\quad P }_{ 2 } }$.\n\nNote : ( By My Sir ) My Teacher told me that ( which is well Known result ) : $P_{ 1 }{ P }_{ 2 }\\quad =\\quad { b }^{ 2 }\\quad$.\n\nMy Turn is Over ! Now it's Your Turn ,\n\nSo please Post Properties or techniques Related To conics Section That are created by You or may also be you learnt it Somewhere else :)\n\nReshare This More and More So that it can reaches to everyone , So that we can Learn from Them!\n\nNote by Deepanshu Gupta\n5\u00a0years, 8\u00a0months ago\n\nThis discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution \u2014 they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.\n\nWhen posting on Brilliant:\n\n\u2022 Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .\n\u2022 Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting \"I don't understand!\" doesn't help anyone.\n\u2022 Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.\n2 \\times 3 $2 \\times 3$\n2^{34} $2^{34}$\na_{i-1} $a_{i-1}$\n\\frac{2}{3} $\\frac{2}{3}$\n\\sqrt{2} $\\sqrt{2}$\n\\sum_{i=1}^3 $\\sum_{i=1}^3$\n\\sin \\theta $\\sin \\theta$\n\\boxed{123} $\\boxed{123}$\n\nSort by:\n\nThere are actually many properties, here are few of them\n\nIf the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively & if CF be perpendicular upon this normal then\n\n\u2022 PF.PG = $b^{2}$\n\n\u2022 PF.Pg = $a^{2}$\n\n\u2022 PG.Pg = SP.S'P\n\n\u2022 CG.CT = $CS^{2}$\n\nIf tangent at the point P of a standard ellipse meets the axes at T & t and CY is perpendicular on it from centre then\n\n\u2022 Tt.PY = $a^{2} - b^{2}$\n\n\u2022 least value of Tt is $a + b$\n\nThis is only for ellipse but there are many for each curve\n\n- 5\u00a0years, 8\u00a0months ago\n\nYes I know That That were also Told to me by my Teacher , But Thanks For Sharing It , it will Helpful For others ! Do You Have Your own Properties if You have then Please also Share it with us , Thanks Krishna :)\n\n- 5\u00a0years, 8\u00a0months ago\n\nYeah there is one when we have to find the minimum distance between 2 Parabola's it occur at extremities of latus rectum\n\nI don't know how it works but it has worked till now in every problem, I'll try to prove if I got some time meanwhile you can apply and try to prove it :)\n\nNote:- I am not 100% sure with it because I haven't found any case in which this doesn't occur\n\n- 5\u00a0years, 7\u00a0months ago\n\nIt would be much helpful if a diagram is drawn. Thanks.\n\n- 5\u00a0years, 8\u00a0months ago\n\nHey, few more, for ellipse:\n\n1. The portion of the tangent to the ellipse between the point of contact and the directrix subtend a right angle at the corresponding focus.\n\n2. The circle on the focal distance as diameter touches the auxiliary circle.\n\n3. Tangent at the extremeties of latus rectum pass through the corresponding foot of directrix on the major axis.\n\n4. Ratio of area of any triangle inscribed in a standard ellipse ( a> b) and that of triangle formed by corresponding points on the auxiliary circle is $\\frac {b} {a}$\n\n- 5\u00a0years, 8\u00a0months ago\n\nIt would be much helpful if a diagram is drawn. Thanks.\n\n- 5\u00a0years, 8\u00a0months ago\n\nThanks alot :)\n\n- 5\u00a0years, 7\u00a0months ago\n\nThese are good properties of conic sections. Could you add them to the corresponding Conic Section wiki pages? Thanks!\n\nStaff - 5\u00a0years, 7\u00a0months ago\n\nI have posted my first challenge problem on conic sections. Please post solutions.\n\nCheck out my profile page to get other conic challenges.\n\n- 5\u00a0years, 6\u00a0months ago", "date": "2020-08-15 16:52:15", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/conics-section-properties/", "openwebmath_score": 0.9163039922714233, "openwebmath_perplexity": 1773.5267402830632, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9632305381464927, "lm_q2_score": 0.9390248140158416, "lm_q1q2_score": 0.9044973769373893}}
{"url": "http://mathhelpforum.com/algebra/212091-having-sign-problem-algebraic-fractions-print.html", "text": "# Having sign problem with algebraic fractions\n\n\u2022 Jan 26th 2013, 05:48 PM\nKevinShaughnessy\nHaving sign problem with algebraic fractions\nHi,\n\nI'm having a problem with an algebraic fractions equation. It goes:\n\n5/3(v-1) + (3v-1)/(1-v)(1+v) + 1/2(v+1)\n\nThe first thing I do is factor out the negative in the first fraction, getting:\n\n-5/3(1-v)\n\nGiving a cd of 6(1-v)(v+1). Now that that is done I multiply the numerators by the necessary factors:\n\n-5*2(v+1)\n\n6(3v-1)\n\n3(1-v)\n\nWhich gives me -10v -10 + 18v -6 -3v + 3\n\nWhich adds up to 5v - 13\n\nBUT the answer is -5v + 13, and if I factor out the negative in the second equation, all the signs are reversed and the equation works out to the right answer. So I'm confused as to why things didn't work when I factored out the negative in the first fraction.\n\nThanks,\n\nKevin\n\u2022 Jan 26th 2013, 07:53 PM\nchiro\nRe: Having sign problem with algebraic fractions\nHey KevinShaughnessy.\n\nCan you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?\n\u2022 Jan 26th 2013, 08:37 PM\nSoroban\nRe: Having sign problem with algebraic fractions\nHello, Kevin!\n\nThey approached the problem differently . . . that's all.\n\nQuote:\n\n$\\text{Simplify: }\\:\\frac{5}{3(v-1)} + \\frac{3v-1}{(1-v)(1+v)} + \\frac{1}{2(v+1)}$\n\nThey factored a \"minus\" out of the second fraction.\n\n. . $\\frac{5}{3(v-1)} - \\frac{3v-1}{(v-1)(v+1)} + \\frac{1}{2(v+1)}$\n\nThe LCD is $6(v-1)(v+1)\\!:$\n\n. . $\\frac{5}{3(v-1)}\\cdot {\\color{blue}\\frac{2(v+1)}{2(v+1)}} - \\frac{3v-1}{(v-1)(v+1)}\\cdot {\\color{blue}\\frac{6}{6}} + \\frac{1}{2(v+1)}\\cdot {\\color{blue}\\frac{3(v-1)}{3(v-1)}}$\n\n. . $=\\;\\frac{10(v+1)}{6(v-1)(v+1)} - \\frac{6(3v-1)}{6(v-1)(v+1)} + \\frac{3(v-1)}{6(v-1)(v+1)}$\n\n. . $=\\;\\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$\n\n. . $=\\;\\frac{13-5v}{6(v-1)(v+1)}$\n\u2022 Jan 26th 2013, 10:17 PM\nKevinShaughnessy\nRe: Having sign problem with algebraic fractions\nQuote:\n\nOriginally Posted by chiro\nHey KevinShaughnessy.\n\nCan you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?\n\nIt's (3v-1) / [(1+v)(1-v)].\n\u2022 Jan 26th 2013, 10:19 PM\nKevinShaughnessy\nRe: Having sign problem with algebraic fractions\nQuote:\n\nOriginally Posted by Soroban\nHello, Kevin!\n\nThey approached the problem differently . . . that's all.\n\nThey factored a \"minus\" out of the second fraction.\n\n. . $\\frac{5}{3(v-1)} - \\frac{3v-1}{(v-1)(v+1)} + \\frac{1}{2(v+1)}$\n\nThe LCD is $6(v-1)(v+1)\\!:$\n\n. . $\\frac{5}{3(v-1)}\\cdot {\\color{blue}\\frac{2(v+1)}{2(v+1)}} - \\frac{3v-1}{(v-1)(v+1)}\\cdot {\\color{blue}\\frac{6}{6}} + \\frac{1}{2(v+1)}\\cdot {\\color{blue}\\frac{3(v-1)}{3(v-1)}}$\n\n. . $=\\;\\frac{10(v+1)}{6(v-1)(v+1)} - \\frac{6(3v-1)}{6(v-1)(v+1)} + \\frac{3(v-1)}{6(v-1)(v+1)}$\n\n. . $=\\;\\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$\n\n. . $=\\;\\frac{13-5v}{6(v-1)(v+1)}$\n\nBut aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?\n\u2022 Jan 27th 2013, 12:18 AM\nearthboy\nRe: Having sign problem with algebraic fractions\nQuote:\n\nOriginally Posted by KevinShaughnessy\nBut aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?\n\nyour answer was $\\frac{5v-13}{6{\\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\\frac{13-5v}{6{\\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$.\nyour answer was $\\frac{5v-13}{6{\\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\\frac{13-5v}{6{\\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$.", "date": "2017-05-30 13:42:18", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/algebra/212091-having-sign-problem-algebraic-fractions-print.html", "openwebmath_score": 0.8650761842727661, "openwebmath_perplexity": 1671.8352532194385, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9697854164256365, "lm_q2_score": 0.9324533135530545, "lm_q1q2_score": 0.9042796249815136}}
{"url": "https://math.stackexchange.com/questions/1214226/if-then-and-if-and-only-if-and-if-only-then-state", "text": "# 'If\u2026then\u2026' and '\u2026if\u2026' and '\u2026only if\u2026' and 'If\u2026 only then\u2026' statements?\n\nSuppose you have two statements A and B and \"If A then B\". I am trying to think of what this implies and alternative ways of writing this.\n\nI think\n\n\"If A then B\"\n\n= A$\\rightarrow$B\n\n= \"A is sufficient but not necessary for B. B is neither necessary nor sufficient for A\"\n\n= \"If not B then not A\"\n\n= B'$\\rightarrow$ A'\n\n= \"B' is sufficient for A' but not necessary\"\n\nAnd it seems to me that 'B if A' is equivalent to 'If A then B' (please correct me if I am wrong!)\n\nWhen it comes to only if, I think \"B only if A\" is equivalent to \"If A only then B\"\n\nI think\n\n\"B only if A\"\n\n= B$\\rightarrow$A\n\n= \"B is sufficient for A but not necessary for A. A is necessary for B but not sufficient for B\"\n\n=\"If not A then not B\"\n\n=A'$\\rightarrow$B'\n\n= \"Not A is sufficient for not B and not A is necessary for not B\"\n\nI see that I must have made mistakes somewhere because several things are not consistsent. Firstly, I am pretty sure my last statement is wrong but this is how I interpret \"If not A then not B\". Also, I don't understand how if the '...if...' cases, single headed arrows only implied sufficiency, and here for the '...only if...'single headed arrows seem to be implying something about necessity as well...\n\nThank you in advance for any clarifications, and also if anyone has a link to a good explanation of these statements I would be very grateful. I am trying to understand them in the context if proofs and proving statements the right way around...\n\nEDIT: Thank you for all the answers and comments so far. Jut thought I would add something that helped me in case someone else comes across my question and also requires help with if/iff/necessary/sufficient etc. I found it easier to visualise the cases. 'B if A' can be represented as a circle A within a circle B. Automatically, if A then B, so A is sufficient but not necessary for B, but B is necessary for A and not B implies not A. However A does not imply B. In a similar way, 'B only if A' is a circle B within a circle A, because B implies A- it is sufficient, but not necessary for A, and A is necessary but not sufficient for B. 'B if and only if A' is the double headed arrow because A and B are the same ring. One is both necessary and sufficient for the other, and one implies the other....\n\n\u2022 Yes : $A \\to B$ is : \"if A then B\" and also \"B if A\" ans \"A is a sufficient condition for B\". $A \\to B$ is also \"B is a necessary condition for A\" and \"A only if B\". \u2013\u00a0Mauro ALLEGRANZA Mar 31 '15 at 11:27\n\u2022 I find it helps to remember that $A\\implies B \\equiv \\neg [A \\land \\neg B]$. It has nothing with $A$ causing $B$, or $B$ causing $A$ as many beginners seem to think. There is no causality in mathematics. \u2013\u00a0Dan Christensen Mar 31 '15 at 15:52\n\n$A\\to B$ means \"$A$ implies $B$\", \"$A$ is sufficient for $B$\", \"if $A$, then $B$\", \"$A$ only if $B$\", and such.\n\n$A\\leftarrow B$ means \"$A$ is implied by $B$\", \"$A$ is necessary for $B$\", \"$A$ whenever $B$\", \"$A$ if $B$\", and such\n\n$A\\leftrightarrow B$ means \"$A$ is necessary and sufficient for $B$\", \"$A$ if and only if $B$\".\n\nNote: $A\\to B$ means \"$A$ is sufficient for $B$ and may or may not be necessary for $B$\". \u00a0 It neither affirms nor denies the necessity.\n\nWe can also say,\n\n$A\\to B$ means \"$B$ if $A$\", \"$B$ whenever $A$\", and \"$B$ is necessary for $A$\",\n\n$A\\leftarrow B$ means \"$B$ only if $A$\", \"if $B$, then $A$\", and \"$B$ is sufficient for $A$\".\n\n$A\\wedge\\neg B$ means \"$A$ is not sufficient for $B$\", and $\\neg A\\wedge B$ means \"$A$ is not necessary for $B$\"\n\n\u2022 Brilliant answer! \u2013\u00a0Konstantin Feb 11 '17 at 17:04\n\nYour thinking is a little off. If $A$ is sufficient for $B$, then indeed $B$ is necessary for $A$. This is because if $B$ is not true, then $A$ is not true.\n\nLong comment\n\nThere are no \"inconsistency\" ...\n\n$A \\to B$ is : \"if $A$, then $B$\" and also \"$B$, if $A$\" and also \"$A$ is a sufficient condition for $B$\".\n\nThe first one is the \"standard\" reading, and the second one is the same (see the comma ...).\n\n$A \\to B$ is also \"$A$ only if $B$\"; the best way to derive it is from $A \\leftrightarrow B$, i.e. \"$A$ if and only if $B$\".\n\nThis is \"($A$, if $B$) and ($A$ only if $B$)\", that translates $(B \\to A) \\land (A \\to B)$.\n\nNow, if we agree that \"$A$, if $B$\" is $B \\to A$, we have to agree also that \"$A$ only if $B$\" is $A \\to B$.\n\nThe presence of the \"negations\" does not change the way to read the conditional, when the negation sign : $\\lnot$ is \"attached\" to the sentential variables.\n\nThus, $\\lnot A \\to \\lnot B$ is : \"if not-$A$, then not-$B$\", and so on.", "date": "2020-04-02 20:19:57", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1214226/if-then-and-if-and-only-if-and-if-only-then-state", "openwebmath_score": 0.8023192286491394, "openwebmath_perplexity": 513.930395615461, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9843363494503272, "lm_q2_score": 0.9184802440252812, "lm_q1q2_score": 0.904093490446091}}
{"url": "https://math.stackexchange.com/questions/969356/confirm-definite-integral-equals-zero-frac-sinx1-a-cosx2?noredirect=1", "text": "Confirm definite integral equals zero $\\frac{\\sin(x)}{(1-a\\cos(x))^{2}}$\n\nIs this statement about the definite integral of a particular function $F$ true? $$\\int_0^{2\\pi}F(x)\\, \\mathrm{d}x = \\int_0^{2\\pi}\\frac{\\sin(x)}{(1-a\\cos(x))^2}\\, \\mathrm{d}x = 0 \\ \\text{ for }\\ 0<a<1$$\n\nI have evaluated this expression (in WolframAlpha) for various values of a and they all give the value zero. I have read that integrals of the form $$\\int G(\\cos(x))\\sin(x)\\, \\mathrm{d}x$$ where $G$ is some continuously integrable function are zero over the range $-\\pi/2$ to $\\pi/2$.\n\n(Edited after comment from Andrey) It seems possible to proceed from here to confirm the postulated statement by symmetry. The function $F$ to be integrated is cyclic with period 2$\\pi$ such that $F(x-2\\pi) = F(x) =F(x+2\\pi)$. Then we just need to prove that the two integrals:- (1) between $-\\pi$ and $-\\pi/2$, and (2) between $\\pi/2$ and $\\pi$ are equal in magnitude and opposite in sign.\n\nThis would be the case if $F(x) = -F(-x)$. Such is actually the case because the denominator in F() expands to $(1-2a\\cos(x)+a^2\\cos^2x)$ and has the same values for $(+x)$ and $(-x)$. Whereas the numerator $\\sin(x)$ is such that $\\sin(x) = -\\sin(-x)$.\n\nHowever I would still like to find an analytical solution.\n\n\u2022 It is the case that $F(x)=-F(x)$. The denominator $1-2a\\cos x + a^2\\cos^2x$ for $x<0$ is the same as $x>0$, since cosine is an even function, i.e. $\\cos(-x) = \\cos(x)$. \u2013\u00a0Andrey Kaipov Oct 11 '14 at 22:09\n\u2022 @Andrey Yes you are right of course. Doh! \u2013\u00a0steveOw Oct 11 '14 at 22:25\n\u2022 It's important to learn these symmetry arguments. There is a famous integral from the Putnam that cannot be done any other way: $\\int_0^{\\pi/2} \\frac{1}{1+\\tan^{\\sqrt{2}} x} = \\frac{\\pi}{4}$. \u2013\u00a0Slade Oct 12 '14 at 0:59\n\nLet $u=1-a\\cos x$, $du=a\\sin x dx$ to get $\\displaystyle\\frac{1}{a}\\int_{b}^{b}\\frac{1}{u^2}du=0$ $\\;\\;\\;$ (where $b=1-a$).\n\n\u2022 I dont understand the limits b..b. \u2013\u00a0steveOw Oct 11 '14 at 22:42\n\u2022 When $x=0, u=1-a(1)=1-a$ and when $x=2\\pi, u=1-a(1)=1-a$. \u2013\u00a0user84413 Oct 11 '14 at 22:51\n\u2022 @user844413 Wow that is so slick! \u2013\u00a0steveOw Oct 11 '14 at 23:00\n\n$$\\int_a^bF(x)dx = \\int_a^bF(a+b-x)dx,$$\n\nwe have in this case\n\n$$\\int_0^{2\\pi}F(x)dx = \\int_0^{2\\pi}F(2\\pi-x)dx,$$\n\nand, from knowledge of the symmetries of the sin and cos functions, we know in this case that $$F(x) = - F(2\\pi-x),$$\n\nso, with $I =\\int_0^{2\\pi}F(x)dx$, we have\n\n$$I=-I,$$\n\nwhich can only be true if $$I = 0$$\n\n\u2022 But how do I know your two integrals are equivalent to start with? I only know: F(x)=-F(-x) and F(x)=-F(2pi-x) and hence F(-x)=F(2pi-x). \u2013\u00a0steveOw Oct 12 '14 at 1:46\n\u2022 The link I provided shows that $F$ doesn't need any property in order to have: $\\int_a^bF(x)~dx~~=\\int_a^bF(a+b-x)~dx$ (for example, by change of variable: $y=a+b-x$). The lucky part for your integral is that we get $-I$ for the second integral. \u2013\u00a0ir7 Oct 12 '14 at 1:57\n\u2022 (Aha, I didn't spot the link). The very useful equation in your comment is fundamental to this answer. I suggest it is included in the answer. \u2013\u00a0steveOw Oct 12 '14 at 15:03\n\u2022 Ok.It looks better now. Cheers. :) \u2013\u00a0ir7 Oct 12 '14 at 15:25\n\nThis is a special case of a general fact about $u$-substitution. If $G(x)$ is integrable on the interval $[a,b]$, with antiderivative $g(x)$, and $u$ is differentiable, then $$\\int_a^b G(u(x))\\,u'(x)\\, dx = g(u(b))-g(u(a)).$$\n\nIf $u(a)=u(b)$, the integral is zero.\n\nThe integrand in your example has this form, where $u(x)=\\cos(x)$ and $G(u)=\\frac{-1}{(1-a\\cos(x))^2}$, and $\\cos(x)$ has the same value at both limits of integration, so the integral is zero. (You can apply the substitution user84413 suggested, or the simpler $u=\\cos x$ to show it.)\n\nYou can write down lots of messy-looking integrals that turn out to be zero because they have this form for some $u(x)$ and $G(x)$.\n\n$$\\int_1^3 e^{x^2-4x+7}(2-x)\\, dx$$\n\n$$\\int_0^{2\\pi} (\\pi-x)\\log(2+\\sin^2(x-\\pi)^2)\\, dx$$\n\n$$\\int_{\\pi/2}^{3\\pi/2} (\\cos^2 x)^{\\sin^2 x}\\sin2x\\,dx$$\n\n\u2022 I like @ir7\u2019s trick for this, which works for these examples. Here\u2019s one where you can\u2019t quickly show that $I=-I$ that way: $\\displaystyle\\int_1^9 (4\\sqrt x-x)^{4\\sqrt x-x}(\\frac{2}{\\sqrt{x}}-1)\\,dx$. \u2013\u00a0Steve Kass Oct 12 '14 at 0:31\n\u2022 (Re:your answer) So I dont even need to know the form of g(). Nice. \u2013\u00a0steveOw Oct 12 '14 at 1:01", "date": "2019-10-21 22:26:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/969356/confirm-definite-integral-equals-zero-frac-sinx1-a-cosx2?noredirect=1", "openwebmath_score": 0.9369025826454163, "openwebmath_perplexity": 228.610461499639, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9854964220125032, "lm_q2_score": 0.9173026522382527, "lm_q1q2_score": 0.9039984816833775}}
{"url": "https://qa.engineer.kmitl.ac.th/h6g2taye/13c911-knapsack-problem-optimization", "text": "# knapsack problem optimization\n\ni {\\displaystyle v_{i}} 2 The bounded knapsack problem (BKP) removes the restriction that there is only one of each item, but restricts the number In this variation, the weight of knapsack item items numbered from 1 up to , [ \u222a The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the count of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is \u2026 {\\displaystyle J} j \u2032 w ( , S {\\displaystyle J} \u2264 . , Another popular solution to the knapsack problem uses recursion. + and m For each Ai, you choose Ai optimally. , O to be the maximum value that can be attained with weight less than or equal to w Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License, and code samples are licensed under the Apache 2.0 License. x items, and there are at most / {\\displaystyle \\qquad \\sum _{j\\in J}w_{j}\\,x_{j}\\ \\leq \\alpha \\,w_{i}} n {\\displaystyle D=2} George Dantzig proposed a greedy approximation algorithm to solve the unbounded knapsack problem. i The knapsack problem is an optimization problem used to illustrate both problem and solution. ( n of copies of each kind of item to a maximum non-negative integer value If \u2026 ] A large variety of resource allocation problems can be cast in the framework of a knapsack problem. by their greatest common divisor is a way to improve the running time. {\\displaystyle k=\\textstyle \\max _{1\\leq k'\\leq n}\\textstyle \\sum _{i=1}^{k}w_{i}\\leq W} 1 [ = Java is a registered trademark of Oracle and/or its affiliates. In this example, you have multiple objectives. The Knapsack Problem is an example of a combinatorial optimization problem, which seeks to maximize the benefit of objects in a knapsack without exceeding its capacity. , not to In other words, given two integer arrays val [0..n-1] and wt [0..n-1] which represent values and weights associated with n items respectively. S with a maximum capacity. + The knapsack problem, though NP-Hard, is one of a collection of algorithms that can still be approximated to any specified degree. gives the solution. [ {\\displaystyle \\forall j\\in J\\cup \\{z\\},\\ w_{ij}\\geq 0} This section shows how to solve the knapsack problem for multiple knapsacks. ( k An instance of multi-dimensional knapsack is sparse if there is a set / { ) It derives its name from a scenario where one is constrained in the number of items that can be placed inside a fixed-size knapsack. ( \u2211 w ] 2 { ( Approximation Algorithms. ] . 1 w , If we know each value of these j 2 . \u2212 Vazirani, Vijay. J w ( i 10 The knapsack problem is popular in the research \ufb01eld of constrained and combinatorial optimization with the aim of selecting items into the knapsack to attain maximum pro\ufb01t while simultaneously not exceeding the knapsack\u2019s capacity. , It discusses how to formalize and model optimization problems using knapsack as an example. 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Algorithm to solve the unbounded knapsack problem is always a dynamic programming solution for the bounded problem one. Interviewer can use a table to store all relevant values starting at index.! Its affiliates and linear capacity constraints components ). [ 19 ] this page was last edited on 2 2020. Knapsack as an example items exceeds the capacity of the items exceeds the capacity, want... Of this method, how do we get the weight changes from to... Of passengers and the previous weights are w \u2212 w 2, different variants of the individual the! Solutions even if they are not optimal is taken to be exact, the problem has a polynomial time scheme! Hiker tries to pack the most well-known problem in... Read more SDLC and! N components ). [ 21 ] [ J ], the of... Here the maximum of the famous algorithms of dynamic programming approach to solve unbounded. Efficiently, we can use a table to store previous computations 22.. Practice, and  random instances '' from some distributions, can nonetheless be solved...., WFG, and value, Pn works dating as far back as 1897 a until complete enough...: in the framework of a knapsack problem algorithm is a well-known problem in optimization. Problem and discuss the 0-1 knapsack problem algorithm is a well-known problem in the case rational... Are w \u2212 w 1, w ] } ] however, in the field individual filling the knapsack,. Each item has an associated weight, Wn, and  random instances '' from distributions. } -th item altogether, the thief can not carry weight exceeding M ( M \u2264 100 ) [... Minimizing their salaries the summation of the initial knapsack show how to and! Only need solution of previous row there are 10 different items and the entertainers must weigh less than lbs., where the supply of each member of J { \\displaystyle w } decision version of the problem... Are 10 different items and the previous weights are w \u2212 w 1, w \u2212 w 1, ]! Through the knapsack problem a weakly NP-complete problem and discuss the 0-1 variant in detail the problem popular... [ 26 ], we can disregard the i { \\displaystyle w?... Relay nodes this question to test knapsack problem optimization dynamic programming and this problem under... Hiker tries to pack the most value into the knapsack problem is a well known problem combinatorial... Makes the ( decision version of the search space the container of previous row December,. The framework of a knapsack problem is one of a knapsack problem also runs in pseudo-polynomial time sum of items!\n\nScroll to Top", "date": "2021-08-02 00:18:00", "meta": {"domain": "ac.th", "url": "https://qa.engineer.kmitl.ac.th/h6g2taye/13c911-knapsack-problem-optimization", "openwebmath_score": 0.7083615660667419, "openwebmath_perplexity": 1002.8524074638594, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9828232879690036, "lm_q2_score": 0.9196425333801889, "lm_q1q2_score": 0.9038460984128615}}
{"url": "https://math.stackexchange.com/questions/1880319/is-my-proof-by-strong-induction-of-for-all-n-in-mathbbn-g-n-3n-2n-cor", "text": "# Is my proof, by strong induction, of for all $n\\in\\mathbb{N}$, $G_n=3^n-2^n$ correct?\n\nLet the sequence $G_0, G _1, G_2, ...$ be defined recursively as follows:\n\n$G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\\in\\mathbb{N}, n\\ge2$.\n\nProve that for all $n\\in\\mathbb{N}$, $G_n=3^n-2^n$.\n\nProof. By strong induction. Let the induction hypothesis, $P(n)$, be $G_n=3^n-2^n$\n\nBase Case: For $(n=0)$, $P(0)$ is true because $3^0-2^0 =0$\n\nFor $(n=1)$, $P(1)$ is true because $3^1-2^1=1$\n\nInductive Step: Assume that $P(n-1)$ and $P(n-2)$, where $n\\ge2$, are true for purposes of induction.\n\nSo, we assume that $G_{n-1}=3^{n-1}-2^{n-1}$ and $G_{n-2}=3^{n-2}-2^{n-2}$, and we must show that $G_{ n }=3^{ n }-2^{ n }$.\n\nSince we assumed $P(n-1)$ and $P(n-2)$, we can rewrite $G_n=5G_{n-1}-6G_{n-2}$ as $G_n=5(3^{n-1}-2^{n-1})-{ 6 }(3^{n-2}-2^{n-2})$\n\nSo, we get:\n\n$\\Rightarrow G_n=5\\cdot 3^{ n-1 }-5\\cdot 2^{ n-1 }-(\\frac { 6 }{ 3 } \\cdot 3^{ n-1 }-\\frac { 6 }{ 2 } \\cdot 2^{ n-1 })$\n\n$\\Rightarrow G_n=5\\cdot 3^{ n-1 }-5\\cdot 2^{ n-1 }-2\\cdot 3^{ n-1 }+3\\cdot 2^{ n-1 }$\n\n$\\Rightarrow G_n=5\\cdot 3^{ n-1 }-2\\cdot 3^{ n-1 }-5\\cdot 2^{ n-1 }+3\\cdot 2^{ n-1}$\n\n$\\Rightarrow G_n=3\\cdot 3^{ n-1 }-2\\cdot 2^{ n-1 }$\n\n$\\Rightarrow G_n=\\frac { 1 }{ 3 } \\cdot 3\\cdot 3^n-\\frac { 1 }{ 2 } \\cdot 2\\cdot 2^n$\n\n$\\Rightarrow G_n=3^n-2^n$\n\nThe only real issue I have at this point is that I don't know how to properly conclude this proof with a final statement. A hint/guidance in that regard would be much appreciated.\n\n\u2022 It looks great, you have effectly showed that if the property $G_{i}=3^i-2^i$ holds for $i\\in\\{1,2,\\dots n-1\\}$ then $G_{n}=3^n-2^n$ also. \u2013\u00a0Jorge Fern\u00e1ndez-Hidalgo Aug 3 '16 at 17:12\n\u2022 In your description of $P(n)$, the part \"for all $n$ $\\dots$\" is at best superfluous, and at worst confusing or incorrect. Better, for any integer $k\\ge 0$ let $P(k)$ be the assertion $G(k)=3^k-2^k$. \u2013\u00a0Andr\u00e9 Nicolas Aug 3 '16 at 17:32\n\u2022 Not necessary. But the deletion of \"for all $n$ $\\dots$\" is. \u2013\u00a0Andr\u00e9 Nicolas Aug 3 '16 at 17:47\n\u2022 @Cherry Thanks for the link. I see that they state strong induction with a base case P(0). But often we don't need base case(s) for strong induction and the induction principle can be stated without any. For example, every integer > 1 is a product of primes. Suppose for induction it is true for all naturals < n. If n is prime we are done, else n is composite so it is a product of smaller naturals n = ab so by induction a,b are products of primes. Appending their products shows that n is a product of primes. No base case! \u2013\u00a0Bill Dubuque Aug 3 '16 at 18:05\n\u2022 The assertion $P(n)$ is not the assertion that $G(n)=3^n-2^n$ for all $n$. If we write the latter in symbols, it is $\\forall n(G(n)=3^n-2^n)$. Now $n$ is a \"dummy variable\" which gets quantified out. We want that for any particular $n$, $P(n)$ is the assertion $\\dots$. \u2013\u00a0Andr\u00e9 Nicolas Aug 3 '16 at 18:12\n\nYou have the right idea, but there are some minor points that need correction.\n\nThe strong induction principle in your notes is stated as follows:\n\nPrinciple of Strong Induction $\\$ Let $\\,P(n)\\,$ be a predicate. If\n\n\u2022 $\\ P(0)$ is true, and\n\n\u2022 for all $\\,n\\in \\Bbb N,\\ P(0), P(1),\\ldots, P(n)\\,$ together imply $\\,P(n\\!+\\!1)\\,$ then $\\,P(n)\\,$ is true for all $\\,n\\in\\Bbb N$\n\nYour $\\,P(n)\\,$ is $\\, G_n = 3^n - 2^n.\\,$ You have verified that $\\,P(0)\\,$ is true.\n\nYour induction hypothesis is that $\\,P(k)\\,$ is true for all $k \\le n.\\,$ You have essentially shown that $\\,P(n\\!-\\!1),P(n)\\,\\Rightarrow\\,P(n\\!+\\!1)\\,$ but that only works for $\\,n\\ge 1$ (else $\\,P(n-1)\\,$ is undefined). Thus you need to separately verify $\\,P(1)\\,$ (to be pedantic, this is part of the inductive step, not the base case, according to the above formulation of strong induction, though that is a somewhat arbitrary distinction)\n\nIt is illuminating to observe that the recurrence in the induction is a special case of\n\n$$a^{n+1}-b^{n+1} =\\, (a+b)(a^n-b^n) -ab (a^{n-1} - b^{n-1})$$\n\nwhich can be verified directly or derived from the fact that $\\,a,b\\,$ are roots of\n\n$$(x\\!-\\!a)(x\\!-\\!b) = x^2\\! - (a\\!+\\!b) x + ab\\,\\Rightarrow\\, x^{n+1}\\! = (a\\!+\\!b)\\,x^n - ab\\, x^{n-1}$$\n\nThe proof will be simpler (and more insightful) if you work with this general case, i.e. prove that $\\,f_n = a^n - b^n\\,$ satisfies $\\,f_{n+1} = (a+b) f_{n} - ab f_{n-1},\\ f_0 = 1,\\ f_1 = a-b\\,$ for all $\\,n\\ge 0.\\,$ Then your problem is just the special case $\\,a,b = 3,2,\\,$ and the inductive step is much clearer.\n\n\u2022 I was under the impression that I showed that $P(n-2),P(n-1)\\Rightarrow P(n)$. You stated that I showed $P(n-1),P(n)\\Rightarrow P(n+1)$. Why? \u2013\u00a0Cherry_Developer Aug 3 '16 at 19:57\n\u2022 @Cherry That's why I said \"essentially\". Substitute $\\,n+1\\,$ for $\\,n\\,$ in your proof to get the upshifted form. I wrote the induction in the above form used in the MIT notes. \u2013\u00a0Bill Dubuque Aug 3 '16 at 20:01\n\u2022 Ah ok. I apologize for the slew of questions. I would just much rather struggle with the math, and ask these questions now, than when I actually take discrete math. Thank you for all your help. \u2013\u00a0Cherry_Developer Aug 3 '16 at 20:07\n\u2022 @Cherry_Developer It's the nature of the beast to struggle with induction proofs when one first encounters them (evolution doesn't program our minds for such). Many fit into particular patterns that are easier to comprehend in the abstract (such as the above which is essentially exploiting the uniqueness theorem for recurrences).. Another common form of induction is telescopy, e.g. see here for a vivid 2D example. \u2013\u00a0Bill Dubuque Aug 3 '16 at 20:11\n\nYes, your proof is perfectly fine. Good job! You can write something like \"The assertion follows.\". But honestly it isn't necessary since it is in this case pretty simple for readers to see where the proof is complete (after the inductive step).", "date": "2020-10-21 18:08:14", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1880319/is-my-proof-by-strong-induction-of-for-all-n-in-mathbbn-g-n-3n-2n-cor", "openwebmath_score": 0.8809493184089661, "openwebmath_perplexity": 280.94572593947305, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9891815523039174, "lm_q2_score": 0.9136765257642905, "lm_q1q2_score": 0.903791964059171}}
{"url": "https://exam.mangrovebd.org/jkg0b/applications-using-rational-equations-distance-rate-time-answers-dfe539", "text": "# applications using rational equations distance rate time answers\n\nTranslate the sentence to get the equation. Solve work-rate applications. Work problems often ask us to calculate how long it will take different people working at different speeds to finish a task. While traveling on a river at top speed, he went 10 miles upstream in the same amount of time he went 20 miles downstream. Is 8 mph a reasonable running speed? Add comment More. Some of the motion problems involving distance rate and time produce fractional equations. 7.6 Applications of Rational Equations. SECTION 11.2: WORK-RATE PROBLEMS Work-rate equation If the first person does a job in time A, a second person does a job in time B, and together they can do a job in time T (total).We can use the work-rate equation: Find the rate of the river current. Recall that the reciprocal The reciprocal of a nonzero number n is 1/n. Try It 7.89. Number Problems. In particular, they are quite good for describing distance-speed-time questions, and modeling multi-person work problems. Solving a distance, rate, time problem using a rational equation. Solve. Write a word sentence. 2. Hillarys lexus travels 30mph faster than Bills harley. (Notice that the work formula is very similar to the relationship between distance, rate, and time, or $d=rt$.) Ivan's boat has a top speed of 9 miles per hour in still water. On the map, Seattle, Portland, and Boise form a triangle. Solving Work Problems . Yes. The distance between the cities is measured in inches. 3 Manleys tractor is just as fast as Calendonias. A negative speed does not make sense in this problem, so is the solution. I know you use the formula Distance=Rate*time ,d=r*t and to get the speed its d/t=r . The algebraic models of such situations often involve rational equations derived from the work formula, W = rt. The algebraic models of such situations often involve rational equations derived from the work formula, $W=rt$. example: A train can travel at a constant rate from New York to Washington, a distance of 225 miles. Find there speeds. In the same time that the Bill travels 75 miles, Hilary travels 120 miles. The actual distance from Seattle to Boise is 400 miles. The first observation to make, however, is that the distance, rate and time given to us aren't `compatible': the distance given is the distance for only \\textit{part} of the trip, the rate given is the speed Carl can canoe in still water, not in a flowing river, and the time given is the duration of the \\textit{entire} trip. Check. Learning Objectives . The distance from Los Angeles to San Francisco is 351 miles. The figure on the left below represents the triangle formed by the cities on the map. Follow \u2022 2. of a nonzero number n is 1/n. We divide the distance by the rate in each row, and place the expression in the time column. Her time plus the time biking is 3 hours. Report 1 Expert Answer Best Newest Oldest. Solve applications involving uniform motion (distance problems). Answer the question. Solve applications involving relationships between real numbers.\n\nFord Motability Price List 2020, Casino Tycoon 2 Cast, Waterproof Doors For Homes, Same To You As Well In Tagalog, Whirlpool Dryer Code F3e2, Examples Of Unenumerated Rights Ireland, 2018 Honda Odyssey Low Battery Warning, Patio Door Locking Mechanism, Test Run Or Test-run, Pea Protein Vs Whey Bodybuilding, Aps Hamza Camp Contact Number, Japanese Honeysuckle Diffuser, Hp Officejet Pro 9010 Review,", "date": "2021-02-27 04:23:38", "meta": {"domain": "mangrovebd.org", "url": "https://exam.mangrovebd.org/jkg0b/applications-using-rational-equations-distance-rate-time-answers-dfe539", "openwebmath_score": 0.3689163625240326, "openwebmath_perplexity": 1890.9402952187716, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9932024682484215, "lm_q2_score": 0.9099069980980297, "lm_q1q2_score": 0.9037218763874748}}
{"url": "https://math.stackexchange.com/questions/1690904/rate-of-water-level-for-cone-shaped-water-tank", "text": "# Rate of water level for cone shaped water tank\n\nA water tank in the form of an inverted cone is being emptied at the rate of $6$ ft$^3$/min. The altitude of the cone is $24$ ft, and the base radius is $12$ ft. Find how fast the water level is lowering when the water is $10$ ft deep.\n\nI am not how to do this problem, but I've tried this using the volume formula for cone:\n\n$$v={1 \\over 3} \\pi r^3 h\\\\ {dv \\over dm} = {1 \\over 3} \\pi (12)^2{dh \\over dm}\\\\ 6 = {1 \\over 3} \\pi 144 \\cdot{dh \\over dm}\\\\ 6 = 48 \\pi \\cdot {dh \\over dm} \\\\ {1 \\over 8 \\pi} = {dh \\over dm}$$\n\nI am pretty sure that I am wrong.\n\nCould someone help me?\n\nThanks\n\nThe answer is ${6 \\over 25 \\pi}$ ft /min according to the answer sheet\n\nYou can't take $r=12$ in $$v={1 \\over 3} \\pi r^2 h\\\\ {dv \\over dm} = {2 \\over 3} \\pi (12){dh \\over dm}$$ because the radius of the water is changing as it drains. What you can do, however, is relate $r$ and $h$, because no matter how much water is left, the cone it forms will be proportional to the original cone. We see (from the given dimensions of the original cone) that $\\frac{r}{h} = \\frac{12}{24} = \\frac{1}{2}$, and $r=\\frac{h}{2}$. Let's substitute this for $r$ right away: $$v={1 \\over 3} \\pi r^2 h$$ $$v={1 \\over 3} \\pi (\\frac{h}{2})^2 h$$ $$v={1 \\over 3} \\pi \\frac{h^3}{4}$$ $$\\frac{dv}{dm} = \\pi (r)\\frac{h^2}{4}\\frac{dh}{dm}$$\n\n$$6 = \\pi (\\frac{h}{2})^2\\frac{dh}{dm}$$ $$6 = \\pi (\\frac{h^2}{4})\\frac{dh}{dm}$$ and plugging in $h=10$: $$6 = \\pi (\\frac{100}{4})\\frac{dh}{dm}$$ We get $\\frac{dh}{dm} = \\frac{6}{25\\pi}$.\n\n\u2022 I added a few more details. Is there anything I should elaborate on? \u2013\u00a0Quinn Greicius Mar 10 '16 at 0:56\n\u2022 Is it $r= {h \\over 2}$ because altitude is 24 and the radius is 12, so ${24 \\over 12} = {1 \\over 2}$? \u2013\u00a0didgocks Mar 10 '16 at 1:00\n\u2022 Exactly. Most related rates problems that involve several variables will use a trick like that to put everything in terms of a single variable to get to a solution. \u2013\u00a0Quinn Greicius Mar 10 '16 at 1:01\n\u2022 Thank you, it was very helpful! \u2013\u00a0didgocks Mar 10 '16 at 1:02\n\u2022 I just noticed that $v = {1 \\over 3} \\pi r^3 h$ is not a correct formula for a cone \u2013\u00a0didgocks Mar 13 '16 at 14:26\n\nRetain symbols till the last plugin step\n\n$$r = h \\cot \\alpha \\; ; \\tan \\alpha = \\frac12 ;\\; V = \\pi r^3/3\\; \\cot \\alpha$$\n\n$$V = \\pi h^3/3\\, \\tan ^2\\alpha$$\n\n$$dV/dt = \\pi h^2 (dh/dt) \\tan ^2\\alpha$$\n\nPlug in given values to get answer tallying with text.", "date": "2019-09-23 05:33:38", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1690904/rate-of-water-level-for-cone-shaped-water-tank", "openwebmath_score": 0.8331156969070435, "openwebmath_perplexity": 258.23426502855233, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9744347875615795, "lm_q2_score": 0.9273632871148287, "lm_q1q2_score": 0.9036550476721462}}
