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77550 | Prove that [imath]\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0[/imath], [imath]x \in \Bbb R[/imath].
Why is [imath]\lim_{n \to \infty} \frac{2^n}{n!}=0\text{ ?}[/imath] Can we generalize it to any exponent [imath]x \in \Bbb R[/imath]? This is to say, is [imath]\lim_{n \to \infty} \frac{x^n}{n!}=0\text{ ?}[/imath] This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions. and here: List of abstract duplicates. | 352360 | How to prove that [imath]\lim_{n\to\infty}\frac{a^n}{n!}=0[/imath]?
What I have got: For any [imath]a\in\mathbb{R}[/imath], we can find an [imath]N[/imath], [imath]N\gt a[/imath], such that, [imath]\lim_{n\to\infty}\frac{a^n}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdot\ldots\cdot\frac{a}{N-1}\cdot\frac{a}{N}\cdot\frac{a}{N+1}\cdot\ldots,[/imath] and the later part [imath]\frac{a}{N}\cdot\frac{a}{N+1}\cdot\ldots\lt\frac{a^m}{n^m}\to 0[/imath] when [imath]m\to\infty[/imath], because [imath]\frac{a}{n}\lt1[/imath]. Then use the definition of convergence, with letting [imath]M=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdot\ldots\cdot\frac{a}{N-1}[/imath], and [imath]\frac{a^m}{n^m}\lt\delta[/imath] for [imath]\delta=\frac{\epsilon}{M}[/imath]. For an n that is large enough, we have [imath]\frac{a^n}{n!}\lt\epsilon[/imath], [imath]\forall\epsilon\gt 0[/imath]. What I want to ask is: Are there any other ways to prove that this limit tends to [imath]0[/imath]? Thanks! |
37647 | If [imath]\sum a_n b_n <\infty[/imath] for all [imath](b_n)\in \ell^2[/imath] then [imath](a_n) \in \ell^2[/imath]
I'm trying to prove the following: If [imath](a_n)[/imath] is a sequence of positive numbers such that [imath]\sum_{n=1}^\infty a_n b_n<\infty[/imath] for all sequences of positive numbers [imath](b_n)[/imath] such that [imath]\sum_{n=1}^\infty b_n^2<\infty[/imath], then [imath]\sum_{n=1}^\infty a_n^2 <\infty[/imath]. The context here is functional analysis homework, in the subject of Hilbert spaces. Here's what I've thought: Let [imath]f=(a_n)>0[/imath]. Then the problem reads: if [imath]\int f\overline{g}<\infty[/imath] for all [imath]g>0,g\in \ell^2[/imath], then [imath]f\in \ell^2[/imath]. This brings the problem into the realm of [imath]\ell^p[/imath] spaces. I know the inner product is defined only in [imath]\ell^2[/imath], but it's sort of like saying: if [imath]\langle f,g\rangle <\infty[/imath] for all [imath]g>0,g\in \ell^2[/imath] then [imath]f\in \ell^2[/imath]. I read this as: "to check a positive sequence is in [imath]\ell^2[/imath], just check its inner product with any positive sequence in [imath]\ell^2[/imath] is finite, then you're done", which I find nice, but I can't prove it :P From there, I don't know what else to do. I thought of Hölder's inequality which in this context states: [imath]\sum_{n=1}^\infty a_nb_n \leq \left( \sum_{n=1}^\infty a_n^2 \right)^{1/2} \left( \sum_{n=1}^\infty b_n^2 \right)^{1/2}[/imath] but it's not useful here. | 359833 | Prove that the sequence is in [imath]\ell^{2}[/imath].
Let [imath](a_{n})[/imath] be a sequence of complex numbers such that for every [imath](b_{n})\in \ell^{2}[/imath]the series [imath]\sum_{1}^{\infty}a_{n}b_{n}[/imath] converges. Prove that [imath](a_{n})\in \ell^{2}.[/imath] What I've tried so far is Let [imath]T(b)=\sum_{n=1}^{\infty}a_{n}b_{n}[/imath] where [imath]b=(b_{n})[/imath] then [imath]T[/imath] is a linear operator on [imath]\ell^{2}[/imath]. So if I can somehow show that [imath]T[/imath] is continuous then I can apply Riez representation theorem for Hilbert spaces can conclude that [imath](a_{n})\in \ell^{2}[/imath]. I'm thinking maybe something along the line of Principle of uniform boundedness but I'm not getting anywhere with that. |
191165 | Proving an equality involving compositions of an integer
Let's consider various representations of a natural number [imath]n \geq 4[/imath] as a sum of positive integers, in which the order of summands is important (i.e. compositions). The task is to prove the number [imath]3[/imath] appears altogether [imath]n2^{n-5}[/imath] times in all of them. I know there're [imath]2^{n-1}[/imath] compositions of [imath]n[/imath]. However, I have no clue as to how to count only those involving the number(s) [imath]3[/imath]. I can't think of any sensible generating function for this. Maybe there's a nice combinatorial interpretation of the given formula, which I can't figure out? Could anyone lend me a hand with handling this problem? | 306541 | Number of times the integer [imath]3[/imath] occurs in all [imath]2^{n-1}[/imath] compositions of [imath]n[/imath]
Suppose [imath]n\ge4[/imath]. Show that in a list of all [imath]2^{n-1}[/imath] compositions of [imath]n[/imath], the integer [imath]3[/imath] occurs exactly [imath]n2^{n-5}[/imath] times. [Hint: Look at ways of drawing lines between n dots.] The number of k-term compositions of [imath]n[/imath] is [imath]\binom{n−1}{k−1}[/imath], for [imath]k\le n[/imath]. We have the two lines [imath]3[/imath] apart and then some combinations on the left of the lines and some combinations on the right of the lines to count but I can't figure out what they are. Any help would be greatly appreciated! |
60590 | Category-theoretic limit related to topological limit?
Is there any connection between category-theoretic term 'limit' (=universal cone) over diagram, and topological term 'limit point' of a sequence, function, net...? To be more precise, is there a category-theoretic setting of some non-trivial topological space such that these different concepts of term 'limit' somehow relate? This question came to me after I saw ( http://www.youtube.com/watch?v=be7rx29eMr4 ) a surprising fact that generalised metric spaces can be seen as categories enriched over preorder [imath]([0,\infty],\leq)[/imath]. | 420753 | Relation between inverse limits (and direct limits) with limits in calculus.
What is the relation between inverse limit (and direct limit) with limits in calculus? Are there some special cases that an inverse limit (or direct limit) is a limit in calculus (for example, the limit of a sequence [imath]a_0, a_1, \ldots[/imath]). Thank you very much. |
47009 | Proof of [imath]\frac{(n-1)S^2}{\sigma^2} \backsim \chi^2_{n-1}[/imath]
It's a standard result that given [imath]X_1,\cdots ,X_n [/imath] random sample from [imath]N(\mu,\sigma^2)[/imath], the random variable [imath]\frac{(n-1)S^2}{\sigma^2}[/imath] has a chi-square distribution with [imath](n-1)[/imath] degrees of freedom, where [imath]S^2=\frac{1}{n-1}\sum^{n}_{i=1}(X_i-\bar{X})^2.[/imath] I would like help in proving the above result. Thanks. | 1773969 | Deriving [imath](n-1)\frac{S^2}{\sigma^2} \sim \chi^2(n-1)[/imath]
I can accept the fact that [imath]Z^2 = \dfrac{\left(X-\mu\right)^2}{\sigma^2} \sim \chi^2(1)[/imath] without knowing too much about this mysterious [imath]\chi[/imath]-function, but I'm wondering how I can show that [imath](n-1)\dfrac{S^2}{\sigma^2} \sim \chi^2(n-1)[/imath] from this? Let [imath]X_1 + X_2+\cdots+X_n[/imath] be independent normally distributed variables: [imath]S = \dfrac{1}{n-1}\sum_{i=1}^n \left(X_i-\bar X\right)^2[/imath] [imath]W^2 = \dfrac{(\sum_{i=0}^n X_i -\bar X)^2}{\sigma^2} = \;? = \dfrac{n-1}{n-1}\dfrac{\sum\left(X_i-\bar X\right)^2}{\sigma^2} = \dfrac{(n-1)S^2}{\sigma^2} \overset{?}{\sim} \chi^2(n-1)[/imath] I'm assuming some of this is wrong, and I don't really understand the argument of [imath]\chi[/imath] (why isn't it [imath]\chi(n)[/imath]?), but I'm mostly interested in seeing a better derivation of the last result, if there is one. Any help is appreciated! |
41303 | Examples and further results about the order of the product of two elements in a group
Let [imath]G[/imath] be a group and let [imath]a,b[/imath] be two elements of [imath]G[/imath]. What can we say about the order of their product [imath]ab[/imath]? Wikipedia says "not much": There is no general formula relating the order of a product [imath]ab[/imath] to the orders of [imath]a[/imath] and [imath]b[/imath]. In fact, it is possible that both [imath]a[/imath] and [imath]b[/imath] have finite order while [imath]ab[/imath] has infinite order, or that both [imath]a[/imath] and [imath]b[/imath] have infinite order while [imath]ab[/imath] has finite order. On the other hand no examples are provided. [imath](\mathbb{Z},+), 1[/imath] and [imath]-1[/imath] give an example of elements of infinite order with product of finite order. I can't think of any example of the other kind! So: What's an example of a group [imath]G[/imath] and two elements [imath]a,b[/imath] both of finite order such that their product has infinite order? Wikipedia then states: If [imath]ab = ba[/imath], we can at least say that [imath]\mathrm{ord}(ab)[/imath] divides [imath]\mathrm{lcm}(\mathrm{ord}(a), \mathrm{ord}(b))[/imath] which is easy to prove, but not very effective. So: What are some similar results about the order of a product, perhaps with some additional hypotheses? | 314850 | Example of a group where [imath]o(a)[/imath] and [imath]o(b)[/imath] are finite but [imath]o(ab)[/imath] is infinite
Let G be a group and [imath]a,b \in G[/imath]. It is given that [imath]o(a)[/imath] and [imath]o(b)[/imath] are finite. Can you give an example of a group where [imath]o(ab)[/imath] is infinite? |
117998 | Finding the error in this proof that 1=2
I have a "proof" that has an error in it and my goal is to figure out what this error is. The proof: If [imath]x = y[/imath], then [imath] \begin{eqnarray} x^2 &=& xy \nonumber \\ x^2 - y^2 &=& xy - y^2 \nonumber \\ (x + y)(x - y) &=& y(x-y) \nonumber \\ x + y &=& y \nonumber \\ 2y &=& y \nonumber \\ 2 &=& 1 \end{eqnarray} [/imath] My best guess is that the error starts with the line [imath]2y = y[/imath]. If we accept that [imath]x + y = y[/imath] is true, then [imath] \begin{eqnarray} x + y &=& y \\ x &=& y - y \\ x &=& y = 0 \end{eqnarray} [/imath] Did I find the error? If not, am I close? | 389180 | Where is wrong in this proof
Suppose [imath]a=b[/imath]. Multiplying by [imath]a[/imath] on both sides gives [imath]a^2 = ab[/imath]. Then we subtract [imath]b^2[/imath] on both sides, and get [imath]a^2-b^2 = ab-b^2[/imath]. Obviously, [imath](a-b)(a+b) = b(a-b)[/imath], so dividing by [imath]a - b[/imath], we find [imath]a+b = b[/imath]. Now, suppose [imath]a=b=1[/imath]. Then [imath]1=2[/imath] :) |
93453 | How to prove that [imath]\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}][/imath], for [imath]p_i[/imath] prime?
This is Exercise 18.14 from Algebra, Isaacs. [imath]p_{1}\ ,\ p_{2}\ ,\ ... p_{n}[/imath] are different prime numbers. How to show that [imath]\mathbb{Q}[\sqrt{p_{1}}, \sqrt{p_{2}}, \ldots, \sqrt{p_{n}} ] = \mathbb{Q}[\sqrt{p_{1}}+ \sqrt{p_{2}}+\cdots + \sqrt{p_{n}}] \quad ?[/imath] First we can note that the Galois group of the first extension over [imath]\mathbb{Q}[/imath] ( which I call [imath]E[/imath] ) is elementary abelian of order [imath]2^{n}[/imath], so we can prove that the orbit of [imath]\sqrt{p_{1}}+\sqrt{p_{2}}+ ... + \sqrt{p_{n}}[/imath] under [imath]\operatorname{Gal}(E/\mathbb{Q})[/imath] contains [imath]2^{n}[/imath] elements, but how to do so? | 1096809 | Is [imath]\mathbb Q(\sqrt{2},\sqrt{3},\sqrt{5})=\mathbb Q(\sqrt{2}+\sqrt{3}+\sqrt{5})[/imath].
