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https://artofproblemsolving.com/wiki/index.php?title=Fermat%27s_Last_Theorem&diff=next&oldid=31193 | # Difference between revisions of "Fermat's Last Theorem"
Fermat's Last Theorem is a recently proven theorem stating that for positive integers with , there are no solutions to the equation .
## History
Fermat's last theorem was proposed by Pierre Fermat in the 1600s in the margin of his copy of the book Arithmetica, by Diophantus. The note in the margin (when translated) read: "It is impossible for a cube to be the sum of two cubes, a fourth power to be the sum of two fourth powers, or in general for any number that is a power greater than the second to be the sum of two like powers. I have discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain."
Many mathematicians today doubt that Fermat actually had a proof for this theorem. If he did have one, he never published it, though he did circulate a proof for the case . It seems unlikely that he would have circulated a proof for the special case when he had a general solution. Some think that Fermat's proof was flawed, and that he saw the flaw after a time.
Some mathematicians have suggested that Fermat had a proof that relied on unique factorization in rings of the form . Unfortunately, this is not often the case. In fact, it has now been known for some time how to solve the problem when this is the case.
Despite Fermat's claim that a simple proof existed, the theorem wasn't proven until Andrew Wiles did so in 1993. Wiles's proof was the culmination of decades of work in number theory. Interestingly enough, Wiles's proof was much more modern than anything Fermat could have produced himself. It exploited connections between modular forms and elliptic curves.
In some sense, Fermat's last theorem is a dead end: it has led to few new mathematical consequences. However, the search for the proof of the theorem generated whole new areas of mathematics. In this sense, it was a good, productive problem.
The ABC Conjecture is a far-reaching conjecture that implies Fermat's Last Theorem for . It is one of the most famous still-open problems in number theory. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9266814589500427, "perplexity": 303.4629095287181}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703519784.35/warc/CC-MAIN-20210119201033-20210119231033-00795.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-10-section-10-6-counting-principles-permutations-and-combinations-exercise-set-page-1103/8 | ## Precalculus (6th Edition) Blitzer
The required solution is $1$
We know that the representation $_{n}{{P}_{r}}$ implies that the number of possible well-organized arrangements of n items is taken r at a time. And the number of possible well-organized arrangements of n items taken r at a time can be evaluated as: $_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ And the provided expression is $_{6}{{P}_{0}}$. Here, $n=6,r=0$. Put the value of n, r in the above formula. Then: \begin{align} & _{6}{{P}_{0}}=\frac{6!}{\left( 6-0 \right)!} \\ & =\frac{6!}{6!} \\ & =1 \end{align} Hence, $_{6}{{P}_{0}}=1$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999353885650635, "perplexity": 408.23457106353584}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039476006.77/warc/CC-MAIN-20210420152755-20210420182755-00631.warc.gz"} |
https://www.piping-designer.com/index.php/properties/571-weight-density | # Weight Density
Written by Jerry Ratzlaff on . Posted in Classical Mechanics
Weight is a force on an object accelerated by gravity. Density is the ratio of the amount of matter in an object compared to its volume. The basic difference between density and weight is that weight is a measure of the amount of matter in an object, whereas density measures the amount of matter in a unit volume.
## weight DENSITY formula
$$\large{ W_{\rho} = \frac{m \; g}{g_c} }$$
### Where:
Units English Metric $$\large{ W_{\rho} }$$ (Greek symbol rho) = weight density $$\large{\frac{lbf-in}{ft^3}}$$ $$\large{\frac{kg-mm}{m^3}}$$ $$\large{ g }$$ = gravity $$\large{\frac{ft}{sec^2}}$$ $$\large{\frac{m}{s^2}}$$ $$\large{ g_c }$$ = gravity conversion constant $$\large{in^4}$$ $$\large{mm^4}$$ $$\large{ m }$$ = mass $$\large{lbm}$$ $$\large{kg}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9893596172332764, "perplexity": 875.0763244651348}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030336674.94/warc/CC-MAIN-20221001132802-20221001162802-00708.warc.gz"} |
http://tex.stackexchange.com/questions/39432/use-a-loop-to-generate-a-list-for-another-foreach-loop?answertab=active | # Use a loop to generate a list for another (foreach) loop
Is it possible to use a loop to generate a list for a foreach loop? More specifically I want the following:
Consider the following answer to a question I asked some time ago: http://tex.stackexchange.com/a/38793/4011
It worked very well except if I define \alist as follows:
\def\alist{
\foreach \i in {1,2,3}{
\i/1,
}4/1
}
I get the following error:
ERROR: Undefined control sequence.