{"url": "https://gmatclub.com/forum/what-is-the-probability-of-getting-at-least-2-heads-in-a-row-130745.html", "text": "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 25 Sep 2018, 02:33\n\nGMAT Club Daily Prep\n\nThank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\nWhat is the probability of getting at least 2 heads in a row\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\nHide Tags\n\nIntern\nJoined: 26 Mar 2012\nPosts: 9\nWhat is the probability of getting at least 2 heads in a row\u00a0 [#permalink]\n\nShow Tags\n\n15 Apr 2012, 22:03\n1\n00:00\n\nDifficulty:\n\n(N/A)\n\nQuestion Stats:\n\n80% (01:23) correct 20% (00:23) wrong based on 10 sessions\n\nHideShow timer Statistics\n\nWhat is the probability of getting at least 2 heads in a row on three flips of a fair coin?\n\nno answers are available, sorry\nIntern\nJoined: 17 Feb 2012\nPosts: 22\nSchools: LBS '14\n\nShow Tags\n\n15 Apr 2012, 22:56\n1\n1\nIf we don't read the question attentively, we will be very likely to make the wrong calculations. In my opinion:\n1/2 X 1/2 X 1 (it doesen't matter) +\n+\n1/2 (the probability of not getting a head after the first flip) X 1/2 (the probability of getting a head) X 1/2( the probability of getting a head again)\n=\n1/4 + 1/8 = 3/8\n_________________\n\nKUDOS needed URGENTLY. Thank you in advance and be ACTIVE!\n\nSenior Manager\nJoined: 13 Mar 2012\nPosts: 277\nConcentration: Operations, Strategy\n\nShow Tags\n\n15 Apr 2012, 23:00\nrovshan85 wrote:\nwhat is the prob. of getting at least 2 heads in a row on three flips of a fair coin?\n\nno answers are available, sorry\n\nprobability for HHT;\n\nTotal number of ways of arranging HHT keeping HH intact is 2!\n\nprobability = 2! * (1/2)^3\n\nprobability for HHH;\n\nno of ways= 1\n\nprobability = (1/2)^3\n\nhence required answer = 2! * (1/2)^3 + (1/2)^3\n= 3* 1/8 = 3/8\n\nHope this helps...!!\n_________________\n\nPractice Practice and practice...!!\n\nIf there's a loophole in my analysis--> suggest measures to make it airtight.\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 49437\nRe: What is the probability of getting at least 2 heads in a row\u00a0 [#permalink]\n\nShow Tags\n\n16 Apr 2012, 01:09\n2\n1\nrovshan85 wrote:\nWhat is the probability of getting at least 2 heads in a row on three flips of a fair coin?\n\nno answers are available, sorry\n\nThe probability of at least 2 heads in a row on three flips is the sum of the probabilities of the following three cases: HHT, THH and HHH.\n\nNow, each case has the probability of $$(\\frac{1}{2})^3$$, so $$P=3*(\\frac{1}{2})^3=\\frac{3}{8}$$.\n\nHope it's clear.\n_________________\nIntern\nJoined: 17 Feb 2012\nPosts: 22\nSchools: LBS '14\nRe: What is the probability of getting at least 2 heads in a row\u00a0 [#permalink]\n\nShow Tags\n\n16 Apr 2012, 01:17\nBanuel's explanation is again the simpliest and the steadiest against mistakes.\n_________________\n\nKUDOS needed URGENTLY. Thank you in advance and be ACTIVE!\n\nTarget Test Prep Representative\nStatus: Head GMAT Instructor\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 2835\nRe: What is the probability of getting at least 2 heads in a row\u00a0 [#permalink]\n\nShow Tags\n\n14 Apr 2017, 05:46\n1\nrovshan85 wrote:\nWhat is the probability of getting at least 2 heads in a row on three flips of a fair coin?\n\nWe can assume the first two flips are heads (H) and the last flip is tails (T). Thus:\n\nP(H-H-T) = 1/2 x 1/2 x 1/2 = \u215b\n\nThe only other way to get two heads in a row would be flipping heads on the second and third flips.\n\nP(T-H-H) = 1/2 x 1/2 x 1/2 = \u215b\n\nThus, the total probability of getting two heads in a row when we flip a coin three times is 1/8 + 1/8 = 2/8.\n\nNext, we need to determine the probability of getting heads on all three flips.\n\nP(H-H-H) = 1/2 x 1/2 x 1/2 = 1/8.\n\nThus, the probability of getting at least two heads in a row is 2/8 + 1/8 = 3/8.\n_________________\n\nJeffery Miller\nHead of GMAT Instruction\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nSenior Manager\nJoined: 22 Feb 2018\nPosts: 326\nRe: What is the probability of getting at least 2 heads in a row\u00a0 [#permalink]\n\nShow Tags\n\n10 Mar 2018, 11:29\ntotal number case =2*2*2 =8\nas listed below\nHHH\nHHT\nTHH\n\nHTH\nTTH\nTHT\nTTT\nHTT\n\nfirst 3 case have alteast 2 consecutive heads, so probability is 3/8\n_________________\n\nGood, good Let the kudos flow through you\n\nRe: What is the probability of getting at least 2 heads in a row &nbs [#permalink] 10 Mar 2018, 11:29\nDisplay posts from previous: Sort by\n\nWhat is the probability of getting at least 2 heads in a row\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\nEvents & Promotions\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.", "date": "2018-09-25 09:33:00", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/what-is-the-probability-of-getting-at-least-2-heads-in-a-row-130745.html", "openwebmath_score": 0.7579609155654907, "openwebmath_perplexity": 1965.4260984372036, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9825575183283514, "lm_q2_score": 0.9196425377849806, "lm_q1q2_score": 0.9036016896751976}}
{"url": "https://sob5050.com/pnpfbmg/article.php?id=619bcf-rank-of-product-of-matrices", "text": "# rank of product of matrices\n\nis a linear combination of the rows of dimension of the linear space spanned by its columns (or rows). Proposition All Rights Reserved. be a the dimension of the space generated by its rows. If , The number of non zero rows is 2 \u2234 Rank of A is 2. \u03c1 (A) = 2. such C. Canadian0469. Find a Basis of the Range, Rank, and Nullity of a Matrix, Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space, Prove a Given Subset is a Subspace and Find a Basis and Dimension, True or False. full-rank matrix with is the : The order of highest order non\u2212zero minor is said to be the rank of a matrix. is the space then. is less than or equal to (a) rank(AB) \u2264 rank(A). ifwhich See the \u2026 Author(s): Heinz Neudecker; Satorra, Albert | Abstract: This paper develops a theorem that facilitates computing the degrees of freedom of an asymptotic \u03c7\u00b2 goodness-of-fit test for moment restrictions under rank deficiency of key matrices involved in the definition of the test. University Math Help. Therefore, by the previous two canonical basis). Enter your email address to subscribe to this blog and receive notifications of new posts by email. is the space As a consequence, also their dimensions (which by definition are As a consequence, the space two We can also In all the definitions in this section, the matrix A is taken to be an m \u00d7 n matrix over an arbitrary field F. Yes. Therefore, there exists an that can be written as linear matrix. is the if. Remember that the rank of a matrix is the The matrix Proof: First we consider a special case when A is a block matrix of the form Ir O1 O2 O3, where Ir is the identity matrix of dimensions r\u00d7r and O1,O2,O3 are zero matrices of appropriate dimensions. Thus, any vector is full-rank, it has less columns than rows and, hence, its columns are Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. . Step by Step Explanation. If is full-rank. Thread starter JG89; Start date Nov 18, 2009; Tags matrices product rank; Home. is full-rank, it has a square vector (being a product of an This implies that the dimension of Let It is left as an exercise (see 38 Partitioned Matrices, Rank, and Eigenvalues Chap. thatThen,ororwhere :where full-rank matrices. Learn how your comment data is processed. : :where . do not generate any vector The proof of this proposition is almost The Adobe Flash plugin is needed to view this content. , givesis haveNow, columns that span the space of all Then prove the followings. Nov 15, 2008 #1 There is a remark my professor made in his notes that I simply can't wrap my head around. we if is a linear combination of the rows of matrix and its transpose. Let then. . . is less than or equal to is impossible because is full-rank, Proving that the product of two full-rank matrices is full-rank Thread starter leden; Start date Sep 19, 2012; Sep 19, 2012 #1 leden. Finally, the rank of product-moment matrices is easily discerned by simply counting up the number of positive eigenvalues. matrix). Rank. As a consequence, the space . matrix. The rank of a matrix is the order of the largest non-zero square submatrix. Matrices. linearly independent rows that span the space of all As a consequence, there exists a PPT \u2013 The rank of a product of two matrices X and Y is equal to the smallest of the rank of X and Y: PowerPoint presentation | free to download - id: 1b7de6-ZDc1Z. Proposition We are going This lecture discusses some facts about If $\\min(m,p)\\leq n\\leq \\max(m,p)$ then the product will have full rank if both matrices in the product have full rank: depending on the relative size of $m$ and $p$ the product will then either be a product of two injective or of two surjective mappings, and this is again injective respectively surjective. This website is no longer maintained by Yu. Let us transform the matrix A to an echelon form by using elementary transformations. we Thus, the space spanned by the rows of :where and are equal because the spaces generated by their columns coincide. vector (being a product of a [Note: Since column rank = row rank, only two of the four columns in A \u2014 c \u2026 can be written as a linear combination of the rows of . is a inequalitiesare be a Let pr.probability matrices st.statistics random-matrices hadamard-product share | cite | improve this question | follow | It is a generalization of the outer product from vectors to matrices, and gives the matrix of the tensor product with respect to a standard choice of basis. , coincide, so that they trivially have the same dimension, and the ranks of the so they are full-rank. Add to solve later Sponsored Links We can also is the identical to that of the previous proposition. Add the \ufb01rst row of (2.3) times A\u22121 to the second row to get (A B I A\u22121 +A\u22121B). Proposition that . We now present a very useful result concerning the product of a non-square thenso is full-rank. such Save my name, email, and website in this browser for the next time I comment. This website\u2019s goal is to encourage people to enjoy Mathematics! are linearly independent and Then, the product equal to the ranks of Since That means,the rank of a matrix is \u2018r\u2019 if i. be two Let 7 0. Find the rank of the matrix A= Solution : The order of A is 3 \u00d7 3. Rank of a Matrix. https://www.statlect.com/matrix-algebra/matrix-product-and-rank. is full-rank, Since is the rank of Being full-rank, both matrices have rank of all vectors Here it is: Two matrices\u2026 The list of linear algebra problems is available here. column vector with coefficients taken from the vector In most data-based problems the rank of C(X), and other types of derived product-moment matrices, will equal the order of the (minor) product-moment matrix. that can be written as linear combinations of the rows of Furthermore, the columns of vector of coefficients of the linear combination. can be written as a linear combination of the columns of Note that if A ~ B, then \u03c1(A) = \u03c1(B) Multiplication by a full-rank square matrix preserves rank, The product of two full-rank square matrices is full-rank. Denote by entry of the matrix and a full-rank coincide. Rank of Product Of Matrices. Rank of the Product of Matrices AB is Less than or Equal to the Rank of A Let A be an m \u00d7 n matrix and B be an n \u00d7 l matrix. Required fields are marked *. If A and B are two equivalent matrices, we write A ~ B. vectors. the space generated by the columns of spanned by the columns of Sum, Difference and Product of Matrices; Inverse Matrix; Rank of a Matrix; Determinant of a Matrix; Matrix Equations; System of Equations Solved by Matrices; Matrix Word Problems; Limits, Derivatives, Integrals; Analysis of Functions Thus, any vector is the This video explains \" how to find RANK OF MATRIX \" with an example of 4*4 matrix. Advanced Algebra. J. JG89. if matrix and Rank of product of matrices with full column rank Get link; Facebook; Twitter; Pinterest the exercise below with its solution). ) The rank of a matrix with m rows and n columns is a number r with the following properties: r is less than or equal to the smallest number out of m and n. r is equal to the order of the greatest minor of the matrix which is not 0. He even gave a proof but it made me even more confused. Column Rank = Row Rank. coincide. writewhere matrix. Advanced Algebra. -th This implies that the dimension of the space spanned by the rows of How to Find Matrix Rank. which implies that the columns of The product of two full-rank square matrices is full-rank An immediate corollary of the previous two propositions is that the product of two full-rank square matrices is full-rank. For example . :where is no larger than the span of the rows of rank of the Oct 2008 27 0. matrix and propositionsBut vector In a strict sense, the rule to multiply matrices is: \"The matrix product of two matrixes A and B is a matrix C whose elements a i j are formed by the sums of the products of the elements of the row i of the matrix A by those of the column j of the matrix B.\" whose dimension is How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, Express a Vector as a Linear Combination of Other Vectors, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Find a Basis for the Subspace spanned by Five Vectors, 12 Examples of Subsets that Are Not Subspaces of Vector Spaces. it, please check the previous articles on Types of Matrices and Properties of Matrices, to give yourself a solid foundation before proceeding to this article. Forums. whose dimension is Keep in mind that the rank of a matrix is . Then, The space Example 1.7. . denotes the for rank. then. multiply it by a full-rank matrix. and two matrices are equal. (1) The product of matrices with full rank always has full rank (for example using the fact that the determinant of the product is the product of the determinants) (2) The rank of the product is always less than or equalto the minimum rank of the matrices being multiplied. To see this, note that for any vector of coefficients Proposition \"Matrix product and rank\", Lectures on matrix algebra. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). An immediate corollary of the previous two propositions is that the product of Let . Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to \u2026 have just proved that any vector Aug 2009 130 16. that A = ( 1 0 ) and B ( 0 ) both have rank 1, but their product, 0, has rank 0 ( 1 ) , the rank of a matrix is the order of a matrix is the dimension of less... Describe a method for finding the product of two full-rank square matrix scalar, we two. A given matrix by applying any of the linear combination of the rows of in particular we! A full-rank matrix we multiply it by a full-rank matrix their dimensions ( which by definition are because! Vector: for any vector note that for any vector rank ; Home published 08/28/2017, email. Suppose that there exists a non-zero vector such thatThusThis means that any is a linear combination of the of... Echelon form by using elementary transformations matrix obtained from a given matrix by applying of. Solution: the order of highest order non\u2212zero minor is said to be equivalent to it an matrix its. Its rows also writewhere is an vector ) by email we can multiply two matrices dimension of less! A to an echelon form by using elementary transformations form by using transformations. Is preserved thus, the space is no larger than the span of columns... ) rank ( a ) = rank ( a ) nor rank ( )... Materials found on this website are now available in a traditional textbook format note that any... The order of a vector ( being a product of block matrices of the vector... So they are full-rank us transform the matrix A= Solution: the order of a is! Matrix C = AB is full-rank an n\u00d7lmatrix now present a very useful result the! Coefficients taken from the vector of coefficients of the space spanned by columns. Non-Zero square submatrix suppose that there exists a non-zero vector such thatThen, denotes... Proved that the rank of a non-square matrix and a square matrix preserves rank, the product of two.... Explains  how to find rank of product-moment matrices is full-rank, as well name email... Does not change when we multiply it by a full-rank matrix a non-square matrix and its transpose easily., with coefficients taken from the vector of coefficients of the elementary row operations is said to the... Is available here is to encourage people to enjoy Mathematics haveThe two inequalitiesare satisfied if only! Matrix product and rank '', Lectures on matrix algebra to it and only if a and be... Scalar, we describe a method for finding the product of two full-rank square matrix inequalitiesare satisfied if only..., by the space is no larger than the span of the previous proposition non-zero row a is 2. (... Does not change when we multiply it by a full-rank square matrix explained solutions if.. Method assumes familiarity with echelon matrices and echelon transformations or rows ) are! Written as a consequence, the space spanned by the columns of do not generate vector! People to enjoy Mathematics ; Start date Nov rank of product of matrices, 2009 ; Tags matrices product ;. Denote by the previous two propositionsBut and are equal to the second row to get ( a ) rank AB! Keep in mind that the rank of any matrix fact is that the matrix a to an form., any vector of coefficients, if thenso that traditional textbook format can be while. Exercise below with its Solution ) matrix, Nullity of a is 3 3... Solution: the order of highest order non\u2212zero minor is said to be the rank of matrices! Any matrix is the rank of a product of two matrices highest order non\u2212zero minor is said to be to. To prove that the rank of a matrix by a full-rank matrix,... Proved that the dimension of the matrices being multiplied is preserved ) are.... Solution: the order of a matrix is the dimension of is than. Non-Square matrix and an vector ) ; Tags matrices product rank ; Home order! Find the rank of any matrix and receive notifications of new posts by.... No larger than the span of the rows of for any vector do not generate any vector: for vector... Independent rows that span the space is no larger than the span of the linear combination of the materials! A matrix is the order of a product of a matrix is the of... Only if algebra problems is available here not change when we multiply it by a scalar, analyze! Is \u2018 r \u2019 if I the elementary row operations is said to the! Than the span of the learning materials found on this website are now available a... S goal is to encourage people to enjoy Mathematics interests us now, email! That any is rank of product of matrices linear combination 2009 ; Tags matrices product rank ; Home product-moment matrices full-rank... Also writewhere rank of product of matrices a linear combination of the columns of, whose dimension is and! 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Two propositionsBut and are, so they are full-rank are equal to are... Vector ) method for finding the rank of, whose dimension is X 0 I ), ( I 0... Such thatThen, ororwhere denotes the -th entry of the columns of and that spanned the! Square matrix easily discerned by simply counting up the number of positive Eigenvalues two is. And ) coincide ( see the exercise below with its rank of product of matrices ) finally, the product of matrices... Start date Nov 18, 2009 ; Tags matrices product rank ; Home website \u2019 s goal to. Two matrices are zero say I have a mxn matrix a to an echelon form by elementary. Of two matrices in addition to multiplying a matrix obtained from a given by... Their columns coincide an example of 4 * 4 matrix two inequalitiesare if... Name, email, and website in this browser for the next proposition provides bound! Another important fact is that the rank of the matrix A= Solution: the order of highest order minor! Echelon matrices and echelon transformations ; Tags matrices product rank ; Home or rows ) in! Of the matrices being multiplied is preserved of new posts by email since is full-rank given matrix by applying of! Date Nov 18, 2009 ; Tags matrices product rank ; Home an immediate corollary of the rows:. Nor rank ( AB ) = 2 a proof but it made me more! Columns of are linearly independent rows that span the space is no larger than the span the... Familiarity with echelon matrices and echelon transformations and echelon transformations problems is here! Multiplication by a scalar, we haveThe two inequalitiesare satisfied if and only if and coincide... Zero rows is 2 \u2234 rank of matrix  with an example of *. With coefficients taken from the vector in particular, we haveThe two inequalitiesare satisfied and...", "date": "2021-04-18 05:32:12", "meta": {"domain": "sob5050.com", "url": "https://sob5050.com/pnpfbmg/article.php?id=619bcf-rank-of-product-of-matrices", "openwebmath_score": 0.7332128286361694, "openwebmath_perplexity": 459.14841579626847, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9848109538667758, "lm_q2_score": 0.9173026561945815, "lm_q1q2_score": 0.9033697038315128}}
{"url": "https://tvrepairtips.org/nicholas-pompeo-rvqte/0lr2d09.php?ee4b06=inverse-of-a-matrix-in-python-without-numpy", "text": "Doing the math to determine the determinant of the matrix, we get, (8) (3)- \u2026 If the generated inverse matrix is correct, the output of the below line will be True. You don\u2019t need to use Jupyter to follow along. Yes! Subtract 1.0 * row 1 of A_M from row 3 of A_M, and \u00a0 \u00a0 Subtract 1.0 * row 1 of I_M from row 3 of I_M, 5. Here are the steps, S, that we\u2019d follow to do this for any size matrix. Those previous posts were essential for this post and the upcoming posts. In this post, we will be learning about different types of matrix multiplication in the numpy \u2026 If you did most of this on your own and compared to what I did, congratulations! \\end{bmatrix} Subtract 2.4 * row 2 of A_M from row 3 of A_M\u00a0 \u00a0 Subtract 2.4 * row 2 of I_M from row 3 of I_M, 7. Now we pick an example matrix from a Schaum's Outline Series book Theory and Problems of Matrices by Frank Aryes, Jr1. Python buffer object pointing to the start of the array\u2019s data. I would even think it\u2019s easier doing the method that we will use when doing it by hand than the ancient teaching of how to do it. The python matrix makes use of arrays, and the same can be implemented. The other sections perform preparations and checks. Plus, if you are a geek, knowing how to code the inversion of a matrix is a great right of passage! NumPy: Determinant of a Matrix. An inverse of a matrix is also known as a reciprocal matrix. The NumPy code is as follows. Python | Numpy matrix.sum() Last Updated: 20-05-2019 With the help of matrix.sum() method, we are able to find the sum of values in a matrix by using the same method. When this is complete, A is an identity matrix, and I becomes the inverse of A. Let\u2019s go thru these steps in detail on a 3 x 3 matrix, with actual numbers. And please note, each S represents an element that we are using for scaling. In Linear Algebra, an identity matrix (or unit matrix) of size $n$ is an $n \\times n$ square matrix with $1$'s along the main diagonal and $0$'s elsewhere. data. Subtract 0.472 * row 3 of A_M from row 2 of A_M\u00a0 \u00a0 Subtract 0.472 * row 3 of I_M from row 2 of I_M. There are also some interesting Jupyter notebooks and .py files in the repo. Returns the (multiplicative) inverse of invertible self. Python statistics and matrices without numpy. Please don\u2019t feel guilty if you want to look at my version immediately, but with some small step by step efforts, and with what you have learned above, you can do it. We start with the A and I matrices shown below. We will see two types of matrices in this chapter. You want to do this one element at a time for each column from left to right. Python Matrix. This means that the number of rows of A and number of columns of A must be equal. Why wouldn\u2019t we just use numpy or scipy? Code faster with the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing. With numpy.linalg.inv an example code would look like that: Python doesn't have a built-in type for matrices. We will be using NumPy (a good tutorial here) and SciPy (a reference guide here). I\u2019ve also saved the cells as MatrixInversion.py in the same repo. It is imported and implemented by LinearAlgebraPractice.py. Also, once an efficient method of matrix inversion is understood, you are ~ 80% of the way to having your own Least Squares Solver and a component to many other personal analysis modules to help you better understand how many of our great machine learning tools are built. The following line of code is used to create the Matrix. See the code below. dtype. Matrix Multiplication in NumPy is a python library used for scientific computing. Below is the output of the above script. Then come back and compare to what we\u2019ve done here. In this post, we create a clustering algorithm class that uses the same principles as scipy, or sklearn, but without using sklearn or numpy or scipy. Let\u2019s start with some basic linear algebra to review why we\u2019d want an inverse to a matrix. If you go about it the way that you would program it, it is MUCH easier in my opinion. A_M and I_M , are initially the same, as A and I, respectively: A_M=\\begin{bmatrix}5&3&1\\\\3&9&4\\\\1&3&5\\end{bmatrix}\\hspace{4em} I_M=\\begin{bmatrix}1&0&0\\\\0&1&0\\\\0&0&1\\end{bmatrix}, 1. You can verify the result using the numpy.allclose() function. In other words, for a matrix [[a,b], [c,d]], the determinant is computed as \u2018ad-bc\u2019. If a is a matrix object, then the return value is a matrix as well: >>> ainv = inv ( np . One way to \u201cmultiply by 1\u201d in linear algebra is to use the identity matrix. To find A^{-1} easily, premultiply B by the identity matrix, and perform row operations on A to drive it to the identity matrix. \\begin{bmatrix} It\u2019s important to note that A must be a square matrix to be inverted. If you found this post valuable, I am confident you will appreciate the upcoming ones. Yes! Let\u2019s simply run these steps for the remaining columns now: That completes all the steps for our 5\u00d75. If at some point, you have a big \u201cAh HA!\u201d moment, try to work ahead on your own and compare to what we\u2019ve done below once you\u2019ve finished or peek at the stuff below as little as possible IF you get stuck. The Numpy module allows us to use array data structures in Python which are really fast and only allow same data type arrays. An identity matrix of size $n$ is denoted by $I_{n}$. GitHub Gist: instantly share code, notes, and snippets. The original A matrix times our I_M matrix is the identity matrix, and this confirms that our I_M matrix is the inverse of A. I want to encourage you one last time to try to code this on your own. which is its inverse. AA^{-1} = A^{-1}A = I_{n} A^{-1}). Let\u2019s first define some helper functions that will help with our work. One of them can generate the formula layouts in LibreOffice Math formats. I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the \u201cprinciples\u201d of a thing, the better I know how to apply it. We will be walking thru a brute force procedural method for inverting a matrix with pure Python. Using the steps and methods that we just described, scale row 1 of both matrices by 1/5.0, 2. \\end{bmatrix} I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the \u201cprinciples\u201d of a thing, the better I know how to apply it. Perform the same row operations on I that you are performing on A, and I will become the inverse of A (i.e. 1 & 0 & 0 & 0\\\\ However, we can treat list of a list as a matrix. In this tutorial, we will make use of NumPy's numpy.linalg.inv() function to find the inverse of a square matrix. So how do we easily find A^{-1} in a way that\u2019s ready for coding? $$If at this point you see enough to muscle through, go for it! The way that I was taught to inverse matrices, in the dark ages that is, was pure torture and hard to remember! 0 & 0 & 1$$ However, we may be using a closely related post on \u201csolving a system of equations\u201d where we bypass finding the inverse of A and use these same basic techniques to go straight to a solution for X. It\u2019s a great right of passage to be able to code your own matrix inversion routine, but let\u2019s make sure we also know how to do it using numpy / scipy from the documentation HERE. 1. The numpy.linalg.det() function calculates the determinant of the input matrix. 0 & 0 & 1 & 0\\\\ Success! left_hand_side_inverse = left_hand_side.I left_hand_side_inverse solution = left_hand_side_inverse*right_hand_side solution Now, we can use that first row, that now has a 1 in the first diagonal position, to drive the other elements in the first column to 0. Let\u2019s get started with Matrices in Python. The first matrix in the above output is our input A matrix. Learning to work with Sparse matrix, a large matrix or 2d-array with a lot elements being zero, can be extremely handy. The second matrix is of course our inverse of A. \\end{bmatrix} My encouragement to you is to make the key mathematical points your prime takeaways. I would not recommend that you use your own such tools UNLESS you are working with smaller problems, OR you are investigating some new approach that requires slight changes to your personal tool suite. Plus, tomorrow\u2026 To find out the solution you have to first find the inverse of the left-hand side matrix and multiply with the right side. In future posts, we will start from here to see first hand how this can be applied to basic machine learning and how it applies to other techniques beyond basic linear least squares linear regression. As per this if i need to calculate the entire matrix inverse it will take me 1779 days. A_M has morphed into an Identity matrix, and I_M has become the inverse of A. $$. I don\u2019t recommend using this. Python provides a very easy method to calculate the inverse of a matrix. Inverse of an identity [I] matrix is an identity matrix [I]. Using this library, we can perform complex matrix operations like multiplication, dot product, multiplicative inverse, etc. In this tutorial, we will learn how to compute the value of a determinant in Python using its numerical package NumPy's numpy.linalg.det() function. This blog is about tools that add efficiency AND clarity. Since the resulting inverse matrix is a 3 \\times 3 matrix, we use the numpy.eye() function to create an identity matrix. The reason is that I am using Numba to speed up the code, but numpy.linalg.inv is not supported, so I am wondering if I can invert a matrix with 'classic' Python code. So hang on! \\begin{bmatrix} In fact, it is so easy that we will start with a 5\u00d75 matrix to make it \u201cclearer\u201d when we get to the coding. PLEASE NOTE: The below gists may take some time to load. To calculate the inverse of a matrix in python, a solution is to use the linear \u2026 We will also go over how to use numpy /scipy to invert a matrix at the end of this post. in a single step. How to do gradient descent in python without numpy or scipy. \\begin{bmatrix} We\u2019ll do a detailed overview with numbers soon after this. Create a Python Matrix using the nested list data type; Create Python Matrix using Arrays from Python Numpy package; Create Python Matrix using a nested list data type. Python is crazy accurate, and rounding allows us to compare to our human level answer. Be sure to learn about Python lists before proceed this article. \\end{bmatrix} I hope that you will make full use of the code in the repo and will refactor the code as you wish to write it in your own style, AND I especially hope that this was helpful and insightful. But it is remarkable that python can do such a task in so few lines of code. There will be many more exercises like this to come. Or, as one of my favorite mentors would commonly say, \u201cIt\u2019s simple, it\u2019s just not easy.\u201d We\u2019ll use python, to reduce the tedium, without losing any view to the insights of the method. I know that feeling you\u2019re having, and it\u2019s great! You can verify the result using the numpy.allclose() function. I encourage you to check them out and experiment with them. Base object if memory is from some other object. This is the last function in LinearAlgebraPurePython.py in the repo. Write a NumPy program compute the inverse of a given matrix. \\end{bmatrix} Would I recommend that you use what we are about to develop for a real project? To work with Python Matrix, we need to import Python numpy module. Kite is a free autocomplete for Python developers. We will see at the end of this chapter that we can solve systems of linear equations by using the inverse matrix. Here, we are going to reverse an array in Python built with the NumPy module. Why wouldn\u2019t we just use numpy or scipy? I_{2} = Creating a Matrix in NumPy; Matrix operations and examples; Slicing of Matrices; BONUS: Putting It All Together \u2013 Python Code to Solve a System of Linear Equations. Get it on GitHub AND check out Integrated Machine Learning & AI coming soon to YouTube. The only really painful thing about this method of inverting a matrix, is that, while it\u2019s very simple, it\u2019s a bit tedious and boring. An object to simplify the interaction of the array with the ctypes module. base. Since the resulting inverse matrix is a 3 \\times 3 matrix, we use the numpy.eye() function to create an identity matrix. Let\u2019s start with the logo for the github repo that stores all this work, because it really says it all: We frequently make clever use of \u201cmultiplying by 1\u201d to make algebra easier. The main thing to learn to master is that once you understand mathematical principles as a series of small repetitive steps, you can code it from scratch and TRULY understand those mathematical principles deeply. Published by Thom Ives on November 1, 2018November 1, 2018. We then divide everything by, 1/determinant. It should be mentioned that we may obtain the inverse of a matrix using ge, by reducing the matrix $$A$$ to the identity, with the identity matrix as the augmented portion.$$. Let\u2019s first introduce some helper functions to use in our notebook work. 0 & 0 & 0 & 1 $$My approach using numpy / scipy is below. The flip() method in the NumPy module reverses the order of a NumPy array and returns the NumPy array object. 1 & 3 & 3 \\\\ When we multiply the original A matrix on our Inverse matrix we do get the identity matrix. We then operate on the remaining rows (S_{k2} to S_{kn}), the ones without fd in them, as follows: We do this for all columns from left to right in both the A and I matrices. When you are ready to look at my code, go to the Jupyter notebook called MatrixInversion.ipynb, which can be obtained from the github repo for this project. >>> import numpy as np #load the Library$$. If the generated inverse matrix is correct, the output of the below line will be True. I want to be part of, or at least foster, those that will make the next generation tools. 0 & 1 \\\\ If you don\u2019t use Jupyter notebooks, there are complementary .py files of each notebook. Data Scientist, PhD multi-physics engineer, and python loving geek living in the United States. It all looks good, but let\u2019s perform a check of A \\cdot IM = I. Matrix Operations: Creation of Matrix. An inverse of a square matrix $A$ of order $n$ is the matrix $A^{-1}$ of the same order, such that, their product results in an identity matrix $I_{n}$. In Python, the \u2026 Python\u2019s SciPy library has a lot of options for creating, storing, and operating with Sparse matrices. \\begin{bmatrix} which clearly indicate that writing one column of inverse matrix to hdf5 takes 16 minutes.", "date": "2021-01-17 06:50:16", "meta": {"domain": "tvrepairtips.org", "url": "https://tvrepairtips.org/nicholas-pompeo-rvqte/0lr2d09.php?ee4b06=inverse-of-a-matrix-in-python-without-numpy", "openwebmath_score": 0.5504042506217957, "openwebmath_perplexity": 495.3180601137551, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.975201841245846, "lm_q2_score": 0.9263037302939515, "lm_q1q2_score": 0.903333103335557}}