Is [imath]\mathbf Q(\sqrt 2,\sqrt 3,\sqrt 5)=\mathbf Q(\sqrt 2+\sqrt 3+\sqrt 5)[/imath]? Say [imath]L=\mathbf Q(\sqrt 2,\sqrt 3,\sqrt 5)[/imath] and [imath]K=\mathbf Q(\sqrt 2+\sqrt 3+\sqrt 5)[/imath]. It is easy to show that [imath]\mathbf Q(\sqrt 2,\sqrt 3)=\mathbf Q(\sqrt 2+\sqrt 3)[/imath]. Also, [imath][L:\mathbf Q]=8[/imath]. If we assume that [imath]L\neq K[/imath], then we will have [imath][K:\mathbf Q]=4[/imath]. The reason for this is that [imath]K(\sqrt 5)=L[/imath]. Now since [imath]K[/imath] is a superfield of [imath]\mathbf Q[/imath], and [imath][\mathbf Q(\sqrt 5):\mathbf Q]=2[/imath], we have [imath][K(\sqrt 5):K]\leq 2[/imath]. Since we have assume that [imath]K\neq L[/imath], we must have [imath][K(\sqrt 5):K]=2[/imath]. By the tower law, [imath][K(\sqrt 5):K][K:\mathbf Q]=[K(\sqrt 5):\mathbf Q]=[L:\mathbf Q]=8[/imath], giving [imath][K:\mathbf Q]=4[/imath]. I am not able to make any further progress. Generalization: Let [imath]p_1,\ldots,p_n[/imath] be pairwise distinct primes. Is [imath]\mathbf Q(\sqrt p_1+\cdots+\sqrt p_n)=\mathbf Q(\sqrt p_1,\ldots,\sqrt p_n)[/imath]? Just something I think might be useful in solving the above: It is known that if [imath]p_1,\ldots,p_n[/imath] are pairwise distinct primes, then [imath][\mathbf Q(\sqrt p_1,\ldots,\sqrt p_n):\mathbf Q]=2^n[/imath]. |
27539 | Proving a special case of the binomial theorem: [imath]\sum^{k}_{m=0}\binom{k}{m} = 2^k[/imath]
I want to know if I can get some help with this proof. I tried, but I just cannot seem to get [imath]2^{k}[/imath]. It states that, For [imath]k \in \mathbb{Z}_{\ge 0}[/imath], [imath]\sum^{k}_{m=0}\binom{k}{m} = 2^k[/imath] Thank You. | 177405 | Prove by induction: [imath]2^n = C(n,0) + C(n,1) + \cdots + C(n,n)[/imath]
This is a question I came across in an old midterm and I'm not sure how to do it. Any help is appreciated. [imath]2^n = C(n,0) + C(n,1) + \cdots + C(n,n).[/imath] Prove this statement is true for all [imath]n \ge 0[/imath] by induction. |
30111 | Unique quadratic subfield of [imath]\mathbb{Q}(\zeta_p)[/imath] is [imath]\mathbb{Q}(\sqrt{p})[/imath] if [imath]p \equiv 1[/imath] [imath](4)[/imath], and [imath]\mathbb{Q}(\sqrt{-p})[/imath] if [imath]p \equiv 3[/imath] [imath](4)[/imath]
I want to prove the assertion: The unique quadratic subfield of [imath]\mathbb{Q}(\zeta_p)[/imath] is [imath]\mathbb{Q}(\sqrt{p})[/imath] when [imath]p \equiv 1 \pmod{4}[/imath], respectively [imath]\mathbb{Q}(\sqrt{-p})[/imath] when [imath]p \equiv 3 \pmod{4}[/imath]. My first attempt is this. In [imath]\mathbb{Z}[\zeta_p][/imath], [imath]1-\zeta_p[/imath] is prime and [imath] p = \epsilon^{-1} (1-\zeta_p)^{p-1} [/imath] where [imath]\epsilon[/imath] is a unit. Since [imath]p[/imath] is an odd prime, [imath](p-1)/2[/imath] is an integer and [imath] \sqrt{\epsilon p } = (1-\zeta_p)^{(p-1)/2} [/imath] makes sense and belongs to [imath]\mathbb{Z}[\zeta_p][/imath]. How do I deal with the [imath]\epsilon[/imath] under the square root? I guess the condition on the congruence class of [imath]p[/imath] comes from that. Is this even the right way to proceed? Uniqueness is not clear to me either. I thought about looking at the valuation [imath]v_p[/imath] on [imath]\mathbb{Q}[/imath], extending it to two possible quadratic extensions beneath [imath]\mathbb{Q}(\zeta_p)[/imath], then seeing how those have to extend to common valuations on [imath]\mathbb{Q}(\zeta_p)[/imath], but I didn't see how to make it work. I would appreciate some help. | 2812412 | [imath]\xi =e^{2*\pi/pi }[/imath] , proof that is only two quadratic extention to [imath]\Bbb Q [/imath] that contained in [imath]Q(\xi_p)[/imath]
[imath]\xi_p =e^{2\pi/p*i }[/imath] , proof that is only two quadratic extention to [imath]\Bbb Q [/imath] that contained in [imath]Q(\xi_p)[/imath] such that [imath]\Bbb Q(\sqrt p)[/imath] if [imath]p=1 mod (4)[/imath] or [imath]\Bbb Q(\sqrt{-p})[/imath] when [imath]p=3mod(4)[/imath] In the class we proof that [imath]p*(-1/p)=z^2 [/imath] when [imath]z=\sum_{a\in \Bbb F^x_p}(a/p)*\xi_p^a[/imath] when (-1/p) = 1 if p=1 mod 4 and =-1 if p=3 mod 4 so it`s how I proof the statment. but how to show that this is the only two that possible according to the statment? I try by contradiction for [imath]z\in \Bbb Q [/imath] such that [imath]\sqrt z \not= \sqrt p [/imath] so [imath]\sum_{i=1}^N a_i*\xi_p^i=\sqrt z[/imath] for some [imath]a_i\in \Bbb Q [/imath] but Its lead me to nowhere... |
134714 | Does [imath]|x|^p[/imath] with [imath]0 satisfy the triangle inequality on \mathbb{R}?[/imath]
I am curious about whether [imath]|x|^p[/imath] with [imath]0<p<1[/imath] satisfy [imath]|x+y|^p\leq|x|^p+|y|^p[/imath] for [imath]x,y\in\mathbb{R}[/imath]. So far my trials show that this seems to be right... So can anybody confirm whether this is right or wrong and give a proof or counterexample? | 1763343 | Root distance function in Metric space
Let [imath]\mathbf X = \Bbb R[/imath] with distance function defined by [imath]d(x,y) = {|x-y|}^\alpha[/imath] , where [imath]\alpha \in \Bbb R[/imath] [imath](0<\alpha\le1)[/imath]. Prove that [imath](\Bbb R , d)[/imath] is a metric space. The first three properties are easy, so I only need the triangle inequality. I tried to use Bernoulli's inequality, but did not lead me to the proof. [imath]d(x,y) \le d(x,z) + d(z,y)[/imath] Please help me, Spanish is my language. |
182888 | Limit of [imath]\frac{1}{x} - \frac{1}{\sin{(x)}}[/imath]
Prove, without using l'Hôpital's Rule, that [imath]\lim\limits_{x \to 0}{\dfrac{1}{x} - \dfrac{1}{\sin{(x)}}} = 0[/imath]. I proved that there exists a [imath]s >0[/imath] such that [imath]\forall x \in (-s,s)[/imath] [imath]\Rightarrow[/imath] [imath]\dfrac{1}{x} - \dfrac{1}{\sin{(x)}} > 0[/imath] if [imath]x<0[/imath] and [imath]\dfrac{1}{x} - \dfrac{1}{\sin{(x)}} < 0[/imath] if [imath]x>0[/imath]. Therefore, this limit exist and it is equal zero, or doesn’t exist. But it is only thing I could do. | 94864 | What is the result of [imath]\lim\limits_{x \to 0}(1/x - 1/\sin x)[/imath]?
Find the limit: [imath]\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)[/imath] I am not able to find it because I don't know how to prove or disprove [imath]0[/imath] is the answer. |
282045 | Three linked question on non-negative definite matrices.
1.a symmetric matrix in [imath]\mathbb{M}_n(\mathbb{R})[/imath] is said to be non-negative definite if [imath]x^Tax≥0[/imath] for all (column) vectors [imath]x\in \mathbb{R}^n[/imath]. Which of the following statements are true? (a) If a real symmetric [imath]n\times n[/imath] matrix is non-negative definite, then all of its eigenvalues are non-negative. (b) If a real symmetric [imath]n\times n[/imath] matrix has all of its eigenvalues are non-negative , then it is non-negative definite. (c) If [imath] A\in \mathbb{M}_n(\mathbb{R})[/imath], then [imath]AA^T[/imath] is non-negetive definite. 2. only one of the following matrices is non-negetive definite. Find it. (a) [imath]\begin{pmatrix} 5 & -3 \\ -3 & 5 \end{pmatrix}[/imath]. (b) [imath]\begin{pmatrix} 1 & -3 \\ -3 & 5 \end{pmatrix}[/imath]. (c) [imath]\begin{pmatrix} 1 & 3 \\ 3 & 5 \end{pmatrix}[/imath]. 3.let [imath]B[/imath] be the real symmetric non-negative definite [imath]2×2[/imath] matrix such that [imath]B^2=A[/imath] where [imath]A[/imath] where is the non-negetive definite matrix in question [imath]2[/imath].write down the characteristic polynomial of [imath]B[/imath]. My thoughts.. For1. (a) & (b) are true but not sure about (c). For 2. (a) is the correct option since it has all positive eigen values.they are [imath]2,8[/imath]. For 3. Eigen values of [imath]B[/imath] will be [imath]\sqrt{2}[/imath] and [imath]\sqrt{8}[/imath]. So the answer is [imath]x^2-3√2x+4=0[/imath]. Can anybody help me to verify the solutions of the above problems. | 311686 | a multiple choice question on non-negative definite matrices
A symmetric matrix in [imath]\mathbb{M}_n(\mathbb{R})[/imath] is said to be non-negative definite if [imath]x^TAx≥0[/imath] for all (column) vectors [imath]x \in \mathbb{R}^n[/imath]. Which of the following statements are true? a. If a real symmetric [imath]n×n[/imath] matrix is non-negative definite, then all of its eigenvalues are non-negative. b. If a real symmetric [imath]n×n[/imath] matrix has all its eigenvalues non-negative, then it is non-negative definite. c. If [imath]A\in\mathbb{M}_n(\mathbb{R})[/imath], then [imath]AA^T[/imath] is non-negative definite. I know that all of the above options are correct but I did not have proof of any one of them.can anybody tell me about the proof of them. |
9776 | How to raise a complex number to the power of another complex number?
How do I calculate the outcome of taking one complex number to the power of another, ie [imath]\displaystyle {(a + bi)}^{(c + di)}[/imath]? | 1027646 | How to calculate [imath]i^i[/imath]
I've been struggling with this problem, actually I was doing a program in python and did 1j ** 1j(complex numbers) (In python a**b = [imath]a^b[/imath] ) and found out the answer to be a real number with value [imath]0.2079[/imath], How to calculate this value of [imath]i^i[/imath]? |
101157 | A commutative ring is a field iff the only ideals are [imath](0)[/imath] and [imath](1)[/imath]
Let [imath]R[/imath] be a commutative ring with identity. Show that [imath]R[/imath] is a field if and only if the only ideals of [imath]R[/imath] are [imath]R[/imath] itself and the zero ideal [imath](0)[/imath]. I can't figure out where to start other that I need to prove some biconditional statement. Any help? | 504192 | Every field has exactly 2 ideals
I want to prove that every field has exactly 2 ideals. I know that the two ideals are [imath]0[/imath] and the whole set itself. That is clear to me. I am a little unsure how to prove it though. |
35598 | Why are addition and multiplication commutative, but not exponentiation?
We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not. My question is why. As background there are plenty of mathematical schemes that can be used to define these operators. One of these is hyperoperation where [imath]H_0(a,b) = b+1[/imath] (successor op) [imath]H_1(a,b) = a+b[/imath] (addition op) [imath]H_2(a,b) = ab [/imath] (multiplication op) [imath]H_3(a,b) = a^b[/imath] (exponentiation op) [imath]H_4(a,b) = a\uparrow \uparrow b[/imath] (tetration op: [imath]a^{(a^{(...a)})}[/imath] nested [imath]b[/imath] times ) etc. Here it is not obvious to me why [imath]H_1(a,b)=H_1(b,a)[/imath] and [imath]H_2(a,b)=H_2(b,a)[/imath] but not [imath]H_3(a,b)=H_3(b,a)[/imath] Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion? Thanks. | 2077612 | Why do we lose the abelian property as soon as we reach exponentiation?
The first operation, addition, is abelian and so is multiplication. However, the next operation, exponentiation is not! Why is this? I understand that [imath]2^3[/imath] and [imath]3^2[/imath] are not equal but why do we suddenly lose this property after adding one other operation? Is there a property that we lose for tetration? Or Pentration (the next operation)? If so what property is lost at the nth operation? Edit: I there anything lost when going from addition to multiplication? |
214871 | Any subgroup of index [imath]p[/imath] in a [imath]p[/imath]-group is normal.
Let [imath]p[/imath] be a prime number and [imath]G[/imath] a finite group where [imath]|G|=p^n[/imath], [imath]n \in \mathbb{Z_+}[/imath]. Show that any subgroup of index [imath]p[/imath] in it is normal in [imath]G[/imath]. Conclude that any group of order [imath]p^2[/imath] have a normal subgroup of order [imath]p[/imath], but without using the Sylow theorems. | 1960875 | Every subgroup of index [imath]p[/imath] is normal in [imath]G[/imath].
Could anyone give me a hint on how to solve the following problem? Prove that if [imath]p[/imath] is a prime and [imath]G[/imath] is a group of order [imath]p^{\alpha}[/imath] for some [imath]\alpha \in \mathbb{Z}^{+}[/imath], then every subroup of index [imath]p[/imath] is normal in [imath]G[/imath]. I don't know where to start to solve this problem |
5248 | Evaluating the integral [imath]\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2[/imath]?
A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: [imath]$$\displaystyle\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$[/imath] Well, can anyone prove this without using Residue theory. I actually thought of doing this: [imath]\int_0^\infty \frac{\sin x} x \, dx = \lim_{t \to \infty} \int_0^t \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm dt[/imath] but I don't see how [imath]\pi[/imath] comes here, since we need the answer to be equal to [imath]$\dfrac{\pi}{2}$[/imath]. | 412847 | How can I evaluate [imath]\int_0^\infty \frac{\sin x}{x} \,dx[/imath]? [may be duplicated]
How can I evaluate [imath]\displaystyle\int_0^\infty \frac{\sin x}{x} \, dx[/imath]? (Let [imath]\displaystyle \frac{\sin0}{0}=1[/imath].) I proved that this integral exists by Cauchy's sequence. However I can't evaluate what is the exact value of this integral. |
9188 | Is [imath]\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})[/imath]?
In this post we saw isomorphism of vector spaces over [imath]\mathbb{Q}[/imath]. Just came across this question: Is [imath]\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})[/imath]? I know these as [imath]\mathbb{Q}[/imath]-Vector spaces, are isomorphic from the linked post. But as fields are they isomorphic? I neither know how to prove it nor how to disprove it. | 759113 | How to show that [imath]\mathbb Q(\sqrt 2)[/imath] is not field isomorphic to [imath]\mathbb Q(\sqrt 3).[/imath]
How to show that [imath]\mathbb Q(\sqrt 2)[/imath] is not field isomorphic to [imath]\mathbb Q(\sqrt 3)?[/imath] My text provides the hint as: Any isomorphism from [imath]\mathbb Q(\sqrt 2)\to\mathbb Q(\sqrt 3)[/imath] is identity when restricted to [imath]\mathbb Q.[/imath] Ya! I can see that: [imath]\mathbb Q(\sqrt 2)\simeq\mathbb Q[x]/\langle x^2-2\rangle[/imath] and [imath]\mathbb Q(\sqrt 3)\simeq\mathbb Q[x]/\langle x^2-3\rangle.[/imath] Now if [imath]\phi[/imath] be such an isomorphism then [imath]\phi:1+\langle x^2-2\rangle\mapsto1+\langle x^2-3\rangle.[/imath] Hence the restriction of [imath]\phi[/imath] over [imath]\{q+\langle x^2-2\rangle:q\in\mathbb Q\}[/imath] remains the identity. Now what? |
438 | Why [imath]\sqrt{-1 \times {-1}} \neq \sqrt{-1}^2[/imath]?
I know there must be something unmathematical in the following but I don't know where it is: \begin{align} \sqrt{-1} &= i \\ \\ \frac1{\sqrt{-1}} &= \frac1i \\ \\ \frac{\sqrt1}{\sqrt{-1}} &= \frac1i \\ \\ \sqrt{\frac1{-1}} &= \frac1i \\ \\ \sqrt{\frac{-1}1} &= \frac1i \\ \\ \sqrt{-1} &= \frac1i \\ \\ i &= \frac1i \\ \\ i^2 &= 1 \\ \\ -1 &= 1 \quad !!! \end{align} | 351929 | Square and square root and negative numbers
Are they equal? -5 = [imath]\sqrt{(-5)^2}[/imath] |
49169 | Why [imath]\sqrt{-1 \times -1} \neq \sqrt{-1}^2[/imath]?