--- TeX said ---
\foreach ...reach \let \pgffor@assign@before@code
=\pgfutil@empty \let \pgff...
l.38 \rectDiv{7}{5}{(1,1)}{(4,3)}{\alist}
Is there any way to fix this?
Edit:
Full example:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}
\def\rectDiv#1#2#3#4#5{%#columns, #rows, rectangle start, rectangle end, list of elements to fill
\begin{tikzpicture}
\draw #3 rectangle #4;
\path #3;
\pgfgetlastxy{\firstx}{\firsty}
\path #4;
\pgfgetlastxy{\secondx}{\secondy}
\pgfmathsetlengthmacro{\xdiff}{\secondx-\firstx}
\pgfmathsetlengthmacro{\ydiff}{\secondy-\firsty}
\pgfmathsetlengthmacro{\myxstep}{\xdiff/#1}
\pgfmathsetlengthmacro{\myystep}{\ydiff/#2}
\foreach \x in {1,...,#1}{
\draw ($#3 +\x*(\myxstep,0)$) -- ($#3 +(0,\ydiff) +\x*(\myxstep,0)$);
}
\foreach \y in {1,...,#2}{
\draw ($#3 +\y*(0,\myystep)$) -- ($#3 +(\xdiff,0) +\y*(0,\myystep)$);
}
\edef\temp{\noexpand\foreach \noexpand\i/\noexpand\j in {#5}}
\temp{
\path[fill=blue!20,draw] ($#3 + (\i*\myxstep,\j*\myystep)$) rectangle ($#3 + (\i*\myxstep,\j*\myystep) + (\myxstep,\myystep)$);
}
\end{tikzpicture}
}
\begin{document}
\rectDiv{7}{5}{(1,1)}{(4,3)}{0/0,1/1,2/0,5/3}
\def\list{1/0}
\rectDiv{7}{5}{(1,1)}{(4,3)}{\list}
\rectDiv{7}{5}{(1,1)}{(4,3)}{\list,2/0,5/3}
\def\alist{
\foreach \i in {1,2,3}{
\i/1,
}4/1
}
\alist
\rectDiv{7}{5}{(1,1)}{(4,3)}{\alist} %this doesn't work
\end{document}
-
Your problem is that you wrote
\edef\temp{\noexpand\foreach \noexpand\i/\noexpand\j in {#5}}
% Here is what #5 is in \temp:
\def\alist{
\foreach \i in {1,2,3}{
\i/1,
}4/1
}
which requires that the stuff in the braces be "fully expandable". Alas, \foreach is not expandable, and it is part of \alist. So when \temp is defined, it expands "as much as possible" and hits some non-expandable part of the code of \foreach, which in this case is \let \pgffor@assign@before@code = ...; \let is an assignment, and not expandable, so is not changed by \edef. Thus, the assignment is not actually performed, and \edef goes on to try to interpret the following control sequence \pgffor@assign@before@code directly. Not surprisingly, it is not defined, since it was the \let that is supposed to define it, so you get an error.
To do this, you have to create a macro that contains the output of \alist. For example,
\gdef\alist{}
\foreach \i in {1,2,3} {
\xdef\alist{\alist \i/1,}
}
\xdef\alist{\alist 4/1}
Then you can use \alist directly instead of \temp. (I have used global assignments for \alist because \foreach executes its contents in a group.)
-
Is there a way to do this without using global macros? – Adam Crume Oct 18 '13 at 4:32
Not with \foreach. It acts in a group, so you have to break out with global macros. – Ryan Reich Oct 18 '13 at 11:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8728861212730408, "perplexity": 3584.277993694464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413558066650.89/warc/CC-MAIN-20141017150106-00124-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://projecteuler.net/problem=368 | You are currently using a secure connection
## A Kempner-like series
### Problem 368
Published on Sunday, 22nd January 2012, 01:00 am; Solved by 339
The harmonic series 1 +
1 2
+
1 3
+
1 4
+ ... is well known to be divergent.
If we however omit from this series every term where the denominator has a 9 in it, the series remarkably enough converges to approximately 22.9206766193.