{"url": "https://amaanswers.com/how-do-you-calculate-arcsin", "text": "# How do you calculate Arcsin?\n\nSamuel Appleberry asked, updated on August 13th, 2021; Topic: arcsin\n\ud83d\udc41 390 \ud83d\udc4d 16 \u2605\u2605\u2605\u2605\u26064.7\n\nUsing arcsine to find an angle First, calculate the sine of \u03b1 by dividng the opposite side by the hypotenuse. This results in sin(\u03b1) = a / c = 52 / 60 = 0.8666. Use the inverse function with this outcome to calculate the angle \u03b1 = arcsin(0.8666) = 60\u00b0 (1.05 radians).\n\nAdd on, is Arctan and tan 1 the same?\n\nThe inverse of tangent is denoted as Arctangent or on a calculator it will appear as atan or tan-1. Note: this does NOT mean tangent raised to the negative one power. ... Sine, cosine, secant, tangent, cosecant and cotangent are all functions however, the inverses are only a function when given a restricted domain.\n\nForbye, how do you find the tangent angle on a calculator? Examples\n\n\u2022 Step 1 The two sides we know are Opposite (300) and Adjacent (400).\n\u2022 Step 2 SOHCAHTOA tells us we must use Tangent.\n\u2022 Step 3 Calculate Opposite/Adjacent = 300/400 = 0.75.\n\u2022 Step 4 Find the angle from your calculator using tan-1\n\u2022 Ever, what is the formula for Arctan?\n\nFrom this given quantity, 1.732 can be written as a function of tan. 60\u00b0 = 60 \\times \\frac{\\pi}{180} = 1.047 radians....Solution:\n\nDimensional Formula Of ResistivityInverse Matrix Formula\n\nHow do you do Arctan on TI 84?\n\nPress the calculator's \"shift,\" \"2nd\" or \"function\" key, and then press the \"tan\" key. Type the number whose arctan you want to find. For this example, type in the number \"0.577.\" Press the \"=\" button.\n\n### What is Arctan 1 in terms of pi?\n\nSince. tan \u03c0/4 = tan 45\u00ba = 1. The arctangent of 1 is equal to the inverse tangent function of 1, which is equal to \u03c0/4 radians or 45 degrees: arctan 1 = tan-1 1 = \u03c0/4 rad = 45\u00ba\n\n### What is the symbol for Arctan?\n\nPrincipal valuesNameUsual notationDefinition\narctangenty = arctan(x)x = tan(y)\narccotangenty = arccot(x)x = cot(y)\narcsecanty = arcsec(x)x = sec(y)\narccosecanty = arccsc(x)x = csc(y)\n\n### What is tan 1x?\n\nThe Function y = tan -1x = arctan x and its Graph: Since y = tan -1x is the inverse of the function y = tan x, the function y = tan -1x if and only if tan y = x.\n\n### What is Arctan of infinity?\n\nThe arctangent is the inverse tangent function. The limit of arctangent of x when x is approaching infinity is equal to pi/2 radians or 90 degrees: The limit of arctangent of x when x is approaching minus infinity is equal to -pi/2 radians or -90 degrees: Arctan\n\n### Where is tan equal to 1?\n\nBasic idea: To find tan-1 1, we ask \"what angle has tangent equal to 1?\" The answer is 45\u00b0. As a result we say that tan-1 1 = 45\u00b0. In radians this is tan-1 1 = \u03c0/4. More: There are actually many angles that have tangent equal to 1.\n\n### Can Arctan be negative?\n\nThe arctangent of a negative number is a negative first quadrant angle, sin-1(-) is in quadrant -I, a clockwise-angle of less than - /2. When you simplify an expression, be sure to use the Arcsine.\n\n### Is Arctan the inverse of tan?\n\nThe arctan function is the inverse of the tangent function. It returns the angle whose tangent is a given number. Try this Drag any vertex of the triangle and see how the angle C is calculated using the arctan() function. Means: The angle whose tangent is 0.577 is 30 degrees.\n\n### What is Arcsin on a calculator?\n\nArcsine definition The arcsine function is the inverse function of y = sin(x). arcsin(y) = sin-1(y) = x + 2k\u03c0\n\n### Does Arcsin cancel out sin?\n\nThe arcsine function is the inverse function for the sine function on the interval [ \u2212 \u03c0 / 2 , \u03c0 / 2 ] . So they \"cancel\" each other under composition of functions, as follows. The notation for inverse functions, f-1(x) is just that: notation, a shorthand way of writing the inverse of a function f.\n\n### What is Arcsin equal to?\n\nThe arcsin function is the inverse of the sine function. It returns the angle whose sine is a given number. ... Means: The angle whose sin is 0.5 is 30 degrees. Use arcsin when you know the sine of an angle and want to know the actual angle.\n\n### How do you use Arctan?\n\nYou can use the arctan to determine an angle measure when the side opposite and the side adjacent to the angle are known....When to Use Arctan\n\u2022 sine = opposite / hypotenuse.\n\u2022 cosine = adjacent / hypotenuse.\n\u2022 tangent = opposite / adjacent.\n\u2022 \ufeff", "date": "2022-08-10 07:35:15", "meta": {"domain": "amaanswers.com", "url": "https://amaanswers.com/how-do-you-calculate-arcsin", "openwebmath_score": 0.8214149475097656, "openwebmath_perplexity": 1383.8321398458863, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9845754497285468, "lm_q2_score": 0.9173026612812897, "lm_q1q2_score": 0.9031536802682186}}
{"url": "https://mathhelpboards.com/threads/x-2-4-versus-x-4-1-2.3005/", "text": "# x^2 = 4 versus x = 4 ^ (1/2)\n\n#### avr5iron\n\n##### New member\nCan someone explain why the solution for x in x^2 = 4 is x = 2, -2\nwhile the solution for x in x = 4 ^ (1/2) is 2\n\n#### MarkFL\n\nStaff member\n1.) $\\displaystyle x^2=4$\n\nNow, using the square root property, we find:\n\n$\\displaystyle x=\\pm\\sqrt{4}=\\pm2$\n\n2.) $\\displaystyle x=4^{\\frac{1}{2}}=\\sqrt{4}=2$\n\nYou see, in the first equation, we have the square of x being equal to a positive value (4), which means x may have two values as the square of a negative is positive.\n\nIn the second equation, we simply have x equal to a positive value, so there is just that one solution.\n\n#### Poly\n\n##### Member\nCan someone explain why the solution for x in x^2 = 4 is x = 2, -2\nwhile the solution for x in x = 4 ^ (1/2) is 2\nI remember being confused about this too, and here is where the confusion comes from I think. You're thinking that the steps in the first statement are $x^2 = 4 \\implies x = 4^{\\frac{1}{2}} = -2, 2$ when in fact they are $x^2 = 4 \\implies x = \\pm 4^{\\frac{1}{2}} = -2, 2$ (as explained above). Now there's no inconsistency.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nnaively, one might think:\n\n$x^2 = 4$\n\ntherefore:\n\n$(x^2)^{\\frac{1}{2}} = 4^{\\frac{1}{2}}$\n\nthat is:\n\n$x^{(2)\\left(\\frac{1}{2}\\right)} = x = 2$.\n\nand we know that if $x = -2$ we have $x^2 = 4$, so what gives?\n\nin general, the rule:\n\n$(a^b)^c = a^{bc}$\n\nonly holds for POSITIVE numbers $a$ (it is a GOOD idea to burn this into your brain). $a = 0$ is a special case, normally it's fine, but problems arise with $0^0$.\n\nwhile it is true that:\n\n$(k^b)^c = k^{bc}$ for INTEGERS $k$, and INTEGERS $b$ and $c$, things go horribly wrong when we try to define things like:\n\n$(-4)^{\\frac{1}{2}}$\n\nand what this means is, when we write:\n\n$x^{\\frac{1}{2}} = y$\n\nwe are already tacitly assuming $x > 0$.\n\nyou can see the graph of $f(x) = \\sqrt{x} = x^{\\frac{1}{2}}$ here:\n\ny = x^(1/2) - Wolfram|Alpha\n\nthe \"orange lines\" mean that the values of $y$ at $x < 0$ are complex-but-not-real (in fact, they are pure imaginary).\n\non a deeper level, what is happening is this:\n\nthe \"squaring function\" is not 1-1, it always converts signs to positive (even if we started with a negative). you can think of this as \"losing information about where we started from\". as a result, we can only \"partially recover\" our beginnings, by taking a square root (we know the size, but we can only guess at the sign).\n\nthe symbol $\\pm$ in the answer to $x^2 = 4$ (that is: $x = \\pm 2$) is the way we indicate this uncertainty.\n\nhowever, the function $y = x^{\\frac{1}{2}}$ is only defined for $x \\geq 0$ (we only get \"the top half of the parabola\" $y^2 = x$), so at $x = 4$, we have a unique value, namely: 2.\n\nthis indicates a peculiarity of functions: they can \"shrink\" or \"collapse\" their domains, but they only give ONE output for ONE input, so they cannot always \"reverse themselves\".\n\n#### soroban\n\n##### Well-known member\nHello, avr5iron!\n\n$\\text{Can someone explain why the solution is }\\pm2\\,\\text{ for }\\,x^2 \\:=\\: 4$\n. . $\\text{while the solution is }2\\,\\text{ for }\\,x \\:=\\: 4^{\\frac{1}{2}}$\n\nThe first is a quadratic equation; it has two roots.\n\n. . $x^2 - 4 \\:=\\:0 \\quad\\Rightarrow\\quad (x-2)(x+2) \\:=\\:0 \\quad\\Rightarrow\\quad x \\:=\\:\\pm2$\n\nThe second is a linear equation; one root.\n\n. . $x \\:=\\:4^{\\frac{1}{2}} \\:=\\:\\sqrt{4} \\quad\\Rightarrow\\quad x \\:=\\:2$\n\n~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~\n\nI have explained this to my students like this . . .\n\nIf they give us a square root,\n. . we assume it has the principal (positive) value.\n\nSo that: .$\\sqrt{9} \\:=\\:3$\n\nIf we introduce a square root,\n. . then we take the responsibility for both values.\n\nSo that: $x^2 \\,=\\,9 \\quad\\Rightarrow\\quad x \\,=\\,\\pm\\sqrt{9} \\quad\\Rightarrow\\quad x \\,=\\,\\pm3$\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\n$\\sqrt{4}$ you are performing the square root operation on a number so the result is unique , it is like usual operations you don't get multiple answers if you add or multiply\nnumbers but when you have $x^2=4 \\,\\,\\Rightarrow \\,\\, \\sqrt{x^2}=\\sqrt{4}$ then $\\pm x=2$ so we are actually performing the square\nroot property on a variable now the result is not unique since a variable might have mutl-\nWhen someone asks for $\\sqrt{4}$ , are they not asking for a value to be determined x , such that\n$4 = x \\cdot x$", "date": "2020-12-03 10:41:23", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/x-2-4-versus-x-4-1-2.3005/", "openwebmath_score": 0.8925647735595703, "openwebmath_perplexity": 454.70053717674534, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9820137942490252, "lm_q2_score": 0.9196425372343816, "lm_q1q2_score": 0.9031016573423356}}
{"url": "https://www.khanacademy.org/math/calculus-2/cs2-integration-techniques/cs2-integrating-with-u-substitution/a/worksheet-u-substitution", "text": "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\n# \ud835\ude36-substitution warmup\n\nAP.CALC:\nFUN\u20116 (EU)\n,\nFUN\u20116.D (LO)\n,\nFUN\u20116.D.1 (EK)\nBefore diving into our practice exercise, gain some risk-free experience performing\u00a0\ud835\ude36-substitution.\nFind each indefinite integral.\n\n# Problem 1\n\nintegral, cosine, left parenthesis, x, squared, right parenthesis, 2, x, d, x, equals\nplus, space, C\n\n# Problem 2\n\nintegral, start fraction, 3, x, squared, divided by, left parenthesis, x, cubed, plus, 3, right parenthesis, squared, end fraction, d, x, equals\nplus, space, C\n\n# Problem 3\n\nintegral, e, start superscript, 4, x, end superscript, d, x, equals\nplus, space, C\n\n# Problem 4\n\nintegral, x, dot, square root of, start fraction, 1, divided by, 6, end fraction, x, squared, plus, 1, end square root, d, x, equals\nplus, space, C\n\n## Want to join the conversation?\n\n\u2022 Well, in the problem #2 what happens to the \"square power\" of \"u\" when you substitue back the equation x^3 + 3 please? I did not catch what happened to get the given answer. Thank you. :)\n\u2022 1/(u^2) == u^(-2). When you integrate, you will increase the power by one (becomes -1) and multiply by the reciprocal of the new power (also -1). Your integral is -1*(u^-1) ==(-1/u).\nThis problem is tricky because of the properties of exponents, just try rewriting the factors to understand where the exponent went to.\n\u2022 =\u222b1/u^2 \u200bdu shoudn't be = ln(|u^2|)?\n\u2022 You are reversing the power rule so the answer is -1/u \u200b+C. However, integral(1/u) =ln(|u|) + C.\n\u2022 In problem 2, why the negative?\n\n1/u^2 * du\n-1/u\n(1 vote)\n\u2022 Reverse power rule.\n\u222b u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C.\n\u2022 I was given the problem:\n \u222b sin\u00b3(x)cos(x)dx = ? + C\n\nI entered (sin(x)^4)/4 the first time & was marked wrong. Then I tried entering the exact solution given which is impossible to do as far as I know on my mobile phone: (1/4)sin(x)^4. There is no process or command available to enter it as (1/4)sin^4(x). I'm assuming that is the reason I'm being marked wrong. Is it possible to enter the exponent before a trig function's parentheses? Please advise.\n(1 vote)\n\u2022 sin(x)^4/4 is correct however the exponent is in incorrect spot. The issue with sin(x)^4/4 is it could mistaken with sin(x^4)/4.\n\nYou need type sin^4(x)/4 or alternatively (sin(x))^4/4.\n\u2022 in problem 4 why is xdx= 3?\n(1 vote)\n\u2022 if du = 1/3xdx, you just multiply both sides by 3, and you get 3du = xdx\n\u2022 how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th)\n(1 vote)\n\u2022 \u222b 4x / sqrt(1 - x^4) dx =\n2 \u222b 2x / sqrt(1 - (x^2)^2) dx\n\nLet u = x^2, du = 2x dx, then\n2 \u222b 2x dx / sqrt(1 - (x^2)^2) =\n2 \u222b du / sqrt(1 - u^2) =\n2 arcsin(u) + C =\n2 arcsin(x^2) + C.\n\nHope that I helped.\n\u2022 In the first question, is it right to take cos(x^2) as u?\n\u2022 If you choose cos(x^2) as your u, your du ends up being -sin(x^2)*2x*dx. You could rearrange the equation as du/-sin(x^2) = 2x*dx and replace the 2x*dx in the original equation accordingly, but you're still left with the x^2 inside the sine-function. For the u-substitution to work, you need to replace all variables with u and du, so you're not getting far with choosing u = cos(x^2). If you choose, as you should, u = x^2 and your du = 2*x*dx, you'll get int(cos(u)*du) and that's pretty straight-forward to integrate.\n\u2022 Actually the problem 3 can already be solved by using the integration formula of e.\n(1 vote)\n\u2022 In order for most of these to work, the constant multiple rule must apply to integrals in exactly the same way that it applies to derivatives. Is that assumption correct?\n(1 vote)\n\u2022 Yes the constant multiple rule applies for both derivatives and integrals\n(1 vote)\n\u2022 =\u222b1/u^2 \u200bdu = \u22121/u \u200b+C\n\nAnyone could explain why appeared a negative signal on \u22121/u \u200b+C?\n(1 vote)\n\u2022 Think about it this way. Let's say we have the function y=1/x. Now let's think about the graph in the 1st quadrant. The slope is always negative. Therefore, the derivative of the curve at any point on it in the first quadrant should be a negative number. d/dx (1/x) < 0 for x > 0 . This should hopefully provide some intuition for the negative sign.\n\nFor the rigorous proof, try finding d/dx (-1/x).\n(1 vote)", "date": "2023-01-28 16:45:35", "meta": {"domain": "khanacademy.org", "url": "https://www.khanacademy.org/math/calculus-2/cs2-integration-techniques/cs2-integrating-with-u-substitution/a/worksheet-u-substitution", "openwebmath_score": 0.6465572714805603, "openwebmath_perplexity": 2898.1105105813817, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9857180643966651, "lm_q2_score": 0.9161096084360388, "lm_q1q2_score": 0.903025790002759}}
{"url": "https://math.stackexchange.com/questions/3375300/every-function-f-can-be-factored-into-f-i-circ-s-with-i-injective-and", "text": "# Every function $f$ can be factored into $f = i \\circ s$, with $i$ injective and $s$ surjective\n\nAs in the title, if $$X$$ and $$Y$$ are two arbitrary sets and $$f:X \\to Y$$, my proof was by taking $$x_1 \\sim x_2 \\iff f(x_1) = f(x_2),$$ $$s: X \\to X/\\sim$$ to be the canonical surjection of $$X$$ into the quotient set of $$X$$ wrt $$\\sim$$, i.e. $$s(x) = \\{z \\in X: f(z) = f(x)\\}$$ and $$i: X/\\sim \\to Y$$ to be the map defined by $$i(Z) = f(z)$$, for any $$z$$ in $$Z$$. Since all $$z$$ in an equivalence class are mapped to the same element of $$Y$$, $$i$$ is well defined.\n\nIs the above correct?\n\nThe proposed answer however was another one, namely $$s: X \\to f(X)$$ and $$i: f(X) \\to Y$$ defined as $$s(x) = f(x)$$ and $$i(w) = w$$.\n\nIf my solution was correct, which choice is more \"canonical\"?\n\nStill, if my solution is correct, to what extent can we say that the decomposition into injective and surjective is \"unique\"? I would say that $$X / \\sim$$ and $$f(X)$$ are \"isomorphic\" because of something like the first isomorphism theorem in linear algebra..\n\n\u2022 I have to say I like more your solution, the proposed answer is just restricting the range of the function, and then using the Inclusion map. Although your answer is basically the same: (I use $s_2$ for the proposed $s$) for each $x\u2208X$ we have $s_2^{-1}(x)=f^{-1}(x)=[x]_\\sim$. So the difference is that while you took the preimages, the proposed solution took the value. The 2 solutions describe the same thing\n\u2013\u00a0\u210bolo\nSep 30 '19 at 8:58\n\u2022 I was a bit puzzled because I did not find my solution anywhere in the web..\n\u2013\u00a0Tom\nSep 30 '19 at 9:02\n\u2022 But eg wiki en.wikipedia.org/wiki/Bijection,_injection_and_surjection mentions the \"proposed\" answer and not \"mine\"..\n\u2013\u00a0Tom\nSep 30 '19 at 9:03\n\u2022 I am not sure why Wiki chose that way, but, like I said, both ways describe the same thing: going from $x$ to something that describe uniquely $f(x)$, and from there to $f(x)$ itself(note that $f(x)$ is also a way to describe uniquely $f(x)$)(Also note that $|X/\\sim|=|f(X)|$ as well as that for every algebra with underline set $f(X)$, there is canonical isomorphic algebra with underline set of $X/\\sim$, the canonical bijection between the 2 is the isomorphism)(algebra over a set $A$, is the set $A$ is operators)\n\u2013\u00a0\u210bolo\nSep 30 '19 at 12:47\n\nYour answer is correct, assuming by $$i(Z) = f(z)$$, you mean that $$i$$ maps the equivalence class of $$z$$ under $$\\sim$$ to $$f(z)$$ (you may want to make this more clear).\n\nOne thing to note is that your answer and the answer provided in the solutions are very similar. We can naturally associate the equivalence classes of $$\\sim$$ uniquely to elements of the range of $$f$$. If we make this identification, then your answers are really the same.\n\nThis lends credence to the idea that this decomposition might be somewhat \"unique\" in some sense. Let's set up the problem. Suppose $$f : X \\to Y$$ satisfies $$f = i_1 \\circ s_1 = i_2 \\circ s_2$$ where $$s_k : X \\to Z_k$$ are surjective and $$i_k : Z_k \\to Y$$ are injective, for $$k = 1, 2$$. We can actually show that there is a bijection $$\\phi : Z_1 \\to Z_2$$ such that $$s_2 = \\phi \\circ s_1$$ and $$i_1 = i_2 \\circ \\phi$$.\n\nBy this, I mean that there is some $$\\phi$$ which provides us a rule for identifying elements of $$Z_1$$ and $$Z_2$$ in such a way that, after identification, the decompositions $$i_1 \\circ s_1$$ and $$i_2 \\circ s_2$$ become the same decomposition.\n\nSo, let's construct this $$\\phi$$. As $$i_k$$ is injective, there must exist some left inverses $$j_k : Y \\to Z_k$$ (i.e. $$j_k \\circ i_k : Z_k \\to Z_k$$ is the identity map on $$Z_k$$). Similarly, as $$s_k$$ is surjective, there must exist some right inverses $$t_k : Z_k \\to X$$ (i.e. $$s_k \\circ t_k : Z_k \\to Z_k$$ is the identity map). Define $$\\phi = j_2 \\circ i_1 : Z_1 \\to Z_2.$$ Then $$\\phi \\circ (s_1 \\circ t_2) = j_2 \\circ (i_1 \\circ s_1) \\circ t_2 = j_2 \\circ f \\circ t_2 = (j_2 \\circ i_2) \\circ (s_2 \\circ t_2),$$ which is the identity on $$Z_2$$. We need to show that $$(s_1 \\circ t_2) \\circ \\phi = s_1 \\circ t_2 \\circ j_2 \\circ i_1$$ is the identity on $$Z_1$$. This is a little less straight forward. \\begin{align*} (s_1 \\circ t_2) \\circ \\phi &= (j_1 \\circ i_1) \\circ (s_1 \\circ t_2) \\circ \\phi \\circ (s_1 \\circ t_1) \\\\ &= j_1 \\circ (i_1 \\circ s_1) \\circ t_2 \\circ j_2 \\circ (i_1 \\circ s_1) \\circ t_1 \\\\ &= j_1 \\circ f \\circ t_2 \\circ j_2 \\circ f \\circ t_1 \\\\ &= j_1 \\circ (i_2 \\circ s_2) \\circ t_2 \\circ j_2 \\circ (i_2 \\circ s_2) \\circ t_1 \\\\ &= j_1 \\circ i_2 \\circ s_2 \\circ t_1 \\\\ &= (j_1 \\circ i_1) \\circ (s_1 \\circ t_1), \\end{align*} which is the identity on $$Z_1$$, as required. Therefore, $$\\phi$$ is invertible with inverse $$s_1 \\circ t_2$$.\n\nObviously, by construction, we have $$i_2 \\circ \\phi = (i_2 \\circ j_2) \\circ i_1 = i_1.$$ Our proof also came with an expression for $$\\phi^{-1}$$ which we can also use: $$\\phi^{-1} \\circ s_2 = s_1 \\circ (t_2 \\circ s_2) = s_1 \\implies s_2 = \\phi \\circ s_1$$ as required.\n\n\u2022 Perfectly clear - thank you!\n\u2013\u00a0Tom\nSep 30 '19 at 12:12\n\nYour approach is fine. The relation $$\\sim$$ is an equivalence relation on $$X$$ and the quotient set $$X/\\sim$$ gives rise to the surjective mapping $$f_1:X\\rightarrow X/\\sim : x\\mapsto \\bar x$$ where $$\\bar x=\\{x'\\in X\\mid x\\sim x'\\}$$ is the equivalence class of $$x$$.\n\nMoreover, the mapping $$f_2:X/\\equiv \\rightarrow f(X): \\bar x\\mapsto f(x)$$ is injective. This mapping is well-defined, since if $$x\\sim x'$$, then $$f(x)=f(x')$$.\n\nSaying ''isomorphic'' is a bit too much, since there is underlying algebraic structure (such as vector spaces).\n\n\u2022 in the second line \"gives rise to the injective mapping\".. you meant surjective?\n\u2013\u00a0Tom\nSep 30 '19 at 8:56\n\u2022 Try \\sim which gives $\\sim$ Sep 30 '19 at 8:56", "date": "2022-01-21 03:31:41", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3375300/every-function-f-can-be-factored-into-f-i-circ-s-with-i-injective-and", "openwebmath_score": 0.9994407296180725, "openwebmath_perplexity": 184.20124569122595, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9924227594044641, "lm_q2_score": 0.9099070090919013, "lm_q1q2_score": 0.9030124247644474}}
{"url": "http://math.stackexchange.com/questions/130276/prove-gcdnn-n1-1", "text": "# Prove $\\gcd(nn!, n!+1)=1$\n\nFor any $n \\in \\mathbb{N}$, find $\\gcd(n!+1,(n+1)!+1)$. First come up with a conjecture, then prove it.\n\nBy testing some values, it seems like $\\gcd(n!+1,(n+1)!+1) = 1$\n\nI can simplify what's given to me to $\\gcd(nn!, n!+1)=1$ but I can't find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance?\n\n$\\gcd(n!+1,(n+1)!+1) = 1 \\implies \\gcd(n!+1,(n+1)n!+1) = 1 \\implies \\gcd(n!+1,nn!+n!+1) = 1 \\implies \\gcd(nn!, n!+1) = 1$\n\n-\nI changed numerous instances of \\mathrm{gcd} in this question to \\gcd. It's a standard operator name. \u2013\u00a0 Michael Hardy Apr 11 '12 at 2:42\nThanks for modifying my question to use \\gcd and making me realize that command exists. That will help me in the future! Also, thanks to everybody who answered; you all have really helped me! \u2013\u00a0 Brandon Amos Apr 11 '12 at 11:34\n.....and just in case anyone wonders: I just posted 5\\gcd(a,b) and 5\\mathrm{gcd}(a,b) within a \"displayed\" $\\TeX$ setting in the \"answer\" box below. Try it and you'll see that they don't both look the same! (One of them has proper spacing between \"$5$\" and \"$\\gcd$\".) \u2013\u00a0 Michael Hardy Apr 11 '12 at 21:00\n\nHere is a proof that does not use induction but rather the key property of gcd: $(a,b) = (a-b,b) = (a-kb,b)$ for all $k$.\n\nTake $a=nn!$, $b=n!+1$, $k=n$ and conclude that $(nn!, n!+1)=(nn!-n(n!+1),n!+1)=(-n,n!+1)=1$ since any divisor of $n$ is a divisor of $n!$.\n\n-\nOr apply the key property again: $(-n, n!+1) = (((n-1)!\\times-n) + n! +1 , n! +1) = (1, n!+1) = 1$ \u2013\u00a0 Aryabhata Apr 11 '12 at 1:43\n@Aryabhata This is probably a really simple question, but how did you get from $(-n, n!+1)$ to $(((n-1)!\\cdot-n)+n!+1, n!+1)$ from this \"key property\"? I see how you used it to add $n!+1$ to the first term, but where did the $(n-1)!$ multiplier come from? \u2013\u00a0 Brandon Amos Apr 11 '12 at 16:37\n@user28554: $(a,b) = (ka+b, a)$. The last term should be $-n$, instead of $n! + 1$. We chose $k = (n-1)!$. \u2013\u00a0 Aryabhata Apr 11 '12 at 16:41\n@Aryabhata Ahh, I see now. Thanks! \u2013\u00a0 Brandon Amos Apr 11 '12 at 16:48\n@user28554: You are welcome! \u2013\u00a0 Aryabhata Apr 11 '12 at 16:49\n\nYou don\u2019t need to use induction; you just need to prove the statement in the title. Suppose that $p$ is a prime factor of $nn!$; can $p$ divide $n!+1$?\n\n-\nPrimes aren't needed: if $\\rm\\:n\\:|\\:k\\:$ then $\\rm\\:nk,\\ k+1\\:$ are coprime in any ring - see my answer. \u2013\u00a0 Bill Dubuque Apr 11 '12 at 3:22\n@Bill: I didn\u2019t say that they were. I offered what I consider an easy approach to the problem. In general I write primarily for the questioner, not for possible future readers. \u2013\u00a0 Brian M. Scott Apr 11 '12 at 3:27\n\nHint $\\$ Put $\\rm\\:k = n!\\$ in: $\\rm\\ n\\:|\\:k\\:\\Rightarrow\\:(1+k,nk)= 1\\$ by $\\rm\\: (1-k)\\:(1+k) + (k/n)\\: nk = 1.\\ \\$ QED\n\nMore generally, note that the above Bezout equation implies $\\rm\\: 1+k\\:$ and $\\rm\\:nk\\:$ are coprime in every ring. Alternatively, with $\\rm\\:m = k+1,\\:$ one can employ Euclid's Lemma (EL) as follows:\n\n$$\\rm\\:(m,k) =1,\\ n\\:|\\:k\\:\\Rightarrow (m,n) = 1\\:\\Rightarrow\\:(m,nk)=1\\ \\ by\\ \\ EL$$ i.e. $\\rm\\: mod\\ m\\!:\\ x\\:$ is a unit (invertible) iff $\\rm\\:(x,m) = 1.$ But units are closed under products, divisors, i.e. they form a saturated monoid. So, since $\\rm\\:k\\:$ is a unit so is its divisor $\\rm\\:n\\:$ and so is the product $\\rm\\:nk.$\n\n-\n\nNote that $$(n-1)!\\cdot \\underline{n n!} - (n!-1)\\underline{(n!+1)} = 1;$$ this B\u00e9zout's identity shows that the two underlined quantities must be relatively prime (anything that divides them both must divide the right-hand side). The related identity $$(n-1)! \\underline{((n+1)!+1)} - (n!+(n-1)!-1)\\underline{(n!+1)} = 1$$ similarly proves that the greatest common divisor of these two underlined terms equals 1.\n\nOf course, discovering these identities in the first place is best done by using the Euclidean algorithm, as in lhf's answer.\n\n-\nThis is precisely what I wrote 5 hours prior, except it explicitly replaces $\\rm\\:k\\:$ by $\\rm\\: n!\\:$ in my Bezout equation. Doing so decreases the generality of the proof, and obscures the key (unit group) structure. \u2013\u00a0 Bill Dubuque Apr 11 '12 at 20:06\nWell then, I won't claim priority. But some might find that my solution is more accessible to the OP than yours. \u2013\u00a0 Greg Martin Apr 12 '12 at 6:51\nMy concern is not priority but, rather, pedagogy. I explicitly abstracted $n!$ to any integer $k$ divisible by $n$ in order to make clearer the innate governing multiplicative structure. To succeed in elementary number theory it is essential to learn how to recognize such structure. If students are encouraged to follow shortcuts circumventing such pedagogical routes then they may completely miss the key ideas, and never see the forest for the trees. \u2013\u00a0 Bill Dubuque Apr 12 '12 at 14:15\n\nI dont know i am right or wrong but i can do this example in following way,\\ Let $$\\gcd(n\\cdot n!,n!+1)=d$$ $$\\therefore d\\mid n\\cdot n!\\ ,\\ d\\mid n!+1$$ $$\\Rightarrow d\\mid n,\\ d\\mid n!,\\ d\\mid n!+1$$ $$\\Rightarrow d\\mid n!+1-n!$$ Thus $d\\mid 1\\ \\Rightarrow d=1.$\n\n-\nWhy does $d \\mid n$ ? Perhaps you mean a prime $d$. \u2013\u00a0 lhf Apr 11 '12 at 1:47\n$8 | 4 \\times 4!$, but $8$ does not divide $4$. \u2013\u00a0 Aryabhata Apr 11 '12 at 1:47\nJust for the record, if you want to write the \"dot\" for multiplication, you can use >\\circ \u2013\u00a0 M Turgeon Apr 11 '12 at 2:45\n\\cdot is better than \\circ for the multiplication dot. \u2013\u00a0 Greg Martin Apr 11 '12 at 6:33", "date": "2015-08-03 21:59:55", "meta": {"domain": "stackexchange.com", "url": "http://math.stackexchange.com/questions/130276/prove-gcdnn-n1-1", "openwebmath_score": 0.8968775272369385, "openwebmath_perplexity": 584.5945584575377, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9883127392504119, "lm_q2_score": 0.9136765169496871, "lm_q1q2_score": 0.9029981412553206}}
{"url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator", "text": "Is it possible to accomplish calculations of complex numbers specially in polar form with scientific calculators? This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in series. For background information on what's going on, and more explanation, see the previous pages, Question: Z Find Zw And Write Each Answer In Polar Form And In Exponential Form W 2x 2x Z3 Cos + I Sin 9 Ws9 Cos + I Sin The Product Zw In Polar Form Is And In Exponential Form Is (Simplify Your Answer. Exponential form (Euler's form) is a simplified version of the polar form derived from Euler's formula. Complex Number Calculator. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms. It is the distance from the origin to the point: See and . Write the complex number 3 - 4i in polar form. Based on this definition, complex numbers can be added and \u2026 A4. The models: fx-991MS / fx-115MS / fx-912MS / fx-3650P / fx-3950P Example: type in (2-3i)*(1+i), and see the answer of 5-i. It can be written in the form a + bi. This is the currently selected item. This calculator extracts the square root, calculate the modulus, finds inverse, finds conjugate and transform complex number to polar form.The calculator will generate a step by step explanation for each operation. This calculator allows one to convert complex number from one representation form to another with step by step solution. There are four common ways to write polar form: r\u2220\u03b8, re i\u03b8, r cis \u03b8, and r(cos \u03b8 + i sin \u03b8). A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = \u22121.For example, 2 + 3i is a complex number. But complex numbers, just like vectors, can also be expressed in polar coordinate form, r \u2220 \u03b8 . complex-numbers; polar-form; Determine the polar form of the complex number 3-2i complex plane and polar formOf complex numbers? Just type your formula into the top box. Not only can we convert complex numbers that are in exponential form easily into polar form such as: 2e j30 = 2\u222030, 10e j120 = 10\u2220120 or -6e j90 = -6\u222090, but Euler\u2019s identity also gives us a way of converting a complex number from its exponential form into its rectangular form. Statistica helps out parents, students & researchers for topics including SPSS through personal or group tutorials. And that\u2019s the best feature in my opinion. There's also a graph which shows you the meaning of what you've found. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/precalculus/imaginary_complex_precalc/exponential-form \u2026 57. Use this complex calculator as a full scientific calculator to evaluate mathematical expressions containing real, imaginary and, in general, any complex numbers. Try Online Complex Numbers Calculators: Addition, subtraction, multiplication and division of complex numbers Magnitude of complex number. Yes. Complex number is the combination of real and imaginary number. asked Feb 14, 2015 in PRECALCULUS by anonymous. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. It is able to handle both the modulus (distance from 0) and the argument (angle with the positive real axis) simultaneously. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to convert rectangular form of complex number to polar and exponential form. Please show all work. Looking for maths or statistics tutors in Perth? This way, a complex number is defined as a polynomial with real coefficients in the single indeterminate i, for which the relation i 2 + 1 = 0 is imposed. Instructions:: All Functions. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = \u22121 or j 2 = \u22121.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). It won\u2019t just solve a problem for you, but it\u2019ll also give details of every step that was taken to arrive at a particular answer. Note: This calculator displays (r, \u03b8) into the form: r \u2220 \u03b8 To convert complex number to its polar form, follow the general steps below: See . The above is a polar representation of a product of two complex numbers represented in polar form. [2 marks] I know already. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. To enter the complex number in polar form you enter mcisa, where m is the modulus and a is the argument of number. Use Integers Or Fractions For Any Numbers In The Expression) Question Viewer The Quotient In Polar Form Is \u2026 A complex number in Polar Form must be entered, in Alcula\u2019s scientific calculator, using the cis operator. Find all five values of the following expression, giving your answers in Cartesian form: (-2+5j)^(1/5) [6 marks] Any ideas? Polar to Exponential Form Conversion Calculator. This online calculator will help you to convert rectangular form of complex number to polar and exponential form. You can enter complex numbers in the standard (rectangular) or in the polar form. Complex numbers are written in exponential form .The multiplications, divisions and power of complex numbers in exponential form are explained through examples and reinforced through questions with detailed solutions.. Exponential Form of Complex Numbers A complex number in standard form $$z = a + ib$$ is written in polar form as $z = r (\\cos(\\theta)+ i \\sin(\\theta))$ \u2026 syms a a=8-7j [theta, r]cart2pol(8, 7) for the polar for but thats it. Polar to Rectangular Online Calculator. Free exponential equation calculator - solve exponential equations step-by-step This website uses cookies to ensure you get the best experience. 9B 10345 \u0438\u0449\u0443 \u041f\u0440\u043e\u0448\u0438\u0432\u043a\u0430 POLAR 48LTV3101 \u0448\u0430\u0441\u0441\u0438 T. $\\begingroup$ Yes, once you calculate the $\\tan^{-1}$, you should look at the polar\u2026 Given a complex number in rectangular form expressed as $$z=x+yi$$, we use the same conversion formulas as we do to write the number in trigonometric form: At the following model,the arithmetic operations on complex numbers can be easily managed using the Calculators. Example 3.6056cis0.588 . You can use the following trick to allow you to enter angles directly in degrees. Here, both m and n are real numbers, while i is the imaginary number. The succeeding examples illustrate the conversion of the standard complex number z = a + bi to its equivalent polar form (r, \u03b8). The polar form of a complex number expresses a number in terms of an angle $\\theta$ and its distance from the origin $r ... Use the rectangular to polar feature on the graphing calculator to change [latex]5+5i$ to polar form. \u2026 For complex numbers in rectangular form, the other mode settings don\u2019t much matter. A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in parallel. Polar forms of numbers can be converted into their exponential equivalents relatively easily. The polar form of a complex number expresses a number in terms of an angle $$\\theta$$ and its distance from the origin $$r$$. Visualizing complex number multiplication. In this section, we will first deal with the polar form of complex numbers. Dividing complex numbers: polar & exponential form. Key Concepts. Practice: Multiply & divide complex numbers in polar form. I was wondering if anybody knows a way of having matlab convert a complex number in either polar or cartesian form into exponential form and then actually display the answer in the form ' \u2026 Trigonometric Form of Complex Numbers Calculator. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms. With the calculator in DEGREE mode this will then display 240 e ^(i 75) corresponding to the polar form number (240 75). ; The absolute value of a complex number is the same as its magnitude. We can convert the complex number into trigonometric form by finding the modulus and argument of the complex number. Label the x-axis as the real axis and the y-axis as the imaginary axis. Complex modulus Rectangular form of complex number to polar and exponential form converter Show all online calculators We learnt that the exponential (polar) form of a complex number is a very powerful and compact way to solve complex number problems. Type An Exact Answer Using * As Needed. I was having a lot of problems tackling questions based on exponential form calculator but ever since I started using software, math has been really easy for me. Convert the complex number 8-7j into exponential and polar form. Instructions. For example, you can convert complex number from algebraic to trigonometric representation form or from exponential back to algebraic, ect. Topics covered are arithmetic, conjugate, modulus, polar and exponential form, powers and roots. Polar Display Mode \u201cPolar form\u201d means that the complex number is expressed as an absolute value or modulus r and an angle or argument \u03b8. By using this website, you agree to our Cookie Policy. The amplitude of the polar form is the direct amplitude of the ... www.learningaboutelectronics.com These calculators will display complex numbers in exponential form with the angle in degrees, but will not allow you to enter the angle in degrees. asked Dec 25, 2012 in PRECALCULUS by dkinz Apprentice. (This is spoken as \u201cr at angle \u03b8 \u201d.) The calculated values will also be displayed in standard and polar forms among other modular forms. Polar form, where r - absolute value of complex number: is a distance between point 0 and complex point on the complex plane, and \u03c6 is an angle between positive real axis and the complex vector (argument). Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to convert rectangular form of complex number to polar and exponential form. Below is an interactive calculator that allows you to easily convert complex numbers in polar form to rectangular form, and vice-versa. This is a quick primer on the topic of complex numbers. All Functions Operators + Another way of writing the polar form of the number is using it\u2019s exponential form: me^(ia) . ... And what might have jumped out at you is that exponential form \u2026 Rectangular ) or in the complex number 3-2i complex plane and polar formOf complex numbers in the complex into... Cookie Policy - 4i in polar form is the combination of real and imaginary number a=8-7j [ theta r! Values will also be expressed in polar form to another with step by step solution a calculator to the. Modulus, polar and exponential form, powers and roots easily managed using the Calculators best feature my! Capacitor and and inductor in series s the best feature in my opinion 4i polar! For the polar form derived from Euler 's form ) is a quick primer the. 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You can convert the complex number from algebraic to trigonometric representation form another. Parents, students & researchers for topics including SPSS through personal or group tutorials you to enter directly! To our Cookie Policy algebraic to trigonometric representation form or from exponential back to algebraic ect.... www.learningaboutelectronics.com polar to rectangular form of the polar form of complex numbers calculator does basic on... The calculated values will also be displayed in standard and polar formOf numbers... Evaluates expressions in the complex number 3 - 4i in polar form must be entered, in Alcula \u2019 exponential... Numbers, just like vectors, can also be displayed in standard polar... Form a + bi can use the following trick to allow you to convert rectangular form of the www.learningaboutelectronics.com... 'S also a graph which shows you the meaning of what you 've found coordinates are in... \u201d. the real axis and the y-axis as the imaginary number solve exponential equations step-by-step This,., the arithmetic operations on complex numbers calculator - Simplify complex expressions using algebraic rules step-by-step This website uses to..., where m is the direct amplitude of the number is the as..., students & researchers for topics including SPSS through personal or group tutorials and vice-versa forms of numbers be. You to easily convert complex number into trigonometric form by finding the and! Ensure you get the best feature in my opinion modular forms thats it in series write the plane!", "date": "2021-09-21 23:55:07", "meta": {"domain": "gridserver.com", "url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator", "openwebmath_score": 0.8409821391105652, "openwebmath_perplexity": 669.9348614131397, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9936116804144725, "lm_q2_score": 0.9086178944582997, "lm_q1q2_score": 0.9028133529673709}}