We know [imath]i^2=-1 [/imath]then why does this happen? [imath] i^2 = \sqrt{-1}\times\sqrt{-1} [/imath] [imath] =\sqrt{-1\times-1} [/imath] [imath] =\sqrt{1} [/imath] [imath] = 1 [/imath] EDIT: I see this has been dealt with before but at least with this answer I'm not making the fundamental mistake of assuming an incorrect definition of [imath]i^2[/imath]. | 755970 | Value of [imath]i^2[/imath] in complex numbers
Please solve this doubt : we know that [imath]\sqrt{a}\sqrt{b}=\sqrt{ab}[/imath] and [imath]i^2 = -1[/imath]. But [imath]i= \sqrt{-1}[/imath] which implies that [imath]i^2 = i \cdot i = \sqrt{-1}\sqrt{-1} = \sqrt{1} = 1[/imath] that is [imath]i^2 = 1[/imath]. So what is the correct value of [imath]i^2[/imath]? Is it [imath]1[/imath] or is it [imath]-1[/imath]? Please explain. |
33215 | What is 48÷2(9+3)?
There is a huge debate on the internet on [imath]48÷2(9+3)[/imath]. I figured if i wanted to know the answer this is the best place to ask. I believe it is 2 as i believe it is part of the bracket operation in BEDMAS. http://www.mathway.com/ agrees with me. I also said if [imath]48÷2*(9+3)[/imath] was asked it would be [imath]288[/imath] which it agrees with as well. However wolframalpha says it is [imath]288[/imath] either way. A friend of mine (who is better at math) said theres no such thing as 'implicit multiplication', only shorthand so that is in fact done after the division (going left to right, not necessarily because division occurs before multiplication. But he didnt explicitly give a reasoning) What is the answer and WHY? | 16502 | Do values attached to integers have implicit parentheses?
Given [imath]5x/30x^2[/imath] I was wondering which is the correct equivalent form. According to BEDMAS this expression is equivalent to [imath]5*\cfrac{x}{30}*x^2[/imath] but, intuitively, I believe that it could also look like: [imath]\cfrac{5x}{30x^2}[/imath] I asked this question on MathOverflow (which was "Off-topic" and closed) and was told it was ambiguous. I was wondering what the convention was or if such a convention exists. According to Wikipedia the order of operations can be different based on the mnemonic used. |
21330 | "Closed" form for [imath]\sum \frac{1}{n^n}[/imath]
Earlier today, I was talking with my friend about some "cool" infinite series and the value they converge to like the Basel problem, Madhava-Leibniz formula for [imath]\pi/4, \log 2[/imath] and similar alternating series etc. One series that popped into our discussion was [imath]\sum\limits_{n=1}^{\infty} \frac{1}{n^n}[/imath]. Proving the convergence of this series is trivial but finding the value to which converges has defied me so far. Mathematica says this series converges to [imath]\approx 1.29129[/imath]. I tried Googling about this series and found very little information about this series (which is actually surprising since the series looks cool enough to arise in some context). We were joking that it should have something to do with [imath]\pi,e,\phi,\gamma[/imath] or at the least it must be a transcendental number :-). My questions are: What does this series converge to? Does this series arise in any context and are there interesting trivia to be known about this series? I am actually slightly puzzled that I have not been able to find much about this series on the Internet. (At least my Google search did not yield any interesting results). | 329613 | How do I find [imath]\sum_{n=1}^\infty \frac{1}{n^n}[/imath]
I stumbled across this problem to find the result of the following expression: [imath]\sum_{n=1}^\infty \frac{1}{n^n}[/imath] but I don't know how to approach it. It was suggested to me that I try this: [imath]\sum_{n=1}^\infty e^{-n\ln n}[/imath] However, I still don't know what to do. How can I find a closed form? |
44113 | What's the value of [imath]\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}[/imath]?
For some series, it is easy to say whether it is convergent or not by the "convergence test", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. The question is motivated from the simple exercise to determining whether the series [imath]\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}[/imath] is convergent. One may immediately get that it is convergent by the ratio test. So here is my question: What's the value of [imath]\sum_{k=1}^{\infty}\frac{k^2}{k!}?[/imath] | 679790 | Value of [imath]\sum\limits_{n= 0}^\infty \frac{n²}{n!}[/imath]
How to compute the value of [imath]\sum\limits_{n= 0}^\infty \frac{n^2}{n!}[/imath] ? I started with the ratio test which told me that it converges but I don't know to what value it converges. I realized I only know how to calculate the limit of a power/geometric series. |
17054 | Group where every element is order 2
Let [imath]G[/imath] be a group where every non-identity element has order 2. If |G| is finite then [imath]G[/imath] is isomorphic to the direct product [imath]\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots \times \mathbb{Z}_{2}[/imath]. Is the analogous result [imath]G= \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots [/imath]. true for the case |G| is infinite? | 1950783 | Let [imath]G[/imath] be a group. Prove that IF [imath]x^2 = e[/imath] for all [imath]x \in G[/imath], then [imath]G[/imath] is abelian.
Let [imath]G[/imath] be a group. Prove that IF [imath]x^2 = e[/imath] for all [imath]x \in G[/imath], then [imath]G[/imath] is abelian. My attempt: [imath]x^2 = e[/imath] [imath]x = x^{-1}[/imath] [imath]xx^{-1}=x^{-1}x^{-1}[/imath] [imath]e = x^{-1}x^{-1}[/imath] [imath]e = (xx)^{-1}[/imath] [imath]e = (x^2)^{-1}[/imath] [imath]e^{-1} = ((x^2)^{-1})^{-1})[/imath] [imath]e = x^2[/imath] Hence, [imath]G[/imath] is abelian. |
95741 | Is there any difference between mapping and function?
I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from [imath]\mathbb R[/imath] to [imath]\mathbb R[/imath]. But I am not ok with this answer. I need a simple way to explain the differences between mapping and function to a lay man together with some illustration (if possible). Thanks for any help. | 1720557 | What is the difference between linear mappings and linear functions?
Let [imath]V[/imath] and [imath]V'[/imath] be vector spaces over a field [imath]K[/imath]. A linear mapping [imath]f:V \to V'[/imath] is a mapping which preserves addition and scalar multiplication. My question is: what is the difference between linear mappings and linear functions? |
10490 | Why is [imath]1^{\infty}[/imath] considered to be an indeterminate form
From Wikipedia: In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form. The indeterminate forms include [imath]0^{0},\frac{0}{0},(\infty - \infty),1^{\infty}, \ \text{etc}\cdots[/imath] My question is can anyone give me a nice explanation of why [imath]1^{\infty}[/imath] is considered to be an indeterminate form? Because, i don't see any justification of this fact. I am still perplexed. | 319764 | 1 to the power of infinity, why is it indeterminate?
I've been taught that [imath]1^\infty[/imath] is undetermined case. Why is it so? Isn't [imath]1*1*1...=1[/imath] whatever times you would multiply it? So if you take a limit, say [imath]\lim_{n\to\infty} 1^n[/imath], doesn't it converge to 1? So why would the limit not exist? |
137277 | Constructing a subset not in [imath]\mathcal{B}(\mathbb{R})[/imath] explicitly
While reading David Williams's "Probability with Martingales", the following statement caught my fancy: Every subset of [imath]\mathbb{R}[/imath] which we meet in everyday use is an element of Borel [imath]\sigma[/imath]-algebra [imath]\mathcal{B}[/imath]; and indeed it is difficult (but possible!) to find a subset of [imath]\mathbb{R}[/imath] constructed explicitly (without the Axiom of Choice) which is not in [imath]\mathcal{B}[/imath]. I am curious to see an example of such a subset. | 1372120 | An example of Lebesgue measurable set but not Borel measurable besides the "subset of Cantor set" example.
The question is to give and example of Lebesgue measurable set but not Borel measurable. I know there exists subset of Cantor set that is not Borel measurable, since the cardinality of all Borel sets in [imath][0,1][/imath] is [imath]\aleph_1[/imath] while the cardinality of the subsets of Cantor set (defined on [imath][0,1][/imath]) is [imath]\aleph_2[/imath], so it must contains subsets that are non-Borel measurable. The problem is, the "cardinality of all Borel sets in [imath][0,1][/imath] is [imath]\aleph_1[/imath]" requires something that is beyond the scope of first-year graduate level real analysis course. I cannot prove it using what I learned. Can anyone give another example? Thank you! |
107617 | An infinite subset of a countable set is countable
In my book, it proves that an infinite subset of a coutnable set is countable. But not all the details are filled in, and I've tried to fill in all the details below. Could someone tell me if what I wrote below is valid? Let [imath]S[/imath] be an infinite subset of a countable set [imath]T[/imath]. Since [imath]T[/imath] is countable, there exists a bijection [imath]f: \mathbb{N} \rightarrow T[/imath]. And [imath]S \subseteq T = \{ f(n) \ | \ n \in \mathbb{N} \}[/imath]. Let [imath]n_{1}[/imath] be the smallest positive integer such that [imath]f(n_{1}) \in S[/imath]. And continue where [imath]n_{k}[/imath] is the smallest positive integer greater than [imath]n_{k-1}[/imath] such that [imath]f(n_{k}) \in S[/imath]. And because [imath]S[/imath] is infinite, we continue forever. Now consider the function [imath]\beta : \mathbb{N} \rightarrow S[/imath] which sends [imath]k \rightarrow f(n_{k})[/imath]. So in order for [imath]S[/imath] to be countable, [imath]\beta[/imath] would have to be a bijection. [imath]\beta[/imath] is injective since if [imath]f(n_{k}) = f(n_{j})[/imath], then [imath]n_{k} = n_{j}[/imath] because [imath]f[/imath] is injective. We also can conclude [imath]k = j[/imath] because the various [imath]n_{i}[/imath] chosen were strictly increasing. Edit: [imath]\beta[/imath] also needs to be surjective. So for every [imath]r \in S[/imath] there exists a [imath]q \in \mathbb{N}[/imath] such that [imath]\beta(q) = f(n_{q}) = r[/imath]. We know that [imath]f^{-1}(r) \in \{n_{1}, n_{2}, ..., n_{k} ... \}[/imath], so we can let [imath]f^{-1}(r) = n_{d}[/imath]. Thus [imath]\beta(d) = f(n_{d}) = r[/imath]. | 775464 | How to prove that this function is surjective?
I try to give a more constructive proof of the following lemma [imath]\qquad[/imath] Let [imath]S[/imath] be countably infinite and [imath]A[/imath] an infinite subset of [imath]S.[/imath] Then [imath]A[/imath] is countable. Here the "constructive proof" means that to prove that [imath]A[/imath] is countable is to construct a proper bijection from the set of positive integers onto [imath]A.[/imath] And I prefer to a proof of without using Zorn's Lemma. I have tried and already got something. Since [imath]S[/imath] is countably infinite, there exists a bijection [imath] f: \mathbb{N} \to S.[/imath] Here we denote the set of positive integers by [imath]\mathbb{N}.[/imath] Then I constructed the following mapping by the method given in Apostol's book Mathematical Analysis, Page 39: [imath]k(1)=\inf\{i\in\mathbb{N}\mid f(i)\in A\}[/imath]. Assume [imath]k(1), k(2), \dots,k(n-1)[/imath] has been constructed, let [imath]k(n)=\inf\{i\in\mathbb{N}\mid f(i)\in A, i>k(n-1)\}, n=2,3,\dots.[/imath] Continue this process on. Then let [imath]h(n)=f(k(n)).[/imath] I have proved that [imath]h: \mathbb{N}\to A[/imath] is injective. Because I have only used the continuity axiom of real numbers and the principle of induction, I think this proof is constructive. But at present I do not know how to prove [imath]h[/imath] is surjective. Can anyone help me to prove that the above [imath]h[/imath] is a surjection from [imath]\mathbb{N} [/imath] to [imath]A?[/imath] |
73348 | Can the Surface Area of a Sphere be found without using Integration?
When we were in school they told us that the Surface Area of a sphere = [imath]4\pi r^2[/imath] Now, when I try to derive it using only high school level mathematics, I am unable to do so. Please help. | 335577 | how to find surface area of a sphere
could any one tell me how to calculate surfaces area of a sphere using elementary mathematical knowledge? I am in Undergraduate second year doing calculus 2. I know its [imath]4\pi r^2[/imath] if the sphere is of radius [imath]r[/imath], I also want to know what is the area of unit square on a sphere. |
30732 | How can I evaluate [imath]\sum_{n=0}^\infty(n+1)x^n[/imath]?
How can I evaluate [imath]\sum_{n=1}^\infty\frac{2n}{3^{n+1}}[/imath]? I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method. In general, how can I evaluate [imath]\sum_{n=0}^\infty (n+1)x^n?[/imath] | 301139 | Sum of a series using derivatives
[imath]1 + 2/2 + 3/4 + \cdots + n/2^{n-1}[/imath] How would find the closed-form expression and also the sum up to 20? I'm not really getting why or the logic behind using derivatives to arrive at an answer. |
303695 | if [imath]f[/imath] is differentiable at a point [imath]x[/imath], is [imath]f[/imath] also necessary lipshitz-continuous at [imath]x[/imath]?
if [imath]f[/imath] is differentiable at a point [imath]x[/imath], is [imath]f[/imath] also necessary Lipshitz at [imath]x[/imath]? Since [imath]f[/imath] is differentiable at [imath]x[/imath], [imath]f[/imath] is also continuous at [imath]x[/imath]. Then we have the [imath]\varepsilon[/imath]-[imath]\delta[/imath] definition of [imath]f[/imath] continuous at [imath]x[/imath], and also we have [imath]f'(x)[/imath] exists. But i have no idea how to connect them together. Since [imath]f[/imath] is only differentiable at a point, i don't think mean value theorem is gonna work here either. | 286715 | Lipschitz condition: if [imath]f[/imath] is differentiable at [imath]b[/imath], then [imath]f[/imath] is Lipschitz of order [imath]1[/imath] at [imath]b[/imath]
I have been trying to solve this, but failing at it. Since [imath]f[/imath] is differentiable at [imath]x[/imath], we have [imath]f'(x)=\lim_{y \to x} \frac{f(y)-f(x)}{y-x}= L[/imath] To prove the proposition, all I need to show is [imath]|f(y)-f(x)|\le C|y-x|[/imath]; but since I cannot apply the MVT at a point, I am lost how to get rid of the [imath]\lim_{y \to x}[/imath] from the defintion of the derivative at [imath]x[/imath]. Any help will be appreciated. |
11 | Is it true that [imath]0.999999999\dots=1[/imath]?
I'm told by smart people that [imath]0.999999999\dots=1[/imath] and I believe them, but is there a proof that explains why this is? | 419866 | Can someone help me solve this problem please.
For the real numbers [imath]x=0.9999999\dots[/imath] and [imath]y=1.0000000\dots[/imath] it is the case that [imath]x^2<y^2[/imath]. Is it true or false? Prove if you think it's true and give a counterexample if you think it's false. |
201906 | Showing that [imath]\frac{\sqrt[n]{n!}}{n}[/imath] [imath]\rightarrow \frac{1}{e}[/imath]
Show:[imath]\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}[/imath] So I can expand the numerator by geometric mean. Letting [imath]C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n[/imath]. Let the numerator be called [imath]a_{n}[/imath] and the denominator be [imath]b_{n}[/imath] Is there a way to use this statement so that I could force the original sequence into the form of [imath]1/\left(1+\frac{1}{n}\right)^n[/imath] | 935490 | The limit of [imath](n!)^{1/n}/n[/imath] as [imath]n\to\infty[/imath]
(Proof necessary) [imath]\lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n}[/imath] I don't have an answer yet, but I know it exists, and is less than [imath]1[/imath]. Edit. Winther's answer is the most correct I don't understand how he is jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). Don't presume, it's wrong, I need to go, and I'll keep looking at it when I get back Any help is appreciated |
189328 | Nonexistence of an injective [imath]C^1[/imath] map between [imath]\mathbb R^2[/imath] and [imath]\mathbb R[/imath]
I am getting bored waiting for the train so I'm thinking whether there can exist a [imath]C^1[/imath] injective map between [imath]\mathbb{R}^2[/imath] and [imath]\mathbb{R}[/imath]. It seems to me that the answer is no but I can't find a proof or a counterexample... Can you help me? | 1079075 | Is there a injective polynomial function from [imath]R^2[/imath] to [imath]R[/imath]?