This modified harmonic series is called the Kempner series.
Let us now consider another modified harmonic series by omitting from the harmonic series every term where the denominator has 3 or more equal consecutive digits. One can verify that out of the first 1200 terms of the harmonic series, only 20 terms will be omitted.
These 20 omitted terms are:
1 111
,
1 222
,
1 333
,
1 444
,
1 555
,
1 666
,
1 777
,
1 888
,
1 999
,
1 1000
,
1 1110
,
1 1111
,
1 1112
,
1 1113
,
1 1114
,
1 1115
,
1 1116
,
1 1117
,
1 1118
and
1 1119
.
This series converges as well.
Find the value the series converges to. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8629895448684692, "perplexity": 1295.4382165285326}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462426.4/warc/CC-MAIN-20150226074102-00281-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/176786/t2-s-implies-textrankt-textranks | $T^2=S$ implies $\text{rank}T=\text{rank}S$?
Is it true, that if $T$ and $S$ are selfadjoint and positive operators in a finite-dimensional Hilbert space, such that $T^2=S$, that then $\text{rank}T=\text{rank}S$ ?
-
This is true more generally if $T$ is diagonalizable (recall that if $T$ is self-adjoint, it is diagonalizable).
If $T$ is diagonalisable, there is an invertible matrix $P$ and a diagonal matrix $D$ such that $T=PDP^{-1}$, the rank of $T$ is the number (counted with multiplicities) of the non-zero eigenvalues of $T$. Now since $T^2=PDP^{-1}PDP^{-1}=PD^2P^{-1}$, the rank of $T^2$ is the number of the non-zero eigenvalues of $D^2$, which is the same as the number of non-zero eigenvalues of $D$.
The ranks of $T$ and $T^2$ are thus equal.
-
Since we are in a finite dimensional vector space, we can diagonalize $T$: there is a hermitian matrix $P$ and a diagonal matrix $D$ such that $T=P^*DP$. The rank of $T$ is the same as the rank of $D$, which is the same as the rank of $D^2$ (the number of non-zero elements of the diagonal), and $S=P^*D^2P$, which proves that $S$ and $T$ have the same rank.
The fact that the operators are self-adjoint is necessary here: if $T^2=0$, $T\neq 0$ and $S=0$, it won't work. Positiveness is not necessary here.
-
Whenever $T$ is a matrix which has the same rank as its square, $\mathrm{i}T$ also is. However it is not possible that $T$ and $\mathrm{i}T$ are both self-adjoined unless $T=0$. Thus your claim that self-adjointness is necessary for this property is disproven. – celtschk Jul 30 '12 at 11:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9636588096618652, "perplexity": 47.939066750987784}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783393093.59/warc/CC-MAIN-20160624154953-00151-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://calculus7.org/2017/09/07/recursive-randomness-of-integers/ | # Recursive randomness of integers
Entering a string such as “random number 0 to 7” into Google search brings up a neat random number generator. For now, it supports only uniform probability distributions over integers. That’s still enough to play a little game.
Pick a positive number, such as 7. Then pick a number at random between 0 and 7 (integers, with equal probability); for example, 5. Then pick a number between 0 and 5, perhaps 2… repeat indefinitely. When we reach 0, the game becomes really boring, so that is a good place to stop. Ignoring the initial non-random number, we got a random non-increasing sequence such as 5, 2, 1, 1, 0. The sum of this one is 9… how are these sums distributed?
Let’s call the initial number A and the sum S. The simplest case is A=1, when S is the number of returns to 1 until the process hits 0. Since each return to 1 has probability 1/2, we get the following geometric distribution
Sum Probability 0 1/2 1 1/4 2 1/8 3 1/16 k 1/2k+1
When starting with A=2, things are already more complicated: for one thing, the probability mass function is no longer decreasing, with P[S=2] being greater than P[S=1]. The histogram shows the counts obtained after 2,000,000 trials with A=2.
The probability mass function is still not too hard to compute: let’s say b is the number of times the process arrives at 2, then the sum is 2b + the result with A=1. So we end up convolving two geometric distributions, one of which is supported on even integers: hence the bias toward even sums.
Sum Probability 0 1/3 1 1/6 2 5/36 3 7/72 k ((4/3)[k/2]+1-1)/2k
For large k, the ratio P[S=k+2]/P[s=k] is asymptotic to (4/3)/4 = 1/3, which means that the tail of the distribution is approximately geometric with the ratio of ${1/\sqrt{3}}$.