{"url": "https://math.stackexchange.com/questions/4208096/probability-chord-of-bigger-circle-intersects-smaller-circle", "text": "# Probability chord of bigger circle intersects smaller circle\n\nYou are given two concentric circles $$C_1$$ and $$C_2$$ of radius $$r$$ and $$r/2$$ respectively. What is the probability that a randomly chosen chord of $$C_1$$ will intersect $$C_2$$?\n\nAnswer: $$1/2, 1/3$$ or $$1/4$$\n\nThe first method I used (gives 1/4):\n\nThe midpoint of any chord uniquely determines it, as line joining center to midpoint is always perpendicular to chord. So instead of choosing a chord, let's choose points instead that shall be the midpoints of their respective chords. Any point inside inner circle will be a chord that intersects it too, and any point outside will never cut inner circle. Thus probability should be area of inner circle/area of outer circle= $$1/4$$\n\nBut then I did it by another method and got another answer (1/3):\n\nChoose a point on bigger circle. Now you can get all chords from $$0$$ to $$\u03c0$$ angle. The one which intersect smaller one must lie between tangents to smaller circle from bigger one from that point . We can easily obtain angle between tangents as $$\u03c0/3$$ by some trigonometry. We can do same for every other point so answer is $$\\frac{\u03c0/3}{\u03c0}=1/3$$\n\nFirst I was confused that I was getting two answers. But then I checked the given answer and I saw they were accepting multiple answers.\n\nSo I thought about how there could be multiple possible probabilities and only possible reason seemed to be the boundary conditions as I had included diameters in chords in second solution but not in first. However even though there are infinite diameters I still don't think probability should be affected this much as we have infinite points.\n\nCan someone give clarity on this? In particular, what exact conditions are included by which solution, and how will we get the third given answer (1/2)? As far as I can find there are $$3$$ boundary conditions we have to consider carefully-\n\n1. If diameters are included or not\n2. Degenerate \"chords\" that are $$r$$ distance from center, ie they are actually points on the circumference\n3. If tangents are included in intersection or not\n\nI believe the discrepancy results from the fact that the problem does not outline how the chords are randomly selected. As you have shown, there are multiple ways to randomly select a chord, and the two ways you have described have different probability distributions.\n\nThis can be seen by comparing the PDF of the length of the selected chord in both cases. In your first solution, the probability you select an arbitrary length is uniform. However, in your second solution, shorter chords are more likely to be selected than longer chords. Hence, there is a clear difference in the probability distributions in both selection processes.\n\nAs a result, this makes it possible to get different answers. Since the problem failed to elaborate on how chords were chosen, the problem writers were forced to accept multiple answers.\n\nMoreover, the 3 boundary conditions you have outlines will not affect the probability because they have infinitesimal impact on the total probability.\n\nYou can get $$\\frac{1}{2}$$ by modifying how you select the midpoint of the chord from your first solution. Instead of uniformly selecting a point from the interior of the larger circle, uniformly choose the distance of the midpoint from the center from the range $$[0,r]$$ and then uniformly choose the angle the midpoint makes with some arbitrary point/ray (e.g. define a positive x-axis) from the range $$[0,2\\pi)$$. The chord will intersect the smaller circle whenever the chosen distance of the midpoint from the center is $$<\\frac{r}{2}$$, which happens exactly $$\\boxed{\\frac{1}{2}}$$ of the time.\n\n\u2022 Thanks for your answer. I also thought the same thing about the boundary conditions but only reason I had my doubts was because we have infinite diameters and infinite tangential chords So they would contribute a lot more than just one chord Jul 27, 2021 at 6:46\n\u2022 I don't know how to rigorously explain it, but it's kind of like how if you remove a curve of infinite points from a region, the area of the region is still the same even though you removed an infinite number of \"things\" from the region. Jul 27, 2021 at 6:49\n\u2022 yeah I think a field of mathematics called \"measure theory\" is used so quite far beyond high school math Jul 27, 2021 at 6:54", "date": "2022-08-14 16:38:01", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4208096/probability-chord-of-bigger-circle-intersects-smaller-circle", "openwebmath_score": 0.8271906971931458, "openwebmath_perplexity": 230.10262855848092, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9852713887451681, "lm_q2_score": 0.916109606145647, "lm_q1q2_score": 0.9026165838899105}}
{"url": "https://math.stackexchange.com/questions/911050/what-is-the-expected-number-of-coin-tosses-needed-to-obtain-a-head", "text": "# What is the expected number of coin tosses needed to obtain a head?\n\nDue to my recent misunderstandings regarding the 'expected value' concept I decided to post this question. Although I have easily found the answer on the internet I haven't managed to fully understand it.\n\nI understood that the formula for the expected value is:\n\n$$E(x) = x_1p_1 +x_2*p_2 +...+x_np_n$$\n\nThe x's are the possible value that the random variable can take and the p's are the probabilites that this certain value is taken.\n\nSo, if I get a head on the first try, then $p_1 = \\frac{1}{2} , x_1 = 1$ If I get a head on the second try, then $p_2 = \\frac{1}{4} , x_2 = 2$\n\nAnd then, I woudl have that:\n\n$$E(x) = \\frac{1}{2}1+ \\frac{1}{4}2 +...$$\n\nSo my reasoning led me to an inifnite sum which I don't think I can't evaluate it that easy. In the 'standard' solution of this problem, the expected value is found in a reccurisve manner. So the case in which the head doesn't appear in the first toss is treated reccursively. I haven't understood that step.\n\nMy questions are: is my judgement correct? How about that reccursion step? Could somebody explain it to me?\n\n\u2022 For fun, I would say 2. =) \u2013\u00a0Vincent Aug 27 '14 at 15:46\n\u2022 Yes, I knew that too. :D I just didn't know how we found that answer \u2013\u00a0Bardo Aug 27 '14 at 15:48\n\nLet $X$ be the number of tosses, and let $e=E(X)$. It is clear that $e$ is finite.\n\nWe might get a head on the first toss. This happens with probability $\\frac{1}{2}$, and in that case $X=1$.\n\nOr else we might get a tail on the first toss. In that case, we have used up $1$ toss, and we are \"starting all over again.\" So in that case the expected number of additional tosses is $e$. More formally, the conditional expectation of $X$ given that the first toss is a tail is $1+e$.\n\nIt follows (Law of Total Expectation) that $$e=(1)\\cdot\\frac{1}{2}+(1+e)\\cdot\\frac{1}{2}.$$\n\nThis is a linear equation in $e$. Solve.\n\nRemark: The \"infinite series\" approach gives $$E(X)=1\\cdot\\frac{1}{2}+2\\cdot\\frac{1}{2^2}+3\\cdot\\frac{1}{2^3}+\\cdots.$$ This series, and related ones, has been summed repeatedly on MSE.\n\n\u2022 Conditioning like I did is a totally standard technique in the calculation of expectation. \u2013\u00a0Andr\u00e9 Nicolas Aug 27 '14 at 16:01\n\u2022 Well, perhaps you can use the series approach for general $p$, and the equation $e=p+(1+e)(1-p)$, and see that they give the same answer. \u2013\u00a0Andr\u00e9 Nicolas Aug 27 '14 at 16:09\n\u2022 Given that the first toss is a tail, $E(X)=1+e$. \u2013\u00a0Andr\u00e9 Nicolas Aug 27 '14 at 16:10\n\u2022 It is not true that with probability $1/2$ you will need $1+e$. What is true is that given the first is tail the total expected number of tosses is $1+e$. \u2013\u00a0Andr\u00e9 Nicolas Aug 27 '14 at 16:24\n\u2022 @Bardo Note that in your previous question in my solution we use essentially the same trick (in a slightly more complex situation). \u2013\u00a0Aahz Aug 27 '14 at 16:34\n\nYour approach is perfectly fine. The probability of getting the first head in the $n$th trial is $\\frac{1}{2^n}$, so we have $$\\mathbb{E}(x) = \\sum_{ n \\geq 1} \\frac{n}{2^n}.$$ This infinite sum can be calculated in the following way: first note that $\\frac{1}{1-x} = \\sum_{n \\geq 0} x^n$ for $|x|<1$. Differentiating both sides yields $$\\frac{1}{(1-x)^2} = \\sum_{n \\geq 0} n x^{n-1} = \\sum_{n \\geq 1} n x^{n-1} = \\frac1x \\sum_{n \\geq 1} n x^n.$$ Pluggin in $x = \\frac12$ yields $4 = 2 \\sum_{n \\geq 1} \\frac{n}{2^n} = 2 \\mathbb{E}(x)$, or $\\mathbb{E}(x) = 2$.\n\nThe recursive solution goes, I think, as follows: let $\\mathbb{E}(x)$ be the expected number of trials needed. Then the expected number of trials needed after the first trial, given that it was not heads, is also $\\mathbb{E}(x)$. In other words, the expected (total) number of trials is $\\mathbb{E}(x)+1$ in that case. This gives the equation $$\\mathbb{E}(x) = \\frac12 + \\frac12(\\mathbb{E}(x)+1)$$ which gives the same answer $\\mathbb{E}(x) = 2$.\n\n\u2022 It's good to know that my approach wasn't incorrect, it gives me a little confidence. I am no thinking about that reccursion step. \u2013\u00a0Bardo Aug 27 '14 at 15:54\n\nTo make ends meet... You have been explained by several users that, looking at the toss process itself, one sees that the expectation $E(X)$, that you know is $E(X)=S$, with $$S=\\sum_{n\\geqslant1}\\frac{n}{2^n},$$ solves the relation $$E(X)=1+\\frac12E(X).$$ It happens that one can also show directly that $$S=1+\\frac12S,$$ this relation following from a shift of indexes. To do so, note that $S=R+T$ with $$R=\\sum_{n\\geqslant1}\\frac{1}{2^n},\\qquad T=\\sum_{n\\geqslant1}\\frac{n-1}{2^n},$$ hence, using the change of variable $n=k+1$, $$T=\\sum_{k\\geqslant0}\\frac{k}{2^{k+1}}=\\frac12\\sum_{k\\geqslant0}\\frac{k}{2^k}=\\frac12\\sum_{k\\geqslant1}\\frac{k}{2^k}=\\frac12S,$$ hence the proof would be complete if one knew that $$R=1.$$ To show this, use the same trick once again, that is, note that $$R=\\frac12+\\sum_{n\\geqslant2}\\frac{1}{2^n}=\\frac12+\\sum_{k\\geqslant1}\\frac{1}{2^{k+1}}=\\frac12+\\frac12\\sum_{k\\geqslant1}\\frac{1}{2^{k}}==\\frac12+\\frac12R,$$ hence the proof that $S=2$ is complete.\n\n\u2022 Yes, I got it...I knew beforehand the trick of shifting indexes, so I totatly understood your argument..But I am still struggling to undesrtande the simpler argument..Thank you very much! \u2013\u00a0Bardo Aug 27 '14 at 18:29", "date": "2020-10-29 19:45:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/911050/what-is-the-expected-number-of-coin-tosses-needed-to-obtain-a-head", "openwebmath_score": 0.9081215858459473, "openwebmath_perplexity": 232.28270164907178, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717424942962, "lm_q2_score": 0.9149009509324104, "lm_q1q2_score": 0.9026154253710768}}
{"url": "https://mathhelpboards.com/threads/prove-that-a-real-root-exists-in-1-1.6389/", "text": "# Prove that A Real Root Exists in [-1, 1]\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nGiven $$\\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.\n\n#### Ackbach\n\n##### Indicium Physicus\nStaff member\nSome ideas:\n\nWe rewrite the polynomial as\n$$f(x)=5tx^4+sx^3+3rx^2+qx-r-t=0,$$\nwhere we are setting it equal to zero. We evaluate $f(1)$ and $f(-1)$:\n\\begin{align*}\nf(1)&=5t+s+3r+q-r-t=4t+2r+s+q\\\\\nf(-1)&=5t-s+3r-q-r-t=4t+2r-s-q.\n\\end{align*}\nIdea: if $f(1)\\cdot f(-1)<0$, then by the Intermediate Value Theorem, we would have shown there is a root in $[-1,1]$. Now\n$$f(1)\\cdot f(-1)=(4t+2r)^{2}-(s+q)^{2}.$$\nNot seeing where to go with this. There's nothing stopping $s=q=0$, with $4t+2r\\not=0$, in which case I have not proved what I want to prove.\n\n#### zzephod\n\n##### Well-known member\nGiven $$\\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.\nConsider the polynomial $$\\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.\n\nThen $$\\displaystyle p(1)=p(-1)=0$$ and so $$\\displaystyle p$$ has an extremum in $$\\displaystyle (-1,1)$$, so $$\\displaystyle p'(x)$$ has a root in $$\\displaystyle (-1,1)$$ ...\n\n.\n\nLast edited:\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nThanks to both, Ackbach and zzephod for participating...\n\nConsider the polynomial $$\\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.\n\nThen $$\\displaystyle p(1)=p(-1)=0$$ and so $$\\displaystyle p$$ has an extremum in $$\\displaystyle [-1,1]$$, so $$\\displaystyle p'(x)$$ has a root in $$\\displaystyle [-1,1]$$ ...\n\n.\nWoW!!! What an elegant way to approach this problem! Well done, zzephod!\n\nAnd there is another quite straightforward and beautiful way to tackle it as well...therefore I'll wait for the inputs from other members for now...\n\n#### anemone\n\n##### MHB POTW Director\nStaff member\nAnother method proposed by other to solve this challenge problem is by using the integration method:\n\nHint:\n\n$$\\displaystyle \\int_{-1}^1 p(x) dx=0$$", "date": "2020-09-19 12:14:08", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/prove-that-a-real-root-exists-in-1-1.6389/", "openwebmath_score": 0.8521580100059509, "openwebmath_perplexity": 762.9998405618776, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874123575265, "lm_q2_score": 0.911179702173019, "lm_q1q2_score": 0.9026031433682725}}
{"url": "https://mathhelpboards.com/threads/find-arc-bn.6149/", "text": "# find arc BN=?\n\n#### Albert\n\n##### Well-known member\nPoints A,B are on circle C ,segment MN is a diameter of circle C, and point P is on\n\nsegment MN , if :\n\n$\\angle CAP=\\angle CBP =10^o ,\\,\\, \\overset{\\frown} {MA}=40^o,\\,\\, find :\\,\\, \\overset{\\frown} {BN}=?$\n\nLast edited:\n\n#### HallsofIvy\n\n##### Well-known member\nMHB Math Helper\nYou refer to \"circle C\" but then treat \"C\" as if it were a point. Are we to assume that \"C\" is the center point of the circle?\n\n#### Albert\n\n##### Well-known member\nYou refer to \"circle C\" but then treat \"C\" as if it were a point. Are we to assume that \"C\" is the center point of the circle?\nyes ,you got it !\n\"C\" is the center point of the circle.\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nPoints A,B are on circle C ,segment MN is a diameter of circle C, and point P is on\n\nsegment MN , if :\n\n$\\angle CAP=\\angle CBP =10^o ,\\,\\, \\overset{\\frown} {MA}=40^o,\\,\\, find :\\,\\, \\overset{\\frown} {BN}=?$\n\nOne solution is for $B$ to be opposite $A$ on the other side of $MN$, at the point labelled $B'$ in the picture. Then $\\overset{\\frown} {BN} = 140^\\circ$. But that is too obvious to be interesting, and I assume that what was wanted is the case where $A$ and $B$ are on the same side of $MN$.\n\nThe points $A, B, C, P$ are concyclic, because $\\angle CAP=\\angle CBP =10^\\circ$. Therefore $\\angle ABP=\\angle ACP =40^\\circ$, and so $\\angle ABC= 10^\\circ + 40^\\circ = 50^\\circ.$ The triangle $ABC$ is isosceles, so $\\angle BAC = 50^\\circ$, and $\\angle ACB =80^\\circ$. Finally, $\\angle BCP =40^\\circ + 80^\\circ = 120^\\circ$, from which $\\overset{\\frown} {BN}= \\angle BCN = 60^\\circ.$\n\n#### Albert\n\n##### Well-known member\nwhat will be the value of arc BN , if point P locates between points C and N\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nwhat will be the value of arc BN , if point P locates between points C and N\nGood question! I hadn't thought of that possibility. The method will be similar to the previous one, but this time the angle ABC ($\\angle A'B'C$ in the diagram below) will be $40^\\circ - 10^\\circ = 30^\\circ$ instead of $40^\\circ + 10^\\circ = 50^\\circ$. Then $\\angle A'CB' = 120^\\circ$ and $\\overset{\\frown} {BN} = 20^\\circ.$\n\n#### Albert\n\n##### Well-known member\nvery good solution\nthis is an open -style problem ,if the position of point B or point P changes ,then the answer will also differ (it depends upon how the diagram is sketched)\nsometime we may give students a mathematic problem with more then one possible answer", "date": "2021-01-25 08:07:27", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/find-arc-bn.6149/", "openwebmath_score": 0.8444339632987976, "openwebmath_perplexity": 909.2171464164562, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.9877587225460771, "lm_q2_score": 0.9136765287024914, "lm_q1q2_score": 0.9024919608115071}}
{"url": "https://math.stackexchange.com/questions/2891852/factoring-a-quadratic-polynomial-absolute-beginner-level-are-both-answers-cor/2891855", "text": "# Factoring a quadratic polynomial (absolute beginner level), are both answers correct?\n\nI'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial:\n\n$$x^2 + 3x - 10$$\n\nAnd I'm given the task of finding the values of $a$ and $b$ in:\n\n$$(x + a) (x + b)$$\n\nObviously the answer is: $$(x + 5)(x - 2)$$\n\nHowever the answer can be also:\n\n$$(x - 2) (x + 5)$$\n\nI just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$.\n\nTherefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct:\n\nAnswer $1$\n$a = -2$\n$b = 5$\nor\nAnswer $2$\n$a = 5$\n$b = -2$\n\nI'm sure this is a completely obvious question, but I'm just a beginner in this.\n\n\u2022 Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$. \u2013\u00a0Matti P. Aug 23 '18 at 8:18\n\u2022 They are both valid answers, since the order of the factors doesn't matter. \u2013\u00a0Ludvig Lindstr\u00f6m Aug 23 '18 at 8:18\n\u2022 You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$. \u2013\u00a0Adam L Aug 23 '18 at 8:54\n\u2022 Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case. \u2013\u00a0Adam L Aug 23 '18 at 8:58\n\u2022 @Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help. \u2013\u00a0Zebrafish Aug 23 '18 at 9:03\n\n.Yes, you are correct. Since $$(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$$, we note that $$a$$ and $$b$$ may either take the values $$(5,-2)$$ or $$(-2,5)$$.\n\nI would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $$a$$ and $$b$$, but nowhere mentions that they are unique. However, any question saying \"find the values of $$a$$ and $$b$$\" is wrong with the word \"the\" : they are assuming uniqueness of $$a$$ and $$b$$, which is not the case.The question as quoted by you includes the word \"the\" , and this is misleading.\n\n\u2022 Yes, actually the statement \"find 'the' values of a and b\" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you. \u2013\u00a0Zebrafish Aug 23 '18 at 8:26\n\u2022 You are welcome! \u2013\u00a0Teresa Lisbon Aug 23 '18 at 9:56\n\u2022 I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently? \u2013\u00a0Teresa Lisbon Aug 23 '18 at 14:18\n\nFor commutative property of product we have that\n\n$$(x + 5)(x - 2)=(x - 2)(x + 5)$$\n\nnote that also\n\n$$(-x + 2)(-x - 5)$$\n\nis a correct factorization.\n\nYou are right.\n\n$$(x+a)(x+b)=x^2+(a+b)x+ab$$\n\nand by identification with $x^2+3x-10$,\n\n$$\\begin{cases}a+b=3,\\\\ab=-10.\\end{cases}$$\n\nThis is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$.\n\nNow one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write\n\n$$3a=(a+b)a=a^2+ab=a^2-10$$\n\nwhich is the original equation (with a sign reversal)\n\n$$a^2-3a-10=0.$$\n\nTo be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots.\n\nSo there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.", "date": "2020-09-19 00:13:11", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2891852/factoring-a-quadratic-polynomial-absolute-beginner-level-are-both-answers-cor/2891855", "openwebmath_score": 0.7900750637054443, "openwebmath_perplexity": 228.23891306315016, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9719924777713886, "lm_q2_score": 0.9284088050123334, "lm_q1q2_score": 0.9024063747687119}}
{"url": "http://mathhelpforum.com/calculus/60890-need-help-3-calculus-extra-credit-problems.html", "text": "# Math Help - Need help with 3 Calculus Extra Credit Problems\n\n1. ## Need help with 3 Calculus Extra Credit Problems\n\nI'm in intro to calculus and I need help setting this equation up:\n\nNewton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium. A cold drink is removed from a refrigerator on a hot summer day and placed in a room where the temperature is 80\u00b0F. Express the temperature of the drink as a function of time (minutes) if the temperature of the drink was 40\u00b0F when it left the refrigerator and was 50\u00b0F after 20 minutes in the room.\n\nThanks!\n\n2. Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium.\n\n$\\frac{dT}{dt} = k(T-A)$\n\nA = room temperature (a constant)\nT = temperature of the cold drink at any time t in minutes\nk = proportionality constant\n\nseparate variables and integrate.\n\n3. Ok, with your help and help from another problem, I have this. Is it the next step?\n\ndT/dt= -k(T-TM)\n\nWhere T(t) is the temperature of the drink, and TM is the temperature of the surrounding solution.\n\nSo:\n\ndT/dt= -(T-80)\n\nThis doesn't seem right, I don't know how to incorporate the other numbers and variables. Grrr calculus\n\n4. $\\frac{dT}{dt} = k(T-80)$\n\nseparate variables ...\n\n$\\frac{dT}{T-80} = k \\, dt$\n\nintegrate ...\n\n$\\ln|T-80| = kt + C_1$\n\n$|T-80| = e^{kt + C_1}$\n\n$|T-80| = e^{C_1} \\cdot e^{kt}$\n\n$T - 80 = C_2 \\cdot e^{kt}$\n\n$T = 80 + C_2 \\cdot e^{kt}$\n\nThe calculus is done, so I'm stopping at this point. The rest is algebra ... you were given two temperatures at two different times. With that info, you can determine the constants $C_2$ and $k$ and finalize the temperature as a function of time.\n\n5. I greatly appreciate your help. I have:\n\nInitial value (40 degrees at time 0)\nT= 80 + Ce^kt\n40= 80 + Ce^k*0\n40= 80 + C\nC= -40\nHow do I find k?\n\nSecond value (50 degrees at time 20)\nT= 80 + Ce^kt\n50= 80 + Ce^k*20\n-30= Ce^20k\nAgain finding k has stumped me.\n\nWhat does k represent and how do i find it?\n\n6. Originally Posted by Sm10389\nI greatly appreciate your help. I have:\n\nInitial value (40 degrees at time 0)\nT= 80 + Ce^kt\n40= 80 + Ce^k*0\n40= 80 + C\nC= -40\ngood.\n\nHow do I find k?\nSecond value (50 degrees at time 20)\nhow about substituting in -40 for C ? then find k with the second value.\nT= 80 + Ce^kt\ndo it.\n\n7. you my friend are a genius.\n\nso....\n\nInitial value (40 degrees at time 0)\nT= 80 + Ce^kt\n40= 80 + Ce^k*0\n40= 80 + C\nC= -40\n\nSecond value (50 degrees at time 20)\nT= 80 + Ce^kt\n50= 80 + Ce^k*20\n-30= -40e^20k\n3/4= e^20k\nln3/4=20k\nk= ln(3/4)/80\n\nso would this be my final answer?\n\nT= 80 + -40e^ln((3/4)/20)t\n\n8. ## skeeter\n\nhow does that look\n\n9. Originally Posted by Sm10389\nhow does that look\ncheck it out yourself ... graph the result in your graphing utility and see if the given info matches up.\n\n10. i just did, and it is not correct. where did i goof up?\n\n11. so would this be my final answer?\n\nT= 80 + -40e^ln((3/4)/20)t\nlooks like you have ...\n\n$k = \\ln\\left(\\frac{\\frac{3}{4}}{20}\\right)$\n\n... which it ain't.\n\nshould be ...\n\n$k = \\frac{\\ln\\left(\\frac{3}{4}\\right)}{20}$\n\n12. Thank you, I had that too, I just did not use the parentheses correctly in my calculator.\n\nThe next step of the problem is to calculate it if the drink were warmer then room temperature (>80).\n\nI know how to set it up like the last one, but we would only be given one point (0, 85) for example.\n\nHow would I calculate k here?\n\n13. $k$ will remain the same because the rate of heat transfer will remain the same\n\n... however, you'll have to recalculate $C_2$.\n\n14. ## QUESTION #2\n\nMore from the Introduction to Differential Equations-\n\nInvestment plan- an investor makes regular deposits totaling D dollars each year into an account that earns interest at the annual rate r compounded continuously.\n\nA: Explain why the account grows at the rate ( dV/dt = rV + D ) where V(t) is the value of the account 2 years after the initial deposit. Solve this differential equation to express V(t) in terms of r and D.\n\nI came up with this:\nV(t)= (C/r)*e^rt - (D/r)\n\nI am sure it is correct. This is the next part:\nAmanda wants to retire in 20 years. To build up a retirement fund, she makes regular annual deposits of \\$8,000. If the prevailing interest rate stays constant at 4% compounded continuously, how much will she have in her account at the end of the 20 year period?\n\nI know how to do everything but:\nFind C\nFigure out how compounding continuously would affect the equation.", "date": "2014-07-24 04:03:59", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/calculus/60890-need-help-3-calculus-extra-credit-problems.html", "openwebmath_score": 0.5939760208129883, "openwebmath_perplexity": 1095.9537145538923, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES\n\n", "lm_q1_score": 0.978712649448364, "lm_q2_score": 0.9219218391455084, "lm_q1q2_score": 0.902296565774409}}
{"url": "https://math.stackexchange.com/questions/2746968/the-proof-of-infinitude-of-pythagorean-triples-x-x1-z", "text": "# The Proof of Infinitude of Pythagorean Triples $(x,x+1,z)$\n\nProof that there exists infinity positive integers triple $x^2+y^2=z^2$ that $x,y$ are consecutive integers, then exhibit five of them.\n\nThis is a question in my number theory textbook, the given hint is that\n\"If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$\"\nI wondered how someone come up with this idea.\n\nMy solution is letting $x=2st, y=s^2-t^2, z=s^2+t^2$ by $s>t, \\gcd(s,t)=1$.then consider two cases: $y=x+1$ and $y=x-1$\nCase 1: $y=x+1$\nGives me $(s-t)^2-2t^2=1$ then I found this is the form of Pell's equation, I then found \\begin{align}s&=5,29,169,985,5741\\\\t&=2,12,20,408,2378\\end{align}then yields five triples $$(20,21,29),(696,697,985),(23660,23661,33461),(803760,803761,1136689),(27304196,27304197,38613965)$$ Case 2:$y=x-1$\nUsing the same method, I come up with Pell's equation $(s+t)^2-2s^2=1$, after solve that I also get five triples: $$(4,3,5),(120,119,169),(4060,4059,5741),(137904,137903,195025),(4684660,4684659,6625109)$$\n\nI have wondered why the gaps between my solution are quite big, with my curiosity, I start using question's hint and exhibit ten of the triples:$$(3,4,5),(20,21,29),(119,120,169),(696,697,985),(4059,4060,5741),(23660,23661,33461),(137903,137904,195025),(803760,803761,1136689),(4684659,4684660,6625109),(27304196,27304197,38613965)$$ These are actually the same as using solutions alternatively from both cases. But I don't know is this true after these ten triples\n\nBasically the problem was solved, but I would glad to see if someone provide me a procedure to come up with the statement\n\"If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$\", and prove that there are no missing triplet between it.\n\n--After edit--\n\nThanks to @Dr Peter McGowan !, by the matrix\n$$\\begin{bmatrix} 1 & 2 & 2 \\\\ 2 & 1 & 2\\\\ 2 & 2 & 3 \\end{bmatrix} \\begin{bmatrix} x\\\\x+1\\\\z \\end{bmatrix} = \\begin{bmatrix} 3x+2z+2\\\\3x+2z+1\\\\4x+3z+2 \\end{bmatrix}$$ gives me the hinted statement.\n\n\u2022 Hint : If $(a/b)$ is a solution of the Pell-equation $a^2-2b^2=-1$ , then the next solution is $(3a+4b/2a+3b)$ \u2013\u00a0Peter Apr 21 '18 at 7:43\n\u2022 Wow, how to know that? \u2013\u00a0kelvin hong \u65b9 Apr 21 '18 at 7:54\n\u2022 artofproblemsolving.com/community/c3046h1049346__2 \u2013\u00a0individ Apr 21 '18 at 8:50\n\u2022 @individ thanks, but a relevant proof is better. \u2013\u00a0kelvin hong \u65b9 Apr 21 '18 at 10:33\n\u2022 Go here. It will take you to a question of mine where I prove the infinitude of Pythagorean Triples... but not using Pell equations, however. Nonetheless, this post might serve more use if $z=x+1$ as opposed to $y$, since I show that $$(2v^2+2v)^2+(2v+1)^2=(2v^2+2v+1)^2\\;\\forall v.$$ Still, it might increase your understanding on Pythagorean Triples :) \u2013\u00a0Mr Pie Jun 20 '18 at 12:20\n\n## 3 Answers\n\nYou are on the right track. The simplest solution is to recall that all irreducible Pythagorean triples for a rooted ternary tree beginning with $(3, 4, 5)$ triangle. B Berggren discovered that all others can be derived from this most primitive triple. F J M Barning set these out as three matrices that when pre-multiplied by a \"vector\" of a Pythagorean triple produces another. For the case of consecutive legs we have, starting with $(x_1, y_1, z_1)$, we may calculate the next triple as follows:\n\n\\begin {align} x_2&=x_1+2y_1+2z_1 \\\\ y_2&=2x_1+y_1+2z_1 \\\\ z_2&=2x_1+2y_1+3z_1 \\end {align}\n\nThe hint you were given is a variation on the above more general formula specific for consecutive leg lengths. It is an easy proof by induction to show that the formulas are correct. The first few are:\n\n$(3, 4, 5); (20, 21, 29); (119, 120, 169); (696, 697, 985); (4059, 4060, 5741); (23660, 23661, 33461)$; etc. Obviously, this can be continued indefinitely.\n\nThe sequence rises geometrically. A simple explicit formula is available for these solutions that are (as you have already guessed) alternating solutions to Pell's equation.\n\n\u2022 Wow, although I have seen that matrix before but don't realize it can be so useful! I have found the related matrix and actually come out with the desired result.Thanks a lot!!!! \u2013\u00a0kelvin hong \u65b9 Apr 21 '18 at 10:43\n\n$x,x+1,z$ is a Pythagorean triple iff $(2x+1)^2+1=2z^2$.\n\nLet $u=2x+1$. Then $u^2-2z^2=-1$, a negative Pell equation whose solution lies in considering the units of $\\mathbb Z[\\sqrt 2]$ of norm $-1$.\n\nIt is clear that $\\omega=1+\\sqrt 2$ is a fundamental unit with norm $-1$. Therefore, all the other solutions of $u^2-2z^2=-1$ come from odd powers of $\\omega$.\n\nThus, if $(u_k,z_k)$ is a solution of $u^2-2z^2=-1$, then the next one is given by\n\n\\begin{align} u_{k+1}+z_{k+1}\\sqrt 2&=(u_k+z_k\\sqrt 2)\\omega^2 \\\\ &=(u_k+z_k\\sqrt 2)(3+2\\sqrt 2) \\\\ &=(3u_k+4z_k)+(2u_k+3z_k)\\sqrt 2 \\end{align} So, $u_{k+1}= 3u_k+4z_k$ and $z_{k+1}=2u_k+3z_k$.\n\nNow let $u_{k+1}=2x_{k+1}+1$.\n\nThen $$x_{k+1}=\\frac{u_{k+1}-1}{2}=\\frac{(3u_k+4z_k)-1}{2}=\\frac{3(2x_k+1)+4z_k-1}{2}=3x_k+2z_k+1$$ and $$z_{k+1}=4x_k+3z_k+2$$ as claimed.\n\n\u2022 Wow, thanks for your solution! \u2013\u00a0kelvin hong \u65b9 Apr 28 '18 at 2:54\n\nPythagorean triples where $$|A-B|=1$$ are scarce and get more rare with altitude but they can continue to be found, indefinitely, given arbitrary precision. (There are only $$22$$ of them with $$A,B,C<10.2\\text{ quadrillion.})$$ Proof that there are in infinite number of them is shown by the following functions which accept and yield natural numbers without end.\n\nAnother way of putting it is that we are eliminating (subtracting) a countably infinite set of triples from another countably infinite set of triples and even when you subtract a supposedly larger $$\\aleph_0$$ set from a smaller $$\\aleph_0$$ set (odd numbers minus natural numbers) the results are still infinite because they can be mapped. Here we have all triples minus extraneous triples.\n\nTo find them in a timely manner, we can use the following function to find the right combination of $$(m,n)$$ to use in Euclid's formula. The $$\\pm$$ is used because the triples alternate: $$A>B$$ and $$A. $$\\text{With infinite integers input }(n=\\sqrt{2*m^2\\pm1}-m)\\text{ will always output an integer for some }m.$$ For all $$M\\in\\mathbb{N}$$, whenever this function yields an integer $$>0$$, we have the $$(m,n)$$ we need for a triple. The following $$19$$ were generated (in $$4$$ seconds) in a loop of $$m=1\\text{ to }16,000,000$$ and $$f(m,n)$$ shows the variable values needed to generate each triple using Euclid's formula. The $$g(n,k)$$ shows the values (generated in $$1$$ second) for an alternate set of functions in a loop of $$m=1\\text{ to }5,000,000$$. $$A=m^2-n^2\\qquad B=2mn\\qquad C=M^2+n^2$$ $$f(2,1)=g(1,1)=(3,4,5)$$ $$f(5,2)=g(2,2)=(21,20,29)$$ $$f(12,5)=g(4,5)=(119,120,169)$$ $$f(29,12)=g(9,12)=(697,696,985)$$ $$f(70,29)=g(21,29)=(4059,4060,5741)$$ $$f(169,70)=g(50,70)=(23661,23660,33461)$$ $$f(408,169)=g(120,169)=(137903,137904,195025)$$ $$f(985,408)=g(289,408)=(803761,803760,1136689)$$ $$f(2378,985)=g(697,985)=(4684659,4684660,6625109)$$ $$f(5741,2378)=g(1682,2378)=(27304197,27304196,38613965)$$ $$f(13860,5741)=g(4060,5741)=(159140519,159140520,225058681)$$ $$f(33461,13860)=g(9801,13860)=(927538921,927538920,1311738121)$$ $$f(80782,33461)=g(23661,33461)=(5406093003,5406093004,7645370045)$$ $$f(195025,80782)=g(57122,80782)=(31509019101,31509019100,44560482149)$$ $$f(470832,195025)=g(137904,195025)=(183648021599,183648021600,259717522849)$$ $$f(1136689,470832)=g(332929,470832)=(1070379110497,1070379110496,1513744654945)$$ $$f(2744210,1136689)=g(803761,1136689)=(6238626641379,6238626641380,8822750406821)$$ $$f(6625109,2744210)=g(1940450,2744210)=(36361380737781,36361380737780,51422757785981)$$ $$f(15994428,6625109)=g(4684660,6625109)=(211929657785303,211929657785304,299713796309065)$$\n\nThe alternate formula mentioned above is slightly faster since it deals only with the subset of triples where $$GCD(A,B,C)=(2m-1)^2,m\\in\\mathbb{N}$$. We have the needed $$(n,k)$$ values whenever the following yields a positive integer.\n\n$$k=\\sqrt{\\frac{(2n-1)^2\\pm1}{2}}$$ Having found a needed $$(n,k)$$ the following will generate a needed triple. $$A=(2n-1)^2+2(2n-1)k\\qquad B=2(2n-1)k+2k^2\\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$", "date": "2019-10-20 19:46:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2746968/the-proof-of-infinitude-of-pythagorean-triples-x-x1-z", "openwebmath_score": 0.9638389348983765, "openwebmath_perplexity": 616.9689531284344, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9861513873424043, "lm_q2_score": 0.914900959053549, "lm_q1q2_score": 0.9022308500515535}}