There is an injective polynomial function from [imath]N^2[/imath] to [imath]N[/imath] (the Cantor-pairing function for example, which is of degree 2), and also one of degree 4 from [imath]Z^2[/imath] to [imath]Z[/imath]. I believe the question is open (at least on this website) whehter there is such a polynomial from [imath]Q^2[/imath] to [imath]Q[/imath]. Not sure if this would be harder, but: is there a injective polynomial function from [imath]R^2[/imath] to [imath]R[/imath]? |
22069 | Is there a name for function with the exponential property [imath]f(x+y)=f(x) \cdot f(y)[/imath]?
I was wondering if there is a name for a function that satisfies the conditions [imath]f:\mathbb{R} \to \mathbb{R}[/imath] and [imath]f(x+y)=f(x) \cdot f(y)[/imath]? Thanks and regards! | 375801 | Is [imath]f(x)f(y)=f(x+y)[/imath] enough to determin [imath]f[/imath]?
I had a discussion with a friend and there it came up the question whether [imath]f(x)f(y)=f(x+y)[/imath], [imath]f(0)=1[/imath] and the existence of [imath]f'(x)[/imath] implies that [imath]f(x)=\exp(a x)[/imath]. This seems very reasonable but I cannot figure this out. Can differentiability be pushed to smoothness in this case? In a similar manner on would expect that [imath]g(x)+g(y)=g(xy)[/imath] and [imath]g(1)=0[/imath] would imply that [imath]g(x)=\log(a x)[/imath]. Any ideas? |
170813 | Prove [imath](-a+b+c)(a-b+c)(a+b-c) \leq abc[/imath], where [imath]a, b[/imath] and [imath]c[/imath] are positive real numbers
I have tried the arithmetic-geometric inequality on [imath](-a+b+c)(a-b+c)(a+b-c)[/imath] which gives [imath](-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3[/imath] and on [imath]abc[/imath] which gives [imath]abc \leq \left(\frac{a+b+c}{3}\right)^3.[/imath] Since both inequalities have the same righthand side, I have tried to deduce something about the lefthand sides, but to no avail. Can somebody help me, please? I am sure it is something simple I have missed. | 2849480 | [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] are sides of a triangle - prove that [imath](x+y-z)(x-y+z)(-x+y+z)\leq xyz[/imath]
[imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] are all sides of a triangle. Prove that [imath](x+y-z)(x-y+z)(-x+y+z)\leq xyz[/imath]. Equality occurs when all sides are the same length. I don't know how to prove that the left side is always less or equal though. Any help is much appreciated! |
13131 | Starting digits of [imath]2^n[/imath].
Prove that for any finite sequence of decimal digits, there exists an [imath]n[/imath] such that the decimal expansion of [imath]2^n[/imath] begins with these digits. | 544214 | Is [imath]2^k = 2013...[/imath] for some [imath]k[/imath]?
I'm wondering if some power of [imath]2[/imath] can be written in base [imath]10[/imath] as [imath]2013[/imath] followed by other digits. Formally, does there exist [imath]k,q,r \in \mathbb N[/imath] such that [imath]2^k=2013 \cdot 10^q+r \,\,\,; \,\,\,r<10^q [/imath] I'm not sure if it's true or not. I would go for a 'no', but I can't prove it. Thanks for your help. |
296101 | An explicit bijection between the power set [imath]\mathcal P \left({\mathbb{N}}\right)[/imath] and [imath]2^\mathbb{N}[/imath].
I know how to show that these two have the same cardinality and from that there must be a bijection between them. Can anyone help with an explicit bijection between these sets? | 41006 | How to show equinumerosity of the powerset of [imath]A[/imath] and the set of functions from [imath]A[/imath] to [imath]\{0,1\}[/imath] without cardinal arithmetic?
How to show equinumerosity of the powerset of [imath]A[/imath] and the set of functions from [imath]A[/imath] to [imath]\{0,1\}[/imath] without cardinal arithmetic? Not homework, practice exercise. |
4467 | How to prove: if [imath]a,b \in \mathbb N[/imath], then [imath]a^{1/b}[/imath] is an integer or an irrational number?
It is well known that [imath]\sqrt{2}[/imath] is irrational, and by modifying the proof (replacing 'even' with 'divisible by [imath]3[/imath]'), one can prove that [imath]\sqrt{3}[/imath] is irrational, as well. On the other hand, clearly [imath]\sqrt{n^2} = n[/imath] for any positive integer [imath]n[/imath]. It seems that any positive integer has a square root that is either an integer or irrational number. How do we prove that if [imath]a \in \mathbb N[/imath], then [imath]\sqrt a[/imath] is an integer or an irrational number? I also notice that I can modify the proof that [imath]\sqrt{2}[/imath] is irrational to prove that [imath]\sqrt[3]{2}, \sqrt[4]{2}, \cdots[/imath] are all irrational. This suggests we can extend the previous result to other radicals. Can we extend 1? That is, can we show that for any [imath]a, b \in \mathbb{N}[/imath], [imath]a^{1/b}[/imath] is either an integer or irrational? | 449974 | Prove [imath]\sqrt{k}[/imath] is not a rational number.
Suppose [imath]k>1[/imath] is an integer, and k is not a square number, then [imath]\sqrt{k}[/imath] is not a rational number. Proof: Let [imath]\sqrt{k}=\frac{p}{q}[/imath], and [imath](p,q)=1[/imath],So [imath]q^2|p^2[/imath], [imath]p\neq 1[/imath], [imath]k[/imath] is not an integer.When [imath]q=p=1[/imath], and [imath]k>1[/imath]. And [imath]q=1[/imath], then [imath]\sqrt{k}=p[/imath], and [imath]k[/imath] is a square number. I do think it's so easy...Where is wrong? How to do it. |
29450 | Self-Contained Proof that [imath]\sum\limits_{n=1}^{\infty} \frac1{n^p}[/imath] Converges for [imath]p > 1[/imath]
To prove the convergence of the p-series [imath]\sum_{n=1}^{\infty} \frac1{n^p}[/imath] for [imath]p > 1[/imath], one typically appeals to either the Integral Test or the Cauchy Condensation Test. I am wondering if there is a self-contained proof that this series converges which does not rely on either test. I suspect that any proof would have to use the ideas behind one of these two tests. | 400604 | Prove the convergence of the series.
Let r > 1 be a real number. Prove that the following series is convergent. [imath]\sum_{n = 1}^{\infty}\frac{1}{n^r}[/imath] |
48161 | In classical logic, why is [imath](p\Rightarrow q)[/imath] True if both [imath]p[/imath] and [imath]q[/imath] are False?
I am studying entailment in classical first-order logic. The Truth Table we have been presented with for the statement [imath](p \Rightarrow q)\;[/imath] (a.k.a. '[imath]p[/imath] implies [imath]q[/imath]') is: [imath]\begin{array}{|c|c|c|} \hline p&q&p\Rightarrow q\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T\\\hline \end{array}[/imath] I 'get' lines 1, 2, and 3, but I do not understand line 4. Why is the statement [imath](p \Rightarrow q)[/imath] True if both p and q are False? We have also been told that [imath](p \Rightarrow q)[/imath] is logically equivalent to [imath](~p || q)[/imath] (that is [imath]\lnot p \lor q[/imath]). Stemming from my lack of understanding of line 4 of the Truth Table, I do not understand why this equivalence is accurate. Administrative note. You may experience being directed here even though your question was actually about line 3 of the truth table instead. In that case, see the companion question In classical logic, why is [imath](p\Rightarrow q)[/imath] True if [imath]p[/imath] is False and [imath]q[/imath] is True? And even if your original worry was about line 4, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines. | 636224 | Why is the implication [imath]P \Rightarrow Q[/imath] false if and only if [imath]P[/imath] is true and [imath]Q[/imath] is false?
Why is the implication [imath]P \Rightarrow Q[/imath] false if and only if [imath]P[/imath] is true and [imath]Q[/imath] is false ? Is this because if [imath]P[/imath] implies [imath]Q[/imath], where [imath]P[/imath] is true and [imath]Q[/imath] is false, then [imath]Q[/imath] is also true by definition of the implication ? The exact definition of the implication [imath]P \Rightarrow Q[/imath]: if [imath]P[/imath] then [imath]Q[/imath] is also true ? So given an implication [imath]P \Rightarrow Q[/imath] where [imath]P[/imath] is true and [imath]Q[/imath] is false it must be false, since no true statement can never imply a false statement, since this would mean the false statement would be true, which it is not ? |
136665 | For every irrational [imath]\alpha[/imath], the set [imath]\{a+b\alpha: a,b\in \mathbb{Z}\}[/imath] is dense in [imath]\mathbb R[/imath]
I am not able to prove that this set is dense in [imath]\mathbb{R}[/imath]. Will be pleased if you help in a easiest way, [imath]\{a+b\alpha: a,b\in \mathbb{Z}\}[/imath] where [imath]\alpha\in\mathbb{Q}^c[/imath] is a fixed irrational. | 1437468 | Denseness of the set [imath]\{ m+n\alpha : m\in\mathbb{N},n\in\mathbb{Z}\}[/imath] with [imath]\alpha[/imath] irrational
How to prove that the set [imath]\{ m+n\alpha : m\in\mathbb{N},n\in\mathbb{Z}\}[/imath], ([imath]\alpha[/imath] is an irrational number) is dense in [imath]\mathbb{R}[/imath]? Using the fact every additive subgroup of [imath]\mathbb{R}[/imath] is either discrete or dense in [imath]\mathbb{R}[/imath], we can prove that the set [imath]\{ m+n\alpha : m,n\in\mathbb{Z}\}[/imath] is dense in [imath]\mathbb{R}[/imath]. But how to prove the denseness of the above set? |
125709 | Example to show the distance between two closed sets can be 0 even if the two sets are disjoint
Let [imath]A[/imath] and [imath]B[/imath] be two sets of real numbers. Define the distance from [imath]A[/imath] to [imath]B[/imath] by [imath]\rho (A,B) = \inf \{ |a-b| : a \in A, b \in B\} \;.[/imath] Give an example to show that the distance between two closed sets can be [imath]0[/imath] even if the two sets are disjoint. | 539687 | I don't understand how sets can be closed, yet disjoint?
What are some closed, disjoint subsets [imath]A, B[/imath] in [imath]R^2[/imath] where [imath]inf\{d(A, B) = 0 \forall a \in A \forall b \in B\}[/imath]? |
72856 | Order of finite fields is [imath]p^n[/imath]
Let [imath]F[/imath] be a finite field. .How do I prove that the order of [imath]F[/imath] is always of order [imath]p^n[/imath] where [imath]p[/imath] is prime? | 317633 | Is there a field with [imath]n[/imath] elements for all [imath]n \in \mathbb{N}[/imath]?
I don't think this is true, but I'm not sure. I certainly know of finite fields with 2,4 and 8 elements, and of course [imath]p^n[/imath] elements where [imath]p[/imath] is prime, for all [imath]n \in \mathbb{N}[/imath]. |
168812 | A compact operator is completely continuous.