I did not feel like computing exact distribution for larger A, resorting to simulations. Here is A=10 (ignore the little bump at the end, an artifact of truncation):
There are three distinct features: P[S=0] is much higher than the rest; the distribution is flat (with a bias toward even, which is diminishing) until about S=n, and after that it looks geometric. Let’s see what we can say for a general starting value A.
Perhaps surprisingly, the expected value E[S] is exactly A. To see this, consider that we are dealing with a Markov chain with states 0,1,…,A. The transition probabilities from n to any number 0,…,n are 1/(n+1). Ignoring the terminal state 0, which does not contribute to the sum, we get the following kind of transition matrix (the case A=4 shown):
${\displaystyle M = \begin{pmatrix}1/2 & 0 & 0 & 0 \\ 1/3 & 1/3 & 0 & 0 \\ 1/4 & 1/4 & 1/4 & 0 \\ 1/5 & 1/5 & 1/5 & 1/5\end{pmatrix} }$
The initial state is a vector such as ${v = (0,0,0,1)}$. So ${vM^j}$ is the state after j steps. The expected value contributed by the j-th step is ${vM^jw}$ where ${w = (1,2,3,4)^T}$ is the weight vector. So, the expected value of the sum is
${\displaystyle \sum_{j=1}^\infty vM^jw = v\left(\sum_{j=1}^\infty M^j\right)w = vM(I-M)^{-1}w}$
It turns out that the matrix ${M(I-M)^{-1}}$ has a simple form, strongly resembling M itself.
${\displaystyle M(I-M)^{-1} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1/2 & 0 & 0 \\ 1 & 1/2 & 1/3 & 0 \\ 1 & 1/2 & 1/3 & 1/4 \end{pmatrix} }$
Left multiplication by v extracts the bottom row of this matrix, and we are left with a dot product of the form ${(1,1/2,1/3,1/4)\cdot (1,2,3,4) = 1 + 1 + 1 + 1 = 4 }$. Neat.
What else can we say? The median is less than A, which is no surprise given the long tail on the right. Also, P[S=0] = 1/(A+1) since the only way to have zero sum is to hit 0 at once. A more interesting question is: what is the limit of the distribution of T = S/A as A tends to infinity? Here is the histogram of 2,000,000 trials with A=50.
It looks like the distribution of T tends to a limit, which has constant density until 1 (so, until A before rescaling) and decays exponentially after that. Writing the supposed probability density function as ${f(t) = c}$ for ${0\le t\le 1}$, ${f(t) = c\exp(k(1-t))}$ for ${t > 1}$, and using the fact that the expected value of T is 1, we arrive at ${c = 2-\sqrt{2} \approx 0.586}$ and ${k=\sqrt{2}}$. This is a pretty good approximation in some aspects: the median of this distribution is ${1/(2c)}$, suggesting that the median of S is around ${n/(4-2\sqrt{2})}$ which is in reasonable agreement with experiment. But the histogram for A=1000 still has a significant deviation from the exponential curve, indicating that the supposedly geometric part of T isn’t really geometric:
One can express S as a sum of several independent geometric random variables, but the number of summands grows quadratically in A, and I didn’t get any useful asymptotics from this. What is the true limiting distribution of S/A, if it’s not the red curve above?