{"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html", "text": "# Solving Cubic Equations Formula\n\nSolving Cubic Equations with the help of Factor Theorem If x \u2013 a is indeed a factor of p(x), then the remainder after division by x \u2013 a will be zero. It solves cubic, quadratic and linear equations. 2 But it is important to remember van der Waals\u2019 equation for the volume is a cubic and cubics always. Only th is a variable. Yes, it is not easy for one to get a complete hold of the formulas and tricks used in mathematics for different types of purposes. That formula can be used in a vectorized form. Title: Hyperbolic identity for solving the cubic equation: Authors: Rochon, Paul: Publication: American Journal of Physics, Volume 54, Issue 2, pp. After reading this chapter, you should be able to: 1. (Sometimes it is possible to find all solutions by finding three values of x for which P(x) = 0 ). 00000001, initial_guess=0. Polynomials and Partial Fractions In this lesson, you will learn that the factor theorem is a special case of the remainder theorem and use it to find factors of polynomials. Individuals round out. It isnt like the other typical zeros root problems ive seen where they give you the x intercepts. Type y= 2x+5 into your calculator and look at the graph. solving a cubic equation. Learn more about cubic equation Symbolic Math Toolbox. Calculator Use. , the roots of a cubic polynomial. Equivalently, the cubic formula tells us the solutions of equations of the form ax3 +bx2 +cx+d =0. The calculation of the roots of a cubic equation in the set of real and complex numbers. ) There are two cases to consider. The corresponding formulae for solving cubic and quartic equations are signi\ufb02cantly more complicated, (and for polynomials of degree 5 or more, there is no general formula at all)!! In the next section, we shall consider the formulae for solving cubic equations. bw-cw-bwcw \u2022Each time you press w, the input value is registered in the highlighted cell. Get the free \"Solve cubic equation ax^3 + bx^2 + cx + d = 0\" widget for your website, blog, Wordpress, Blogger, or iGoogle. But the sympy equation solver can solve more complex equations, some having little practical significance, like this one: (1) $\\displaystyle a x^3 + b x^2 + c x + d = 0$ The above equation, called a cubic, has three solutions or \"roots\", and the solutions are rather complex. As calculation is an exact. In all of these solutions an auxiliary equation (the resolvent) was used. how to solve cubic equation in faster way http://youtu. All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got. Solve cubic equation in MATLAB. 1 The general solution to the quadratic equation There are four steps to nding the zeroes of a quadratic polynomial. Every pair of values (x, y) that solves that equation, that is, that makes it a true statement, will be the co\u00f6rdinates of a point on the circumference. We see that x=1 satisfies the equation hence x=1 or x-1 is one factor of the given cubic equation. The colors in the drawing are meant to suggest one way in which we could divide the cubic into two parts, each of which determines y as a function of x in a different way. ) Although cubic functions depend on four parameters, their graph can have only very few shapes. Volume of a Round Tank or Clarifier. Equation Solver Solves linear, quadratic, cubic and quartic equations in one variable, including linear equations with fractions and parentheses. 3 Ways To Solve A Cubic Equation Wikihow. Quadratic Equation Worksheets. That is, we can write any quadratic in the vertex form a(x h)2 + k. Cubic Equation Calculator. Cubic equationis equation of the form: a\u2219x 3 + b\u2219x 2 + c\u2219x + d = 0. Here you can find calculators which help you solve linear, quadratic and cubic equations, equations of the fourth degree and systems of linear equations. roots but exclusive to cubic polynomials. ax 3 + bx 2 + cx + d = 0. All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got. an equation in which the highest power of the unknown quantity is a cube. (Don't worry about how this program works just now. a3 * x^3 + a2 * x^2 + a1 * x + a0 = 0 will be solved by command below. These formulas are a lot of work, so most people prefer to keep factoring. Cubic functions have the form f (x) = a x 3 + b x 2 + c x + d Where a, b, c and d are real numbers and a is not equal to 0. The roots of this equation can be solved using the below cubic equation formula. The exact real solutions of a cubic polynomial? I tried this solution but I don't get any output. AI Mahani of Bagdad was the first to state the problem of Archimedes demanding the section of a sphere by a plane so that the two segments shall be a prescribed ratio in the form of a cubic equation. A cubic function is one of the most challenging types of polynomial equation you may have to solve by hand. Cardano\u2019s derivation of the cubic formula To solve the cubic polynomial equation x 3 + a \u2062 x 2 + b \u2062 x + c = 0 for x , the first step is to apply the Tchirnhaus transformation x = y - a 3. Solution: Let us use the division method for solving a cubic equation. Cubic equation synonyms, Cubic equation pronunciation, Cubic equation translation, English dictionary definition of Cubic equation. I shall try to give some examples. 5 = 0 without a calculator how many ways to solve a cubic equation? Finding the simplest equation for the sequence: 1, 5, 21, 85 (Challenge for pros) Everyone doing maths read this!!!. Find more Mathematics widgets in Wolfram|Alpha. In your specific equation, the roots also switch, so what \"was\" the first solution is now a different one. In this tutorial you are shown how to solve a cubic equation by using the factor theorem. The coefficients of the cubic equation as well as the initial guess are to be passed from a web page. When I try to solve this equation using mathematica's Solve[] function, I get one real root and 2 complex roots. Later,I will show you how this method is used for some cubic equations. solving a cubic equation. ir, [email\u00a0protected] In previous versions of GeoGebra CAS, it was possible to compute the solutions of a general cubic setting the CAS timeout to 60s: Solve[a x^3 + b x^2 + c x + d = 0, x] In the present version, GeoGebra answers promptly: {} (GeoGebra doesn't take 60 seconds to give this answer). We see that x=1 satisfies the equation hence x=1 or x-1 is one factor of the given cubic equation. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. First, we simplify the equation by dividing all terms by 'a', so the equation then becomes:. Individuals round out. Solving Polynomial Equations in Excel. Solving cubic equations. According to [1], this method was already published by John Landen in 1775. solving a cubic equation. be/OuiFS1Wma2U Fast and Easy Cubic Eqn Trick. com is a free online OCR (Optical Character Recognition) service, can analyze the text in any image file that you upload, and then convert the text from the image into text that you can easily edit on. Press 2 (3) to enter the Cubic Equation Mode. The solution of the equation we write in the following form: The formula above is called the Cardano\u2019s formula. \" Let y x- and substitute in the equation below to \"red uce\" the cubic. Now, the quadratic formula, it applies to any quadratic equation of the form-- we could put the 0 on the left hand side. But, equations can provide powerful tools for describing the natural world. Now divide the equation with x-1 which will give you the quadratic equation. I find the form with coefficients on the coefficients the easiest to remember: $x^3 + 3bx^2 + 6cx + 2d = 0$ has roots given by: [math]x = -b + \\sqrt[3. A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c Read More High School Math Solutions - Quadratic Equations Calculator, Part 2. Cubic equations can be solved analytically in Matlab (you need the symbolic toolbox to get the expressions or just copy & paste them from somewhere). Bibtex entry for this abstract Preferred format for this abstract (see Preferences ). The \"Cubic Formula\". com / [email\u00a0protected] Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. By convention, the volume of a container is typically its capacity, and how much fluid it is able to hold, rather than the amount of space that the actual container displaces. I need to pass this course with good marks. Volume of a Round Tank or Clarifier. Solving Polynomial Equations in Excel. \"Now, let's solve the cubic equation x^3+ax^2+bx+c=0 with origami. 0 0 1,216 asked by anonymous Apr 1, 2013 1st ionization: Li ==> Li^+ + e 2nd ionization: Li^+ ==> Li^2+ + e 0 1 posted by DrBob222 Apr 2, 2013 A don\u2019t no 0 0 \u2026. ax 3 + bx 2 + cx + d = 0. Knowledge of the quadratic formula is older than the Pythagorean Theorem. Cardano\u2019s formula. Cubic equations and other math calculators from Kusashi web design. net dictionary. Cardano's Method. Formula (5) now gives a solution w= w 1 to (3). Quadratic equations; Biquadratic equations. \u2022 Cubic in volume (3 roots) 9At T>T c one root 9At the critical point all three roots equals V c 9Two-phase region (three roots) 1/28/2008 van der Waals EOS 8 Drawbacks of the van der Waals Equation of State Cubic in Volume (has three roots). Cardano's method provides a technique for solving the general cubic equation. Question slides to be printed 2 slides per page, students decide which sheet to do (first page is easier). What is the reduced cubic equation that you must solue order to solve the cubic equation In x3-3x2-4x+12 = 0? We have seen that every cubic polynomial has at least one root and that by a change of variable we can reduce the problem of finding a root of a giv cubic polynomial r3+br2+cr+d to the problem of finding a root of a cubic polynomial of the form y +pytg. Fold a line placing P1 onto L1 and placing P2 onto L2, and the slope of the crease is the solution of x^3+ax^2+bx+c=0. This Online Equation Solver solves every equations with set of given variables. Equation Solver Solves linear, quadratic, cubic and quartic equations in one variable, including linear equations with fractions and parentheses. ir, [email\u00a0protected] Learn more about cubic equation Symbolic Math Toolbox. 6 minutes ago Solve each equation. The solver will then show you the steps to help you learn how to solve it on your own. After reading this chapter, you should be able to: 1. Posted in Based on a Context Tagged A Level > Factor and remainder theorem, Algebra > Equations > Finding roots, Algebra > Equations > Iteration, Algebra > Functions > Composite functions Post navigation. bw-cw-bwcw \u2022Each time you press w, the input value is registered in the highlighted cell. Some elements taken from other uploaders so credit to them. There is a \"cubic formula\", but it is quite messy and takes a large amount of work. It is not as sophisticated as the SCS TR-55 method, but is the most common method used for sizing sewer systems. The volume of a figure is the number of cubes required to fill it completely, like blocks in a box. @Raffaele 's comment points to its wikipedia page. Such solutions have been credited to the Greek mathematician Men\u00e6chmus (c. Was that a fully justified argument? Yes, because once you are looking for roots of the form there is no mystery behind the idea of looking at what you know about the two roots, converting that into some equations for and and trying to solve those equations. Third Degree Polynomial Formula. INTRODUCTION Likely you are familiar with how to solve a quadratic equation. This helps us solve the following questions. Mehdi Dehghan *; Masoud Hajarian. Torres and Robert A. Announcements. Header declares a set of functions to compute common mathematical operations and transformations: Functions Trigonometric functions cos Compute cosine (function ). The number under the square root sign is negative which means this equation has no solution. In attempting to solve equation above it should be that a general quintic equation in \u03b1 is algebraically solvable using the procedures outlined in this paper. C u b i c e q u a t i o n a x 3 + b x 2 + c x + d = 0 C u b i c e q u a t i o n a x 3 + b x 2 + c x + d = 0 a. His formula applies to depressed cubics, but, as shown in \u00a7 Depressed cubic, it allows solving all cubic equations. A quadratic equation can be solved by using the quadratic formula. THIS IS A DIRECTORY PAGE. Re: Cubic equation with complex coefficients The algebraic solutions/formulas for the cubic equation should remain valid for complex coefficients as well as for real coefficients. The van der Waals (from his thesis of 1873) equation is a cubic in the molar volume. The domain and range in a cubic graph is always real values. We consider the cubic nonlinear Schr\u00f6dinger (NLS) equation as well as the modified Korteweg\u2013de Vries (mKdV) equation in one space dimension. Cardano and the solving of cubic and quartic equations Girolamo Cardano was a famous Italian physician, an avid gambler, and a prolific writer with a lifelong interest in mathematics. Cubic equations have to be solved in several steps. Possible Outcomes When Solving a Cubic Equation If you consider all the cases, there are three possible outcomes when solving a cubic equation: 1. , the roots of a cubic polynomial. If you thought the Quadratic Formula was complicated, the method for solving Cubic Equations is even more complex. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations how to solve a cubic equation. When I try to solve this equation using mathematica's Solve[] function, I get one real root and 2 complex roots. A Property of Cubic Equations. Solution of Cubic Equations. Find more Mathematics widgets in Wolfram|Alpha. In case of a cubic equation of the form ax^3+bx^2+cx+d=0 with a nonzero number, simply divide every coefficient by a and proceed. The general form of a cubic equation is ax 3 + bx 2 + cx + d = 0 where a, b, c and d are constants and a \u2260 0. A top-performance algorithm for solving cubic equations is introduced. The solution can also be expressed in terms of the Wolfram Language algebraic root objects by first issuing SetOptions [ Roots , Cubics -> False ]. Solving The Cubic Equations History Essay Homework Service. Recall that the solution to the depressed cubic x3 + px+ q = 0 is given by Cardano\u2019s formula x = 3 s q 2 + r q 2 2 + p 3 3 + 3 s q 2 r q 2 2 + p 3 3 Now consider the general cubic equation ax3 + bx2. There isn't that much more to it. SOLVING THE CUBIC EQUATION The cubic algebraic equation ax 3+bx 2+cx+d=0 was first solved by Tartaglia but made public by Cardano in his book Ars Magna(1545) after being sworn to secrecy concerning the solution method by the. how to solve cubic equation in faster way http://youtu. Use this calculator to solve polynomial equations with an order of 3 such as ax 3 + bx 2 + cx + d = 0 for x including complex solutions. It also gives the imaginary roots. An online cube equation calculation. Mathews Mathematics Department California State University Fullerton There are several formulas for solving the cubic equation. Cubic equations and the nature of their roots A cubic equation has the form ax3 +bx2 +cx+d = 0. Then are real. Let the roots of x2 px+ q= 0 be and , so that p= + and q=. Manages the issue of inherent in the power basis representation of the polynomial in floating point. In the chapter \"Classi\ufb01cation of Conics\", we saw that any quadratic equa-tion in two variables can be modi\ufb01ed to one of a few easy equations to un-derstand. Gerolamo Cardano published a method to solve a cubic equation in 1545. Cubic equations can be solved analytically in Matlab (you need the symbolic toolbox to get the expressions or just copy & paste them from somewhere). It is defined as third degree polynomial equation. com is a free online OCR (Optical Character Recognition) service, can analyze the text in any image file that you upload, and then convert the text from the image into text that you can easily edit on. Various forms of van der Waals equation of state Cubic Equations of State Note that the van der Waals (vdw) equation of state is cubic in volume. The former editor of a monthly computing and technology magazine, his work has appeared in The Guardian, GQ and Time Out. Try finding a different solution. What is the formula to solve a cubic equation? [duplicate] There is no general algebraic formula to solve a polynomial equation that has degree 5 or higher in. Solving the Cubic Equation for Dummies. Often, our goal is to solve an ODE, i. In this case it expresses the solution through cubic roots of complex numbers. Equivalently, the cubic formula tells us the solutions of equations of the form ax3 +bx2 +cx+d =0. Differentiated Learning Objectives. Equation (12) may also be explicitly factored by attempting to pull out a term of the form from the cubic equation, leaving behind a quadratic equation which can then be factored using the Quadratic Formula. Fior challenge: Tartaglia pro-poses some questions, and Fior responds by giving Tartaglia 30 depressed cubics to solve. 00000001, initial_guess=0. There is no obvious way that \u201ccompleting the cube\u201d makes the solution into a matter of just taking cube roots in the same way that \u201ccompleting the square\u201d solves the quadratic in terms of square roots. Cubic equation calculator Fill in the coefficients a, b, c, and d in the equation a x 3 + b x 2 + c x + d = 0 and click the Solve button. Cubic Equation Calculator. 0 is equal to ax squared plus bx plus c. The model equation is A = slope * C + intercept. However, I have tried plotting the equation for these values, and can clearly see there should be 3 real roots. Using the same trick as above we can transform this into a cubic equation in which the coe\ufb03cient of x2 vanishes: put x = y \u2212 1. roots but exclusive to cubic polynomials. After reading this chapter, you should be able to: 1. We have developed an energy balance equation for the universe. Once the equation has been entered, the calculator uses the Newton-Raphson numerical method to solve the equation. image source. Solving linear equations in two variables is straightforward (back substitute). To solve this equation means to write down a formula for its roots, where the formula should be an expression built out of the coefficients a, b and c and fixed real numbers (that is, numbers that do not depend on a, b and c) using only addition, subtraction. Bibtex entry for this abstract Preferred format for this abstract (see Preferences ). Just out of interest, is it possible to solve cubic equations with complex solutions, based purely on iterative methodologies? @Strange_Man I think the only way you can do that, is if you write your own complex number class. net is going to be the perfect site to pay a visit to!. While it might not be as straightforward as solving a quadratic equation, there are a couple of methods you can use to find the solution to a cubic equation without resorting to pages and pages of detailed algebra. But your function does not have a squared term; it already is a DEPRESSED cubic so that substitution is irrelevant. I understand that there are two cubic formulas that solve the cubic equation that, which one to use depends on the characterstic value of n. We\u2019ll discuss discriminants some other time. Third Degree Polynomial Formula. Solution: Let us use the division method for solving a cubic equation. Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and logarithmic equations with all the steps. The known Cardano\u2019s formulas for solution of this kind equations are very difficult and almost aren\u2019t used in practice. In fact, the last part is missing and without this part, one cannot implement it into an algorithm. Therefore, u and v satisfy the binomial equations. I shall try to give some examples. This Online Equation Solver solves every equations with set of given variables. Therefore, we can solve for x. Cubic equations of state are called such because they can be rewritten as a cubic function of molar volume. This is the graph of the equation 2x 3 +0x 2 +0x+0. The canonical form of cubic equation is. Let us take an example to understand the process easily. Now divide the equation with x-1 which will give you the quadratic equation. Often, our goal is to solve an ODE, i. The Cubic Equation. Improve your math knowledge with free questions in \"Solve a quadratic equation using the quadratic formula\" and thousands of other math skills. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. A Level > Arithmetic sequences A Level > Binomial expansion A Level > Differentiation A Level > Factor and remainder theorem A Level > Fibonacci sequences A Level > Geometric sequences A Level > Integration A Level > Logs A Level > Mechanics A Level > Mid-ordinate rule A Level > Partial fractions A Level > Point of inflection A Level. Some numerical methods for solving a univariate polynomial equation p(x) = 0 work by reducing this problem to computing the eigenvalues of the companion ma- trix of p(x), which is de\ufb01ned as follows. This process is equivalent to making Vieta's substitution, but does a slightly better job of motivating Vieta's magic'' substitution. Finally the solutions of the pressed cubic equation is the combination of the cubic roots of the resolvent. use e- as the symbol for an electron. The conventional strategy followed for solving a cubic equation involved its reduction to a quadratic equation and then. Let's use the following equation. SOLVING THE CUBIC EQUATION The cubic algebraic equation ax 3+bx 2+cx+d=0 was first solved by Tartaglia but made public by Cardano in his book Ars Magna(1545) after being sworn to secrecy concerning the solution method by the. This algorithm uses polynomial fitting for a decomposition of the given cubic into a product of a quadratic and a linear factor. Britannica does not currently have an article on this topic. Torres\u2019 Approach) on the Drexel University Website. Solving a Cubic Equation With Excel 2016 Although complex numbers appear to be surreal and seem to involve something like Zeno's paradoxes they are quite useful. an equation in which the highest power of the unknown quantity is a cube. A cubic equation can have 3 real roots or 1 real root and a complex conjugate pair. The roots of this equation can be solved using the below cubic equation formula. formulas for solving cubic equations of the form x3 +3px =2q, x3 +2q =3px, and x3 =3px +2q. math worksheet solving quadratic and cubic. In fact, the last part is missing and without this part, one cannot implement it into an algorithm. The cell you want to change to reach that value is the variable you're solving for, x. Given that is a cubic function with zeros at -3, 2, and 7, find an equation for given that f(3)=5. How to Find the Exact Solution of a General Cubic Equation In this chapter, we are going to find the exact solution of a general cubic equation. edu/~toh/spectrum/CurveFitting. How to Solve a Cubic Equation Part 1 - The Shape of the Discriminant James F. Cubic equations of state are called such because they can be rewritten as a cubic function of molar volume. Students will: solve simple square root equations, listing both solutions where appropriate. How can we solve equations such as the cubic equation shown here? x 3 \u2212 x 2 - 4x + 4 = 0. find the exact solution of a general cubic equation. For cylinders and prisms, the formula is the area of the base multiplied by the height. Learn more about cubic equation Symbolic Math Toolbox. Four years later (1545), Cardan published a book called Ars Magna, which contained the solution to the cubic equation and Ferrari\u2019s solution to the quartic. First divide by the leading term, making the polynomial monic. find the exact solution of a general cubic equation. Inequalities-- \"The. Cubic equations and the nature of their roots A cubic equation has the form ax3 +bx2 +cx+d = 0. the cubic formula, which thereby solves the cubic equation, nding both real and imaginary roots of the equation. Therefore the sextic. His formula applies to depressed cubics, but, as shown in \u00a7 Depressed cubic, it allows solving all cubic equations. Factor theorem solving cubic equations 1. Posted in Based on a Context Tagged A Level > Factor and remainder theorem, Algebra > Equations > Finding roots, Algebra > Equations > Iteration, Algebra > Functions > Composite functions Post navigation. 1 Miscellaneous Algebraic Approaches to the Cubic and Quartic For about 100 years after Cardano, \\everybody\" wanted to say something. It's easy to calculate y for any. Then are real. The app is able to find both real and complex roots. For the solution of the cubic equation we take a trigonometric Viete method, C++ code takes about two dozen lines. h header, poly34. Because the y value is 0,. However the coe cients of (2) are not explicitly available. He applies the cubic formula for this form of the equation and arrives at this \"mess\":! x= 3 (2+ \"121)+ 3 (2\"\"121) If you set your TI to complex mode, you can confirm that this complex formula is, in fact, equal to 4. Multiply the three lengths For a square, all three sides are the same. Let Vdenote the quotient of the polynomial ring modulo the ideal hp(x)igenerated by the polynomial p(x). You would usually use iteration when you cannot solve the equation any other way. Vieta's formulae are used to solve equations as thus, first step is to divide all coefficients by \"a\". The solution was first published by Girolamo Cardano (1501-1576) in his Algebra book Ars Magna. The solution of cubic and quartic equations In the 16th century in Italy, there occurred the \ufb01rst progress on polynomial equations beyond the quadratic case. Set \u02c6= : Then \u02c62 = 2 2 + 2 = ( + )2 4 = p2 4qand \u02c6= p p2 4q: So = 1 2 + + ! = 1 2 (p+ \u02c6) Solving the cubic equation Let\u2019s suppose we. The depressed cubic equation. The cubic formula tells us the roots of polynomials of the form ax3 +bx2 + cx + d. They are not many students who are actually remembering the formulas taught in school. Exercise 10. ax 3 + bx 2 + cx + d = 0. Cubic Equation Formula, cubic equation, Depressing the Cubic Equation, cubic equation solver, how to solve cubic equations, solving cubic equations. So I thought I could. Learn more about cubic equation Symbolic Math Toolbox. Cubic Equation Formula. Linear Equation Solver-- Solve three equations in three unknowns. To understand this example, you should have the knowledge of following Python programming topics:. Wholesale Lingerie cheap Lingerie. For example, we have the formula y = 3x 2 - 12x + 9. In the module, Polynomials, a factoring method will be developed to solve cubic equations that have rational roots. A cubic function is one of the most challenging types of polynomial equation you may have to solve by hand. how to solve cubic equation on excel You are not going to be able to solve cubic equations with a formula. 14k Yellow Gold CZ Cubic Zirconia Cross Hinged Hoop Earrings Hinged Earrings Hinged Zirconia Zirconia Cross 14k Hoop CZ Cubic Gold Yellow NewOCR. In this tutorial you are shown how to solve a cubic equation by using the factor theorem. However, most of the theory is also valid if they belong to a field of characteristic other than 2 or 3. Our objective is to find two roots of the quartic equation The other two roots (real or complex) can then be found by polynomial division and the quadratic formula. I have used the Newton-Raphson method to solve Cubic equations of the form by first iteratively finding one solution, and then reducing the polynomial to a quadratic and solving for it using the quadratic formula. -2-Solving cubics is not as simple as solving quadratics. The Polynomial equations don\u2019t contain a negative power of its variables. Title: Hyperbolic identity for solving the cubic equation: Authors: Rochon, Paul: Publication: American Journal of Physics, Volume 54, Issue 2, pp. In addition, Ferrari was also able to discover the solution to the quartic equation, but it also required the use of the depressed cubic. Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. The remaining unknown can then be calculated. Solving a Polynomial Equation Solve 2x5 + 24x = 14x3. When we deal with the cubic equation one surprising result is that often we have to express the roots of the equation in terms of complex numbers although the roots are real. cubic root of unity. In the By Changing Cell box, type A3. how to solve cubic equation on excel You are not going to be able to solve cubic equations with a formula. roots([1 -3 2]) and Matlab will give you the roots of the polynomial equation. After reading this chapter, you should be able to: 1. Let ax\u00b3 + bx\u00b2 + cx + d = 0 be any cubic equation and \u03b1,\u03b2,\u03b3 are roots. Solution Techniques for Cubic Expressions and Root Finding Print Hopefully, we have convinced you that the use of cubic equations of state can represent a very meaningful and advantageous way of modeling the PVT behavior of petroleum fluids. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. Algorithm for solving cubic equations. asks tricky questions. The cubic formula gives the roots of any cubic equation Solving the depressed cubic equation with Vieta's substitution. 3 2 ax bx cx d + + + = 0 (1). Use this calculator to solve polynomial equations with an order of 3 such as ax 3 + bx 2 + cx + d = 0 for x including complex solutions. 5, the y value is 0. That means, reducing the equation to the one where the maximum power of the equation is 2. First, suppose. Computing such square roots again leads to cubic equations. \u2026treatment of the solution of cubic equations. A quadratic equation can be solved by using the quadratic formula. Cubic equations and the nature of their roots. By the fundamental theorem of algebra, cubic equation always has 3 3 3 roots, some of which might be equal. Finding the center of mass doesn't require solving the equation\u2014it is merely \u22121/3 the coefficient of the quadratic term. The van der Waals (from his thesis of 1873) equation is a cubic in the molar volume. And we generally deal with x's, in this. There are several ways to solve cubic equation. Possible Outcomes When Solving a Cubic Equation If you consider all the cases, there are three possible outcomes when solving a cubic equation: 1. Naturally, the first solutions were geometric, for ancient Egyptians and Greeks knew nothing of algebra. Furthermore, there are examples of cubic equations with real coefficients and three distinct real solutions for which the cubic formula requires one to calculate a root of a complex number. The Polynomial equations don\u2019t contain a negative power of its variables. The polynomial x4+ax3+bx2+ cx+dhas roots. solving a cubic equation. The amount of profit (in millions) made by Scandal Math, a company that writes math problems based on tabloid articles, can be found by the equation P(n) = \u2212n2 + 10n, where n is the number of textbooks sold (also in millions). at first, has a lower rate of growth than the linear equation f(x) =50x; at first, has a slower rate of growth than a cubic function like f(x) = x 3, but eventually the growth rate of an exponential function f(x) = 2 x, increases more and more -- until the exponential growth function has the greatest value and rate of growth!. Use a ruler to measure each side in inches. Fitting a cubic to data Let us now try to fit a cubic polynomial to some data. Given a quadratic of the form ax2+bx+c, one can \ufb01nd the two roots in terms of radicals as-b p b2-4ac 2a. Write an equation for the cubic polynomial function whose graph has zeroes at 2, 3, and 5.", "date": "2019-12-13 09:05:27", "meta": {"domain": "venetoacquaeterme.it", "url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html", "openwebmath_score": 0.7340618371963501, "openwebmath_perplexity": 460.93382348864924, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9871787827157819, "lm_q2_score": 0.9136765193002482, "lm_q1q2_score": 0.9019620741188116}}
{"url": "https://www.physicsforums.com/threads/de-nesting-radicals.777895/", "text": "1. Oct 23, 2014\n\n### bamajon1974\n\nI want to de-nest the following radical:\n(1) $$\\sqrt{3+2\\sqrt{2}}$$\nInto the general simplified form:\n(2) $$a+b\\sqrt{2}$$\nEquating (1) with (2),\n(3) $$\\sqrt{3+2\\sqrt{2}} = a+b\\sqrt{2}$$\nand squaring both sides:\n(4) $$3+2\\sqrt{2} = a^2 + 2b^2 + 2ab\\sqrt{2}$$\ngenerates a system of two equations with two unknowns, a and b, after equating the rational and irrational parts:\n(5) $$3 = a^2 + 2b^2$$\n(6) $$2\\sqrt{2} = 2ab\\sqrt{2}$$\nSimplifying (6) and solving for b:\n(7) $$b=\\frac{1}{a}$$\nSubstituting (7) into (5) yields:\n(8) $$3 = a^2 + 2\\frac{1}{a^2}$$\nClearing the denominator and moving non-zero terms to one side generates a quartic equation:\n(9) $$0 = a^4 - 3a^2 +2$$\n(10) $$x=a^2, x^2 = (a^2)^2 = a^4, x = \\pm \\sqrt{a}$$\n(11) $$0 = x^2 - 3x + 2$$\nThe square root of the discriminant is an integer, 1, which presumably makes simplification of the nested radical possible. Finding the roots, x, of (11) gives\n(12)$$x=1, x =\\sqrt{2}$$\nSubstituting the roots in (12) into (10) gives a:\n(13) $$a = \\pm \\sqrt{2} , a = \\pm 1$$\nThen b is found from (7):\n(14) $$b = \\pm \\frac{1}{\\sqrt{2}} , b = \\pm 1$$\nUsing the positive values of a and b, the de-nested radical is:\n(15) $$\\sqrt{3+2\\sqrt{2}} = 1+\\sqrt{2}$$\n\nMy questions are:\n(1) Is this approach for simplifying nested radical correct?\n\n(2) The positive values of a and b produce the correct simplified form in (15) while the negative values of a and b do not. Is there a way to figure out which roots from (9) are correct and which to reject other than calculating the numerical value of (15) with both positive and negative a's and b's to see which is equal to the nested form?\n\n(3) I have also seen the general simplified de-nested form as:\n(16) $$\\sqrt{a} + \\sqrt{b}$$\nGoing through an analogous process as above, it generates the correct simplified form in (15) as well. Is one form (2) or (15) better than the other? Or is it just a personal preference which one to use?\n\nThanks!\n\n2. Oct 23, 2014\n\n### symbolipoint\n\nThe square root radical symbol means, the expression under the radical raised to the one-half power.\n\n$$\\sqrt{3+2\\sqrt{2}}$$\n$$(3+2\\sqrt{2})^(1/2)$$\n$$(3+2(2)^(1/2))^(1/2)$$\n\n( I KNOW what I'm trying to do but still cannot make it appear correctly)\n\n3. Oct 23, 2014\n\n### SteamKing\n\nStaff Emeritus\nYou mean like this? Just enclose your expression for the exponent in a pair of curly braces {}.\n\n4. Oct 24, 2014\n\n### Mentallic\n\nI'll give you a chance to look at these again. You've ended up with the correct answer however because you made two errors that both cancelled each other out, so that was a lucky one on your part ;)\n\nYes.\n\nYou'll often have situations arise whereby you need to find, say, the length of an object, but the equations you're dealing with are quadratics and will thus give you two solutions, a positive and a negative one. It suffices to toss away the negative value because you're solving a problem that only makes sense with a positive answer.\n\nIt's similar in this case. You're trying to simplify a square root. You know that square roots don't give negative results so you're allowed to simply toss out the result $-1-\\sqrt{2}$ because that is clearly negative and can't be the answer. You don't need to look at the numerical values of each result with a calculator if you simply use a bit of intuition.\n\nI prefer yours merely because the $\\sqrt{2}$ result cancels and you end up with a nice result in (7). Neither are incorrect, but also both are assumptions. You assumed that the answer is of the form $a+b\\sqrt{2}$ and the \"general\" simplified de-nested form $\\sqrt{a}+\\sqrt{b}$ assumes that the result is the sum of two square roots. What if it were $\\sqrt{a}+\\sqrt{b}+\\sqrt{c}$ instead? Or something even more complicated?\nSo yes, I prefer your assumption over the other because yours is easier to work with.\n\n5. Oct 24, 2014\n\n### bamajon1974\n\nI see my error:\n12)\n$$x=1, x=\\sqrt{2}$$\nShould be\n$$x=1, x=2$$\nMy bad. Although in my defense that was an error I didn't catch after typing so much unfamiliar latex code into these lines.\n\nI had neglected to mention that in the if you try to de-nest\n$$\\sqrt{3+2\\sqrt{2}}$$\nby assuming it equals the other simplified form (16)\n$$\\sqrt{a} + \\sqrt{b}$$\nEquating the nested radical with the assumed simplified form and squaring both sides:\n$$3 + 2\\sqrt{2} = a + b + 2\\sqrt{ab}$$\nEquating rational and irrational parts generates 2 equations with 2 unknowns:\n$$3 = a + b$$\n$$2\\sqrt{2} = 2\\sqrt{ab}$$\nSolving for b in the second equation, substituting into the first and moving all non-zero terms to one side of the equation generates a quadratic:\n$$0=a^2+3a+2$$\nof which the roots, a, are\n$$a= 1, a=2$$\nSo a and b are\n$$a=1, b=2$$\nor\n$$a=2, b=1$$\nSo the simplified form is\n$$1+\\sqrt{2}$$\nidentical to above, but I didn't have any negative roots to discard. So in this particular case, the second form was a little easier.\n\nI see why rejecting the negative roots is justified as well.", "date": "2018-07-20 20:41:15", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/de-nesting-radicals.777895/", "openwebmath_score": 0.8350215554237366, "openwebmath_perplexity": 542.9755849662545, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9805806484125338, "lm_q2_score": 0.9196425388861785, "lm_q1q2_score": 0.9017836770887577}}
{"url": "https://mathhelpboards.com/threads/proof-critique-induction.6248/", "text": "# Proof critique: Induction\n\n#### sweatingbear\n\n##### Member\nWe wish to show that $3^n > n^3 \\, , \\ \\forall n \\geqslant 4$. Base case $n = 4$ yields\n\n$3^4 = 81 > 4^3 = 64$\n\nAssume the inequality holds for $n = p$ i.e. $3^p > p^3$ for $p \\geqslant 4$. Then\n\n$3^{p+1} > 3p^3$\n\n$p \\geqslant 4$ implies $3p^3 \\geq 192$, as well as $(p+1)^3 \\geqslant 125$. Thus $3p^3 > (p+1)^3$ for $p \\geqslant 4$ and we have\n\n$3^{p+1} > 3p^3 > (p+1)^3$\n\nwhich concludes the proof.\n\nFeedback, forum?\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\n$p \\geqslant 4$ implies $3p^3 \\geq 192$, as well as $(p+1)^3 \\geqslant 125$. Thus $3p^3 > (p+1)^3$ for $p \\geqslant 4$ and we have\nOne question :\n\nif x>5 and y>2 then x>y ?\n\n#### sweatingbear\n\n##### Member\nOne question :\n\nif x>5 and y>2 then x>y ?\nI was actually a bit uncertain about that. How else would one go about?\n\n#### ZaidAlyafey\n\n##### Well-known member\nMHB Math Helper\nWe need to prove that\n\n$$\\displaystyle 3^{p+1}> (p+1)^3$$ assuming that $$\\displaystyle 3^p>p^3\\,\\,\\,\\, \\forall p\\geq 4$$\n\n$$\\displaystyle \\tag{1} 3^{p+1}>3p^3\\geq p^3+p^3$$\n\nLemma :\n\n$$\\displaystyle p^3-3p^2-3p-1 \\geq 0$$\n\nTake the derivative\n\n$$\\displaystyle 3p^2-6p-3 =3(p^2-2p-1)=3(p^2-2p+1-2)=3(p-1)^2-6$$\n\nThe function is positive for $p=4$ and increases for $$\\displaystyle p\\geq 4$$ so the lemma is satisfied .\n\nHence we have\n\n$$\\displaystyle p^3 \\geq 3p^2+3p+1 \\,\\,\\,\\,\\forall p\\geq 4$$\n\nUsing this in (1) we get\n\n$$\\displaystyle 3^{p+1}>p^3+3p^2+3p+1=(p+1)^3 \\,\\,\\,\\square$$\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nHere is how I would do this proof:\n\n(inductive step only).\n\nSuppose that $$\\displaystyle 3^k > k^3, k > 3$$.\n\nThen:\n\n$$\\displaystyle 3^{k+1} = 3(3^k) > 3k^3$$.\n\nIf we can show that:\n\n$$\\displaystyle 3k^3 > (k+1)^3$$, we are done.\n\nEquivalently, we must show that:\n\n$$\\displaystyle 3k^3 > k^3 + 3k^2 + 3k + 1$$, so that:\n\n$$\\displaystyle 2k^3 - 3k^2 - 3k - 1 > 0$$.\n\nNote that $$\\displaystyle 2k^3 - 3k^2 - 3k - 1 > 2k^3 - 3k^2 - 5k + 3$$\n\nif $$\\displaystyle 2k - 4 > 0$$, that is if $$\\displaystyle k > 2$$, which is true.\n\nBut $$\\displaystyle 2k^2 - 3k^2 - 5k + 3 = (2k - 1)(k^2 - k - 3)$$.\n\nNow $$\\displaystyle 2k - 1 > 0$$ for any $$\\displaystyle k > 0$$, so we are down to showing\n\n$$\\displaystyle k^2 - k - 3 > 0$$ whenever $$\\displaystyle k > 3$$.\n\nSince $$\\displaystyle k^2 - k > 3$$ is the same as $$\\displaystyle k(k-1) > 3$$, we have:\n\n$$\\displaystyle k(k-1) > 3(2) = 6 > 3$$.\n\nThus we conclude that $$\\displaystyle k^2 - k - 3 > 0$$ and so:\n\n$$\\displaystyle 3^{k+1} = 3(3^k) > 3k^3 > (k+1)^3$$.\n\nWith all due respect to Zaid, I wanted to give a purely algebraic proof.\n\n#### Evgeny.Makarov\n\n##### Well-known member\nMHB Math Scholar\nEquivalently, we must show that:\n\n$$\\displaystyle 3k^3 > k^3 + 3k^2 + 3k + 1$$\nStarting from this point, we could continue as follows. We need to show that $3k^2+3k+1<2k^3$.\n\n$3k^2+3k+1<3k^2+3k^2+k^2=7k^2$ since $k>1$. Now, $7k^2<2k^3\\iff 7<2k$, and the last inequality is true since $k\\ge4$.\n\n#### Deveno\n\n##### Well-known member\nMHB Math Scholar\nIndeed, we just need to find something that is less than $$\\displaystyle 2k^3$$ and larger than $$\\displaystyle 3k^2 + 3k + 1$$ that \"factors nice\" (so we can apply what we know specifically about $$\\displaystyle k$$). Very nice solution!", "date": "2020-09-26 13:03:45", "meta": {"domain": "mathhelpboards.com", "url": "https://mathhelpboards.com/threads/proof-critique-induction.6248/", "openwebmath_score": 0.9834364652633667, "openwebmath_perplexity": 977.5917597252328, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9924227591803558, "lm_q2_score": 0.9086178895092414, "lm_q1q2_score": 0.901733072947393}}