I have a question. If [imath]X[/imath] and [imath]Y[/imath] are Banach spaces, we have to prove that a compact linear operator is completely continuous. A mapping [imath]T \colon X \to Y[/imath] is called completely continuous, if it maps a weakly convergent sequence in [imath]X[/imath] to a strongly convergent sequence in [imath]Y[/imath] , i.e., [imath]x_n\underset{n\to +\infty}\rightharpoonup x[/imath] implies [imath]\lVert Tx_n- Tx\rVert_Y\to 0[/imath]. | 811219 | Problems proving that a compact operator is completely continuous
I would like to prove that if [imath]T:X\rightarrow Y[/imath] is a compact operator, then for every weak convergent sequence [imath](x_n)_{n\in\mathbb N}[/imath] with [imath]x_n\rightharpoonup x[/imath] for some [imath]x\in X[/imath] it follows that [imath]Tx_n\rightarrow Tx[/imath] with respect to the norm on [imath]Y[/imath]. I started the proof with, Since [imath](x_n)_{n\in\mathbb N}[/imath] is weak convergent there exists a positive constan [imath]C[/imath] sucht that [imath]||x_n||\leq C[/imath] for all [imath]n[/imath]. Hence by compactness it follows that [imath]Tx_n[/imath] contains a convergent subsequence i.e [imath]Tx_{n_k}\rightarrow y[/imath] for some [imath]y\in Y[/imath]. Now it should be somehow possible to deduce that [imath]y=Tx[/imath]. But I am not able to move on. Could someone give me hint how to proceed? or even how to find a better start for my proof? Thanks in advance! |
236578 | Recurrence relation, Fibonacci numbers
[imath](a)[/imath] Consider the recurrence relation [imath]a_{n+2}a_n = a^2 _{n+1} + 2[/imath] with [imath]a_1 = a_2 = 1[/imath]. [imath](i)[/imath] Assume that all [imath]a_n[/imath] are integers. Prove that they are all odd and the integers [imath]a_n[/imath] and [imath]a_{n+1}[/imath] are coprime for [imath]n \in \mathbb N[/imath] [imath](ii)[/imath] Assume that the set [imath]\{a_n , a_{n+1} , a_{n+2}\}[/imath] is pairwise coprime for [imath]n \in \mathbb N[/imath]. Prove that all [imath]a_n[/imath] are integers by induction. [imath](b)[/imath] Consider the recurrence relation [imath]a_{n+2}a_n = a^2_{n+1} + 1[/imath] with [imath]a_1 = 1, a_2 = 2[/imath] and compare this sequence to the Fibonacci numbers. What do you find? Formulate it as a mathematical statement and prove it. | 241663 | Recurrence relation, induction and Fibonacci numbers
1.(a) Consider the recurrence relation [imath]a_{n+2}a_n = a^2_{n+1} + 2[/imath] with [imath]a_1 = a_2 = 1[/imath]. (i) Assume that all [imath]a_n[/imath] are integers. Prove that they are all odd and the integers [imath]a_n[/imath] and [imath]a_{n+1}[/imath] are coprime for [imath]n \in \mathbb{N}[/imath]. (ii) Assume that the set [imath]\left\{ a_n,\ a_{n+1},\ a_{n+2}\right\}[/imath] is pairwise coprime for [imath]n \in \mathbb{N}[/imath]. Prove that all [imath]a_n[/imath] are integers by induction. (b) Consider the recurrence relation [imath]a_{n+2}a_n = a^2_{n+1} + 1[/imath] with [imath]a_ 1= 1,\ a_2 = 2[/imath] and compare this sequence to the Fibonacci numbers. What do you find? Formulate it as a mathematical statement and prove it. I have no idea where to start with 1(ai) but with 1(aii) i have started with: Given that [imath]\left\{a_n,\ a_{n+1},\ a_{n+2}\right\}[/imath] is pairwise coprime you get [imath]gcd(a_n,a_{n+1})=1,\ gcd(a_n,a_{n+2})=1,\ gcd(a_{n+1},a_{n+2})=1[/imath] Using the initial terms, you can do base induction on [imath]a_{n+2} = \frac{a^2_{n+1}+2}{a_n}[/imath] to prove whether the next terms will be integers. [imath]a_3 = \frac{1^2+2}{1}[/imath] [imath]a_4 = \frac{3^2+2}{1}[/imath] [imath]a_5 = \frac{11^2+2}{3}[/imath] gives: [imath]a_1=1,\ a_2 = 1,\ a_3 = 3,\ a_4 = 11,\ a_5 = 41[/imath] which are all integers. So the base induction is correct. Now for the inductive step [imath]n=k, k \rightarrow k+1[/imath] [imath]a_{k+3} = \frac{a^2_{k+2}+2}{a_{k+1}}[/imath]. I am not sure how to prove this for the inductive step. Also for (b) how do you formulate a mathematical statement and prove it? |
278330 | Consider the series [imath] ∑_{n=1}^∞ x^2+ n/n^2[/imath] . Pick out the true statements:
Consider the series [imath] \sum_{n=1}^\infty x^2+ n/n^2[/imath] . Pick out the true statements: (a) The series converges for all real values of [imath]x[/imath]. (b) The series converges uniformly on [imath]\mathbb{R}[/imath]. (c) The series does not converge absolutely for any real value of [imath]x[/imath]. stuck on this problem totally. please help me to solve this problem.thanks . | 641336 | different convergence of the series [imath]\sum_1^\infty{(-1)^n\frac{x^2+n}{n^2}}[/imath]
Consider the series [imath]\sum_1^\infty{(-1)^n\frac{x^2+n}{n^2}}[/imath] Pick out the true statements: (a) The series converges for all real values of [imath]x[/imath]. (b) The series converges uniformly on [imath]\mathbb{R}[/imath]. (c) The series does not converge absolutely for any real value of [imath]x[/imath]. (a) is true by alternating series test. (c) is not true as [imath]\sum_1^\infty{\frac{x^2+n}{n^2}} \ge \sum_1^\infty{\frac{1}{n}}[/imath]. (b) I have no idea. seek your help for (b) |
8337 | Different methods to compute [imath]\sum\limits_{k=1}^\infty \frac{1}{k^2}[/imath] (Basel problem)
As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) [imath]\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.[/imath] However, Euler was Euler and he gave other proofs. I believe many of you know some nice proofs of this, can you please share it with us? | 302310 | Show that [imath]\sum_{n=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}[/imath]
Show that [imath]\sum_{n=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}[/imath] Anyone can help ? |
207395 | Limit of a continuous function
Suppose that [imath]f[/imath] is a continuous and real function on [imath][0,\infty][/imath]. How can we show that if [imath]\lim_{n\rightarrow\infty}(f(na))=0[/imath] for all [imath]a>0[/imath] then [imath]\lim_{x\rightarrow+\infty} f(x)=0[/imath]? | 1935620 | Example: Continuous function having a limit along every arithmetic sequence but having no limit over the reals
The following question seems to be very elementary and must be a folklore, but we are not able to find an answer. Let [imath]f: [0,\infty)\to \mathbb R[/imath] be a continuous function such that for every [imath]a>0[/imath], [imath] \lim_{n\to+\infty} f(na) = 0, [/imath] where [imath]n[/imath] is restricted to be integer. Does this imply that [imath] \lim_{t\to+\infty} f(t) = 0, [/imath] where [imath]t[/imath] is allowed to be real? |
3852 | If [imath]AB = I[/imath] then [imath]BA = I[/imath]
If [imath]A[/imath] and [imath]B[/imath] are square matrices such that [imath]AB = I[/imath], where [imath]I[/imath] is the identity matrix, show that [imath]BA = I[/imath]. I do not understand anything more than the following. Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix. If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem? P.S.: Please avoid using the transpose and/or inverse of a matrix. | 462983 | Prove that a square matrix commutes with its inverse
The Question: This is a very fundamental and commonly used result in linear algebra, but I haven't been able to find a proof or prove it myself. The statement is as follows: let [imath]A[/imath] be an [imath]n\times n[/imath] square matrix, and suppose that [imath]B=\operatorname{LeftInv}(A)[/imath] is a matrix such that [imath]BA=I[/imath]. Prove that [imath]AB=I[/imath]. That is, prove that a matrix commutes with its inverse, that the left-inverse is also the right-inverse My thoughts so far: This is particularly annoying to me because it seems like it should be easy. We have a similar statement for group multiplication, but the commutativity of inverses is often presented as part of the definition. Does this property necessarily follow from the associativity of multiplication? I've noticed that from associativity, we have [imath] \left(A\operatorname{LeftInv}(A)\right)A=A\left(\operatorname{LeftInv}(A)A\right) [/imath] But is that enough? It might help to talk about generalized inverses. |
28438 | Alternative proof that [imath](a^2+b^2)/(ab+1)[/imath] is a square when it's an integer
Let [imath]a,b[/imath] be positive integers. When [imath]k = \frac{a^2 + b^2}{ab+1}[/imath] is an integer, it is a square. Proof 1: (Ngô Bảo Châu): Rearrange to get [imath]a^2-akb+b^2-k=0[/imath], as a quadratic in [imath]a[/imath] this has two values: [imath]a[/imath] and [imath]kb - a = (b^2-k)/a[/imath]. (The second root is determined in two different ways from the expansion [imath](x-r_1)(x-r_2) = x^2 - (r_1 + r_2)x + r_1 r_2[/imath].) Now suppose we have [imath]a,b[/imath] such that [imath]k[/imath] is an integer but not a square, by the investigation about roots we have that the second root is a nonzero integer since [imath]k,b,a[/imath] are integers and [imath]k \not = b^2[/imath], futhermore it is positive which is easily seen from its defining equation. WLOG assume [imath]a \ge b[/imath] so that the second root is strictly smaller than [imath]a[/imath]. This leads to a decent, replacing [imath]a[/imath] with the second root. Proof 2 (Don Zagier): Apply reduction theory (specifically, Sätze 1 and 2 of Section 13 of my book on quadratic fields) to the quadratic form [imath]x^2 + kxy + y^2[/imath], which is the unique reduced quadratic form in its equivalence class. Note that Proof 2 is pretty much the same as Proof 1 when written out in explicit detail, but I could not read Zagier's book because I cannot read German. I would like to know more approaches to this and other alternative proofs of this result if possible! Thanks in advance. I would also be interested in related problems (especially easier ones of a similar nature) and texts which cover the reduction theory in English. | 372200 | To prove [imath]\frac {a^2+b^2}{ab+1}[/imath] is a perfect square , without geometry or induction.
Let [imath]a[/imath] and [imath]b[/imath] be positive integers such that [imath]ab+1[/imath] divides [imath]a^2+b^2[/imath] ; then prove that [imath]\frac {a^2+b^2}{ab+1}[/imath] is a perfect square (this problem came in [imath]\Bbb {IMO}[/imath] [imath]1988[/imath]). How to prove it without using geometry or induction ? |
54210 | Is [imath]x^x=y[/imath] solvable for [imath]x[/imath]?
Given that [imath]x^x = y[/imath]; and given some value for [imath]y[/imath] is there a way to expressly solve that equation for [imath]x[/imath]? | 351046 | Solve [imath]x^x = a[/imath] for known [imath]a[/imath]?
For example if you have [imath]x^x = 2[/imath], can you express [imath]x[/imath] as a numerical expression containing only the addition, multiplication and exponentiation operators? |
300904 | Prove that G and G' are isomorphic.
Let [imath]G=[/imath] the set of [imath]2\times 2[/imath] matrices [imath]\left\{\begin{bmatrix} a& b\\ 0& a-2b\end{bmatrix}\;\middle\vert\; a,b \in\mathbb{R} \text{ and }A^2\neq 2ab\right\},[/imath] where the group operation is matrix multiplication. Let [imath]G'=\{(c,d) \mid c,d \in\mathbb{R} \text{ and } c,d \neq 0\}[/imath] with group operation [imath](c_1,d_1)(c_2,d_2)=(c_1c_2,d_1d_2)[/imath]. Prove that [imath]G[/imath] and [imath]G'[/imath] are isomorphic. | 300338 | Need help is defining an isomorphism.
Let [imath]A = \left\{ \begin{bmatrix} a & b \\ 0 & {a-2b} \end{bmatrix} \mid a,b \in \mathbb{R}, a^2 \ne 2ab \right\}[/imath] where the group operation is matrix multiplication. Let [imath]A'= \left\{(c,d) \mid c,d\in \mathbb{R}, c,d \ne 0\right\}[/imath] with group operation [imath](c_1,d_1)(c_2,d_2) =(c_1c_2,d_1d_2)[/imath]. I need to prove that these are isomorphic, so i need to show that A is isomorphic to some group of permutations and A' is isomorphic to that same group and therefore A and A' are isomorphic. I am not sure how to find the isomorphism to a permutation group in either case so any guidance as to how I would go about doing this would be greatly appreciated. Thanks you |
14721 | Three variable, third degree Diophantine equation
I haven't found any useful method to solve the following problem: Prove that if [imath]x,y,z\in\mathbb{Z}[/imath] and [imath]x^3+y^3=3z^3[/imath] then [imath]xyz=0[/imath]. Source: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=382377 | 473610 | a diophantine equation from Stewart and Tall
This is from Stewart and Tall from the chapter on Kummer's Theorem. Show that there are no non trivial (non-zero) solutions to [imath]x^3 + y^3=3z^3[/imath] |
15591 | Number of even and odd subsets
Suppose we have the following two identities: [imath]\displaystyle \sum_{k=0}^{n} \binom{n}{k} = 2^n[/imath] [imath]\displaystyle \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0[/imath] The first says that the number of subsets of an [imath]n[/imath]-set is [imath]2^n[/imath]. The second says that the number of subsets of even size equals the number of subsets of odd size (of an [imath]n[/imath]-set). Thus there are [imath]2^{n-1}[/imath] subsets of even length and [imath]2^{n-1}[/imath] subsets of odd length? To combinatorially prove the second identity, let [imath]A[/imath] be a [imath]k[/imath]-subset of [imath][n][/imath]. Then note whether [imath]k[/imath] is odd or even? | 94514 | Proving [imath]\sum_{k=0}^n(-1)^k\binom nk=0[/imath]
Show that [imath]\sum_{k=0}^n(-1)^k\binom nk=0[/imath] So for odd [imath]n[/imath] we have an even number of terms. So [imath]\binom nk=\binom n{n-k}[/imath] which have opposite signs. Thus the sum is 0. For even [imath]n[/imath] we have that [imath]\sum_{k=0}^n(-1)^k\binom nk= \binom n0+\sum_{k=1}^{n-1}(-1)^k\binom nk+\binom nn[/imath] Now [imath]\sum_{k=1}^{n-1}(-1)^k\binom nk= \sum_{k=1}^{n-1}(-1)^k\left[\binom{n-1}k+\binom{n-1}{k-1}\right][/imath] What would that sum be in the square brackets? |
16374 | Universal Chord Theorem
Let [imath]f \in C[0,1][/imath] and [imath]f(0)=f(1)[/imath]. How do we prove [imath]\exists a \in [0,1/2][/imath] such that [imath]f(a)=f(a+1/2)[/imath]? In fact, for every positive integer [imath]n[/imath], there is some [imath]a[/imath], such that [imath]f(a) = f(a+\frac{1}{n})[/imath]. For any other non-zero real [imath]r[/imath] (i.e not of the form [imath]\frac{1}{n}[/imath]), there is a continuous function [imath]f \in C[0,1][/imath], such that [imath]f(0) = f(1)[/imath] and [imath]f(a) \neq f(a+r)[/imath] for any [imath]a[/imath]. This is called the Universal Chord Theorem and is due to Paul Levy. Note: the accepted answer answers only the first question, so please read the other answers too, and also this answer by Arturo to a different question: https://math.stackexchange.com/a/113471/1102 This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions. and here: List of abstract duplicates. | 371571 | Prove that there exists [imath]x_n[/imath] such that [imath]0 \leq x_n \leq 1-\frac{1}{n}[/imath] and [imath]f(x_n)=f(x_n+\frac{1}{n})[/imath].
Suppose that the function [imath]f:[0,1] \to \mathbb{R}[/imath] is continuous on [imath][0,1][/imath] and [imath]f(0)=f(1)[/imath]. Prove that for each natural number [imath]n[/imath], there exists [imath]x_n \in \mathbb{R}[/imath] such that [imath]0 \leq x_n \leq 1-\frac{1}{n}[/imath] and [imath]f(x_n)=f(x_n+\frac{1}{n})[/imath]. Though I don't know how the proof would look like, I have a strong feeling that it has something to do with the Intermediate Value Theorem, judging by the continuity of [imath]f[/imath] and the existence of such a [imath]x_n[/imath]. So I guess I'm supposed to define a function [imath]g(x)=f(x)-f(x+\frac{1}{n})[/imath] on [imath][0,\frac{1}{n}][/imath] and try to claim that [imath]g(x_n)=0[/imath] for some [imath]x_n \in [0,\frac{1}{n}][/imath]. Unfortunately I don't know how to proceed further from here, maybe because I haven't made use of the fact that [imath]f(0)=f(1)[/imath]. Any hint and suggestion is much appreciated. Thank you! |
61087 | How would one go about proving that the rationals are not the countable intersection of open sets?
I'm trying to prove that the rationals are not the countable intersection of open sets, but I still can't understand why [imath]\bigcap_{n \in \mathbf{N}} \left\{\left(q - \frac 1n, q + \frac 1n\right) : q \in \mathbf{Q}\right\}[/imath] isn't a counter-example. Any ideas? Thanks! | 2765171 | Set of rational number in (0,1) can not be expressed as intersection of countable collection of open sets
Set S=Q[imath]\cap[/imath](0,1) I wanted to show that this can not be expressed as intersection of countable collection of open sets. I know Any closed set can be shown as intersection of countable collection of open set .I had hint show this by Cantor intersection property which says that infinite intersection of nested sequence of nonempty compact set is nonempty . I could not able to link this hint with problem till now .Any help will be appreciated Without using Baire Category Theorem Is it possible to solve this problem? As this problem occur in exercise where there in no mention of that theorem in text. |
151864 | Continuous images of compact sets are compact
Let [imath]X[/imath] be a compact metric space and [imath]Y[/imath] any metric space. If [imath]f:X \to Y[/imath] is continuous, then [imath]f(X)[/imath] is compact (that is, continuous functions carry compact sets into compact sets). Proof: Consider an open cover of [imath]f(X)[/imath]. Then [imath]f(X) \subset \bigcup_{\alpha \in A}V_\alpha[/imath] where each [imath]V_\alpha[/imath] is open in [imath]Y[/imath]. [imath]X \subset f^{-1}(f(X)) \subset f^{-1}\left(\bigcup_{\alpha \in A}V_\alpha\right) = \bigcup_{\alpha \in A}f^{-1}(V_\alpha)[/imath]. Hence [imath]\bigcup_{\alpha \in A}f^{-1}(V_\alpha)[/imath] is an open cover of [imath]X[/imath]. Since [imath]X[/imath] is compact then we can choose a finite subcover [imath]\{V_i\}_{i=1}^n[/imath] such that [imath]X \subset \bigcup_{i=1}^n f^{-1}(V_i)[/imath]. So then [imath]f(X) \subset f\left(\bigcup_{i=1}^n f^{-1}(V_i)\right) = \bigcup_{i=1}^n f\left(f^{-1}(V_i)\right) \subset \bigcup_{i=1}^n V_i[/imath], a finite subcover of [imath]f(X)[/imath]. [imath]\therefore f(X)[/imath] is compact. Does this proof have an error? Thanks for your help. | 2047150 | If [imath]h:X \to Y[/imath] is a homeomorphism of metric spaces, how do I prove that [imath]X[/imath] is compact if and only if [imath]Y[/imath] is compact?