## 3 thoughts on “Recursive randomness of integers”
1. Rahul says:
I’m almost certain I’ve seen the continuous version of this problem before (i.e pick a random number x0 uniformly in [0, 1], pick x1 uniformly in [0, x0], …). If I recall correctly, the distribution was piecewise polynomial: constant on [0, 1], linear on [1, 2], quadratic on [2, 3], and so on. I can’t find the page where I saw it though — it’s really hard to search online for this problem. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 18, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.956184983253479, "perplexity": 342.60199749997435}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646176.6/warc/CC-MAIN-20180318204522-20180318224522-00779.warc.gz"} |
http://mathhelpforum.com/calculus/134153-d-2y-dx-2-a-print.html | # d^2y/dx^2
• Mar 16th 2010, 04:58 PM
bigwave
d^2y/dx^2
find $d^2y/dx^2$ by implicit differentiation
$3x^2-4y^2=7$
$6x-8yy'=0$
$y'=\frac{6x}{8y}=-\frac{3x}{4y}$
$\frac{d^2y}{dx^2} = \frac{3x(4y')-4y(3)}{16y^2}$
i don't see this moving towards $-\frac{21}{16y^3}$which is the answer(Crying)
• Mar 16th 2010, 05:01 PM
Moo
In your formula for $\frac{d^2y}{dx^2}$, you have a y', which you can replace by what you found just above ! : $y'=-\frac{3x}{4y}$ :)
And then finish it off with the original equation : $3x^2-4y^2=7 \Rightarrow x^2=\frac{7+4y^2}{3}$
Plus, you applied the quotient rule incorrectly, the derivative of $\frac uv$ is $\frac{u'v-uv'}{v^2}$, but you did $uv'-u'v$
• Mar 16th 2010, 05:42 PM
bigwave
Quote:
Plus, you applied the quotient rule incorrectly, the derivative of $\frac {u}{v}$ is $\frac{u'v-uv'}{v^2}$, but you did $uv'-u'v$
then
$\left(-\frac{3x}{4y}\right)' = -\frac{(3)4y-3x(4y')}{16y^2}$
• Mar 17th 2010, 12:07 AM
Moo
Quote:
Originally Posted by bigwave
then
$\left(-\frac{3x}{4y}\right)' = -\frac{(3)4y-3x(4y')}{16y^2}$
So sorry, there was indeed the minus sign...
Did you manage to go through all the calculations to get the desired answer ?
• Mar 17th 2010, 11:19 AM
Moo
We have $3x^2-4y^2=7$ and $y'=-\frac{3x}{4y}$
$\frac{d^2y}{dx^2}=\frac{3x(4y')-4y(3)}{16y^2}=\frac{12(xy'-y)}{16y^2}=\frac 34\cdot\frac{xy'-y}{y^2}$
But $y'=-\frac{3x}{4y}\Rightarrow xy'=-\frac{3x^2}{4y}\Rightarrow xy'-y=-\frac{3x^2}{4y}-y=-\frac{3x^2-4y^2}{4y}$
But the very first equation tells us that $3x^2-4y^2=7$
Hence $xy'-y=-\frac{7}{4y}$
Therefore, $\frac{d^2y}{dx^2}=\frac 34\cdot\left(-\frac{7}{4y^3}\right)=-\frac{21}{16y^3}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9670714139938354, "perplexity": 1489.4543550911883}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917120844.10/warc/CC-MAIN-20170423031200-00149-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://analytixon.com/2021/08/27/if-you-did-not-already-know-1488/ | Observable Operator Model (OOM)
Observable Operator Models (OOMs) were introduced by Jaeger as a generalization of hidden Markov models (HMMs). The theory of OOMs makes use of both probabilistic and linear algebraic tools, which has an important advantage: using the tools of linear algebra a very simple and efficient learning algorithm can be developed for OOMs. This seems to be better than the known algorithms for HMMs.
A widely used class of models for stochastic systems is hidden Markov models. Systems that can be modeled by hidden Markov models are a proper subclass of linearly dependent processes, a class of stochastic systems known from mathematical investigations carried out over the past four decades. This article provides a novel, simple characterization of linearly dependent processes, called observable operator models . The mathematical properties of observable operator models lead to a constructive learning algorithm for the identification of linearly dependent processes. The core of the algorithm has a time complexity of O(N + nm3), where N is the size of training data, n is the number of distinguishable outcomes of observations, and m is model state-space dimension.