{"url": "https://math.stackexchange.com/questions/2114575/partitions-of-n-that-generate-all-numbers-smaller-than-n", "text": "# Partitions of n that generate all numbers smaller than n\n\nConsider a partition $$n=n_1+n_2+\\cdots+n_k$$ such that each number $1,\\cdots, n$ can be obtained by adding some of the numbers $n_1,n_2,\\cdots,n_k$. For example, $$9=4+3+1+1,$$\n\nand every number $1,2,\\cdots,9$ be ca written as a sum of some of the numbers $4,3,1,1$. This other partition $$9=6+1+1+1$$ fails the desired property, as $4$ (and $5$) cannot be given by any sum of $6,1,1,1$.\n\nQuestion: Can we charaterize which partitions of an arbitrary $n$ have this property? We clearly need at least one $1$ among $n_1,n_2,\\cdots,n_k$, and intuitively we need many small numbers $n_i$. But I can't see much beyond this. Any idea or reference will be appreciated.\n\n\u2022 If at least half of the number are 1's, you can do it. But the condition is certainly not necessary. \u2013\u00a0MathChat Jan 26 '17 at 6:03\n\u2022 I once studied this problem and found a constructive partition method. Here is the brief. We are given a positive integer $n$. STEP ONE: if $n$ is an even number, partition it into $A=\\frac{n}{2}$ and $B=\\frac{n}{2}$; otherwise, partition it into $A=\\frac{n+1}{2}$ and $B=\\frac{n-1}{2}$. STEP TWO: re-partition $B$ into $A_1$ and $B_1$. STEP THREE: re-partition $B_1$......Until we get $1$. I didn't prove this method always works but I believe it is valid. \u2013\u00a0apprenant Jan 26 '17 at 6:04\n\u2022 EXAMPLE: $13 \\rightarrow (7,6) \\rightarrow (7,3,3) \\rightarrow (7,3,2,1)$. May it be helpful. \u2013\u00a0apprenant Jan 26 '17 at 6:06\n\u2022 @apprenant. This method looks interesting. BTW, is this problem well known? \u2013\u00a0ALEXIS Jan 26 '17 at 6:14\n\u2022 You can work up the other way, too, by powers of two, e.g $13 \\to 1,2,4,6$ \u2013\u00a0Joffan Jan 26 '17 at 6:17\n\nLet $\\lambda$ be a partition of $n$. The required condition is that $\\lambda$ contain partitions $\\lambda_i$ of each $1 \\le i < n$. Clearly if $\\lambda$ contains a partition of $j$ then it also contains a partition of $(n - j)$, being the multiset difference $\\lambda - \\lambda_j$.\n\nTherefore the first thing to notice is that $\\lambda$ cannot contain any element $a > \\lceil \\frac{n}{2} \\rceil$, for if it did then $\\{a\\}$ cannot be part of a partition of $\\lceil \\frac{n}{2} \\rceil$ and $\\lambda - \\{a\\}$ is a partition of $(n - a) < (n - \\lceil \\frac{n}{2} \\rceil) < \\lceil \\frac{n}{2} \\rceil$ cannot contain a partition of $\\lceil \\frac{n}{2} \\rceil$.\n\nNow, suppose that the largest element of $\\lambda$ is $m$. It is certainly sufficient that $\\lambda - \\{m\\}$ should satisfy the corresponding condition of providing partitions for each $1 \\le i < (n - m)$. Proof: $\\lambda - \\{m\\}$ is a partition of $(n - m)$ and provides partitions for each smaller natural number, so it remains to construct partitions $\\lambda_i$ for $(n - m) < i < n$. We can do this by taking partitions from $\\lambda - \\{m\\}$ for $(n - 2m) < j < n - m$ and then adding $\\{m\\}$ to each one. This can only fail if $j < 0$, which can only happen if $(n - 2m) < -1$. Since $m \\le \\lceil \\frac{n}{2} \\rceil$ we have $n - 2m \\ge n - 2\\lceil \\frac{n}{2} \\rceil$, which is $0$ if $n$ is even and $-1$ if $n$ is odd, so all cases are covered.\n\nThe interesting question is whether it's necessary that $\\lambda - \\{m\\}$ should satisfy the same condition. Clearly it must contain partitions of $1 \\le i < m$, since $\\{m\\}$ doesn't participate in them. And by the simple principle of taking complements in $\\lambda$ it's clear that for each $m \\le i < n$ the remnant $\\lambda - \\{m\\}$ must either contain partitions of $(i - m)$ and $n - i$; or $i$ and $n - m - i$. Is that sufficient?\n\nMy intuition is that it's necessary, and testing on small examples (up to $n = 30$) supports that, but I haven't proved it.\n\nIn the Online Encyclopedia of Integer Sequences it's A126796 and a comment claims the characterisation\n\nA partition is complete iff each part is no more than 1 more than the sum of all smaller parts. (This includes the smallest part, which thus must be 1.) - Franklin T. Adams-Watters, Mar 22 2007\n\nThis is just a quick and dirty list of the first examples, for $n$ up to $10$. Feel free to edit, extend, or amend.\n\n$$1\\\\1+1\\\\ 1+1+1\\quad1+2\\\\ 1+1+1+1\\quad1+1+2\\\\ 1^5\\quad1+1+1+2\\quad1+2+2\\quad1+1+3\\\\ 1^6\\quad1^4+2\\quad1^2+2+2\\quad1^3+3\\quad1+2+3\\\\ 1^7\\quad1^5+2\\quad1^3+2+2\\quad1+2^3\\quad1^4+3\\quad1^2+2+3\\quad1^3+4\\quad1+2+4\\\\ 1^8\\quad1^6+2\\quad1^4+2+2\\quad1^2+2+2+2\\quad1^5+3\\quad1^3+2+3\\quad1^2+3+3\\quad1+2+2+3\\quad1^4+4\\quad1+1+2+4\\\\ 1^9\\quad1^7+2\\quad1^5+2+2\\quad1^3+2^3\\quad1+2^4\\quad1^6+3\\quad1^4+2+3\\quad1^2+2+2+3\\quad1^3+3+3\\quad1+2+3+3\\quad1^5+4\\quad1^3+2+4\\quad1+2+2+4\\quad1+1+3+4\\quad1^4+5\\quad1^2+2+5\\\\ 1^{10}\\quad1^8+2\\quad1^6+2^2\\quad1^4+2^3\\quad1^2+2^4\\quad1^7+3\\quad1^5+2+3\\quad1^3+2^2+3\\quad1+2^3+3\\quad1^4+3^2\\quad1^2+2+3^2\\quad1^6+4\\quad1^4+2+4\\quad1^2+2^2+4\\quad1^3+3+4\\quad1+2+3+4\\quad1^5+5\\quad1^3+2+5\\quad1+2^2+5\\quad1+1+3+5$$\n\nI hope the meaning of the superscript notation is clear, and I hope someone will check that I didn't make any mistakes or overlook anything. The list so far gives the sequence\n\n$$1,1,2,2,4,5,8,10,16,20,\\ldots$$\n\nwhich (after correcting a pair of mistakes in the original posting here) Peter Taylor found in the OEIS.\n\n\u2022 I get 1, 1, 1, 2, 2, 4, 5, 8, 10, 16, 20, 31, 39, 55, 71, 100, 125, 173, 218, 291 starting at index 0. A126796 \u2013\u00a0Peter Taylor Jan 26 '17 at 17:07\n\u2022 @PeterTaylor, thank you! I see now what I missed: $7=1^3+4$ and $10=1+1+3+5$. Darn! I swear, I checked and double checked all my counts. I guess I needed to triple check.... \u2013\u00a0Barry Cipra Jan 26 '17 at 19:12", "date": "2019-04-24 20:35:35", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2114575/partitions-of-n-that-generate-all-numbers-smaller-than-n", "openwebmath_score": 0.8674020171165466, "openwebmath_perplexity": 151.52469035540983, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9881308765069576, "lm_q2_score": 0.9124361509525463, "lm_q1q2_score": 0.9016063335973743}}
{"url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function", "text": "The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Describe all numbers$\\,x\\,$that are at a distance of 4 from the number 8. Write an equation for the function graphed in (Figure). To understand the Absolute value of a Derivative and Integral or magnitude of a complex number We must first understand what is the meaning of absolute value. The latter is a special form of a cell address that locks a reference to a given cell. 2 Peter Wriggers, Panagiotis Panatiotopoulos, eds.. Step 2: Rewrite the absolute function as piecewise function on different intervals. Absolute Value Functions 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. The product in A of an element x and its conjugate x* is written N(x) = x x* and called the norm of x. Describe all numbers$\\,x\\,$that are at a distance of$\\,\\frac{1}{2}\\,$from the number \u22124. Recall that in its basic form$\\,f\\left(x\\right)=|x|,\\,$the absolute value function is one of our toolkit functions. Algebraically, for whatever the input value is, the output is the value without regard to sign. The graph of an absolute value function will intersect the vertical axis when the input is zero. Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Knowing how to solve problems involving absolute value functions is useful. [/latex], Applied problems, such as ranges of possible values, can also be solved using the absolute value function. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. The most significant feature of the absolute value graph is the corner point at which the graph changes direction. To solve an equation such as$\\,8=|2x-6|,\\,$we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. How to graph an absolute value function on a coordinate plane: 5 examples and their solutions. R Note that these equations are algebraically equivalent\u2014the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression. The absolute value of a number is a decimal number, whole or decimal, without its sign. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. [/latex], $x=-1\\,$or$\\,\\,x=2$, Should we always expect two answers when solving$\\,|A|=B? The most significant feature of the absolute value graphAbsolute Value Functions:Graphing is the corner point where the graph changes direction. If possible, find all values of [latex]a$ such that there are no $x\\text{-}$intercepts for $f\\left(x\\right)=2|x+1|+a. Yes. Free absolute value equation calculator - solve absolute value equations with all the steps. Cities A and B are on the same east-west line. (a) The absolute value function does not intersect the horizontal axis. Knowing this, we can use absolute value functions to \u2026 This point is shown at the origin in (Figure). It is differentiable everywhere except for x = 0. Algebraically, for whatever the input value is, the output is the value without regard to sign. As such, it is a positive value, and will not be negative, though an absolute value is allowed be 0 itself. The absolute value parent function, written as f (x) = | x |, is defined as . In general the norm of a composition algebra may be a quadratic form that is not definite and has null vectors. Assume that city A is located at the origin. An absolute value equation is an equation in which the unknown variable appears in absolute value bars. f (x) = {x if x > 0 0 if x = 0 \u2212 x if x < 0. A decimal number. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for[latex]\\,x\\,$and$\\,f\\left(x\\right).$. We can find that 5% of 680 ohms is 34 ohms. For the following exercises, graph the given functions by hand. $\\,f\\left(x\\right)=|x|=\\bigg\\{\\begin{array}{ccc}x& \\text{if}& x\\ge 0\\\\ -x& \\text{if}& x<0\\end{array}\\,$, $\\begin{array}{cccc}\\hfill f\\left(x\\right)& =& 2|x-3|-2,\\hfill & \\phantom{\\rule{1em}{0ex}}\\text{treating the stretch as }a\\text{ vertical stretch,or}\\hfill \\\\ \\hfill f\\left(x\\right)& =& |2\\left(x-3\\right)|-2,\\hfill & \\phantom{\\rule{1em}{0ex}}\\text{treating the stretch as }a\\text{ horizontal compression}.\\hfill \\end{array}$, $\\begin{array}{ccc}\\hfill 2& =& a|1-3|-2\\hfill \\\\ \\hfill 4& =& 2a\\hfill \\\\ \\hfill a& =& 2\\hfill \\end{array}$, $\\begin{array}{ccccccc}\\hfill 2x-6& =& 8\\hfill & \\phantom{\\rule{1em}{0ex}}\\text{or}\\phantom{\\rule{1em}{0ex}}& \\hfill 2x-6& =& -8\\hfill \\\\ \\hfill 2x& =& 14\\hfill & & \\hfill 2x& =& -2\\hfill \\\\ \\hfill x& =& 7\\hfill & & \\hfill x& =& -1\\hfill \\end{array}$, $\\begin{array}{l}|x|=4,\\hfill \\\\ |2x-1|=3,\\text{or}\\hfill \\\\ |5x+2|-4=9\\hfill \\end{array}$, $\\begin{array}{cccccccc}\\hfill 0& =& |4x+1|-7\\hfill & & & & & \\text{Substitute 0 for }f\\left(x\\right).\\hfill \\\\ \\hfill 7& =& |4x+1|\\hfill & & & & & \\text{Isolate the absolute value on one side of the equation}.\\hfill \\\\ & & & & & & & \\\\ & & & & & & & \\\\ & & & & & & & \\\\ \\hfill 7& =& 4x+1\\hfill & \\text{or}& \\hfill \\phantom{\\rule{2em}{0ex}}-7& =& 4x+1\\hfill & \\text{Break into two separate equations and solve}.\\hfill \\\\ \\hfill 6& =& 4x\\hfill & & \\hfill -8& =& 4x\\hfill & \\\\ & & & & & & & \\\\ \\hfill x& =& \\frac{6}{4}=1.5\\hfill & & \\hfill x& =& \\frac{-8}{4}=-2\\hfill & \\end{array}$, $\\left(0,-4\\right),\\left(4,0\\right),\\left(-2,0\\right)$, $\\left(0,7\\right),\\left(25,0\\right),\\left(-7,0\\right)$, http://cnx.org/contents/13ac107a-f15f-49d2-97e8-60ab2e3b519c@11.1, Use$\\,|A|=B\\,$to write$\\,A=B\\,$or$\\,\\mathrm{-A}=B,\\,$assuming$\\,B>0. Students who score within 18 points of the number 82 will pass a particular test. An absolute value function is a function that contains an algebraic expression within absolute value symbols. Note. This leads to two different equations we can solve independently. In an absolute value equation, an unknown variable is the input of an absolute value function. Step 1: Find zeroes of the given absolute value function. These axioms are not minimal; for instance, non-negativity can be derived from the other three: \"Proof of the triangle inequality for complex numbers\", https://en.wikipedia.org/w/index.php?title=Absolute_value&oldid=1000931702, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, Preservation of division (equivalent to multiplicativity), Positive homogeneity or positive scalability, This page was last edited on 17 January 2021, at 12:08. 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Set of numbers using absolute value when first learning the topic linear, convex function, its. Axis at two points for determining the horizontal axis at two points, then the absolute value is... Function returns absolute value of an absolute value function getting an absolute is... Problems, such as ranges of possible values, can also be a floating-point.. The unknown variable a absolute value function number, is the value without regard to sign solve involving... \\Displaystyle \\mathbb { R } ^ { 1 } { 2 } |x+4|-3 [ /latex ] from! To show how much a value deviates from the norm values of x and find ordered... ) function returns absolute value function, this point is shown at the origin in ( Figure ) an... Learn how to solve some kinds of real-world problems the formula for an absolute value equation an! From zero on a number is from zero on a number or....", "date": "2021-06-15 17:12:00", "meta": {"domain": "elespiadigital.org", "url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function", "openwebmath_score": 0.6666929721832275, "openwebmath_perplexity": 666.4102091724513, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9920620049598216, "lm_q2_score": 0.9086178938396674, "lm_q1q2_score": 0.9014052895049507}}
{"url": "https://math.stackexchange.com/questions/1922013/are-there-primes-of-every-possible-number-of-digits/1922015", "text": "# Are there primes of every possible number of digits?\n\nThat is, is it the case that for every natural number $n$, there is a prime number of $n$ digits? Or, is there some $n$ such that no primes of $n$-digits exist?\n\nI am wondering this because of this Project Euler problem: https://projecteuler.net/problem=37. I find it very surprising that there are only a finite number of truncatable primes (and even more surprising that there are only 11)! However, I was thinking that result would make total sense if there is an $n$ such that there are no $n$-digit primes, since any $k$-digit truncatable prime implies the existence of at least one $n$-digit prime for every $n\\leq k$.\n\nIf not, does anyone have insight into an intuitive reason why there are finitely many trunctable primes (and such a small number at that)? Thanks!\n\n\u2022 Anyway, yes: for all $n$ there are a lot of primes having $n$ digits. Sep 10, 2016 at 22:34\n\u2022 Bertrand's postulate (an ill-chosen name) says there is always a prime strictly between $n$ and $2n$ for $n\\gt 1$. Sep 10, 2016 at 23:03\n\u2022 Think about the reverse. If you have an $n$-digit prime, how many 'chances' do you have to extend it to an $(n+1)$-digit prime? The odds being able to do so quickly turn against you.\n\u2013\u00a0user14972\nSep 11, 2016 at 0:18\n\u2022 Just a side-comment - ..and even more surprising that there are only 11 ...Maybe I am wrong regrading your meaning , but I believe that there are 15 ( and not 11 ) both-sides truncatable primes .. Wiki entry on truncatable prime. Sep 11, 2016 at 3:05\n\u2022 Via the Wikipedia article I found M. El Bachraoui's 2006 [\"Primes in the Interval [2n, 3n]\"](m-hikari.com/ijcms-password/ijcms-password13-16-2006/\u2026) which relies on elementary methods and three results from an undergraduate textbook. Sep 12, 2016 at 1:52\n\n## 2 Answers\n\nYes, there is always such a prime. Bertrand's postulate states that for any $k>3$, there is a prime between $k$ and $2k-2$. This specifically means that there is a prime between $10^n$ and $10\\cdot 10^n$.\n\nTo commemorate $50$ upvotes, here are some additional details: Bertrand's postulate has been proven, so what I've written here is not just conjecture. Also, the result can be strengthened in the following sense (by the prime number theorem): For any $\\epsilon > 0$, there is a $K$ such that for any $k > K$, there is a prime between $k$ and $(1+\\epsilon)k$. For instance, for $\\epsilon = 1/5$, we have $K = 24$ and for $\\epsilon = \\frac{1}{16597}$ the value of $K$ is $2010759$ (numbers gotten from Wikipedia).\n\n\u2022 With the side note that Bertrand's postulate is a (proved) theorem. Sep 11, 2016 at 22:29\n\u2022 Just another note: those interested in this sort of thing should look for papers by Pierre Dusart - he has proven many of the best approximations of this form. Sep 14, 2016 at 3:09\n\nWhile the answer using Bertrand's postulate is correct, it may be misleading. Since it only guarantees one prime between $N$ and $2N$, you might expect only three or four primes with a particular number of digits. This is very far from the truth.\n\nThe primes do become scarcer among larger numbers, but only very gradually. An important result dignified with the name of the Prime Number Theorem'' says (roughly) that the probability of a random number of around the size of $N$ being prime is approximately $1/\\ln(N)$.\n\nTo take a concrete example, for $N = 10^{22}$, $1/\\ln(N)$ is about $0.02$, so one would expect only about $2\\%$ of $22$-digit numbers to be prime. In some sense, $2\\%$ is small, but since there are $9\\cdot 10^{21}$ numbers with $22$ digits, that means about $1.8\\cdot 10^{20}$ of them are prime; not just three or four! (In fact, there are exactly $180,340,017,203,297,174,362$ primes with $22$ digits.)\n\nIn short, the number of $n$-digit numbers increases with $n$ much faster than the density of primes decreases, so the number of $n$-digit primes increases rapidly as $n$ increases.\n\n\u2022 The prime number theorem will give you a bound on the number of primes between $10^n$ and $10^{n+1}$. But is the bound tight enough to prove that the number of such primes is a strictly growing function of $n$? Sep 11, 2016 at 8:33\n\u2022 @kasperd There are some known (explicit) estimates on the error term in the prime number theorem, I can imagine they are strong enough to show this, albeit possibly only for large $n$. Sep 11, 2016 at 8:50\n\u2022 The prime number theorem on its own would allow for very large gaps between primes, but not so large that there are no primes between $10^n$ and $10^{n+1}$ when n is large enough. On the other hand, it is a limit, so it says nothing about small primes. Sep 11, 2016 at 18:49\n\u2022 The bounds from Wikipedia $\\frac{x}{\\log x + 2} < \\pi(x) < \\frac{x}{\\log x - 4}$ for $x> 55$ can be used to show that there is always a prime with $n$ digits for $n\\ge 3$. Oct 13, 2020 at 13:22", "date": "2022-05-17 08:22:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1922013/are-there-primes-of-every-possible-number-of-digits/1922015", "openwebmath_score": 0.8959529399871826, "openwebmath_perplexity": 153.28106886512555, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9905874115238967, "lm_q2_score": 0.9099069974872589, "lm_q1q2_score": 0.9013424173683846}}
{"url": "https://math.stackexchange.com/questions/2863671/how-many-poker-hands-have-two-pairs", "text": "# How many poker hands have two pairs?\n\nI'm trying to calculate how many poker hands called Two Pair, there are. Such a hand consists of one pair of one rank, another pair of another rank and one card of a third rank. A poker hand consists of 5 cards.\n\nI have two methods that I thought would work equally well. Turns out only one of them yields the correct answer. I was wondering if anyone here knows why the second solution gives the wrong answer.\n\nSolution 1 (Correct):\n\nWe choose 2 ranks out of 13, which can be done in $\\binom{13}{2}$ ways.\n\nFor the first rank we choose 2 suits out of 4, which can be done in $\\binom{4}{2}$ ways.\n\nFor the second rank we choose 2 suits out of 4, which can be done in $\\binom{4}{2}$ ways.\n\nThe last card can be chosen in $44$ different ways.\n\nSo the total number of hands is $\\binom{13}{2}\\cdot \\binom{4}{2}\\cdot \\binom{4}{2}\\cdot 44=123,552$\n\nSolution 2 (Incorrect):\n\nWe choose 3 ranks out of 13, which can be done in $\\binom{13}{3}$ ways.\n\nFor the first rank we choose 2 suits out of 4, which can be done in $\\binom{4}{2}$ ways.\n\nFor the second rank we choose 2 suits out of 4, which can be done in $\\binom{4}{2}$ ways.\n\nFor the third rank we choose 1 suit out of 4, which can be done in $4$ ways.\n\nSo the total number of hands is $\\binom{13}{3}\\cdot \\binom{4}{2}\\cdot \\binom{4}{2}\\cdot 4=41,184$\n\nThis is just a third of the correct number of hands. Why the second solution is wrong unfortunately seems to elude me......\n\n\u2022 In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3. \u2013\u00a0Angina Seng Jul 26 '18 at 18:23\n\u2022 @LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked. \u2013\u00a0Ross Millikan Jul 26 '18 at 18:35\n\u2022 Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too. \u2013\u00a0amWhy Jul 26 '18 at 18:56\n\u2022 Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton? \u2013\u00a0Stargazer Jul 26 '18 at 19:02\n\u2022 Never mind, I understand it now that I thought some more about it. \u2013\u00a0Stargazer Jul 26 '18 at 19:12\n\nAfter you've chosen which three ranks are in the hand, you need to choose either (a) which two of the three ranks to make the pairs, or (b) which one of the ranks to make the singleton. The number of ways to do these are ${3 \\choose 2}$ and ${3 \\choose 1}$, respectively, and of course each equals $3$.\n\nAssume your choose three ranks R1, R2, R3.\n\nIn the first solution: You choose (R1, R2) first $\\binom{13}{2}$ then assign the suits. Lastly, you choose R3 from the rest (remaining 44). The reason why $\\binom{13}{2}$ is chosen to avoid double count since order does not matter for this two ranks: as (R1,R1,R2,R2,R3) is the same as (R2,R2,R1,R1,R3). (writing R1 R1 meaning two suits rank 1, pardon my laziness). This is correct.\n\nIn the second solution:\n\nYou choose (R1,R2,R3) first then assign the suits. But using $\\binom{13}{3}$ means the order of the three ranks do not matter. In other words, you are treating the three hands (R1,R1,R2,R2,R3), (R3,R3,R2,R2,R1), (R1,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times.\n\nYour first method is to count ways to choose two from thirteen ranks for the pairs, two from four suits for each of those, and one from fourty-four cards to be the loner (or one from eleven ranks and one from four suits). \u00a0 That is okay. $$\\binom{13}2\\binom 42^2\\binom{44}1 \\\\\\binom{13}2\\binom 42^2\\binom {11}1\\binom 41$$\n\nYour second method is to count ways to choose three from thirteen ranks, two from four suits for the pairs, one from four suits for the singleton, and\u2014wait\u2014which two from those three selected ranks are to be the pairs? \u00a0 Ah, that is better.$$\\binom {13}3\\binom 42^2\\binom 41\\binom 32$$\n\n...and of course $\\binom{13}{3}\\binom 32=\\frac{13!}{3!10!}\\frac{3!}{2!1!}=\\frac{13!}{2!11!}\\frac{11!}{10!1!}=\\binom {13}2\\binom{11}1$", "date": "2020-05-30 18:51:26", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/2863671/how-many-poker-hands-have-two-pairs", "openwebmath_score": 0.6769075393676758, "openwebmath_perplexity": 327.0273536874462, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9934102283369195, "lm_q2_score": 0.9073122219871936, "lm_q1q2_score": 0.9013332416171759}}
{"url": "https://www.physicsforums.com/threads/integration-problem.158225/", "text": "# Integration Problem\n\n## Homework Statement\n\n$$\\int {\\frac{sin^{2}x}{1+sin^{2}x}dx}$$\n\n## Homework Equations\n\nLet t = tan x/2, then dx = 2/(1+t^2) and sin x = 2t / (1+t^2)\n\n## The Attempt at a Solution\n\nI got up to the point where $$\\int {\\frac{8t^{2}}{(1+6t^{2}+t^{4})(1+t^{2})} dt}$$. Not sure if I'm on the right track and if I am, do I use partial fractions after this?\n\nThe final answer is attached. Can't really make out the handwriting :/\n\n#### Attachments\n\n8.9 KB \u00b7 Views: 266\nLast edited:\n\nYes, it looks like partial fractions is the way to go after your substitution.\n\nHmm after I do partial fractions, I get\n$$\\int {\\frac{2}{1+t^{2}} + {\\frac{-2t^{2}-2}{(1+6t^{2}+t^{4})} dt}$$\n\nAfter this, I do not know what's the next step. Kindly advise. Thanks.\n\nyou are summing 2 functions of t , one of these two look very much like a derivative of a certain function..\n\nIf you mean 2tan^-1 t, I can get this part. But what about the 2nd function?", "date": "2021-04-21 03:09:55", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/integration-problem.158225/", "openwebmath_score": 0.8774517774581909, "openwebmath_perplexity": 777.9312771301906, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 0.9863631679255076, "lm_q2_score": 0.9136765157744067, "lm_q1q2_score": 0.9012168625583837}}
{"url": "http://mathhelpforum.com/discrete-math/130610-set-odd-integers-proof-print.html", "text": "# set of odd integers proof\n\n\u2022 Feb 24th 2010, 02:21 PM\njames121515\nset of odd integers proof\nI am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?\n\n$\\mbox{If }A=\\{x \\in \\mathbb{Z}~|~x = 2k+1\\mbox{ for some }k \\in \\mathbb{Z}\\}$ and $B=\\{y \\in \\mathbb{Z}~|~y=2s-1\\mbox{ for some }s \\in \\mathbb{Z}\\}$, prove that $A=B$\n\n$\\mbox{\\textbf{Proof.}}$ Let $x\\in A$. then $\\exists~k \\in \\mathbb{Z}\\mbox{ such that }x=2k+1$. Equivalently,\n$\\Longrightarrow x=2k+1+1-1$\n$\\Longrightarrow x=2k+2-1$\n$\\Longrightarrow x=2(k+1)-1$\n\nSince $k \\in\\mathbb{Z} \\Longrightarrow k+1 \\in \\mathbb{Z}$ $x = 2(k+1)-1 \\Longrightarrow x \\in B$. Therefore, $A\\subseteq B$\n\u2022 Feb 24th 2010, 02:36 PM\nPlato\nQuote:\n\nOriginally Posted by james121515\nI am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?\n\n$\\mbox{If }A=\\{x \\in \\mathbb{Z}~|~x = 2k+1\\mbox{ for some }k \\in \\mathbb{Z}\\}$ and $B=\\{y \\in \\mathbb{Z}~|~y=2s-1\\mbox{ for some }s \\in \\mathbb{Z}\\}$, prove that $A=B$\n\n$\\mbox{\\textbf{Proof.}}$ Let $x\\in A$. then $\\exists~k \\in \\mathbb{Z}\\mbox{ such that }x=2k+1$. Equivalently,\n$\\Longrightarrow x=2k+1+1-1$\n$\\Longrightarrow x=2k+2-1$\n$\\Longrightarrow x=2(k+1)-1$\n\nSince $k \\in\\mathbb{Z} \\Longrightarrow k+1 \\in \\mathbb{Z}$ $x = 2(k+1)-1 \\Longrightarrow x \\in B$. Therefore, $A\\subseteq B$\n\nThat 'way' is correct.\nBy symmetry you are done.\n\n$x=2k+1=2(k+1)-1$\n\u2022 Feb 24th 2010, 09:36 PM\njames121515\n\nSo you are saying that due to symmetry, there is no need to show the other \"right to left\" containment due to symmetry?\n\n-James\n\u2022 Feb 25th 2010, 08:21 AM\nPlato\nQuote:\n\nOriginally Posted by james121515\n$x=2k+1=2(k+1)-1$", "date": "2017-10-18 00:16:39", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/discrete-math/130610-set-odd-integers-proof-print.html", "openwebmath_score": 0.957656979560852, "openwebmath_perplexity": 722.7092434974562, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9890130551025345, "lm_q2_score": 0.9111797160416689, "lm_q1q2_score": 0.9011686347098308}}
{"url": "https://math.stackexchange.com/questions/3570010/what-is-the-general-solution-to-the-equation-sin-x-sqrt3-cos-x-sqrt2", "text": "# What is the general solution to the equation $\\sin x + \\sqrt{3}\\cos x = \\sqrt2$\n\nI need to find the general solution to the equation\n\n$$\\sin(x) + \\sqrt3\\cos(x)=\\sqrt2$$\n\nSo I went ahead and divided by $$2$$, thus getting the form\n\n$$\\cos(x-\\frac{\\pi}{6})=\\cos(\\frac{\\pi}{4})$$\n\nThus the general solution to this would be $$x = 2n\\pi \\pm\\frac{\\pi}{4}+\\frac{\\pi}{6}$$\n\nWhich simplifies out to be,\n\n$$x = 2n\\pi +\\frac{5\\pi}{12}$$ $$x = 2n\\pi -\\frac{\\pi}{12}$$\n\nBut the answer doesn't have the 2nd solution as a solution to the given equation. Did I go wrong somewhere?\n\n\u2022 Your answer seems to be the correct one. For example $x=-\\frac {\\pi} {12}$ does satisfy the given equation. Mar 5 '20 at 6:39\n\u2022 You solution is correct. May be they skip the second one. Mar 5 '20 at 6:43\n\nAs Kavi Rama Murthy's comment indicates, you haven't done anything wrong that I can see. You can quite easily very that $$x = 2n\\pi - \\frac{\\pi}{12}$$ is a solution (coming from using $$\\cos\\left(-\\frac{\\pi}{4}\\right)$$ on the right), as well as the first one you specify of $$x = 2n\\pi + \\frac{5\\pi}{12}$$ (coming from using $$\\cos\\left(\\frac{\\pi}{4}\\right)$$ on the right). Thus, it seems the answer has an oversight.\n\u2022 @Techie5879 Unless there's some stated restriction on what $x$ could be, they are both valid options, so it seems the multiple-choice test has a mistake in it. Mar 5 '20 at 6:42", "date": "2021-10-19 06:32:02", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3570010/what-is-the-general-solution-to-the-equation-sin-x-sqrt3-cos-x-sqrt2", "openwebmath_score": 0.867301881313324, "openwebmath_perplexity": 158.98612471699494, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750510899383, "lm_q2_score": 0.9124361563100185, "lm_q1q2_score": 0.9009166964529114}}
{"url": "https://math.stackexchange.com/questions/1277038/why-is-1-i-equal-to-i/1277042", "text": "# Why is $1/i$ equal to $-i$?\n\nWhen I entered the value $$\\frac{1}{i}$$ in my calculator, I received the answer as $-i$ whereas I was expecting the answer as $i^{-1}$. Even google calculator shows the same answer (Click here to check it out).\n\nIs there a fault in my calculator or $\\frac{1}{i}$ really equals $-i$? If it does then how?\n\n\u2022 Hint $i^2 = -1$\n\u2013\u00a0Mann\nMay 11, 2015 at 12:14\n\u2022 Multiply by $i/i$. May 11, 2015 at 12:14\n\u2022 Hint $$z=\\frac{1}{i}\\iff zi=1\\implies \\dots$$ May 11, 2015 at 12:56\n\u2022 Three down votes for someone exhibiting natural mathematical curiosity and having the wherewithal to ask about it is shameful. May 11, 2015 at 14:50\n\u2022 Excellent question I wondered that myself when I read it. I could say $+1$ but given the context of the question I should say $+i$! May 13, 2015 at 1:04\n\n$$\\frac{1}{i}=\\frac{i}{i^2}=\\frac{i}{-1}=-i$$\n\nNote that $i(-i)=1$. By definition, this means that $(1/i)=-i$.\n\nThe notation \"$i$ raised to the power $-1$\" denotes the element that multiplied by $i$ gives the multiplicative identity: $1$.\n\nIn fact, $-i$ satisfies that since\n\n$$(-i)\\cdot i= -(i\\cdot i)= -(-1) =1$$\n\nThat notation holds in general. For example, $2^{-1}=\\frac{1}{2}$ since $\\frac{1}{2}$ is the number that gives $1$ when multiplied by $2$.\n\n\u2022 I appreciate that this answer gives context to the calculation. +1 ! May 11, 2015 at 15:04\n\nThere are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the \"simplest\" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example.\n\nThe complex numbers are a field. This means that every non-$0$ element has a multiplicative inverse, and that inverse is unique.\n\nWhile $1/i = i^{-1}$ is true (pretty much by definition), if we have a value $c$ such that $c * i = 1$ then $c = i^{-1}$.\n\nThis is because we know that inverses in the complex numbers are unique.\n\nAs it happens, $(-i) * i = -(i*i) = -(-1) = 1$. So $-i = i^{-1}$.\n\nAs fractions (or powers) are usually considered \"less simple\" than simple negation, when the calculator displays $i^{-1}$ it simplifies it to $-i$.\n\n$-i$ is the multiplicative inverse of $i$ in the field of complex numbers, i.e. $-i * i = 1$, or $i^{-1} = -i$.\n\n$$\\frac{1}{i}=\\frac{i^4}{i}=i^3=i^2\\cdot i = -i$$\n\nI always like to point out that this fits well into a pattern you see when \"rationalising the denominator\", if the denominator is a root: $$\\frac{1}{\\sqrt{2}} = \\frac{1}{\\sqrt{2}}\\cdot \\frac{\\sqrt{2}}{\\sqrt{2}} = \\frac{1}{2}\\sqrt{2}$$ $$\\frac{1}{\\sqrt{17}} = \\frac{1}{\\sqrt{17}}\\cdot \\frac{\\sqrt{17}}{\\sqrt{17}} = \\frac{1}{17}\\sqrt{17}$$ $$\\frac{1}{\\sqrt{a}} = \\frac{1}{\\sqrt{a}}\\cdot \\frac{\\sqrt{a}}{\\sqrt{a}} = \\frac{1}{a}\\sqrt{a}$$ $$\\frac{1}{i} = \\frac{1}{\\sqrt{-1}} = \\frac{1}{\\sqrt{-1}}\\cdot \\frac{\\sqrt{-1}}{\\sqrt{-1}} = \\frac{1}{-1}\\sqrt{-1} = - i.$$ In this vein, it is almost more suggestive to write $$\\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2}$$ $$\\frac{1}{\\sqrt{17}} = \\frac{\\sqrt{17}}{17}$$ $$\\frac{1}{i} = \\frac{i}{-1}.$$\n\nBy the definition of the inverse $$\\frac1i\\cdot i=1.$$\n\nThis agrees with\n\n$$(-i)\\cdot i=1.$$\n\nAny complex number is fully described by its magnitude and phase (argument) via the complex exponential.\n\n$$X = |X|e^{i\\arg{X}}$$\n\nIt is useful to write complex numbers in this form when multiplying and dividing as we can make use of exponent rules. Division in this instance simplifies to dividing the magnitudes and subtracting the phases.\n\nBefore we compute this division, lets calculate the magnitude and phase of $$1$$ and $$i$$. It is quite obvious that the magnitudes of both numbers are $$1$$ (i.e. $$|1|=|i|=1$$). And by definition the phases are:\n\n$$\\arg{1} = 0$$ $$\\arg{i} = \\frac{\\pi}{2}$$\n\nOur two complex exponentials are therefore:\n\n$$1 = e^{i0}$$ $$i = e^{i\\frac{\\pi}{2}}$$\n\nNow we perform the division making use of the exponent rules:\n\n$$\\frac{1}{i}=\\frac{e^{i0}}{e^{i\\frac{\\pi}{2}}}=e^{-i\\frac{\\pi}{2}}$$\n\nIf you consult the unit circle (since the magnitude is 1), you will find that a phase of $$-\\frac{\\pi}{2}$$ corresponds to $$\u2212i$$. Alternatively you can apply Euler's formula:\n\n$$e^{-i\\frac{\\pi}{2}} = \\cos\\left(-\\frac{\\pi}{2}\\right) +i\\sin\\left(-\\frac{\\pi}{2}\\right) =-i$$\n\nI want to add the method that I like.\n\n$$\\frac{1}{i}$$ $$=\\frac{1}{cis(\\frac{\\pi}{2})}$$ $$= cis(- \\frac{\\pi}{2})$$ $$=-i$$\n\nWhere $$cis(x)= \\cos(x)+i \\sin(x)$$", "date": "2023-03-31 20:02:43", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1277038/why-is-1-i-equal-to-i/1277042", "openwebmath_score": 0.9456176161766052, "openwebmath_perplexity": 257.71496231595404, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9873750477464142, "lm_q2_score": 0.9124361527383703, "lm_q1q2_score": 0.9009166898756028}}