If [imath]h:X \to Y[/imath] is a homeomorphism of metric spaces, how do I prove that [imath]X[/imath] is compact if and only if [imath]Y[/imath] is compact? I can't find the complete formal proof for it anywhere. My idea: I assume I would take an open cover [imath]\mathcal U[/imath] of [imath]f(X)[/imath] by sets open in [imath]Y[/imath]? Then since [imath]f[/imath] is continuous, [imath]f^{-1}(U)[/imath] is open in [imath]X[/imath] for all [imath]U \in \mathcal U[/imath]? Or is that the wrong place to start? Step by step help/solutions appreciated! |
139659 | Total number of solutions of an equation
What is the total number of solutions of an equation of the form [imath]x_1 + x_2 + \cdots + x_r = m[/imath] such that [imath]1 \le x_1 < x_2 < \cdots < x_r < N[/imath] where [imath]N[/imath] is some natural number and [imath]x_1, x_2, \cdots, x_r, m[/imath] are integers? Also, what would be the solution in the relaxed case where [imath]x_1, x_2, x_3, \dots, x_r[/imath] could be equal? | 1653130 | How many solutions are [imath] a+b+c+d = 30 ,(a\leq b\leq c\leq d) [/imath]?
I would appreciate if somebody could help me with the following problem: Q: How many solutions are there to the equation [imath] a+b+c+d = 30 ,(a\leq b\leq c\leq d) [/imath] where [imath]a,b,c,d\in \{1,2,\cdots,30\}[/imath] |
15423 | Optimal algorithm for finding the odd sphere with a balance scale
Say we have [imath]N[/imath] spheres indexed as [imath]1,2,3,\dotsc, N[/imath] such that all of them have identical weight apart from one, and we don't know if that one is heavier or lighter. We have to determine which sphere has the odd weight using just a balance scale. We could solve this problem by weighing repeatedly, but I am interested in a solution involving weighing as few times as possible, so my question is what is the optimal algorithm for this task? | 2711627 | Balls and scales (generalization)
A classical brain teaser is : You have twelve balls identical in size and appearance but one ball is an odd weight (could be either light or heavy). You have a set of balance scales. How many steps are necessary to identify the odd ball and to say whether it is light or heavy? The answer is [imath]3[/imath] steps, and the the solution is easily found on the web. But what if twelve is replaced by any number [imath]N > 2[/imath]? Is there a general way to find the optimal number of steps? |
26722 | calculating [imath]a^b \!\mod c[/imath]
What is the fastest way (general method) to calculate the quantity [imath]a^b \!\mod c[/imath]? For example [imath]a=2205[/imath], [imath]b=23[/imath], [imath]c=4891[/imath]. | 81228 | How do I compute [imath]a^b\,\bmod c[/imath] by hand?
How do I efficiently compute [imath]a^b\,\bmod c[/imath]: When [imath]b[/imath] is huge, for instance [imath]5^{844325}\,\bmod 21[/imath]? When [imath]b[/imath] is less than [imath]c[/imath] but it would still be a lot of work to multiply [imath]a[/imath] by itself [imath]b[/imath] times, for instance [imath]5^{69}\,\bmod 101[/imath]? When [imath](a,c) \neq 1[/imath], for instance [imath]6^{103}\,\bmod 14[/imath]? Are there any other tricks for evaluating exponents in modular arithmetic? This is being asked in an effort to cut down on duplicates, see here: Coping with *abstract* duplicate questions. and here: List of Generalizations of Common Questions |
254865 | Simple binomial theorem proof: [imath]\sum_{j=0}^{k} \binom{a+j}j = \binom{a+k+1}k[/imath]
I am trying to prove this binomial statement: For [imath]a \in \mathbb{C}[/imath] and [imath]k \in \mathbb{N_0}[/imath], [imath]\sum_{j=0}^{k} {a+j \choose j} = {a+k+1 \choose k}.[/imath] I am stuck where and how to start. My steps are these: [imath]{a+j \choose j} = \frac{(a+j)!}{j!(a+j-j)!} = \frac{(a+j)!}{j!a!}[/imath] but now I dont know how to go further to show the equality. Or I said: [imath]\sum_{j=0}^{k} {a+j \choose j} = {a+k \choose k} +\sum_{j=0}^{k-1} {a+j \choose j} = [help!] = {a+k+1 \choose k}[/imath] Thanks for help! | 866046 | Identity with binomials
Does there exist a closed formula for [imath]\underset{n=1}{\overset{N-1}{\sum}}\dbinom{N+n}{n}?[/imath] I've searching on wikipedia but I haven't found this kind of sum. |
91087 | Subfields of finite fields
We know that if a finite field [imath]F[/imath] has characteristic [imath]p[/imath] (prime), then [imath]F[/imath] has cardinality [imath]p^r[/imath] where [imath]r = [F:\mathbb{F}_p][/imath]. I'm now trying to say something about the possible cardinalities of subfields of [imath]F[/imath]. I can see that there is a subfield of cardinality [imath]p^s[/imath] for each [imath]s[/imath] that divides [imath]r[/imath], given by the fixed field of the group generated by [imath]\phi^s[/imath], where [imath]\phi[/imath] is the Frobenius automorphism. Now suppose [imath]K[/imath] is a subfield of [imath]F[/imath]. Then (since both are additive groups), Lagrange gives us that [imath]|K|[/imath] divides [imath]|F|[/imath], so [imath]|K| = p^t[/imath] for some [imath]1 \leq t \leq r [/imath] (alternatively, [imath]K[/imath] contains [imath]\mathbb{F}_p[/imath] and so is a vector space over [imath]\mathbb{F}_p[/imath] and is thus isomorphic to [imath]\mathbb{F}_p^t[/imath], where [imath]t = [K:\mathbb{F}_p][/imath]). By considering the multiplicative group of units of [imath]K[/imath] and [imath]F[/imath] respectively, we get that [imath] p^t - 1[/imath] divides [imath]p^r -1[/imath]. I want to make the leap to [imath]t|r[/imath], but I'm failing to see why this needs to be true. Any help would be appreciated. | 1377867 | I have to show no proper intermediate fields exist between [imath]Z_2[/imath] and [imath]GF(2^3)[/imath]
I have to show no proper intermediate fields exist between [imath]Z_2[/imath] and its overfield [imath]GF(2^3)[/imath], Can any one help? |
96739 | Do continuous linear functions between Banach spaces extend?
Just wondering... Let [imath]E[/imath], [imath]G[/imath] be Banach spaces, let [imath]U\subset E[/imath] be a subset of [imath]E[/imath], and let [imath]f:U\rightarrow G[/imath] be a continuous linear function. Can [imath]f[/imath] be extended to a continuous linear function on [imath]E[/imath], [imath]F:E\rightarrow G[/imath]? For which cases does this happen? | 1182253 | Reference request: linear operators into [imath]L^\infty[/imath] can be extended presrving the norm.
Suppose [imath]X[/imath] is a normed linear space and [imath]Y\subset X[/imath] a linear subspace. I remember that any linear map [imath]L\colon Y\to L^\infty(\Omega)[/imath] can be extended to a linear map [imath]\tilde{L}\colon X\to L^\infty(\Omega)[/imath] that has the same operator norm. (Under some assumptions, I guess [imath]\sigma[/imath]-finiteness, on the measure that on [imath]\Omega[/imath]) Does anyone have a reference for that? |
296745 | There exist a function such that [imath]f\circ f(x)=e^x[/imath]?
Based on this question: How to calculate [imath]f(x)[/imath] in [imath]f(f(x)) = e^x[/imath]? I would like to know if I can get a function such that [imath]f:\mathbb R \to \mathbb R^+[/imath], defined by [imath]f\circ f(x)=e^x[/imath]. My guess is no, but I can't prove it, I need help. Note that different than the previous question, the function is from [imath]\mathbb R[/imath] to [imath]\mathbb R^+[/imath]. Thanks a lot | 59023 | How to calculate [imath]f(x)[/imath] in [imath]f(f(x)) = e^x[/imath]?
How would I calculate the power series of [imath]f(x)[/imath] if [imath]f(f(x)) = e^x[/imath]? Is there a faster-converging method than power series for fractional iteration/functional square roots? |
15129 | how can be prove that [imath]\max(f(n),g(n)) = \Theta(f(n)+g(n))[/imath]
how can be prove that [imath]\max(f(n),g(n)) = \Theta(f(n)+g(n))[/imath] though the big O case is simple since [imath]\max(f(n),g(n)) \leq f(n)+g(n)[/imath] edit : where [imath]f(n)[/imath] and [imath]g(n)[/imath] are asymptotically nonnegative functions. | 267252 | How to prove that [imath]\max(f(n), g(n)) = \Theta(f(n) + g(n))[/imath]?
Using the basic definition of theta notation prove that [imath]\max(f(n), g(n)) = \Theta(f(n) + g(n))[/imath] I came across two answer to this question on this website but the answers weren't clear to me. Would you mind to elaborate how this can be proven? I am first year student of computer sciences. Thank you! Edit: What exactly does [imath]\max(f(n), g(n))[/imath] return? |
102476 | [imath]f\geq 0[/imath], continuous and [imath]\int_a^b f=0[/imath] implies [imath]f=0[/imath] everywhere on [imath][a,b][/imath]
This is problem 6.2 from the 3rd edition of Principles of Mathematical Analysis. Problem 6.2: Suppose [imath]f\geq 0[/imath], f is continuous on [imath][a, b][/imath], and [imath]\int_a^b f(x) \, dx = 0[/imath]. Prove that [imath]f(x)=0[/imath] for all [imath]x \in [a, b][/imath]. I'm looking for a critique of my proof. It's a pretty easy problem, but I am always wary of making too bold of assumptions, especially on these low level/fundamental proofs. I'll be using Rudin's notation and refer to theorems from the text (If I should include the text of each theorem, feel free to leave a comment... I'm lazy but could probably use the TeX practice :p) Proof: Assume, for contradiction, that [imath]f>0[/imath]. Then, for any partition [imath]P[/imath] we have the Lower Riemann Sum: [imath]L(P, f)=\sum_{i=1}^n m_i \, \Delta x_i[/imath]. At least one [imath]\Delta x_i[/imath] must be positive, since [imath]a < b[/imath], and each [imath]m_i[/imath] must be positive since we have [imath]f>0[/imath] by assumption, so certainly [imath]\sup f > 0[/imath]. That means [imath]L(P, f)>0[/imath]. Thus, we have: [imath]0 = 0(b-a) < L(P,f)\leq \sup L(P, f) = \inf U(P,f) =L[/imath] where the last string of equalities holds because our function is continuous on a compact interval, so is integrable by theorems 6.8 and 6.6. So our integral has value [imath]L>0[/imath]. This is in contradiction to our given assumption that [imath]\int_a^b f(x) \, dx = 0[/imath], so we must have that [imath]f=0[/imath] on [imath][a, b][/imath]. [imath]\Box[/imath] So, I am wondering if my proof is correct (and is presented well). Also if someone could enlighten me as to what Rudin means when he say "Compare this with exercise 1," I'd be appreciative. Is it a hint or is there something else he expects you to notice? there are a lot of things I could compare :)... Exercise 6.1: Suppose [imath]\alpha[/imath] increases on [imath][a, b][/imath], [imath]a \leq x_0 \leq b[/imath], [imath]\alpha[/imath] is continuous at [imath]x_0[/imath], [imath]f(x_0)=1[/imath], and [imath]f(x)=0[/imath] if [imath]x\neq x_0[/imath]. Prove that [imath]f[/imath] is Riemann-Stieltjes Integrable and that [imath]\int f \, d\alpha = 0[/imath] | 1750372 | Show that [imath]f[/imath] is identically zero if and only if [imath]\int_a^b f(x)dx = 0[/imath]
Assume [imath]f[/imath] is continuous and nonnegative over [imath][a,b][/imath]. Show that [imath]f[/imath] is identically zero if and only if [imath]\displaystyle \int_{a}^b f(x)dx = 0[/imath]. Proving the first direction is easy: If [imath]f(x) = 0[/imath], then obviously [imath]\displaystyle \int_{a}^b f(x)dx = 0[/imath]. On the other hand if [imath]\displaystyle \int_{a}^b f(x)dx = 0[/imath], then since [imath]f[/imath] is continuous on [imath][a,b][/imath] and [imath]f(x) \geq 0[/imath] we must have [imath]f(x) = 0[/imath]. |
197393 | Why does [imath]\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi[/imath]?
Playing around on wolframalpha shows [imath]\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi[/imath]. I know [imath]\tan^{-1}(1)=\pi/4[/imath], but how could you compute that [imath]\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi[/imath] to get this result? | 1657856 | proofs on trigonometric identities involving complex numbers
Provide a reason for each step of the proof. Prove the identity [imath]\arctan(1)+\arctan(2)+\arctan(3)=180^{\circ}[/imath]Proof:\begin{align*}\arctan(1)+\arctan(2)+\arctan(3)&=\text{arg}(1+i)+\text{arg}(1+2i)+\text{arg}(1+3i)\\&=\text{arg}\left[(1+i)(1+2i)(1+3i)\right]\\&=\text{arg}(-10)=180^{\circ} \hspace{4cm}\blacksquare\end{align*} |
18983 | why is [imath]\sum\limits_{k=1}^{n} k^m[/imath] a polynomial with degree [imath]m+1[/imath] in [imath]n[/imath]
why is [imath]\sum\limits_{k=1}^{n} k^m[/imath] a polynomial with degree [imath]m+1[/imath] in [imath]n[/imath]? I know this is well-known. But how to prove it rigorously? Even mathematical induction does not seem so straight-forward. Thanks. | 731047 | Sum of series: [imath]1^k+2^k+3^k+...+n^k =?[/imath]
Is there any better algorithm to solve this equation other than brute force. [imath]1^k+2^k+3^k+...+n^k=[/imath] formula? Here [imath]k[/imath] is a natural number, [imath]n[/imath] is a natural number. |
84392 | Why do the [imath]n \times n[/imath] non-singular matrices form an "open" set?
Why is the set of [imath]n\times n[/imath] real, non-singular matrices an open subset of the set of all [imath]n\times n[/imath] real matrices? I don't quite understand what "open" means in this context. Thank you. | 430488 | Topology of Lie groups.