A short introduction to observable operator models of discrete stochastic processes
Observable Operator Models for Discrete Stochastic Time Series
A Consistent Method for Learning OOMs from Asymptotically Stationary Time Series Data Containing Missing Values
“Partially Observable Markov Decision Process”
Propheticus
Due to recent technological developments, Machine Learning (ML), a subfield of Artificial Intelligence (AI), has been successfully used to process and extract knowledge from a variety of complex problems. However, a thorough ML approach is complex and highly dependent on the problem at hand. Additionally, implementing the logic required to execute the experiments is no small nor trivial deed, consequentially increasing the probability of faulty code which can compromise the results. Propheticus is a data-driven framework which results of the need for a tool that abstracts some of the inherent complexity of ML, whilst being easy to understand and use, as well as to adapt and expand to assist the user’s specific needs. Propheticus systematizes and enforces various complex concepts of an ML experiment workflow, taking into account the nature of both the problem and the data. It contains functionalities to execute all the different tasks, from data preprocessing, to results analysis and comparison. Notwithstanding, it can be fairly easily adapted to different problems due to its flexible architecture, and customized as needed to address the user’s needs. …
Missing Value PC (MVPC)
Missing data are ubiquitous in many domains such as healthcare. Depending on how they are missing, the (conditional) independence relations in the observed data may be different from those for the complete data generated by the underlying causal process and, as a consequence, simply applying existing causal discovery methods to the observed data may lead to wrong conclusions. It is then essential to extend existing causal discovery approaches to find true underlying causal structure from such incomplete data. In this paper, we aim at solving this problem for data that are missing with different mechanisms, including missing completely at random (MCAR), missing at random (MAR), and missing not at random (MNAR). With missingness mechanisms represented by missingness Graph (m-Graph), we analyze conditions under which addition correction is needed to derive conditional independence/dependence relations in the complete data. Based on our analysis, we propose missing value PC (MVPC), which combines additional corrections with traditional causal discovery algorithm, in particular, PC. Our proposed MVPC is shown in theory to give asymptotically correct results even using data that are MAR and MNAR. Experiment results illustrate that the proposed algorithm can correct the conditional independence for values MCAR, MAR and rather general cases of values MNAR both with synthetic data as well as real-life healthcare application. …
Probabilistic Robustness
Neural networks are becoming increasingly prevalent in software, and it is therefore important to be able to verify their behavior. Because verifying the correctness of neural networks is extremely challenging, it is common to focus on the verification of other properties of these systems. One important property, in particular, is robustness. Most existing definitions of robustness, however, focus on the worst-case scenario where the inputs are adversarial. Such notions of robustness are too strong, and unlikely to be satisfied by-and verifiable for-practical neural networks. Observing that real-world inputs to neural networks are drawn from non-adversarial probability distributions, we propose a novel notion of robustness: probabilistic robustness, which requires the neural network to be robust with at least $(1 – \epsilon)$ probability with respect to the input distribution. This probabilistic approach is practical and provides a principled way of estimating the robustness of a neural network. We also present an algorithm, based on abstract interpretation and importance sampling, for checking whether a neural network is probabilistically robust. Our algorithm uses abstract interpretation to approximate the behavior of a neural network and compute an overapproximation of the input regions that violate robustness. It then uses importance sampling to counter the effect of such overapproximation and compute an accurate estimate of the probability that the neural network violates the robustness property. … | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8690068125724792, "perplexity": 593.3347743205966}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00251.warc.gz"} |
https://www.nag.com/numeric/nl/nagdoc_27/flhtml/e02/e02aff.html | # NAG FL Interfacee02aff (dim1_cheb_glp)
## 1Purpose
e02aff computes the coefficients of a polynomial, in its Chebyshev series form, which interpolates (passes exactly through) data at a special set of points. Least squares polynomial approximations can also be obtained.
## 2Specification
Fortran Interface
Subroutine e02aff ( f, a,
Integer, Intent (In) :: nplus1 Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: f(nplus1) Real (Kind=nag_wp), Intent (Out) :: a(nplus1)
#include <nag.h>
void e02aff_ (const Integer *nplus1, const double f[], double a[], Integer *ifail)
The routine may be called by the names e02aff or nagf_fit_dim1_cheb_glp.
## 3Description
e02aff computes the coefficients ${a}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,n+1$, in the Chebyshev series
$12a1T0x¯+a2T1x¯+a3T2x¯+⋯+an+1Tnx¯,$
which interpolates the data ${f}_{r}$ at the points
$x¯r=cosr-1π/n , r=1,2,…,n+1.$
Here ${T}_{j}\left(\overline{x}\right)$ denotes the Chebyshev polynomial of the first kind of degree $j$ with argument $\overline{x}$. The use of these points minimizes the risk of unwanted fluctuations in the polynomial and is recommended when the data abscissae can be chosen by you, e.g., when the data is given as a graph. For further advantages of this choice of points, see Clenshaw (1962).
In terms of your original variables, $x$ say, the values of $x$ at which the data ${f}_{r}$ are to be provided are
$xr=12xmax-xmincosπr-1/n+12xmax+xmin, r=1,2,…,n+1$
where ${x}_{\mathrm{max}}$ and ${x}_{\mathrm{min}}$ are respectively the upper and lower ends of the range of $x$ over which you wish to interpolate.