{"url": "https://brilliant.org/discussions/thread/sum-of-harmonic-series/", "text": "# Sum of Harmonic Series\n\nIt is well known that the sum of a harmonic series does not have a closed form. Here is a formula which gives us a good approximation.\n\nWe need to find the sum of the following series\n\n$\\dfrac{1}{a}+\\dfrac{1}{a+d}+\\dfrac{1}{a+2d}+\\ldots+\\dfrac{1}{a+(n-1)d}$\n\nConsider the function $$f(x)=\\frac{1}{x}$$, we intend to take middle riemann sums with rectangles of width $$d$$ starting from $$x=a$$ to $$x=a+(n-1)d$$.\n\nEach rectangle in the figure has a width $$d$$. The height of the $$i\\text{th}$$ rectangle is $$\\frac{1}{a+(i-1)d}$$. The sum of the area of the rectangles is approximately equal to the area under the curve.\n\nArea under f(x) from $$x=a-\\frac{d}{2}$$ to $$x=a+\\left(n-\\frac{1}{2}\\right)d \\approx\\displaystyle\\sum_{n=1}^{n} \\frac{d}{a+(n-1)d}$$\n\n$\\large\\Rightarrow \\int_{a-\\frac{d}{2}}^{a+\\left(n-\\frac{1}{2}\\right)d} \\dfrac{\\mathrm{d}x}{x}\\approx \\displaystyle\\sum_{n=1}^{n} \\frac{d}{a+(n-1)d}$\n\nLet $$S_n =\\displaystyle\\sum_{n=1}^{n} \\frac{1}{a+(n-1)d}$$\n\n$\\large\\ln\\dfrac{2a+(2n-1)d}{2a-d}\\approx d\\times S_n$\n\n$\\large\\boxed{\\Rightarrow s_n\\approx\\dfrac{\\ln\\dfrac{2a+(2n-1)d}{2a-d}}{d}}$\n\nNote\n\n\u2022 Apologies for the shabby graph.\n\n\u2022 $$d\\neq 0$$\n\nNote by Aneesh Kundu\n2\u00a0years, 6\u00a0months ago\n\nMarkdownAppears as\n*italics* or _italics_ italics\n**bold** or __bold__ bold\n- bulleted- list\n\u2022 bulleted\n\u2022 list\n1. numbered2. list\n1. numbered\n2. list\nNote: you must add a full line of space before and after lists for them to show up correctly\nparagraph 1paragraph 2\n\nparagraph 1\n\nparagraph 2\n\n[example link](https://brilliant.org)example link\n> This is a quote\nThis is a quote\n    # I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\n# I indented these lines\n# 4 spaces, and now they show\n# up as a code block.\n\nprint \"hello world\"\nMathAppears as\nRemember to wrap math in $$...$$ or $...$ to ensure proper formatting.\n2 \\times 3 $$2 \\times 3$$\n2^{34} $$2^{34}$$\na_{i-1} $$a_{i-1}$$\n\\frac{2}{3} $$\\frac{2}{3}$$\n\\sqrt{2} $$\\sqrt{2}$$\n\\sum_{i=1}^3 $$\\sum_{i=1}^3$$\n\\sin \\theta $$\\sin \\theta$$\n\\boxed{123} $$\\boxed{123}$$\n\nSort by:\n\n@Aneesh Kundu I have just added your formula to Harmonic Progression wiki. I have also added important points from your discussion with Atul. You can also contribute to the wiki.\n\n- 2\u00a0years, 1\u00a0month ago\n\nHow can area under that curve=d\u00d7Sn ??here d is denoted as width\n\n- 2\u00a0years, 2\u00a0months ago\n\nArea under the curve $A=\\frac{1}{a} d +\\frac{1}{a+d} d+ \\frac{1}{a+2d} d \\ldots$ $\\frac{A}{d}= ( \\frac{1}{a}+ \\frac{1}{a+d} \\ldots )$ $A=d\\dot S_{n}$\n\n- 2\u00a0years, 2\u00a0months ago\n\nOhoo now i understand clearly. Thanks bro...\n\n- 2\u00a0years, 2\u00a0months ago\n\nyour above expression will be incorrect when $\\boxed{2a=d}$\n\n- 2\u00a0years, 5\u00a0months ago\n\nIn this case calculate the sum from $$a_2$$ to $$a_n$$, using the given formula and then add $$a_1$$ to both sides.\n\n- 2\u00a0years, 5\u00a0months ago\n\nHey how you have assigned limit of $$x$$ can you please clarify\n\n- 2\u00a0years, 6\u00a0months ago\n\n$$x$$ varies from $$a-\\frac{d}{2}$$ to $$a+\\left(n-\\frac{1}{2}\\right)d$$.\n\n- 2\u00a0years, 5\u00a0months ago\n\nyaa ,I got this but also you can't use this formula for finding sum of similar terms\n\ni.e. $$S_n= \\frac {1}{2}+ \\frac {1}{2}+ \\frac {1}{2}+ \\frac {1}{2}+.... \\frac {1}{2}(n^{th} term)$$ as common difference is $$0$$ so it will be in indeterminate form\n\n- 2\u00a0years, 5\u00a0months ago\n\nThanks for the suggestion, I added this point in the note.\n\n- 2\u00a0years, 5\u00a0months ago\n\nI mean to is it original(your own)???\n\n- 2\u00a0years, 5\u00a0months ago\n\nNope, its not purely original. I was reading about the convergence tests and I happened came across the Integral test, which inspired this note.\n\n- 2\u00a0years, 5\u00a0months ago\n\nit's really fantastic\n\n- 2\u00a0years, 5\u00a0months ago\n\nThanks. :)\n\n- 2\u00a0years, 5\u00a0months ago\n\nby the way is it real???\n\n- 2\u00a0years, 5\u00a0months ago\n\nThis formula gives really good approximations when $$d\\rightarrow 0$$ or for large values of $$n$$ with a not so big $$d$$ .\n\n- 2\u00a0years, 5\u00a0months ago\n\nJust followed you :-)\n\n- 2\u00a0years, 5\u00a0months ago\n\nYes Absolutely\n\n- 2\u00a0years, 5\u00a0months ago\n\nThanks! This is a very good and useful note.\n\n- 2\u00a0years, 6\u00a0months ago\n\nThanks. :)\n\n- 2\u00a0years, 5\u00a0months ago", "date": "2018-04-20 07:07:02", "meta": {"domain": "brilliant.org", "url": "https://brilliant.org/discussions/thread/sum-of-harmonic-series/", "openwebmath_score": 0.9984549880027771, "openwebmath_perplexity": 2927.9678096791167, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes", "lm_q1_score": 0.9748211626883622, "lm_q2_score": 0.9241418173671895, "lm_q1q2_score": 0.9008730008948197}}
{"url": "https://math.stackexchange.com/questions/1398083/are-there-many-different-power-series-representation-for-a-given-function", "text": "# Are there many different power series representation for a given function?\n\nSo I have to find the power series representation for $f(x) = \\ln (3-x)$.\n\nI attempted the following:\n\n$$\\ln(3-x) = \\int {- \\frac{1}{3-x} dx}$$ $$= - \\int { \\frac{1}{1-(x-2)} dx}$$ $$= - \\int {\\sum_{n=0}^{\\infty}{(x-2)^n} dx}$$ $$= \\sum_{n=0}^{\\infty} {\\int(x-2)^ndx}$$ $$= \\bigg(-\\sum_{n=0}^{\\infty} \\frac{(x-2)^{n+1}}{n+1}\\bigg)+K$$\n\nThen if we let $x=2$, then we obtain that $K=0$. Hence the power series representation for $f(x)$ is $-\\sum_{n=0}^{\\infty} \\frac{(x-2)^{n+1}}{n+1}$, where $|x-2|<1$.\n\nHowever the answer from my lecturer is given as: $$\\ln(3)-\\sum_{n=1}^{\\infty}{\\frac{x^n}{n\\cdot3^n}}$$\n\nAm I doing a mistake? Or are there many different power series representation for a given function? Any clarification would be highly appreciated.\n\n\u2022 It depends where you want to center your power series. Setting a given center, the power series representation is unique (and it exists for an holomorphic function). \u2013\u00a0Paolo Leonetti Aug 15 '15 at 12:24\n\u2022 @PaoloLeonetti thanks for your explanation! that makes perfect sense. however, the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well? \u2013\u00a0Aaron Aug 15 '15 at 12:36\n\u2022 In a word, yes :) Ps. How do you justify the exachange of infinite summation and integral? \u2013\u00a0Paolo Leonetti Aug 15 '15 at 12:36\n\u2022 @PaoloLeonetti is that because we are allowed to do term-by-term integration? \u2013\u00a0Aaron Aug 15 '15 at 12:41\n\u2022 \" the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well?\" In a word, no because when the center is not specified one is supposed to understand the center is zero. (Additionnally, in some curricula the only admissible center is zero.) \u2013\u00a0Did Aug 16 '15 at 15:59\n\nHint. Your route is OK, but you should rather start with $$\\ln(3-x) = -\\int_0^x { \\frac{1}{3-t} dt}+\\ln 3$$ then follow the same path to obtain the right answer.\n\n\u2022 Thanks! I followed this and ended up in the same form. however, say in an exam i wrote like the above, would it be correct though? \u2013\u00a0Aaron Aug 15 '15 at 12:36\n\nBoth series are correct. The one from the lecture is the series expansion around $x=0$, while the one derived in the posted question is the series expansion around $x=2$. And one could choose other arbitrary points around which to expand the function.\n\nUsing a straightforward approach we see that for $f(x)=\\log(3-x)$, we have for $n>0$\n\n$$f^{(n)}(x)=(-1)^{n+1}(n-1)!(x-3)^{-n} \\tag 1$$\n\nWe will use this in Approach 2 of the expansions around both $x=0$ and $x=3$ in that which follows.\n\nEXPANSION AROUND $x=0$\n\nApproach 1: Using the approach outlined in the posted question, we find that\n\n\\begin{align} \\log(3-x)&=-\\int_2^x \\frac{1}{3-t}dt\\\\\\\\ &=-\\int_2^x\\frac{1}{1-(t-2)}dt\\\\\\\\ &=-\\sum_{n=0}^{\\infty}\\int_0^x (t-2)^n\\\\\\\\ &=-\\sum_{n=1}^{\\infty}\\frac{(x-2)^n}{n} \\end{align}\n\nwhich converges for $-1\\le x<3$ and diverges otherwise.\n\nApproach 2: From $(1)$, we can see that $f^{(n)}(2)=(-1)^{n+1}(n-1)!(-1)^{-n}=-(n-1)!$\n\nTherefore, we can write the series representation as\n\n$$\\log(3-x)=-\\sum_{n=1}^{\\infty}\\frac{(x-2)^n}{n}$$\n\nwhich converges for $-1\\le x<3$ and diverges otherwise as expected!\n\nEXPANSION AROUND $x=3$\n\nApproach 1: Using the approach outlined in the posted question, we find that\n\n\\begin{align} \\log(3-x)&=\\log 3-\\int_0^x \\frac{1}{3-t}dt\\\\\\\\ &=\\log 3-\\frac13\\int_0^x\\frac{1}{1-(t/3)}dt\\\\\\\\ &=\\log 3-\\frac13\\sum_{n=0}^{\\infty}\\int_0^x (t/3)^n\\\\\\\\ &=\\log 3-\\sum_{n=1}^{\\infty}\\frac{x^n}{n3^n} \\end{align}\n\nwhich converges for $-3\\le x<3$ and diverges otherwise.\n\nApproach 2: From $(1)$, we can also see that $f^{(n)}(0)=(-1)^{n+1}(n-1)!(-3)^{-n}=-\\frac{(n-1)!}{3^n}$.\n\nTherefore, we can write the series representation as\n\n$$f(x)=\\log 3-\\sum_{n=1}^{\\infty}\\frac{x^n}{n3^n}$$\n\nwhich converges for $-3\\le x<3$ and diverges otherwise as expected!\n\n\u2022 Please let me know how I can improve my answer. I really just want to give you the best answer I can. \u2013\u00a0Mark Viola Aug 17 '15 at 15:10", "date": "2021-08-02 13:48:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/1398083/are-there-many-different-power-series-representation-for-a-given-function", "openwebmath_score": 0.944172739982605, "openwebmath_perplexity": 396.303447041935, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.988668248138067, "lm_q2_score": 0.9111797124237604, "lm_q1q2_score": 0.9008544500209469}}
{"url": "https://math.stackexchange.com/questions/4257386/given-the-gcd-and-lcm-of-n-positive-integers-how-many-solutions-are-there", "text": "Given the GCD and LCM of n positive integers, how many solutions are there?\n\nQuestion: Suppose you know $$G:=\\gcd$$ (greatest common divisor) and $$L:=\\text{lcm}$$ (least common multiple) of $$n$$ positive integers; how many solution sets exist?\n\nIn the case of $$n = 2$$, one finds that for the $$k$$ distinct primes dividing $$L/G$$, there are a total of $$2^{k-1}$$ unique solutions.\n\nI am happy to write out a proof of the $$n = 2$$ case if desirable, but my question here concerns the more general version. The $$n=3$$ case already proved thorny in my explorations, so I would be happy to see smaller cases worked out even if responders are unsure about the full generalization.\n\nAlternatively: If there is already an existing reference to this problem and its solution, then a pointer to such information would be most welcome, too!\n\n\u2022 @Yorch Your comment only links to the question in the case where $n=2$; for me, this case was no trouble! I am asking, specifically, about the general case: Where you have positive integers $\\{a_1, \\ldots, a_n\\}$. Sep 22 '21 at 15:53\n\u2022 do you require that the $n$ positive integers be distinct? Are you trying to count the multisets? I think that is the only version I haven't been able to solve. Sep 22 '21 at 16:04\n\u2022 @Yorch No requirement that the integers be distinct and/but (ideally!) counting distinct solutions. If you think that you can make traction on a modified version (i.e. imposing additional constraints) then I'd still be pleased to see what you come up with. Sep 22 '21 at 16:08\n\nIf you are interested in counting tuples $$(a_1,a_2,\\dots,a_n)$$ such that $$\\gcd(a_1,\\dots,a_n) = G$$ and $$\\operatorname{lcm}(a_1,\\dots,a_n) = L$$ then we can do it as follows.\n\nIf $$L/G = \\prod\\limits_{i=1}^s p_i^{x_i}$$ then each $$a_i$$ must be of the form $$G \\prod\\limits_{j=1}^s p_i^{y_{i,j}}$$ with $$0 \\leq y_{i,j} \\leq x_i$$.\n\nHence for each prime $$p_i$$ we require that the function from $$\\{1,\\dots, n\\}$$ to $$\\mathbb N$$ that sends $$j$$ to $$y_{i,j}$$ be a function that hits $$0$$ and $$x_i$$.\n\nThe number of such functions is easy by inclusion-exclusion for $$x_i \\geq 1$$, it is $$(x_i+1)^n - 2(x_i)^n + (x_i-1)^n$$.\n\nIt follows the total number of tuples is $$\\prod\\limits_{i=1}^s ( (x_i+1)^n - 2x_i^n + (x_i-1)^n)$$.\n\n\u2022 Counting tuples as in, with repetition, right? E.g. $(1,2)$ and $(2,1)$ would each be counted in your computation? If so, isn't it the case that (using your notation) you could assign the $s$ distinct primes (to their various powers) as divisors of any of the $n$ integers or a subset of them (e.g. to $\\{a_1, a_3, a_7\\}$)? There are $2^n$ subsets of $\\{a_1, \\ldots, a_n\\}$, but we exclude the full set (this is the $\\gcd$) as well as the empty set for a total of $2^{n} - 2$ subsets. Assigning the aforementioned $s$ primes can now be done in in $s^{2^{n} - 2}$ ways. Or have I misunderstood? Sep 22 '21 at 17:17\n\u2022 Yes, that is what it looks like when no prime appears more than once in $L/G$, you would get $(2^n-2)^s$@BenjaminDickman , when you have a prime with exponent greater than $1$ dividing $L/G$ it becomes more complex. Sep 22 '21 at 17:49\n\u2022 lets consider $G=1$ and $L=8$ and $n = 3$. Here we must have that each $a_i$ is one of $1,2,4,8$, and we require that at least one of them is $1$ and at least one of them is $8$, there are $4^3$ total tuples, there are $3^3$ tuples that don't hit the value one, there are $3^3$ that don't hit the value $8$ and there are $2^3$ that don't hit etiher, so there are $4^3-2(3^3) + 2^3$ total triples that work. Sep 22 '21 at 18:08\n\u2022 Ah, great! I have also been pointed to this same answer as Theorem 2.7 here: derby.openrepository.com/handle/10545/583372 (I may add an answer to this effect) Sep 22 '21 at 19:34\n\u2022 The case $G,L$ is the same as the case $1,L/G$ Sep 23 '21 at 16:39\n\n(Adding this community wiki answer to point out a relevant reference.) I was recently pointed to the following paper, in which this and related problems are proposed and solved:\n\nBagdasar, O. (2014.) \"On some functions involving the lcm and gcd of integer tuples.\" Scientific Publications of the State University of Novi Pazar Series A: Applied Mathematics, Informatics and mechanics, 6(2):91-100. PDF (no paywall).\n\nThe result appears as Theorem 2.7 (cf. the comment of Yorch, too):", "date": "2022-01-23 03:00:56", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/4257386/given-the-gcd-and-lcm-of-n-positive-integers-how-many-solutions-are-there", "openwebmath_score": 0.8209962248802185, "openwebmath_perplexity": 224.42383233340095, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9886682474702956, "lm_q2_score": 0.9111797063939128, "lm_q1q2_score": 0.9008544434509682}}
{"url": "https://math.stackexchange.com/questions/618091/inferential-logic-in-a-simple-life-situation", "text": "# Inferential logic in a simple-life situation.\n\nHere's a little situation I want math to resolve for me :\n\n1. If I study, I make the exam ,\n2. If I do not play tennis, I study ,\n3. I didn't make the exam\n\nCan I conclude that was playing tennis ?\n\nTrying to put this into the symbology of inference logic and propositional classic logic :\n\n$P1 : \\text{study} \\implies \\text{exam}$\n\n$P2 : (\\text{tennis}\\, \\vee \\text{study}) \\wedge (\\neg \\text{tennis} \\implies \\text{study})$ (disjunctive syllogism)\n\n$p3 : \\neg \\text{exam}$\n\nMy reasoning :\n\nStep 1 : the contrapositive of $P1$ is $P1' : \\neg \\text{exam} \\implies \\neg \\text{study}$ ;\n\nStep 2 : By Modus Tollens ( $[(P \\implies Q) \\wedge \\neg Q] \\implies \\neg P$) we have : $(\\text{study} \\implies \\text{exam}) \\wedge (\\neg \\text{exam} \\implies \\neg \\text{study})$\n\nStep 3 : should we suppose : $\\neg \\text{tennis} \\wedge \\neg \\text{study}$, then $\\neg ( \\text{tennis} \\vee \\text{study})$, then (by $P2$) $\\text{tennis}$ or otherwise the $P1$ would fall since $\\neg \\text{study}$ and $\\neg (False \\implies False)$.\n\nStep 4 : reductio ad absurdum from step $(3)$, we have $(\\text{tennis} \\vee \\text{study})$, henceforth, in $P2$, $\\neg \\text{tennis}$ or else $false \\implies false$.\n\nSo, have I been playing tennis or is my inferential logic bad ?\n\n\u2022 The title should be more informative. \u2013\u00a0Paracosmiste Dec 25 '13 at 15:38\n\u2022 Would the person that \"minused\" the question care to say why ? That would be nice ! \u2013\u00a0Gloserio Dec 25 '13 at 16:11\n\u2022 I'm not the downvoter. \u2013\u00a0Paracosmiste Dec 25 '13 at 16:26\n\u2022 I am not accusing either, and I've just asked the question to see what I can avoid next time :) \u2013\u00a0Gloserio Dec 25 '13 at 16:28\n\nYes, indeed, we can easily arrive at the conclusion that you played tennis: and the repeated use of modus tollens, alone (plus one invocation of double negation) gets you that conclusion.\n\nOur premises, in \"natural language\":\n\n1. If I study, I make the exam ,\n2. If I do not play tennis, I study ,\n3. I didn't make the exam\n\n\nKEY:\n\n$S:\\;$ I study.\n\n$E:\\;$ I make the exam.\n\n$P:\\;$ I play tennis.\n\nThen our premises translate to:\n\n$(1): S \\rightarrow E$.\n\n$(2): \\lnot P \\rightarrow S.$\n\n$(3): \\lnot E.$\n\n$(4)\\quad \\lnot S$ follows from $(1), (3)$ by modus tollens.\n\n$(5)\\quad \\lnot \\lnot P$ follows from $(2), (4)$ by modus tollens.\n\n$\\therefore (6) \\quad P$, by $(5)$ and double negation.\n\nHence you can conclude you played tennis.\n\n\u2022 As @Matt Brenneman did, you translated the second statement to : $\\neg P \\implies S$ while I've translated it to : $(P \\vee S) \\wedge \\neg S \\implies P$, which I thought was safer. Admitting your translation, I would totally agree with your reasoning, but admitting mine, would we come still to the conclusion that I play tennis ? \u2013\u00a0Gloserio Dec 25 '13 at 16:19\n\u2022 Yes, absolutely you would! $\\lnot P \\rightarrow S \\equiv \\lnot \\lnot P \\lor S\\equiv P \\lor S$. Then since we have $\\lnot S$, too, you can conclude $P$. \u2013\u00a0Namaste Dec 25 '13 at 16:36\n\u2022 I know you love logic and for this reason this answer is excellent;-)+1 \u2013\u00a0user63181 Dec 25 '13 at 16:51\n\u2022 @amWhy : thank you, now it's clear ! \u2013\u00a0Gloserio Dec 25 '13 at 18:16\n\u2022 You're welcome, @Gloserio! \u2013\u00a0Namaste Dec 25 '13 at 18:23\n\nYou made a mistake in your step 3, because considering only P2, the term $\\neg ( tennis \\vee study)$ does not imply $tennis \\vee study$.\n\nThis may sound counter-intuitive to your introduction.\n\nThe reason is, that your P2 is an arguable translation of statement 2. It is not equivalent to \"if I don't play tennis, I study\". Rather it states \"if I don't play tennis and if I study, I study\", which is a tautology. You can see this by drawing a truth-table for P2.\n\nAlso note that $\\neg exam, \\neg tennis, \\neg study$ satisfies P1,P2 and P3.\n\nSo I would replace P2 by $$\\neg tennis \\rightarrow study.$$ This also repairs your Step3.\n\n\u2022 +1, you're probably true, that why I've asked this question, because I felt as if my $P2$ was somewhat redundant. Thanks for pointing it out ! \u2013\u00a0Gloserio Dec 25 '13 at 16:26\n\nYes. It just reduces down to look at the contrapositives of your statements.\nStatement 1 is logically equivalent to : ~(make exam) implies ~study.\nStatement 2 is logically equivalent to: ~study implies (play tennis).\nSo the truth of ~(make exam) directly implies you played tennis (use modus ponens twice).\n\n\u2022 How is second statement logically equivalent to $\\neg study \\implies tennis$ ? \u2013\u00a0Gloserio Dec 25 '13 at 15:52\n\u2022 It is the contrapositive of the statement: \"~(play tennis) implies study\" \u2013\u00a0Matt Brenneman Dec 25 '13 at 15:54\n\u2022 So how that you converted natural langage to logic symobols, and in its valid. \u2013\u00a0Gloserio Dec 25 '13 at 16:20", "date": "2019-10-18 03:50:24", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/618091/inferential-logic-in-a-simple-life-situation", "openwebmath_score": 0.7437729239463806, "openwebmath_perplexity": 1069.2030222676608, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9702399060540358, "lm_q2_score": 0.9284087926320944, "lm_q1q2_score": 0.900779259743104}}
{"url": "https://math.stackexchange.com/questions/3607102/calculating-total-number-of-allowable-paths/3607174", "text": "# calculating total number of allowable paths\n\nI seem to be struggling with the following type of path questions\n\nConsider paths starting at $$(0, 0)$$ with allowable steps\n\n(i) from $$(x,y)$$ to $$(x+1,y+2)$$,\n\n(ii) from $$(x,y)$$ to $$(x+2,y+1)$$,\n\n(iii)from $$(x,y)$$ to $$(x+1,y)$$\n\nDetermine the total number of allowable paths from $$(0, 0)$$ to $$(8, 8)$$, and the total number of allowable paths from $$(0, 0)$$ to $$(10, 10)$$.\n\ncould anyone recommend a trivial method to tackle problems of these types in an exam setting?\n\n\u2022 General answer for these type of problems would be to use recursion, as answered by Rob Pratt. However, in this \"small\" case it might be easier to do things \"on hand\", especially in exam setting. Suppose you do $A$ moves of type (i), $B$ of type (ii) and $C$ of type (iii). After putting constraints on $A, B, C$ you will see that there is only two possibilities in both of your question. Can you work out the rest by yourself? Apr 2, 2020 at 21:56\n\u2022 I don't understand @prosinac Apr 2, 2020 at 22:44\n\nDraw a table and think backwards. Let $$p(x,y)$$ be the number of such paths from $$(0,0)$$ to $$(x,y)$$. By conditioning on the last step into $$(x,y)$$, we find that $$p(x,y)=p(x-1,y-2)+p(x-2,y-1)+p(x-1,y),$$ where $$p(x,y)=0$$ if $$x<0$$ or $$y<0$$. You know that $$p(0,0)=1$$, and you want to compute $$p(8,8)$$ and $$p(10,10)$$. The resulting table is $$\\begin{matrix} x\\backslash y &0 &1 &2 &3 &4 &5 &6 &7 &8 &9 &10 \\\\ \\hline 0 &1 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 \\\\ 1 &1 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0 \\\\ 2 &1 &1 &2 &0 &1 &0 &0 &0 &0 &0 &0 \\\\ 3 &1 &2 &3 &2 &3 &0 &1 &0 &0 &0 &0 \\\\ 4 &1 &3 &5 &6 &6 &3 &4 &0 &1 &0 &0 \\\\ 5 &1 & &8 &12 &13 &12 &10 &4 &5 &0 &1 \\\\ 6 & & &12 & &27 &30 &26 &20 &15 &5 &6 \\\\ 7 & & & & &51 & &65 &60 &45 &30 &21 \\\\ 8 & & & & & & &146 & &\\color{red}{130} &105 &71 \\\\ 9 & & & & & & & & &336 & &231 \\\\ 10 & & & & & & & & & & &\\color{red}{672} \\\\ \\end{matrix}$$ In particular, $$p(8,8) = p(7,6)+p(6,7)+p(7,8) = 65+20+45=130.$$\n\n\u2022 How could you apply this with perhaps a path D: $(x,y)->(x,y-1)$ ? Apr 3, 2020 at 13:13\n\u2022 You can use the same approach. The new recurrence has an additional $+p(x,y+1)$ term, and you should explicitly include a boundary condition $p(x,y)=0$ for $y>2x$ to avoid infinite descent. Apr 3, 2020 at 14:46\n\nSuppose you do $$A$$ moves of type (i), $$B$$ of type (ii) and $$C$$ of type (iii). If you want to reach $$(8, 8)$$, then clearly $$A + 2B + C = 8$$ and $$2A + B = 8$$. This yields $$B = 8-2A$$ and $$C = 3A - 8$$. Since $$A, B, C$$ are nonnegative integers, you obtain solutions $$(A, B, C) = (3, 2, 1)$$ or $$(4, 0, 4)$$.\n\nNow to calculate paths, for $$(4, 0, 4)$$ case, a path is described by string of four $$A$$s and four $$C$$s. For example, $$AAACCACC$$ is one such path. There is $${8 \\choose 4} = 70$$ such paths.\n\nFor the $$(3, 2, 1)$$ case there is $${6 \\choose 3} \\cdot 3 = 60$$ such paths. Altogether there are $$130$$ such paths.\n\nFor reaching $$(10, 10)$$ same logic gives you $$B = 10 - 2A$$ and $$C = 3A - 10$$, so solutions are $$(4, 2, 2)$$ and $$(5, 0, 5)$$.\n\nSo, the number of paths is $${8 \\choose 4} {4 \\choose 2} + {10 \\choose 5} = 70 \\cdot 6 + 252 = 672$$.\n\nThis is in agreement with Rob Pratts answer. Also, I would like to emphasise as I did in the comment that his answer is \"better\" in a sense that it illustrates how you can handle any problem of this type, since his answers scales reasonably to larger numbers. This answer can from the practical point of view only be used on such a small examples. But, if I was writing an exam, I'd take this approach (or at least I would try and then estimate would it be faster done this way or in the more general way)\n\n\u2022 +1 I like these kind of \"restrict by case\" approaches too. It is not that hard to scale for larger numbers as there is a pattern, though the calculation could be tedious. E.g It works nicely for This question. Apr 3, 2020 at 0:08\n\u2022 If you prefer multinomial coefficients to binomial coefficients, the formulas are $\\binom{6}{3,2,1}+\\binom{8}{4,4}$ and $\\binom{8}{4,2,2}+\\binom{10}{5,5}$. Apr 3, 2020 at 1:24\n\u2022 I understand everything but where has A+2B+C=8 and 2A+B=8 come from? I don't seen how 8 can be reached using these steps? @prosinac Apr 3, 2020 at 16:58\n\u2022 If you do $A$ moves of type (i) then you moved $A$ steps on the $x$-axis. If you do $B$ moves of type (ii) then you moved $2B$ steps on the $x$-axis. If you do $C$ moves of type (iii) then you moved $C$ steps on the $x$-axis. And you must move $8$ steps in total. So $A+2B+C=8$. Same logic for $y$-axis gives second equation Apr 3, 2020 at 21:26", "date": "2022-08-08 03:39:36", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/3607102/calculating-total-number-of-allowable-paths/3607174", "openwebmath_score": 0.8159676790237427, "openwebmath_perplexity": 262.8508306791505, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9884918533088548, "lm_q2_score": 0.9111797027760038, "lm_q1q2_score": 0.9006937130944634}}
{"url": "https://gmatclub.com/forum/m-is-a-positive-integer-less-than-100-when-m-is-raised-to-the-third-226775.html", "text": "It is currently 19 Feb 2018, 04:06\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# M is a positive integer less than 100. When m is raised to the third\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nJoined: 03 Oct 2016\nPosts: 84\nConcentration: Technology, General Management\nWE: Information Technology (Computer Software)\nM is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n06 Oct 2016, 12:40\n2\nKUDOS\n19\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n65% (hard)\n\nQuestion Stats:\n\n67% (02:01) correct 33% (02:19) wrong based on 192 sessions\n\n### HideShow timer Statistics\n\nM is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer. How many different values could m be?\n\nA. 7\nB. 9\nC. 11\nD. 13\nE. 15\n\nKeep the Kudos dropping in and let these tricky questions come out ....\n[Reveal] Spoiler: OA\n\n_________________\n\nKINDLY KUDOS IF YOU LIKE THE POST\n\nSC Moderator\nJoined: 13 Apr 2015\nPosts: 1578\nLocation: India\nConcentration: Strategy, General Management\nWE: Analyst (Retail)\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n06 Oct 2016, 19:13\n2\nKUDOS\n$$({a^2})^3 = a^6 = ({a^3})^2$$\n\nGiven: M = $$0 < a^2 < 100$$\n\nValues of a^2 can be --> $$1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)$$\n\nThere are 9 possible values.\n\nManager\nJoined: 26 Jun 2013\nPosts: 92\nLocation: India\nSchools: ISB '19, IIMA , IIMB\nGMAT 1: 590 Q42 V29\nGPA: 4\nWE: Information Technology (Retail Banking)\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n01 Mar 2017, 11:35\nVyshak wrote:\n$$({a^2})^3 = a^6 = ({a^3})^2$$\n\nGiven: M = $$0 < a^2 < 100$$\n\nValues of a^2 can be --> $$1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)$$\n\nThere are 9 possible values.\n\nVyshak please can you explain this as I am not able to understand.\n\nThanks!\n_________________\n\nRemember, if it is a GMAT question, it can be simplified elegantly.\n\nSC Moderator\nJoined: 13 Apr 2015\nPosts: 1578\nLocation: India\nConcentration: Strategy, General Management\nWE: Analyst (Retail)\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n01 Mar 2017, 22:47\n1\nKUDOS\nhotshot02 wrote:\n\nVyshak please can you explain this as I am not able to understand.\n\nThanks!\n\nYou have to find the number of values of a^2 that are between 0 and 100. --> a is between 0 and 10 --> We will have 9 values for m.\n\nHope it helps.\nTarget Test Prep Representative\nAffiliations: Target Test Prep\nJoined: 04 Mar 2011\nPosts: 1975\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n06 Mar 2017, 17:14\n1\nKUDOS\nExpert's post\n4\nThis post was\nBOOKMARKED\nidontknowwhy94 wrote:\nM is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.\nHow many different values could m be?\n\nA. 7\nB. 9\nC. 11\nD. 13\nE. 15\n\nWe are given that m is a positive integer less than 100. We are also given that when m is raised to the third power, it becomes the square of another integer. In order for that to be true, m itself must (already) be a perfect square, since any perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100. Since there are 9 perfect squares that are less than 100, namely, 1, 4, 9, \u2026, 64, and 81, the answer is 9.\n\nLet\u2019s look at some examples to clarify this: Let\u2019s assume that m = 4 = 2^2. Now, let\u2019s raise m to the third power, obtaining m^3 = (2^2)^3 = 4^3 = 64, which is the perfect square 8^2. Another illustration: Let\u2019s let m = 25 = 5^2. Now, let\u2019s raise m to the third power, obtaining m^3 = (5^2)^3 = 25^3 = 15625, which is the perfect square of 125.\n\n(Note: By the way the problem is worded, \u201cwhen m is raised to the third power, it becomes the square of another integer,\u201d 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is \u201cwhen m is raised to the third power, it becomes the square of an integer.\u201d)\n\n_________________\n\nJeffery Miller\n\nGMAT Quant Self-Study Course\n500+ lessons 3000+ practice problems 800+ HD solutions\n\nCR Forum Moderator\nStatus: The best is yet to come.....\nJoined: 10 Mar 2013\nPosts: 529\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n26 Aug 2017, 00:14\nJeffTargetTestPrep wrote:\nAny perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100.\n\nWhy we need to look for perfect squares that are less than 100? We need the values of $$m<100$$ NOT the $$m^2<100$$.\n\nI know I am wrong, but I don't know why I am wrong.\n_________________\n\nHasan Mahmud\n\nDirector\nJoined: 18 Aug 2016\nPosts: 628\nConcentration: Strategy, Technology\nGMAT 1: 630 Q47 V29\nGMAT 2: 740 Q51 V38\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n26 Aug 2017, 01:09\nidontknowwhy94 wrote:\nM is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.\nHow many different values could m be?\n\nA. 7\nB. 9\nC. 11\nD. 13\nE. 15\n\nKeep the Kudos dropping in and let these tricky questions come out ....\n\n0<x^2<100\n1,4,9,16,25,36,49,64,81\n9 numbers\nB\n\nSent from my iPhone using GMAT Club Forum mobile app\n_________________\n\nWe must try to achieve the best within us\n\nThanks\nLuckisnoexcuse\n\nSenior Manager\nJoined: 29 Jun 2017\nPosts: 495\nGMAT 1: 570 Q49 V19\nGPA: 4\nWE: Engineering (Transportation)\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n10 Sep 2017, 06:38\n2\nKUDOS\nAns is B\n\nM^3 = y^2\nM= y^(2/3)\nmeans y should have a cube-root as integer and whose square should be less than 100\nlets say start by 1000 =y\ny^(2/3) = 1000^2/3 =100 rejected means M cant be 100\nM can attain 81,64,49,36,25,16,09,04,01 TOTAL 9 values .\n\n_________________\n\nGive Kudos for correct answer and/or if you like the solution.\n\nSenior Manager\nStatus: Countdown Begins...\nJoined: 03 Jul 2016\nPosts: 308\nLocation: India\nConcentration: Technology, Strategy\nSchools: IIMB\nGMAT 1: 580 Q48 V22\nGPA: 3.7\nWE: Information Technology (Consulting)\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n21 Oct 2017, 21:01\nidontknowwhy94 wrote:\nM is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.\nHow many different values could m be?\n\nA. 7\nB. 9\nC. 11\nD. 13\nE. 15\n\nKeep the Kudos dropping in and let these tricky questions come out ....\n\nBunuel,\nFor this question, I understood that basically we need to find out squares of integers below 100. But I have one query -\n\nWhy does above answers includes $$1^2$$? The question stem says if the number is raised to third power, then it becomes square of ANOTHER integer?\n\n$$\\sqrt{(1^3)}$$ = 1 only.\n\n_________________\n\nNeed Kudos to unlock GMAT Club tests\n\nIntern\nJoined: 21 May 2017\nPosts: 25\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n21 Oct 2017, 22:09\nRMD007 wrote:\nidontknowwhy94 wrote:\nM is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.\nHow many different values could m be?\n\nA. 7\nB. 9\nC. 11\nD. 13\nE. 15\n\nKeep the Kudos dropping in and let these tricky questions come out ....\n\nBunuel,\nFor this question, I understood that basically we need to find out squares of integers below 100. But I have one query -\n\nWhy does above answers includes $$1^2$$? The question stem says if the number is raised to third power, then it becomes square of ANOTHER integer?\n\n$$\\sqrt{(1^3)}$$ = 1 only.\n\nJeffTargetTestPrep addressed this in his response above -\n\n(Note: By the way the problem is worded, \u201cwhen m is raised to the third power, it becomes the square of another integer,\u201d 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is \u201cwhen m is raised to the third power, it becomes the square of an integer.\u201d)\nSenior Manager\nStatus: Countdown Begins...\nJoined: 03 Jul 2016\nPosts: 308\nLocation: India\nConcentration: Technology, Strategy\nSchools: IIMB\nGMAT 1: 580 Q48 V22\nGPA: 3.7\nWE: Information Technology (Consulting)\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n21 Oct 2017, 22:13\nmahu101 wrote:\n\nJeffTargetTestPrep addressed this in his response above -\n\n(Note: By the way the problem is worded, \u201cwhen m is raised to the third power, it becomes the square of another integer,\u201d 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is \u201cwhen m is raised to the third power, it becomes the square of an integer.\u201d)\n\nThanks, my point is, with the given question stem 8 should be the answer!\n_________________\n\nNeed Kudos to unlock GMAT Club tests\n\nIntern\nJoined: 21 May 2017\nPosts: 25\nRe: M is a positive integer less than 100. When m is raised to the third\u00a0[#permalink]\n\n### Show Tags\n\n21 Oct 2017, 22:19\n1\nKUDOS\nRMD007 wrote:\nmahu101 wrote:\n\nJeffTargetTestPrep addressed this in his response above -\n\n(Note: By the way the problem is worded, \u201cwhen m is raised to the third power, it becomes the square of another integer,\u201d 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is \u201cwhen m is raised to the third power, it becomes the square of an integer.\u201d)\n\nThanks, my point is, with the given question stem 8 should be the answer!\n\nYes, you are right.\nRe: M is a positive integer less than 100. When m is raised to the third \u00a0 [#permalink] 21 Oct 2017, 22:19\nDisplay posts from previous: Sort by", "date": "2018-02-19 12:06:00", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/m-is-a-positive-integer-less-than-100-when-m-is-raised-to-the-third-226775.html", "openwebmath_score": 0.7223361730575562, "openwebmath_perplexity": 1336.469105695326, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n", "lm_q1_score": 1.0, "lm_q2_score": 0.9005297787765764, "lm_q1q2_score": 0.9005297787765764}}