I do not understand the topology of a Lie group clearly. Let [imath]G[/imath] be a Lie group and [imath]T_eG[/imath] be its tangent space at the identity [imath]e \in G[/imath]. Why [imath]Aut(T_eG)[/imath] is an open subset of the vector space of endomorphisms of [imath]T_eG[/imath] (i.e. [imath]End(T_eG)[/imath])? What does "open" mean? Thank you very much. |
298614 | General term of [imath]a_n = 2a_{n-1} + 1[/imath]
Find the general term of the sequence defined by: [imath]a_n = 2a_{n-1} + 1[/imath] where [imath]a_1[/imath] is given Thank You | 183599 | can one derive the [imath]n^{th}[/imath] term for the series, [imath]u_{n+1}=2u_{n}+1[/imath],[imath]u_{0}=0[/imath], [imath]n[/imath] is a non-negative integer
derive the [imath]n^{th}[/imath] term for the series [imath]0,1,3,7,15,31,63,127,255,\ldots[/imath] observation gives, [imath]t_{n}=2^n-1[/imath], where [imath]n[/imath] is a non-negative integer [imath]t_{0}=0[/imath] |
78560 | How do you solve the Initial value probelm [imath]dp/dt = 10p(1-p), p(0)=0.1[/imath]?
The problem is... [imath] \frac{dp}{dt} = 10p(1-p),[/imath] [imath]p(0)=0.1[/imath]. Solve and show that [imath]p(t) \to 1[/imath] as [imath]t\to \infty.[/imath] I know this is probably really simple, I was trying to go down the line of finding a general solution and then imposing the boundary condition. But I can't even see how to find the general solution... Then for the second bit taking the limit of the general solution [imath]p(t)[/imath]? | 1070953 | Differential equation problem. Integrating the logistic equation.
I would like to know how to integrate or rather solve this: [imath] \frac{dP}{dt} = kP(L-P). [/imath] I have the solution, but I would like to know how to arrive at it. I have been told it involves separation of variables and partial fractions. |
61828 | Proof of Frullani's theorem
How can I prove the Theorem of Frullani? I did not even know all the hypothesis that [imath]f[/imath] must satisfy, but I think that this are Let [imath]\,f:\left[ {0,\infty } \right) \to \mathbb R[/imath] be a a continuously differentiable function such that [imath] \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 0, [/imath] and let [imath] a,b \in \left( {0,\infty } \right)[/imath]. Prove that [imath] \int\limits_0^{\infty} {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}} {x}}dx = f\left( 0 \right)\left[ {\ln \frac{b} {a}} \right] [/imath] If you know a more general version please give it to me )= I can´t prove it. | 1046074 | Show that [imath]\int_0^{\infty}\frac{f(ax)-f(bx)}{x}dx=[f(0)-L]\ln\frac{b}{a}[/imath]
Let [imath]f:[0,\infty)\to\mathbb{R}[/imath] be continuous and [imath]\lim_{x\to\infty}f(x)=L[/imath]. Show that [imath]\int_0^{\infty}\frac{f(ax)-f(bx)}{x}dx=[f(0)-L]\ln\frac{b}{a}[/imath] where [imath]0<a<b[/imath]. I don't even know how to start. |
9286 | Proving [imath]\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \frac{\sqrt \pi}{2}[/imath]
How to prove [imath]\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}[/imath] | 355024 | How to calculate the following integral?
How to calculate the following integral? [imath]\int_{- \infty}^{\infty} \mathrm{e}^{- \frac{x^2}{2}} \mathrm{d} x[/imath] |
187729 | Evaluating [imath]\int_0^\infty \sin x^2\, dx[/imath] with real methods?
I have seen the Fresnel integral [imath]\int_0^\infty \sin x^2\, dx = \sqrt{\frac{\pi}{8}}[/imath] evaluated by contour integration and other complex analysis methods, and I have found these methods to be the standard way to evaluate this integral. I was wondering, however, does anyone know a real analysis method to evaluate this integral? | 1142514 | Evaluate [imath]\int_{0}^{\infty} \cos(x^2)dx [/imath]
Prove that the above integral is equal to [imath]\frac{\sqrt{2\pi}}{2}[/imath] I have already tried expanding using [imath]\cos[/imath] identity and also taking Laplace for it. I am getting nowhere with this. |
14666 | Number of permutations of [imath]n[/imath] elements where no number [imath]i[/imath] is in position [imath]i[/imath]
I am trying to figure out how many permutations exist in a set where none of the numbers equal their own position in the set; for example, [imath]3,1,5,2,4[/imath] is an acceptable permutation where [imath]3,1,2,4,5[/imath] is not because 5 is in position 5. I know that the number of total permutations is [imath]n![/imath]. Is there a formula for how many are acceptable given the case that no position holds its own number? | 408341 | A basic probability doubt on derangment
Is there any implication that the probability that a random permutation is a derangment is [imath]\frac{1}{e}[/imath] when [imath]n->\infty[/imath] ? |
95799 | Why [imath]\gcd(qb+r,b)=\gcd(b,r)[/imath]?
Given: [imath]a = qb + r[/imath]. Then it holds that [imath]\gcd(a,b)=\gcd(b,r)[/imath]. That doesn't sound logical to me. Why is this so? Addendum by LePressentiment on 11/29/2013: (in the interest of http://meta.math.stackexchange.com/a/4110/53259 and averting a duplicate) What's the intuition behind this result? I only recognise the proof and examples solely due to algebraic properties and formal definitions; I'd like to apprehend the result naturally. | 488661 | If [imath]a = r \pmod b[/imath] when [imath]0 \le r < b[/imath], then [imath]\gcd(a, b) = \gcd (b, r)[/imath]
If [imath]a = r \pmod b[/imath] when [imath]0 \le r < b[/imath], then [imath]\gcd(a, b) = \gcd (b, r)[/imath]. I do not understand why this is, can somebody explain? I've also looked over this thread: Why is [imath]\gcd(a,b)=\gcd(b,r)[/imath] when [imath]a = qb + r[/imath]? , but I still can't seem to see how that applies to [imath]a = r \pmod b[/imath]. |
51502 | If [imath]f_k \to f[/imath] a.e. and the [imath]L^p[/imath] norms converge, then [imath]f_k \to f[/imath] in [imath]L^p[/imath]
Let [imath]1\leq p < \infty[/imath]. Suppose that [imath]\{f_k\} \subset L^p[/imath] (the domain here does not necessarily have to be finite), [imath]f_k \to f[/imath] almost everywhere, and [imath]\|f_k\|_{L^p} \to \|f\|_{L^p}[/imath]. Why is it the case that [imath]\|f_k - f\|_{L^p} \to 0?[/imath] A statement in the other direction (i.e. [imath]\|f_k - f\|_{L^p} \to 0 \Rightarrow \|f_k\|_{L^p} \to \|f\|_{L^p}[/imath] ) follows pretty easily and is the one that I've seen most of the time. I'm not how to show the result above though. | 592586 | Basic [imath]L_1[/imath] convergence question
I think this is quite a simple question, but for some reason am finding it difficult to answer. The question is: If [imath](f_n:n\in\mathbb{N})[/imath] is a sequence of integrable functions, with [imath]f_n \to f[/imath] a.e. for some integrable [imath]f[/imath], then is it true that if [imath]\|f_n\|_1 \to \|f\|_1 [/imath], then [imath]\|f_n-f\|_1\to 0[/imath]. Any help is very good! |
18690 | Algebraic Proof that [imath]\sum\limits_{i=0}^n \binom{n}{i}=2^n[/imath]
I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subsets of a set of [imath]n[/imath] elements. I'm curious if there's a series of algebraic manipulations that can lead from [imath]\sum\limits_{i=0}^n \binom{n}{i}[/imath] to [imath]2^n[/imath]. | 59554 | Algebraic proof that collection of all subsets of a set (power set) of [imath]N[/imath] elements has [imath]2^N[/imath] elements
In other words, is there an algebraic proof showing that [imath]\sum_{k=0}^{N} {N\choose k} = 2^N[/imath]? I've been trying to do it some some time now, but I can't seem to figure it out. |
63870 | A classical problem about limit of continuous function at infinity and its connection with Baire Category Theorem
When I google "baire category theorem", I get a link to Ben Green's website. And at the end of the paper, he mentioned such a classic problem: Suppose that [imath]f:\mathbb{R}^+\to\mathbb{R}^+[/imath] is a continuous function with following property: for all [imath]x\in\mathbb{R}^+[/imath], the sequence [imath]f(x), f(2x), f(3x),\dots[/imath] tends to [imath]0[/imath]. Prove that [imath]\lim_{t\to\infty}f(t)=0[/imath]. I find the problem 1.17 on P.27 of the book "Selected problems of real analysis", and on P.169 gives the answer by prove the following lemma: If [imath]G[/imath] is a unbounded open set of [imath]\mathbb{R}^+[/imath], then for any closed interval [imath][p,q]\ (0<p<q)[/imath], there exist a [imath]x_0\in [p,q][/imath] such that [imath]G[/imath] contains infinitely many points of the form [imath]nx_0\ (n\in\mathbb{N})[/imath]. But, on the above book, it also says: If [imath]\lim_{n\to\infty}f(nx)[/imath] exists only for points [imath]x[/imath] in a nonempty closed set without isolated points, then [imath]\lim_{x\to\infty}f(x)[/imath] also exist. I didn't find a proof for this result, I want to know whether for any nonempty closed set with no isolated points, the above lemma is true? and could someone tell me why Ben Green mentioned this problem on his paper (see the hint of his paper)? | 101086 | [imath]\lim_{n\to \infty}f(nx)=0[/imath] implies [imath]\lim_{x\to \infty}f(x)=0[/imath]
Can anyone help me with this problem? Let [imath]f:[0,\infty)\longrightarrow \mathbb R[/imath] be a continuous function such that for each [imath]x>0[/imath], we have [imath]\lim_{n\to \infty}f(nx)=0[/imath]. Then prove that [imath]\lim_{x\to \infty}f(x)=0[/imath]. Our teacher told first to prove Baire's theorem, and then show that this is a consequence of that theorem. I proved Baire's theorem, and I spend a few hours thinking on how Baire's theorem is related to this problem, but I couldn't find anything. I'd really appreciate your help. |
24060 | Uniform continuity on (0,1) implies boundedness
I need to prove that if [imath]f: (0,1) \rightarrow \mathbb{R}[/imath] is Uniformly continuous then it is bounded. Thank you. | 567577 | Why does uniform continuity of a function imply that the function is bounded?
As the title states, I'm wondering why: If [imath]A[/imath] is a bounded subset of [imath]\mathbb{R}[/imath] and [imath]f:A\to \mathbb{R}[/imath] is uniformly continuous on [imath]A[/imath], then [imath]f[/imath] must be bounded on [imath]A[/imath]. Proof: Since it is uniformly continuous, the function is a Lipschitz function. [imath]|f(x)-f(y)| \leq L|x-y|[/imath]. Since [imath]A[/imath] is bounded, [imath]|x-y|[/imath] does not get arbitrarily big and that too is bounded by a constant. Let [imath]|x-y| \leq M[/imath]. Then we have a Lipschitz condition where [imath]|f(x)-f(y)| \leq LM[/imath]. The function is then bounded by the product of two constants, [imath]LM[/imath], which means that it is bounded. Can someone check this? |
67620 | Set of continuity points of a real function
I have a question about subsets [imath] A \subseteq \mathbb R [/imath] for which there exists a function [imath]f : \mathbb R \to \mathbb R[/imath] such that the set of continuity points of [imath]f[/imath] is [imath]A[/imath]. Can I characterize this kind of sets? In a topological,measurable or in some way? For example, does there exist a function continuous on [imath]\mathbb Q[/imath] and discontinuous on the irrationals? | 64677 | characteristic function of the rationals
Let [imath]\chi[/imath] be the characteristic function of the rational numbers in [imath][0,1][/imath]. Does there exist a sequence [imath]\{f_n\}[/imath] of continuous functions on [imath][0,1][/imath] that converges pointwise to [imath]\chi[/imath]? |
141774 | Choice of [imath]q[/imath] in Baby Rudin's Example 1.1
First, my apologies if this has already been asked/answered. I wasn't able to find this question via search. My question comes from Rudin's "Principles of Mathematical Analysis," or "Baby Rudin," Ch 1, Example 1.1 on p. 2. In the second version of the proof, showing that sets A and B do not have greatest or lowest elements respectively, he presents a seemingly arbitrary assignment of a number [imath]q[/imath] that satisfies equations (3) and (4), plus other conditions needed to show that [imath]q[/imath] is the right number for the proof. As an exercise, I tried to derive his choice of [imath]q[/imath] so that I may learn more about the problem. If we write equations (3) as [imath]q = p - (p^2 - 2)x[/imath], we can write (4) as [imath] q^2 - 2 = (p^2 - 2)[1 - 2px + (p^2 - 2)x^2]. [/imath] Here, we need a rational [imath]x > 0[/imath], chosen such that the expression in [imath][...][/imath] is positive. Using the quadratic formula and the sign of [imath](p^2 - 2)[/imath], it can be shown that we need [imath] x \in \left(0, \frac{1}{p + \sqrt{2}}\right) \mbox{ for } p \in A, [/imath] or, for [imath]p \in B[/imath], [imath]x < 1/\left(p + \sqrt{2}\right)[/imath] or [imath]x > 1/\left(p - \sqrt{2}\right)[/imath]. Notice that there are MANY solutions to these equations! The easiest to see, perhaps, is letting [imath]x = 1/(p + n)[/imath] for [imath]n \geq 2[/imath]. Notice that Rudin chooses [imath]n = 2[/imath] for his answer, but it checks out easily for other [imath]n[/imath]. The Question: Why does Rudin choose [imath]x = 1/(p + 2)[/imath] specifically? Is it just to make the expressions work out clearly algebraically? Why doesn't he comment on his particular choice or the nature of the set of solutions that will work for the proof? Is there a simpler derivation for the number [imath]q[/imath] that I am missing? | 511228 | Intuition in Rudin's Proof on Page 2
Rudin has a proof in which he is proving that [imath]A=\{ p \in \mathbb{Q}\;|\; p^2<2\}[/imath] has no maximum element (or in other words, an element which is greater than every other element). For this he creates a rational [imath]q=p-\dfrac{p^2-2}{p+2}[/imath] and uses this rational [imath]q[/imath] to show that it is greater than [imath]p[/imath] and still in [imath]A[/imath]. Since [imath]p[/imath] was arbitrary element of [imath]A[/imath], this finishes the proof. I follow the proof but I don't understand where the formula for [imath]q[/imath] magically came from. Any help? |
199026 | Don't we need the axiom of choice to choose from a non-empty set?
I recently read a proof that had the following in it: "since [imath]A[/imath] is non-empty, we can find an element [imath]x[/imath] in [imath]A[/imath]." This proof did not mention the axiom of choice, but it seems to me that it would be required to make the proof formal. Would I not require a choice function to allow me to find/pick some element [imath]x[/imath] from [imath]A[/imath] after noting that A is non-empty? Thanks | 2786279 | Schröder–Bernstein theorem for finite sets
I want to prove Schröder–Bernstein theorem for 2 finite sets [imath]A, B[/imath] of the same cardinal. I do it with induction on the cardinal number of [imath]A, B[/imath]. In the inducative step, I write: Let [imath]A, B[/imath] sets such that [imath]|A|=|B|=n+1[/imath] for some [imath]n\in\omega[/imath]. Here, I want to use the the inducative assumption for the sets [imath]A-\{a\}, B-\{b\}[/imath] for some [imath]a\in A, b\in B[/imath]. For choosing the above [imath]a,b[/imath], do I need to use the axiom of choice? |
668 | What's an intuitive way to think about the determinant?