Truncation of the resulting series after the term involving ${a}_{i+1}$, say, yields a least squares approximation to the data. This approximation, $p\left(\overline{x}\right)$, say, is the polynomial of degree $i$ which minimizes
$12ε12+ε22+ε32+⋯+εn2+12εn+12,$
where the residual ${\epsilon }_{\mathit{r}}=p\left({\overline{x}}_{\mathit{r}}\right)-{f}_{\mathit{r}}$, for $\mathit{r}=1,2,\dots ,n+1$.
The method employed is based on the application of the three-term recurrence relation due to Clenshaw (1955) for the evaluation of the defining expression for the Chebyshev coefficients (see, for example, Clenshaw (1962)). The modifications to this recurrence relation suggested by Reinsch and Gentleman (see Gentleman (1969)) are used to give greater numerical stability.
For further details of the algorithm and its use see Cox (1974) and Cox and Hayes (1973).
Subsequent evaluation of the computed polynomial, perhaps truncated after an appropriate number of terms, should be carried out using e02aef.
## 4References
Clenshaw C W (1955) A note on the summation of Chebyshev series Math. Tables Aids Comput. 9 118–120
Clenshaw C W (1962) Chebyshev Series for Mathematical Functions Mathematical tables HMSO
Cox M G (1974) A data-fitting package for the non-specialist user Software for Numerical Mathematics (ed D J Evans) Academic Press
Cox M G and Hayes J G (1973) Curve fitting: a guide and suite of algorithms for the non-specialist user NPL Report NAC26 National Physical Laboratory
Gentleman W M (1969) An error analysis of Goertzel's (Watt's) method for computing Fourier coefficients Comput. J. 12 160–165
## 5Arguments
1: $\mathbf{nplus1}$Integer Input
On entry: the number $n+1$ of data points (one greater than the degree $n$ of the interpolating polynomial).
Constraint: ${\mathbf{nplus1}}\ge 2$.
2: $\mathbf{f}\left({\mathbf{nplus1}}\right)$Real (Kind=nag_wp) array Input
On entry: for $r=1,2,\dots ,n+1$, ${\mathbf{f}}\left(r\right)$ must contain ${f}_{r}$ the value of the dependent variable (ordinate) corresponding to the value
$x¯r=cosπr-1/n$
of the independent variable (abscissa) $\overline{x}$, or equivalently to the value
$xr=12xmax-xmincosπr-1/n+12xmax+xmin$
of your original variable $x$. Here ${x}_{\mathrm{max}}$ and ${x}_{\mathrm{min}}$ are respectively the upper and lower ends of the range over which you wish to interpolate.
3: $\mathbf{a}\left({\mathbf{nplus1}}\right)$Real (Kind=nag_wp) array Output
On exit: ${\mathbf{a}}\left(\mathit{j}\right)$ is the coefficient ${a}_{\mathit{j}}$ in the interpolating polynomial, for $\mathit{j}=1,2,\dots ,n+1$.
4: $\mathbf{ifail}$Integer Input/Output
On entry: ifail must be set to $0$, . If you are unfamiliar with this argument you should refer to Section 4 in the Introduction to the NAG Library FL Interface for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this argument, the recommended value is $0$. When the value is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
On entry, ${\mathbf{nplus1}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{nplus1}}\ge 2$.
${\mathbf{ifail}}=-99$
An unexpected error has been triggered by this routine. Please contact NAG.
See Section 7 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 9 in the Introduction to the NAG Library FL Interface for further information.
## 7Accuracy
The rounding errors committed are such that the computed coefficients are exact for a slightly perturbed set of ordinates ${f}_{r}+\delta {f}_{r}$. The ratio of the sum of the absolute values of the $\delta {f}_{r}$ to the sum of the absolute values of the ${f}_{r}$ is less than a small multiple of $\left(n+1\right)\epsilon$, where $\epsilon$ is the machine precision.
## 8Parallelism and Performance
e02aff is not threaded in any implementation.
The time taken is approximately proportional to ${\left(n+1\right)}^{2}+30$.
For choice of degree when using the routine for least squares approximation, see Section 3.2 in the E02 Chapter Introduction.