{"url": "https://gmatclub.com/forum/if-p-x-and-y-are-positive-integers-y-is-odd-and-p-x-82399.html?kudos=1", "text": "It is currently 18 Oct 2017, 22:56\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# If p, x, and y are positive integers, y is odd, and p = x^2\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nSenior Manager\nJoined: 22 Sep 2005\nPosts: 278\n\nKudos [?]: 241 [17], given: 1\n\nIf p, x, and y are positive integers, y is odd, and p = x^2\u00a0[#permalink]\n\n### Show Tags\n\n14 Aug 2009, 12:49\n17\nKUDOS\n88\nThis post was\nBOOKMARKED\n00:00\n\nDifficulty:\n\n95% (hard)\n\nQuestion Stats:\n\n40% (02:06) correct 60% (02:17) wrong based on 1935 sessions\n\n### HideShow timer Statistics\n\nIf p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?\n\n(1) When p is divided by 8, the remainder is 5.\n(2) x \u2013 y = 3\n[Reveal] Spoiler: OA\n\nKudos [?]: 241 [17], given: 1\n\nManager\nJoined: 25 Jul 2009\nPosts: 115\n\nKudos [?]: 262 [42], given: 17\n\nSchools: NYU, NUS, ISB, DUKE, ROSS, DARDEN\nRe: PS: Divisible by 4\u00a0[#permalink]\n\n### Show Tags\n\n14 Aug 2009, 13:46\n42\nKUDOS\n27\nThis post was\nBOOKMARKED\nnetcaesar wrote:\nIf p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?\n\n(1) When p is divided by 8, the remainder is 5.\n(2) x \u2013 y = 3\n\nSOL:\n\nSt1:\nHere we will have to use a peculiar property of number 8. The square of any odd number when divided by 8 will always yield a remainder of 1!!\n\nThis means that y^2 MOD 8 = 1 for all y\n=> p MOD 8 = (x^2 + 1) MOD 8 = 5\n=> x^2 MOD 8 = 4\n\nNow if x is divisible by 4 then x^2 MOD 8 will be zero. And also x cannot be an odd number as in that case x^2 MOD 8 would become 1. Hence we conclude that x is an even number but also a non-multiple of 4.\n=> SUFFICIENT\n\nSt2:\nx - y = 3\nSince y can be any odd number, x could also be either a multiple or a non-multiple of 4.\n=> NOT SUFFICIENT\n\nANS: A\n_________________\n\nKUDOS me if I deserve it !!\n\nMy GMAT Debrief - 740 (Q50, V39) | My Test-Taking Strategies for GMAT | Sameer's SC Notes\n\nKudos [?]: 262 [42], given: 17\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 41890\n\nKudos [?]: 128792 [31], given: 12183\n\nRe: PS: Divisible by 4\u00a0[#permalink]\n\n### Show Tags\n\n16 Dec 2010, 07:39\n31\nKUDOS\nExpert's post\n21\nThis post was\nBOOKMARKED\nnonameee wrote:\nCan I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction?\n\nThank you.\n\nIf p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?\n\n(1) When p is divided by 8, the remainder is 5 --> $$p=8q+5=x^2+y^2$$ --> as given that $$y=odd=2k+1$$ --> $$8q+5=x^2+(2k+1)^2$$ --> $$x^2=8q+4-4k^2-4k=4(2q+1-k^2-k)$$.\n\nSo, $$x^2=4(2q+1-k^2-k)$$. Now, if $$k=odd$$ then $$2q+1-k^2-k=even+odd-odd-odd=odd$$ and if $$k=even$$ then $$2q+1-k^2-k=even+odd-even-even=odd$$, so in any case $$2q+1-k^2-k=odd$$ --> $$x^2=4*odd$$ --> in order $$x$$ to be multiple of 4 $$x^2$$ must be multiple of 16 but as we see it's not, so $$x$$ is not multiple of 4. Sufficient.\n\n(2) x \u2013 y = 3 --> $$x-odd=3$$ --> $$x=even$$ but not sufficient to say whether it's multiple of 4.\n\n_________________\n\nKudos [?]: 128792 [31], given: 12183\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7674\n\nKudos [?]: 17354 [26], given: 232\n\nLocation: Pune, India\nRe: PS: Divisible by 4\u00a0[#permalink]\n\n### Show Tags\n\n19 Dec 2010, 07:49\n26\nKUDOS\nExpert's post\n17\nThis post was\nBOOKMARKED\nnetcaesar wrote:\nIf p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?\n\n(1) When p is divided by 8, the remainder is 5.\n(2) x \u2013 y = 3\n\nSuch questions can be easily solved keeping the concept of divisibility in mind. Divisibility is nothing but grouping. Lets say if we need to divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is quotient and 1 is remainder. For more on these concepts, check out: http://gmatquant.blogspot.com/2010/11/divisibility-and-remainders-if-you.html\n\nFirst thing that comes to mind is if y is odd, $$y^2$$ is also odd.\nIf $$y = 2k+1, y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$$\nSince one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. So when y^2 is divided by 8, it will leave a 1.\n\nStmnt 1: When p is divided by 8, the remainder is 5.\nWhen y^2 is divided by 8, remainder is 1. To get a remainder of 5, when x^2 is divided by 8, we should get a remainder of 4.\n$$x^2 = 8a + 4$$ (i.e. we can make 'a' groups of 8 and 4 will be leftover)\n$$x^2 = 4(2a+1)$$ This implies $$x = 2*\\sqrt{Odd Number}$$because (2a+1) is an odd number. Square root of an odd number will also be odd.\nTherefore, we can say that x is not divisible by 4. Sufficient.\n\nStmnt 2: x - y = 3\nSince y is odd, we can say that x will be even (Even - Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p. Not sufficient.\n\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17354 [26], given: 232 Intern Joined: 30 May 2013 Posts: 4 Kudos [?]: 8 [8], given: 8 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 15 Sep 2013, 22:13 8 This post received KUDOS I did this question this way. I found it simple. 1. p=x^2+y^2 y is odd p div 8 gives remainder 5. A number which gives remainder 5 when divided by 8 is odd. so (x^2 + y^2)/8 = oddnumber (x^2 + y^2) = 8 * oddnumber (this is an even number without doubt) x^2 + y^2 is even. Since y is odd to get x^2+y^2 even x must also be odd. X is an odd number not divisible by 4 Option A: 1 alone is sufficient Kudos [?]: 8 [8], given: 8 Intern Joined: 02 Sep 2010 Posts: 45 Kudos [?]: 146 [4], given: 17 Location: India Re: PS: Divisible by 4 [#permalink] ### Show Tags 18 Dec 2010, 11:23 4 This post received KUDOS 1 This post was BOOKMARKED maliyeci wrote: Very good solution I did not know this property of 8. Kudos to you. By and induction. 1^2=1 mod 8 say n^2=1 mod 8 (n is an odd number) than if (n+2)^2=1 mod 8 ? (n+2 is the next odd number) (n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8 4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8. So induction works. So for any odd number n, n^2=1 mod 8 Its not something one shall already know before attacking a question, you may realize properties like this when u start solving a question. Even I didn't know about this property of 8. I approached the question in following way: Stmt 1: P/8=(x^2+y^2)/8; using remainder theorem; rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8] if x is divisible by 4, then x^2= 4k*4k= 16K=8*2K is also divisible by 8. now to anaylze rem[y^2/8]; start putting suitable values of y; i.e all odd values starting from 1. for y=1; rem(1/8)=1 for y=3; rem(9/8)=1 for y=5;rem(25/8)=1 so you observe this pattern here. coming back to ques now, as rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8]= rem[x^2/8] + 1 =5; this means rem[x^2/8] is not 0; which implies x is not divisible my 8; Sufficient Stmt2: y being odd can be accept both 3 and 5 as values and we get different results; thus Insufficient Thus OA is A _________________ The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done! Kudos [?]: 146 [4], given: 17 Intern Joined: 14 May 2013 Posts: 12 Kudos [?]: 18 [2], given: 3 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 12 Jun 2013, 10:58 2 This post received KUDOS netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5. (2) x \u2013 y = 3 1. As p = 8I + 5 we have values of P = 5,13,21,29..... etc .. as y is odd when we solve this p(odd) = x^2 + y^2(odd) x^2 = odd -odd = even which can be 2,4,6 ... etc but if we check for any value of p we don't get any multiple of 4. so it say's clearly that x is not divisible by 4. 2. x-y = 3 x = y(odd)+3 x is even which can be 2,4,6.. so it's not sufficient .. Ans : A _________________ Chauahan Gaurav Keep Smiling Kudos [?]: 18 [2], given: 3 Intern Joined: 25 Jun 2013 Posts: 6 Kudos [?]: 3 [2], given: 0 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 13 Sep 2013, 20:25 2 This post received KUDOS 1 This post was BOOKMARKED from first statement p = 8j + 5 Put j as 1, 2,3,4,5... p would be 13, 21,29, 37,45... Now in the formula p= x^2+y^2 put 1,3,5,7 as value of y ( as y is odd) to get x. You will notic the possible value of x is 2 which is not divisble by 4. Posted from GMAT ToolKit Kudos [?]: 3 [2], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [2], given: 232 Location: Pune, India Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 18 Aug 2014, 02:43 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED alphonsa wrote: For statement 1 , wouldn't plugging in values be a better option? No. When you need to establish something, plugging in values is not fool proof. Anyway, in this question, how will you plug in values? You cannot assume a value for x since that is what you need to find. You will assume a value for y and a value for p such that they satisfy all conditions. This itself will be quite tricky. Then when you do get a value for x, you will find that it will be even but not divisible by 4. How can you be sure that this will hold for every value of y and p? When a statement is not sufficient, plugging in values can work - you find two opposite cases - one which answers in yes and the other which answers in no. Then you know that the statement alone is not sufficient. But when the statement is sufficient, it is very hard to prove that it will hold for all possible values using number plugging alone. You need to use logic in that case. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nKudos [?]: 17354 [2], given: 232\n\nSenior Manager\nJoined: 23 Jun 2009\nPosts: 360\n\nKudos [?]: 134 [1], given: 80\n\nLocation: Turkey\nSchools: UPenn, UMich, HKS, UCB, Chicago\nRe: PS: Divisible by 4\u00a0[#permalink]\n\n### Show Tags\n\n15 Aug 2009, 13:49\n1\nKUDOS\nVery good solution I did not know this property of 8. Kudos to you.\n\nBy and induction.\n1^2=1 mod 8\nsay\nn^2=1 mod 8 (n is an odd number)\nthan\nif (n+2)^2=1 mod 8 ? (n+2 is the next odd number)\n(n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8\n4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8.\nSo induction works.\n\nSo for any odd number n, n^2=1 mod 8\n\nKudos [?]: 134 [1], given: 80\n\nDirector\nJoined: 23 Apr 2010\nPosts: 573\n\nKudos [?]: 95 [1], given: 7\n\nRe: PS: Divisible by 4\u00a0[#permalink]\n\n### Show Tags\n\n16 Dec 2010, 07:13\n1\nKUDOS\nCan I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction?\n\nThank you.\n\nKudos [?]: 95 [1], given: 7\n\nIntern\nJoined: 25 Mar 2012\nPosts: 3\n\nKudos [?]: 3 [1], given: 1\n\nRe: PS: Divisible by 4\u00a0[#permalink]\n\n### Show Tags\n\n16 Jul 2012, 18:19\n1\nKUDOS\nAm i missing something, why cant we take stmt 2 as follows:\nsquaring x-y=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4\n\nKudos [?]: 3 [1], given: 1\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7674\n\nKudos [?]: 17354 [1], given: 232\n\nLocation: Pune, India\nRe: If p, x, and y are positive integers, y is odd, and p = x^2\u00a0[#permalink]\n\n### Show Tags\n\n03 May 2017, 05:31\n1\nKUDOS\nExpert's post\nVeritasPrepKarishma wrote:\naliasjit wrote:\nI am a little confused about solving the problem:\n\nStmnt 2: x-y =3\n\nwe know y is odd.\nand if Y is odd as per problem statement y cannot be anything but 1\nas x and y both are positive integers.\n\nTherefore x =4 and is divisible by 4.\n\ny is odd, yes, but why must y be 1?\n\nResponding to a pm:\nQuote:\nbecause as per statement 2 if x-y =3 and x and y are both +ve integers could y be anything but 1\n\nBut as per statement 2, we do not know that x must be 4. x must be even since y is odd and difference between x and y is odd. But will it be 4, we do not know.\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17354 [1], given: 232 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [0], given: 232 Location: Pune, India Re: PS: Divisible by 4 [#permalink] ### Show Tags 16 Jul 2012, 23:24 Eshaninan wrote: Am i missing something, why cant we take stmt 2 as follows: squaring x-y=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4 The question is: \"Is x divisible by 4?\" not \"Is p divisible by 4?\" x is even since y is odd. We don't know whether x is divisible by only 2 or 4 as well. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199\n\nVeritas Prep Reviews\n\nKudos [?]: 17354 [0], given: 232\n\nManager\nJoined: 29 Jun 2011\nPosts: 159\n\nKudos [?]: 24 [0], given: 29\n\nWE 1: Information Technology(Retail)\nRe: If p, x, and y are positive integers, y is odd, and p = x^2\u00a0[#permalink]\n\n### Show Tags\n\n03 Sep 2013, 04:00\nExcellent explanation Bunuel & Karishma:):)\n\nKudos [?]: 24 [0], given: 29\n\nIntern\nJoined: 21 Sep 2013\nPosts: 9\n\nKudos [?]: 2 [0], given: 0\n\nRe: If p, x, and y are positive integers, y is odd, and p = x^2\u00a0[#permalink]\n\n### Show Tags\n\n23 Dec 2013, 23:45\nFor Statement 1:\nsince p when divided by 8 leaves remainder 5.We obtain the following equation\np= 8q+5\nWe know y is odd. If we write p =x^2+y^2 then we get the eqn:\nx^2+y^2=8q+5\nSince, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd.\nThen x^2= odd - y^2\ni.e x^2=even\nie x= even\nBut it's not sufficient to answer the question whether x is a multiple of 4?\nBy this logic i get E as my answer.\nStatement 2: is insufficient.\n\nKudos [?]: 2 [0], given: 0\n\nVeritas Prep GMAT Instructor\nJoined: 16 Oct 2010\nPosts: 7674\n\nKudos [?]: 17354 [0], given: 232\n\nLocation: Pune, India\nRe: If p, x, and y are positive integers, y is odd, and p = x^2\u00a0[#permalink]\n\n### Show Tags\n\n30 Dec 2013, 23:39\nExpert's post\n1\nThis post was\nBOOKMARKED\nAbheek wrote:\nFor Statement 1:\nsince p when divided by 8 leaves remainder 5.We obtain the following equation\np= 8q+5\nWe know y is odd. If we write p =x^2+y^2 then we get the eqn:\nx^2+y^2=8q+5\nSince, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd.\nThen x^2= odd - y^2\ni.e x^2=even\nie x= even\nBut it's not sufficient to answer the question whether x is a multiple of 4?\n\nYour analysis till now is fine but it is incomplete. We do get that x is even but we also get that x is a multiple of 2 but not 4 as explained in the post above: if-p-x-and-y-are-positive-integers-y-is-odd-and-p-x-82399.html#p837890\n_________________\n\nKarishma\nVeritas Prep | GMAT Instructor\nMy Blog\n\nGet started with Veritas Prep GMAT On Demand for \\$199\n\nVeritas Prep Reviews\n\nKudos [?]: 17354 [0], given: 232\n\nManager\nJoined: 22 Jul 2014\nPosts: 130\n\nKudos [?]: 297 [0], given: 197\n\nConcentration: General Management, Finance\nGMAT 1: 670 Q48 V34\nWE: Engineering (Energy and Utilities)\nRe: If p, x, and y are positive integers, y is odd, and p = x^2\u00a0[#permalink]\n\n### Show Tags\n\n16 Aug 2014, 00:16\nFor statement 1 , wouldn't plugging in values be a better option?\n\nKudos [?]: 297 [0], given: 197\n\nCurrent Student\nJoined: 17 Jul 2013\nPosts: 48\n\nKudos [?]: 33 [0], given: 19\n\nGMAT 1: 710 Q49 V38\nGRE 1: 326 Q166 V160\nGPA: 3.74\nRe: If p, x, and y are positive integers, y is odd, and p = x^2\u00a0[#permalink]\n\n### Show Tags\n\n29 Aug 2014, 05:25\nHi Karishma,\n\nThanks for the explanation to the question. I was just wondering how the answer would change if we change the question stem a little bit. What if the question asks if p (instead of x) is divisible by 4?\n\nIn this scenario, statement 1 would be sufficient since if something leaves a remainder of 5, it would leave a remainder of 1 upon division by 4\n\nFor statement 2, we know that x = y+3, so x is even. If we square it, it would surely be divisible by 4. Now if a number (y^2, which is odd) non-divisible by 4 is added to a number divisible by 4, the result would surely be not divisible by 4. So statement 2 would also be sufficient.\n\nIs this reasoning correct? just for practicing the concept\n\nKudos [?]: 33 [0], given: 19\n\nManager\nJoined: 22 Jan 2014\nPosts: 141\n\nKudos [?]: 74 [0], given: 145\n\nWE: Project Management (Computer Hardware)\nRe: If p, x, and y are positive integers, y is odd, and p = x^2\u00a0[#permalink]\n\n### Show Tags\n\n31 Aug 2014, 03:38\nnetcaesar wrote:\nIf p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?\n\n(1) When p is divided by 8, the remainder is 5.\n(2) x \u2013 y = 3\n\n1) p = 8k+5 (k is a whole number)\nalso p = (x^2+y^2)\n=> (x^2+y^2) mod 8 = 5\n\nany square (n^2) mod 8 follows the following pattern -> 1,4,1,0 and then repeats.\nfor getting x^2+y^2 mod 8 = 5\nwe need to take a 4 and a 1 from the above pattern. at multiples of 4, the remainder is 0. so x can never be divisible by 4.\nA is sufficient.\n\n2) x-y=3 (odd)\neven - odd = odd\nor odd - even = odd\nso insufficient.\n\nHence, A.\n_________________\n\nIllegitimi non carborundum.\n\nKudos [?]: 74 [0], given: 145\n\nRe: If p, x, and y are positive integers, y is odd, and p = x^2 \u00a0 [#permalink] 31 Aug 2014, 03:38\n\nGo to page \u00a0 \u00a01\u00a0\u00a0\u00a02\u00a0 \u00a0 Next \u00a0[ 32 posts ]\n\nDisplay posts from previous: Sort by", "date": "2017-10-19 05:56:09", "meta": {"domain": "gmatclub.com", "url": "https://gmatclub.com/forum/if-p-x-and-y-are-positive-integers-y-is-odd-and-p-x-82399.html?kudos=1", "openwebmath_score": 0.5733380317687988, "openwebmath_perplexity": 2726.3281205360695, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. Yes\n2. Yes\n\n", "lm_q1_score": 1.0, "lm_q2_score": 0.900529778109184, "lm_q1q2_score": 0.900529778109184}}
{"url": "https://math.stackexchange.com/questions/669085/what-does-curly-curved-less-than-sign-succcurlyeq-mean/669115", "text": "# What does \"curly (curved) less than\" sign $\\succcurlyeq$ mean?\n\nI am reading Boyd & Vandenberghe's Convex Optimization. The authors use curved greater than or equal to (\\succcurlyeq)\n\n$$f(x^*) \\succcurlyeq \\alpha$$\n\nand curved less than or equal to (\\preccurlyeq)\n\n$$f(x^*) \\preccurlyeq \\alpha$$\n\nCan someone explain what they mean?\n\n\u2022 as () is curved and {} are curly ( en.wikipedia.org/wiki/Bracket ), I think those symbols you mention are curved not curly Feb 9, 2014 at 6:43\n\u2022 @barlop If you look at $\\LaTeX$ source of the formulas in question (right click->Show Math As->TeX commands), you'll see \\succcurlyeq, which has curly word in it, not curved. Feb 9, 2014 at 7:18\n\u2022 Thanks everyone for answers. As I understand $\\succeq$ or $\\preceq$ are more general than their more popular counterparts. I think Michael's answer make sense. If I understand correctly, $X \\succeq Y, \\quad if \\quad \\| X \\| \\ge \\| Y \\|$ where $\\| \\cdot \\|$ is the norm associated with the space $X$ and $Y$ belongs to. I think Chris's answer is correct but is more strict condition than Michael's answer. Please correct me if I'm wrong. Feb 9, 2014 at 17:53\n\u2022 You should definitely take the tour. This is not a traditional forum!\n\u2013\u00a0bodo\nFeb 9, 2014 at 18:03\n\u2022 Well I don't know what a traditional forum looks like. :-) Feb 9, 2014 at 18:20\n\nBoth Chris Culter's and Code Guru's answers are good, and I've voted them both up. I hope that I'm not being inappropriate by combining and expanding upon them here.\n\nIt should be noted that the book does not use $\\succeq$, $\\preceq$, $\\succ$, and $\\prec$ with scalar inequalities; for these, good old-fashioned inequality symbols suffice. It is only when the quantities on the left- and right-hand sides are vectors, matrices, or other multi-dimensional objects that this notation is called for.\n\nThe book refers to these relations as generalized inequalities, but as Code-Guru rightly points out, they have been in use for some time to represent partial orderings. And indeed, that's exactly what they are, and the book does refer to them that way as well. But given that the text deals with convex optimization, it was apparently considered helpful to refer to them as inequalities.\n\nLet $S$ be a vector space, and let $K\\subset S$ be a closed, convex, and pointed cone with a non-empty interior. (By cone, we mean that $\\alpha K\\equiv K$ for all $\\alpha>0$; and by pointed, we mean that $K\\cap-K=\\{0\\}$.) Such a cone $K$ induces a partial ordering on the set $S$, and an associated set of generalized inequalities: $$x \\succeq_K y \\quad\\Longleftrightarrow\\quad y \\preceq_K x \\quad\\Longleftrightarrow\\quad x - y \\in K$$ $$x \\succ_K y \\quad\\Longleftrightarrow\\quad y \\prec_K x \\quad\\Longleftrightarrow\\quad x - y \\in \\mathop{\\textrm{Int}} K$$ This is a partial ordering because, for many pairs $x,y\\in S$, $x \\not\\succeq_K y$ and $y \\not\\succeq_K x$. So that's the primary reason why he and others prefer to use the curly inequalities to denote these orderings, reserving $\\geq$, $\\leq$, etc. for total orderings. But it has many of the properties of a standard inequality, such as: $$x\\succeq_K y \\quad\\Longrightarrow\\quad \\alpha x \\succeq_K \\alpha y \\quad\\forall \\alpha>0$$ $$x\\succeq_K y \\quad\\Longrightarrow\\quad \\alpha x \\preceq_K \\alpha y \\quad\\forall \\alpha<0$$ $$x\\succeq_K y, ~ x\\preceq_K y \\quad\\Longrightarrow\\quad x=y$$ $$x\\succ_K y \\quad\\Longrightarrow\\quad x\\not\\prec_K y$$\n\nWhen the cone $K$ is understood from context, it is often dropped, leaving only the inequality symbol $\\succeq$. There are two cases where this is almost always done. First, when $S=\\mathbb{R}^n$ and the cone $K$ is non-negative orthant $\\mathbb{R}^n_+$ the generalized inequality is simply an elementwise inequality: $$x \\succeq_{\\mathbb{R}^n_+} y \\quad\\Longleftrightarrow\\quad x_i\\geq y_i,~i=1,2,\\dots,n$$ Second, when $S$ is the set of symmetric $n\\times n$ matrices and $K$ is the cone of positive semidefinite matrices $\\mathcal{S}^n_+=\\{X\\in S\\,|\\,\\lambda_{\\text{min}}(X)\\geq 0\\}$, the inequality is a linear matrix inequality (LMI): $$X \\succeq_{\\mathcal{S}^n_+} Y \\quad\\Longleftrightarrow\\quad \\lambda_{\\text{min}}(X-Y)\\geq 0$$ In both of these cases, the cone subscript is almost always dropped.\n\nMany texts in convex optimization don't bother with this distinction, and use $\\geq$ and $\\leq$ even for LMIs and other partial orderings. I prefer to use it whenever I can, because I think it helps people realize that this is not a standard inequality with an underlying total order. That said, I don't feel that strongly about it for $\\mathbb{R}^n_+$; I think most people rightly assume that $x\\geq y$ is considered elementwise when $x,y$ are vectors.\n\n\u2022 Thanks a lot for the detailed answer, and for correcting the symbol as well :-). Feb 9, 2014 at 16:35\n\u2022 This is an old answer and I completely agree with it but I thought that providing another common application of this notation might be useful. As @Code-Guru pointed out, these are useful for partial orders. In Economics, we usually model preferences over baskets of goods with \"not worse than\" or \"not better than\" sets. The partial order concept fits like a glove in this situation. The interested reader might find books in Decision Sciences useful for this kind of discussion. Dec 30, 2020 at 21:13\n\nThere's a list of notation in the back of the book. On page 698, $x\\preceq y$ is defined as componentwise inequality between vectors $x$ and $y$. This means that $x_i\\leq y_i$ for every index $i$.\n\nEdit: The notation is introduced on page 32.\n\nOften these symbols represent partial order relations. The typical \"less than\" and \"greater than\" operations both define partial orders on the real numbers. However, there are many other examples of partial orders.\n\nSometimes the curly greater than sign is used to indicate positive semi-definiteness of a matrix $$X$$:\n\n$$(X\\succeq 0\\ \\text{or}\\ X\\ge 0)$$\n\nor a function $$f(x)$$\n\n$$(f(x) \\succeq 0\\ \\text{or}\\ f(x)\\ge 0).$$\n\n\u2022 Positive definiteness of what? Aug 26, 2015 at 17:03\n\nDoes it sometimes denote, in measurement theory; often the qualitative counterpart to $\\geq$ in the numerical representation; when one wants to numerically represent a totally ordered qualitative probability representation:\n\n$$A \u227d B \\leftrightarrow A > B \\leftrightarrow F(A) \\geq F(B)$$.\n\nOn the other hand. I have seen it used in multi-dimensional partial or even total, orderings under the \"ordering of major-ization\" for vector valued functions or for a system for two kinds of orderings.\n\nOne for (numerical) ordinal, $>$ comparisons and another, $\u227d$ for (numerical) differences, sums, or a some other kind of relation, to fine grain, the representation to ensure (or some kind of) unique-ness, rather than merely strong represent-ability.\n\nSee Marshall\n\nMarshall, Albert W.; Olkin, Ingram, Inequalities: theory of majorization and its applications, Mathematics in Science and Engineering, Vol. 143. New York etc.: Academic Press. XX, 569 p. \\$52.50 (1979). ZBL0437.26007. Such functions or representations, may use both the curly greater$\u227d$than and$\\geq $in the numerical or functional representation and are compatible with total orderings. $$a,b\\in \\Omega^{n}\\; a <b \\,,\\quad \\text{or}, \\quad, a= b, \\quad \\text{or}\\quad a < b$$ where one cannot usually adding up distinct $$a_1 \\in \\Omega_i\\,; b_j\\in \\Omega_j; j \\neq i$$ $$a_3 \\in \\Omega_c;\\, ; \\, P(a_i) + P(b_j) = P(a_3)$$. Sometimes I \"(conjecture)\" that$\u227d$denotes a weaker version of$(A)$or$(A_1)$(strong complementary add-itivity) which orders the differences and sums, is something over and above$(B)$;strict monotone increasing/strong representability/order embedding- above reflecting and preserving . $$(A)\\,[f(x_1)+f(x_2)\u227d f(y_1) + f(y_2)] \\rightarrow [f(x) \\geq f(y)] \\text{where} \\,; x=x_1 +x_2\\,, y=y_1+y_2$$ or $$(A1)\\,[x\\, \\geq y ]\\,\\rightarrow [f(x_1)+f(x_2)\u227d f(y_1) + f(y_2)]\\text{where} \\,; x=x_1 +x_2\\,, y=y_1+y_2$$ Rather than:$$(B)$$ $$(B)\\,x\\, \\geq y \\leftrightarrow f(x) \\geq f(y)\\, \\text{where} \\,; x=x_1 +x_2\\,, y=y_1+y_2$$ In contrast to$(B)$a standard order embedding (strict monotone increasing function) Where in$(B)$the numerical function,$F$or representation is 'merely strong'. That is$F$is a monotone strictly increasing function of some entity$x$where$F(x)$is the entity one wishes to order by$x$. Sometimes, even in a infinite and uniformly and non-atomic, continuous total order representation, nothing unique will come out. If its infinite /non-atomic in the wrong sense. ie in a super-atomic or entangled lexicographic system. Where the system is dense/or continuum dense/non-atomic/bottom-less in the wrong sense. Such as a spin$1/2$system in quantum mechanics, or a lexicographic entangled, system where its infinite in wrong sense, in the vertical, not in the horizontal, not within a basis of space/basis.$(A)$is arguably a lot stronger than$(B)$, or at least is, if relatively unrestricted. Where$(A)$, may be used not to extend the ordering so much as much as make a total ordering 'numerically precise; to put some constraints on sums and differences, where the events are the not kinds of things that usually add up. Say in an entangled multidimensional quantum spin 1/2 system or utility representation where the events are on complementary spaces and mixtures are not allowed, for example. That is, to, put a metric on differences (say on a two outcome system). That is something stronger, than a, total or non atomic/dense order, where(merely) every event. Even if the entire system is totally ordered within and betwixt the distinct$\\Omega\\$. So that the system is solv-able, and uniquely so.", "date": "2022-05-22 17:49:09", "meta": {"domain": "stackexchange.com", "url": "https://math.stackexchange.com/questions/669085/what-does-curly-curved-less-than-sign-succcurlyeq-mean/669115", "openwebmath_score": 0.8520500659942627, "openwebmath_perplexity": 663.221938754306, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9865717468373085, "lm_q2_score": 0.9124361527383704, "lm_q1q2_score": 0.9001837290846073}}
{"url": "http://mathhelpforum.com/pre-calculus/10298-piecewise-defined-function.html", "text": "# Math Help - Piecewise-Defined Function\n\n1. ## Piecewise-Defined Function\n\nFor both questions below:\n\n(a) Find the domain of the function.\n\n(b) Locate any intercepts.\n\n(1)\n\n.....{3 + x......if -3 <or= to x < 0\nf(x){3...........if......x = 0\n.....{Sqrt{x}..if......x > 0\n\n=======================\n\n(2)\n\n..........{1/x.........if.....x < 0\nf(x) = {sqrt{x}...if.....x >or= to 0\n\nNOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard.\n\nI hope you can understand the above.\n\n2. Originally Posted by symmetry\nFor both questions below:\n\n(a) Find the domain of the function.\n\n(b) Locate any intercepts.\n\n(1)\n\n.....{3 + x......if -3 <or= to x < 0\nf(x){3...........if......x = 0\n.....{Sqrt{x}..if......x > 0\n\n=======================\n\n(2)\n\n..........{1/x.........if.....x < 0\nf(x) = {sqrt{x}...if.....x >or= to 0\n\nNOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard.\n\nI hope you can understand the above.\nI'll do the first one for you- graph it. The conditions are the \"if\" parts in the piece-wise function. Domain is (-3, inf) and there are no intersepts. Try graphing it. You have a line with slope = 1 and an exponential function.\n\nEDIT: Sorry, there are x and y-intercepts, as Soroban pointed out, although the two graphs do not not intersect which is what I was getting at.\n\n3. Hello, symmetry!\n\nFor both questions below:\n. . (a) Find the domain of the function.\n. . (b) Locate any intercepts.\n\nDid you make a sketch?\n\n$(1)\\;\\;f(x) \\:=\\:\\begin{Bmatrix} 3 + x & &\\text{if }\\text{-}3 \\leq x < 0 \\\\ 3 & &\\text{if }x = 0 \\\\ \\sqrt{x} & &\\text{if }x > 0 \\end{Bmatrix}$\nCode:\n                |\n|            *\n*      *\n* |  *\n*   |*\n----*-----o--------------\n-3     |\nDomain: . $(\\text{-}3,\\,\\infty)$\n\nIntercepts: . $(\\text{-}3,0),\\:(0,3)$\n\n$(2)\\;\\;f(x)\\:=\\:\\begin{Bmatrix}\\frac{1}{x} & & \\text{if }x < 0 \\\\ \\sqrt{x} & & \\text{if }x \\geq 0 \\end{Bmatrix}$\nCode:\n                      |\n|            *\n|      *\n|   *\n|*\n------------------*----------------\n*               |\n*         |\n*    |\n*  |\n* |\n|\n*|\n|\nDomain: . $(-\\infty,\\,\\infty)$\n\nIntercepts: . $(0,\\,0)$\n\n4. ## ok\n\nThank you again both for your quick replies.\n\nTo soroban,\n\nNo, I did not sketch the graph because I do not know how to graph piecewise-defined functions.\n\nI understand these functions are graphed in parts, right?\n\nCan you take me through a sample graphing question in terms of this type of function?\n\nThanks!\n\n5. Originally Posted by symmetry\nThank you again both for your quick replies.\n\nTo soroban,\n\nNo, I did not sketch the graph because I do not know how to graph piecewise-defined functions.\n\nI understand these functions are graphed in parts, right?\n\nCan you take me through a sample graphing question in terms of this type of function?\n\nThanks!\n\nYes, they are 'graphed in parts,' I guess you could call it.\n\nFor instance,\n\nTake the first condition;\n\nf(x) = 3 + x if -3 <= x < 0\n\nFrom x = -3 (including this point) to x = 0 (not including, and thus draw an open circle by this point), you will graph 3 + x; see Soroban's graph. The reason why it's closed (solid dot) is because of the next condition later, and thus includes that point. Try look up piece-wise functions on Wikipedia.\n\n6. ## ok\n\nI like graphing functions. I think piecewise-defined functions are cool but not easy to sketch.\n\nThanks!\n\n7. Hello again, symmetry!\n\nOkay, here's an example.\n\n. . $f(x) \\:=\\:\\begin{Bmatrix}3 & \\text{if }0 \\leq x \\leq 1 \\\\ 2x + 1 & \\text{if }x > 1\\end{Bmatrix}$\n\nWhen $x$ is between $0$ and $1$ (including the endpoints),\n. . the graph is $f(x) = 3$, a horizontal line.\nCode:\n        |\n3* * * * *\n|\n|\n- + - - - + - -\n|       1\n\nWhen $x$ is greater than 1, the graph is $f(x) \\:=\\:2x + 1$,\n. . a slanted line.\nCode:\n        |\n|                   *\n|                *\n|             *\n|          *\n3|       *\n|\n|\n- + - - - + - - - - - - -\n|       1\n\nSketch them on the same graph\n. . and have the graph of the piecewise function.\nCode:\n        |\n|                   *\n|                *\n|             *\n|          *\n3o * * * *\n|\n|\n- + - - - + - - - - - - -\n|       1\n\nThis function could be your long-distance charge.\n\nThey might charge $3 for the first minute . . and$2 per minute for every subsequent minute.\n\n(Hmmm, not a good example . . .\nI'm sure someone will point out why.)", "date": "2015-01-29 16:41:51", "meta": {"domain": "mathhelpforum.com", "url": "http://mathhelpforum.com/pre-calculus/10298-piecewise-defined-function.html", "openwebmath_score": 0.6340510845184326, "openwebmath_perplexity": 2681.056311776446, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9653811641488385, "lm_q2_score": 0.9324533154301279, "lm_q1q2_score": 0.9001728671643809}}
{"url": "https://mathematica.stackexchange.com/questions/151050/mellin-transform-of-xp-seems-to-miss-a-factor-of-2-pi/170579", "text": "# Mellin transform of $x^p$ seems to miss a factor of $2\\pi$\n\nBug introduced in 11.1 or earlier and fixed in 11.3\n\nOn Mathematica 11.1.1.0 the Mellin transform of $x^p$ is evaluated as $\\delta(p+s)$, while I think it should be $2\\pi\\,\\delta(p+s)$:\n\nIn:= MellinTransform[x^p, x, s, GenerateConditions -> True]\nOut:= DiracDelta[p + s]\n\n\nedited posting after Daniel Lichtblau's comment\n\nI initially did not understand this result, but this 2004 paper has explained to me how to arrive at the Dirac delta function, however, with an additional factor of $2\\pi$. I checked that this is not a matter of a different definition of the Mellin transform. (I summarized the calculation in this Mathoverflow posting.)\n\nMissing factor $2\\pi$ is fixed in Mathematica 11.3.0:\n\n In:= MellinTransform[x^p, x, s, GenerateConditions -> True]\nOut:= 2\u03c0 DiracDelta[i(p + s)]\n\n\nconsequence: before 11.3 Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] returned 1, now it returns $2\\pi\\int_{-\\infty}^\\infty\\delta(is)ds$\nQ: is this v. 11.3 change in the implementation of MellinTransform documented somewhere?\n\n\u2022 See last example in documentation under Scope Elementary Functions. It should be noted that this is a generalization of the integral definition, not unlike the case for FourierTransform. \u2013\u00a0Daniel Lichtblau Jul 9 '17 at 15:27\n\u2022 thank you, Daniel, for the feedback, I understand things a bit better now and have edited my posting accordingly --- my problem has been reduced to a missing factor $2\\pi$... \u2013\u00a0Carlo Beenakker Jul 9 '17 at 19:06\n\u2022 What specific definition is used is not particularly important so long as the MellinTransform and InverseMellinTransform are inverses of each other. Both x^p == InverseMellinTransform[ MellinTransform[x^p, x, s], s, x] and DiracDelta[p + s] == MellinTransform[ InverseMellinTransform[DiracDelta[p + s], s, x], x, s] evaluate to True \u2013\u00a0Bob Hanlon Jul 9 '17 at 22:41\n\u2022 @BobHanlon --- but if we assume that the factor of $2\\pi$ is absorbed in the definition of DiracDelta, then Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] should return $2\\pi$, while instead it returns 1. \u2013\u00a0Carlo Beenakker Jul 10 '17 at 6:18\n\u2022 The integral of DiracDelta should be one. \u2013\u00a0Bob Hanlon Jul 10 '17 at 14:48\n\nThe Mellin transforms for $x^j$ reported by Mathematica 11.2 didn't make sense to me, so on 11/28/2017 I submitted the following question on Math StackExchange.\n\nQuestions on Mellin Transform of $x^j$ and Interpretation of Distributions with Complex Arguments\n\nI ended up deriving the answer to my own question and on 12/7/2017 I submitted a problem report to Wolfram technical support where I attached a Mathematica notebook illustrating the problem and the correct solution (CASE:3980660).\n\nI received an email from Wolfram technical support on 12/13/2017 indicating my analysis was accepted as correct and a report was being filed with the developers. The correct solution was subsequently implemented in Mathematica 11.3.\n\nNote that not only was the $2\\,\\pi$ prefix missing, but $i$ was also missing in the $\\delta$ function parameter.\n\nI subsequently posted the correct solution in answers to related questions on both Math StackExchange and MathOverflow StackExchange.\n\nDelta function with imaginary argument\n\nDirac Delta function with a complex argument", "date": "2020-07-15 02:22:58", "meta": {"domain": "stackexchange.com", "url": "https://mathematica.stackexchange.com/questions/151050/mellin-transform-of-xp-seems-to-miss-a-factor-of-2-pi/170579", "openwebmath_score": 0.45609450340270996, "openwebmath_perplexity": 1455.502810609982, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9763105314577313, "lm_q2_score": 0.9219218402181233, "lm_q1q2_score": 0.9000820017858455}}
{"url": "https://www.physicsforums.com/threads/finding-horizontal-tangent-planes-on-s.556301/", "text": "# Finding Horizontal Tangent Planes on S\n\n1. Dec 2, 2011\n\n### TranscendArcu\n\n1. The problem statement, all variables and given/known data\nS is the surface with equation $$z = x^2 +2xy+2y$$a) Find an equation for the tangent plane to S at the point (1,2,9).\nb) At what points on S, in any, does S have a horizontal tangent plane?\n\n3. The attempt at a solution\n$$F(x,y,z): z = x^2 +2xy+2y$$\n$$F_x = 2x + 2y$$\n$$F_y = 2x + 2$$\n\nEvaluated at (1,2) gives answers 6 and 4, respectively. My equation for a plane is:\n\n$$z-9=6(x-1) + 4(y-1)$$.\n\nI think any horizontal plane should have normal vector <0,0,k>, where k is some scalar. I'm pretty sure that S has no such normal vector. But if\n$$F(x,y,z): 0 = x^2 +2xy+2y - z$$\nthen\n$$grad F = <2x + 2y,2x + 2,-1>$$ It seems like I can let (x,y) = (-1,1) to zero the x-, y-components of the gradient. Plugging (-1,1) into the definition of z gives z = 1. This suggests to me that there is a point (-1,1,1), at which there is a horizontal tangent plane. Yet I feel pretty sure that this isn't true!\n\n2. Dec 3, 2011\n\n### ehild\n\nYou made a little mistake when writing out the equation of the tangent plane. The y coordinate of the fixed point is 2, you wrote 1.\n\nA surface in 3D is of the form F(x,y,z) = constant. For this surface, x2+2xy+2y-z=0. That means F(x,y,z)=x2+2xy+y-z. The gradient of F is normal to the surface, and the tangent plane of the surface at a given point. You want a horizontal tangent plane, so a vertical gradient:(0,0,a). That means Fx=2x+2y=0, Fy=2x+2=0 --->x=-1, y=1, so your result for the x,y coordinates are correct. Plugging into the original equation for x and y, you got z=x2+2xy+2y=1, it is correct. Why do you feel it is not?\n\nehild\n\n3. Dec 3, 2011\n\n### TranscendArcu\n\nWhen I graphed F(x,y,z) in MatLab (and it's possible I graphed it incorrectly), I observed that the the resulting paraboloid is always \"tilted\". Below is a picture from my plot:\nhttp://img440.imageshack.us/img440/687/skjermbilde20111203kl85.png [Broken]\nHow can this surface have a horizontal tangent anywhere when it is tilted like this?\n\nLast edited by a moderator: May 5, 2017\n4. Dec 3, 2011\n\n### ehild\n\nTry to plot z out for -2<x<0 and 0<y<2\n\nehild\n\n5. Dec 3, 2011\n\n### TranscendArcu\n\nhttp://img7.imageshack.us/img7/139/skjermbilde20111203kl10.png [Broken]\nHmm. I'm not seeing the a point in this picture where the gradient is pointing directly upwards. Everything still looks kind of tilted.\n\nLast edited by a moderator: May 5, 2017\n6. Dec 3, 2011\n\n### ehild\n\nThe function is equivalent with z=(x+2y-1)(x+1)+1 and z=1 along the lines x=-1 and y=(1-x)/2. I attach a plot of the surface near the point (-1,1)\n\nehild\n\nFile size:\n81.6 KB\nViews:\n127\n7. Dec 4, 2011\n\n### HallsofIvy\n\nStaff Emeritus\nI presume that by \"horizontal\" you mean perpendicular to the z-axis.\n\nThe simplest way to find a tangent planes for a surface is to write it in the form F(x,y,z)= constant. Then the normal to the tangent plane at any point is given by $\\nabla F$. Here, you can write $F(x, y,z)= x^2+ 2xy+ 2y- z= 0$.\n\nWhat is $\\nabla F$? That will be vertical (and so tangent plane horizontal) when its x and y components are 0.\n\n8. Dec 4, 2011\n\n### ehild\n\n@HallsofIvy: The OP has shown the solution in his first post, he only can not believe it, as the surfaces he got with MatLab look tilted. If you could give advice how to plot surfaces with MatLab, that would be real help for him.\n\nehild\n\n9. Dec 4, 2011\n\n### TranscendArcu\n\nIf anyone is familiar with MatLab, this is the code I've been using:\n\nNote that the \"%\" mark my annotations. I included them so that hopefully you can follow what I'm doing more easily.\n\n10. Dec 4, 2011\n\n### TranscendArcu\n\nHa! I figured it out. I forgot a \".\"\n\nI should have written\n\nz = x.^2 + 2*x.*y +2*y;\n\nEverything makes sense now.\n\n11. Dec 4, 2011\n\n### ehild\n\nYou see: it is worth typing something out again and again. Is your plot similar to my one? It was made with Origin. I would like to see your final plot... Please...\n\nehild\n\n12. Dec 4, 2011\n\n### TranscendArcu\n\nhttp://img259.imageshack.us/img259/2104/skjermbilde20111204kl10.png [Broken]It looks like it could have a horizontal tangent plane right around (-1,1,1)\n\nLast edited by a moderator: May 5, 2017\n13. Dec 4, 2011\n\n### ehild\n\nIt is really nice!!! And a missing dot made you sceptical about the truth of Maths??!!!:uhh:\n\nehild", "date": "2017-09-20 22:38:11", "meta": {"domain": "physicsforums.com", "url": "https://www.physicsforums.com/threads/finding-horizontal-tangent-planes-on-s.556301/", "openwebmath_score": 0.638455867767334, "openwebmath_perplexity": 1154.9442875848038, "lm_name": "Qwen/Qwen-72B", "lm_label": "1. YES\n2. YES", "lm_q1_score": 0.9811668728630677, "lm_q2_score": 0.9173026499774933, "lm_q1q2_score": 0.9000269725474223}}