In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a [imath]2\times 2[/imath] matrix by the formula. Our teacher showed us how to compute the determinant of an [imath]$n \times n$[/imath] matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related? | 580754 | Physical meaning of the determinant .
In my undergraduate mechanics class, we just finished the section on coupled oscillators where we use eigenvalues to find solutions to the differential equations describing the motion of our spring mass system. When we finished, one of my fellow students started going on a rant about how he "knew the determinant had a physical meaning," and he went on to talk about how frustrated he was with his differential equations professor who said that the determinant was just an operation and only had physical meaning when it was given one (ie- showing the system is linearly independent by setting the determinant to [imath]0[/imath]). I sided with the professor, but I wasn't sure how to explain it to my classmate. What do you guys think? |
13801 | What's the thing with [imath]\sqrt{-1} = i[/imath]
What's the thing with [imath]\sqrt{-1} = i[/imath]? Do they really teach this in the US? It makes very little sense, because [imath]-i[/imath] is also a square root of [imath]-1[/imath], and the choice of which root to label as [imath]i[/imath] is arbitrary. So saying [imath]\sqrt{-1} = i[/imath] is plainly false! So why do people say [imath]\sqrt{-1} = i[/imath]? Is this how it's taught in the US? | 1097134 | why is [imath]\sqrt{-1} = i[/imath] and not [imath]\pm i[/imath]?
this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE. Why is [imath]\sqrt{-1} = i[/imath] and not [imath]\sqrt{-1}=\pm i[/imath]? With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers? |
2260 | Proof for formula for sum of sequence [imath]1+2+3+\ldots+n[/imath]?
Apparently [imath]1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2[/imath]. How? What's the proof? Or maybe it is self apparent just looking at the above? PS: This problem is known as "The sum of the first [imath]n[/imath] positive integers". | 658705 | What is the sum of integers from [imath]1[/imath] to [imath]789999[/imath] ? asks the professor
How to resolve it? How to find that sum? |
264980 | How to prove that [imath] \sum_{n \in \mathbb{N} } | \frac{\sin( n)}{n} | [/imath] diverges?
It is stated as a problem in Spivak's Calculus and I can't wrap my head around it. | 1897860 | How to prove that [imath]\sum_{n=1}^{\infty} \frac{|\sin n|}{n}[/imath] and [imath]\sum_{n=1}^{\infty} \frac{\sin^2 n}{n}[/imath] both diverge?
How can I prove that [imath]\sum_{n=1}^{\infty} \frac{|\sin n|}{n}[/imath] and [imath]\sum_{n=1}^{\infty} \frac{\sin^2 n}{n}[/imath] both diverge? I thought of using Comparison test, but I couldn't find any sequence to compare with. This question is from the book 'Real Analysis and Foundations' by Steven G. Krantz. |
92105 | [imath]f[/imath] uniformly continuous and [imath]\int_a^\infty f(x)\,dx[/imath] converges imply [imath]\lim_{x \to \infty} f(x) = 0[/imath]
Trying to solve [imath]f(x)[/imath] is uniformly continuous in the range of [imath][0, +\infty)[/imath] and [imath]\int_a^\infty f(x)dx [/imath] converges. I need to prove that: [imath]\lim \limits_{x \to \infty} f(x) = 0[/imath] Would appreciate your help! | 397214 | uniformly continuous and [imath]\int_0^\infty f(t)\,\mathrm dt[/imath] exists [imath]\implies \lim_{x\to\infty}f(x) = 0 [/imath]
I appreciate your help with this one. Let [imath]f \colon[0,\infty)\rightarrow \mathbb{R}[/imath] be uniformly continuous and let the integral [imath]\int_0^\infty f(t)\,\mathrm dt[/imath] exist and be final. I need to show that [imath]\lim_{x\to\infty}f(x) = 0. [/imath] Thank you very much. |
305023 | Is the graph of every real function a null set?
This question popped to my mind during an analysis lecture: Let [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] be a (general) function. Is there an [imath]N\subset \mathbb{R}^2[/imath] with [imath]\lambda^2(N)=0[/imath], such that [imath]\{(x,f(x)):x\in \mathbb{R}\}[/imath] [imath]\subset N[/imath] ? If [imath]f[/imath] is measurable or even continuous, we could apply classical calculus, but what if f is not measurable? | 35606 | Lebesgue Measure of the Graph of a Function
Let [imath]f:R^n \rightarrow R^m[/imath] be any function. Will the graph of [imath]f[/imath] always have Lebesgue measure zero? [imath](1)[/imath] I could prove that this is true if [imath]f[/imath] is continuous. [imath](2)[/imath] I suspect it is true if [imath]f[/imath] is measurable, but I'm not sure. (My idea was to use Fubini's Theorem to integrate the indicator function of the graph, but I don't know if I'm using the Theorem properly). If [imath](2)[/imath] is incorrect, what would be a counterexample where the graph of [imath]f[/imath] has positive measure? If [imath](2)[/imath] is correct, can we prove the existence of a non-measurable function whose graph has positive outer measure? |
4551 | How can I prove [imath]\sup(A+B)=\sup A+\sup B[/imath] if [imath]A+B=\{a+b\mid a\in A, b\in B\}[/imath]
If [imath]A,B[/imath] non empty, upper bounded sets and [imath]A+B=\{a+b\mid a\in A, b\in B\}[/imath], how can I prove that [imath]\sup(A+B)=\sup A+\sup B[/imath]? | 509661 | Taking suprema on a set-equality
Suppose we have the following: [imath]X = A + B [/imath] where [imath]X, A,[/imath] and [imath]B[/imath] are any sets. [imath]A + B = \{ a + b : a \in A , \; \; \; b \in B \} [/imath] Can we conclude that [imath]\sup X = \sup A + \sup B [/imath] ? |
13490 | Proving that the sequence [imath]F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}[/imath] is boundedly convergent on [imath]\mathbb{R}[/imath]
Here is an exercise, on analysis which i am stuck. How do I prove that if [imath]F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}[/imath], then the sequence [imath]\{F_{n}(x)\}[/imath] is boundedly convergent on [imath]\mathbb{R}[/imath]? | 1528099 | Problem dealing with [imath]\sum \frac{\sin(n)}{n}[/imath] and its convergence
[imath]\text{If} \ S=\displaystyle\sum_{n=1}^{\infty}\dfrac{\sin (n)}{n}, \ \text{then what is} \ 2S+1[/imath] I know that [imath]\sum \frac{\sin(n)}{n}[/imath] converges. But now what do I do? |
21792 | Norms Induced by Inner Products and the Parallelogram Law
Let [imath] V [/imath] be a normed vector space (over [imath]\mathbb{R}[/imath], say, for simplicity) with norm [imath] \lVert\cdot\rVert[/imath]. It's not hard to show that if [imath]\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}[/imath] for some (real) inner product [imath]\langle \cdot, \cdot \rangle[/imath], then the parallelogram equality [imath] 2\lVert u\rVert^2 + 2\lVert v\rVert^2 = \lVert u + v\rVert^2 + \lVert u - v\rVert^2 [/imath] holds for all pairs [imath]u, v \in V[/imath]. I'm having difficulty with the converse. Assuming the parallelogram identity, I'm able to convince myself that the inner product should be [imath] \langle u, v \rangle = \frac{\lVert u\rVert^2 + \lVert v\rVert^2 - \lVert u - v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u\rVert^2 - \lVert v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u - v\rVert^2}{4} [/imath] I cannot seem to get that [imath]\langle \lambda u,v \rangle = \lambda \langle u,v \rangle[/imath] for [imath]\lambda \in \mathbb{R}[/imath]. How would one go about proving this? | 370697 | Parallelogram law, dot product
Prove that if [imath]||\cdot||[/imath] satisfies [imath]||u-v||^2 + ||u+v||^2 = 2(||u||^2 + ||v||^2)[/imath] , then [imath]u \cdot v = \frac{1}{2} (||u+v||^2 - ||u||^2 - ||v||^2)[/imath] is dot product and [imath]||u||^2 = u \cdot u[/imath]. I've already shown that [imath](u+w)\cdot v = u \cdot v + w \cdot v[/imath], but I have serious troubles showing that [imath](\lambda v)\cdot(w) = \lambda (v\cdot w)[/imath]. Could you help me with that? |
29023 | Value of [imath]\sum\limits_n x^n[/imath]
Why does the following hold: \begin{equation*} \displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3 ? \end{equation*} Can we generalize the above to [imath]\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}[/imath] ? Are there some values of [imath]x[/imath] for which the above formula is invalid? What about if we take only a finite number of terms? Is there a simpler formula? [imath]\displaystyle \sum_{n=0}^{N} x^n[/imath] Is there a name for such a sequence? This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions. and here: List of abstract duplicates. | 68572 | Show that the geometric series [imath]a + ar +ar^2 + \cdots + ar^{n-1} + \cdots[/imath] converges if and only if [imath]|r| < 1[/imath]
Show, rigorously, that the geometric series [imath]a + ar +ar^2 + \cdots + ar^{n-1} + \cdots[/imath] converges if and only if |r| < 1. Also, show that if |r| < 1, the sum is given by [imath]S = \frac{a}{1-r}[/imath]. Note(s): Question 1, First part: if |r| < 1, show (rigorously) that the geometric series above converges. Update: I'm reposting this question as the accepted answer on the possible duplicate question doesn't provide a rigorous [[imath]\epsilon[/imath]-[imath]\delta[/imath]] derivation of an important step, namely, [imath]\lim_{N\to\infty}r^{N+1} = 0[/imath] if [imath]|r|\lt 1[/imath]. [Translated to the current question's context, [imath]\lim_{n\to\infty}ar^n = 0[/imath] if [imath]|r|\lt 1[/imath]]. Update 2: The problem of the above issue has been addressed. Question 1, Second part: if the geometric series above converges, then show (rigorously) that |r| < 1. -- Any hints would be appreciated. -- |
6979 | Areas versus volumes of revolution: why does the area require approximation by a cone?
Suppose we rotate the graph of [imath]y = f(x)[/imath] about the [imath]x[/imath]-axis from [imath]a[/imath] to [imath]b[/imath]. Then (using the disk method) the volume is [imath]\int_a^b \pi f(x)^2 dx[/imath] since we approximate a little piece as a cylinder. However, if we want to find the surface area, then we approximate it as part of a cone and the formula is [imath]\int_a^b 2\pi f(x)\sqrt{1+f'(x)^2} dx.[/imath] But if approximated it by a circle with thickness [imath]dx[/imath] we would get [imath]\int_a^b 2\pi f(x) dx.[/imath] So my question is how come for volume we can make the cruder approximation of a disk but for surface area we can't. | 325547 | Why does the surface area integral need the arc length differential but the volume doesn't?
When calculating the surface area of a revolution you need to use the arc length differential [imath]\sqrt{1 + y'^2}[/imath] but you don't need to use that when calculating the volume. Why is that? Thanks! |
29366 | Do sets, whose power sets have the same cardinality, have the same cardinality?
Is it generally true that if [imath]|P(A)|=|P(B)|[/imath] then [imath]|A|=|B|[/imath]? Why? Thanks. | 842789 | How do we prove that, if [imath]\mathcal{P}(A) \sim \mathcal{P}(B)[/imath], then [imath]A \sim B[/imath]?
The converse--if [imath]\ A \sim B[/imath] then [imath] \mathcal{P}(A) \sim \mathcal{P}(B)[/imath]--is very easy to prove. I can't see an immediate, simple proof for the converse case. It seems like a potentially good strategy would be to prove that there is a bijection [imath]f[/imath] from [imath]\mathcal{P}(A)[/imath] to [imath]\mathcal{P}(B)[/imath] such that, if [imath]X \in \mathcal{P}(A)[/imath], then [imath]X \sim f(X)[/imath], but that seems non-trivial as well. In case it isn't clear, the notation [imath]C \sim D[/imath] means that there us a bijection from [imath]C[/imath] to [imath]D[/imath]. |
191453 | How to show that the modulus of [imath]\frac{z-w}{1-\bar{z}w}[/imath] is always [imath]1[/imath]?
Let's suppose that [imath]|z|<1[/imath] and [imath]|w|=1[/imath]. Show that the modulus of [imath]\displaystyle \frac{z-w}{1-\bar{z}w}[/imath] is always [imath]1[/imath]. Some hint. | 1613426 | Prove that [imath]|\frac{a-b}{1-\bar ab}|=1[/imath] if [imath]|a|=1[/imath] or [imath]|b|=1[/imath]
Prove that [imath]|\frac{a-b}{1-\bar ab}|=1[/imath] if [imath]|a|=1[/imath] or [imath]|b|=1[/imath] I assumed [imath]|a|=1[/imath]. Then tried to show that our statement holds. I wrote [imath]a=a_1+ia_2[/imath] and [imath]b=b_1+ib_2[/imath] and [imath]\bar a=a_1-ia_2[/imath] Also [imath]|a|=|\bar a|=a_1^2+a_2^2=1[/imath] However, after multiplying it all out it doesn't get me anywhere. I figured maybe I have to use triangle inequality, in particular I know that: [imath]|a+b|\leq |a|+|b|[/imath] and [imath]|a+b| \geq |a|-|b|[/imath] Should I take [imath]-b[/imath] instead of [imath]b[/imath]? Then the inequality becomes: [imath]|a-b| \leq |a|+|-b|[/imath] and [imath]|a-b| \geq |a|-|-b|[/imath] which looks more similar to what I need. |
6244 | Is there a quick proof as to why the vector space of [imath]\mathbb{R}[/imath] over [imath]\mathbb{Q}[/imath] is infinite-dimensional?
It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this? | 868858 | Dimension of R over Q without cardinality argument.
I am looking for the easiest (elementary) proof that [imath]\mathbb R[/imath] is infinite dimensional as a [imath]\mathbb Q[/imath]-vector space, without using cardinality. It should be understandable at highschool level. So I guess the question could be reformulated as: what is the easiest infinite family of reals that can be showed to be independent ? So far square roots of primes seems a good candidate, but the proof is still a little intricate, is it the easiest possible ? The goal of this is to show students that we can prove the result by different ways, and see that understanding cardinals is useful. But I still want the students to be able to understand the other proof. |
147642 | If a group satisfies [imath]x^3=1[/imath] for all [imath]x[/imath], is it necessarily abelian?
I know that any group satisfying [imath]x^2=1[/imath] for all [imath]x[/imath] is abelian. Is the same true if [imath]x^3=1[/imath]? I don't think it is, but I can't find a basic counterexample. | 678570 | Non-abelian group in which [imath]\forall_{a,b\in G} (ab)^3=a^3b^3[/imath]
Give an example of a non-abelian group, in which [imath](ab)^3=a^3b^3[/imath] for every element [imath]a,b[/imath] in [imath]G[/imath]. I understand that such a group should be of order divisible by 3 (see Problem from Herstein on group theory). Also, it is easily seen that [imath](ab)^3=a^3b^3 \iff (ba)^2=a^2b^2[/imath]. But I can't come up with one single example. |