## 10Example
Determine the Chebyshev coefficients of the polynomial which interpolates the data ${\overline{x}}_{\mathit{r}},{f}_{\mathit{r}}$, for $\mathit{r}=1,2,\dots ,11$, where ${\overline{x}}_{r}=\mathrm{cos}\left(\pi ×\left(r-1\right)/10\right)$ and ${f}_{r}={e}^{{\overline{x}}_{r}}$. Evaluate, for comparison with the values of ${f}_{\mathit{r}}$, the resulting Chebyshev series at ${\overline{x}}_{\mathit{r}}$, for $\mathit{r}=1,2,\dots ,11$.
The example program supplied is written in a general form that will enable polynomial interpolations of arbitrary data at the cosine points $\mathrm{cos}\left(\pi ×\left(\mathit{r}-1\right)/n\right)$, for $\mathit{r}=1,2,\dots ,n+1$, to be obtained for any $n$ ($\text{}={\mathbf{nplus1}}-1$). Note that e02aef is used to evaluate the interpolating polynomial. The program is self-starting in that any number of datasets can be supplied.
### 10.1Program Text
Program Text (e02affe.f90)
### 10.2Program Data
Program Data (e02affe.d)
### 10.3Program Results
Program Results (e02affe.r) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 69, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9270707368850708, "perplexity": 1052.0724204063736}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00520.warc.gz"} |
http://umj.imath.kiev.ua/article/?lang=en&article=8761 | 2019
Том 71
№ 7
# Functions of shift operator and their applications to difference equations
Chaikovs'kyi A. V.
Abstract
We study the representation for functions of shift operator acting upon bounded sequences of elements of a Banach space. An estimate is obtained for the bounded solution of a linear difference equation in the Banach space. For two types of differential equations in Banach spaces, we present sufficient conditions for their bounded solutions to be limits of bounded solutions of the corresponding difference equations and establish estimates for the rate of convergence.
English version (Springer): Ukrainian Mathematical Journal 62 (2010), no. 10, pp 1635-1648.
Citation Example: Chaikovs'kyi A. V. Functions of shift operator and their applications to difference equations // Ukr. Mat. Zh. - 2010. - 62, № 10. - pp. 1408–1419.
Full text | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9051069021224976, "perplexity": 488.7276324340172}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027319470.94/warc/CC-MAIN-20190824020840-20190824042840-00535.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/135956-poisson-distribution.html | # Math Help - The Poisson Distribution?
1. ## The Poisson Distribution?
Customer arrivals at a checkout counter have a Poisson distribution with an average of 7 per hour. For a given hour, find the probabilities of the following events:
a) Exactly seven customers arrive
b) No more than two customers arrive
c) At least two customers arrive
Can someone help me figure this out? I don't know how to approach this problem.
2. These can be calculated directly.
$p(7) = \frac{e^{-7}\cdot 7^{7}}{7!}$
p(No more than 2) = p(0)+p(1)+p(2)
p(at least 2) = 1- (p(0)+p(1))
Let's see what you get.
3. Originally Posted by TKHunny
These can be calculated directly.
$p(7) = \frac{e^{-7}\cdot 7^{7}}{7!}$
p(No more than 2) = p(0)+p(1)+p(2)
p(at least 2) = 1- (p(0)+p(1))
Let's see what you get. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9089488387107849, "perplexity": 1034.989857583124}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042986806.32/warc/CC-MAIN-20150728002306-00136-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/determine-the-mass-moment-of-inertia-of-a-quarter-of-an-annulus.600344/ | Homework Help: Determine the mass moment of inertia of a quarter of an annulus
1. Apr 25, 2012
jaredogden
1. The problem statement, all variables and given/known data
B.1
The quarter ring has mass m and was cut from a thin uniform plate. Knowing that r1 = 1/2r2 determine the mass moment of inertia of the quarter ring with respect to A. axis AA' B. The centroidal axis CC' that is perpendicular to the plane of the quarter ring.
CC' is located possibly where the center of mass is? (See attached file). EDIT: obviously it is the centroid.. the problem says centroidal axis, I'm dumb..
2. Relevant equations
IAA' = 1/4mr2
ICC' = 1/2mr2
Parallel Axis Theorem
I = I(bar) + md2
3. The attempt at a solution
I originally tried to just use the mass moments of inertia to calculate it. I then realized that the center of mass of the quarter annulus will not be at the origin O in this case so I probably will have to use the parallel axis theorem.
I am really lost on this and I originally calculated
IAA' = (1/16)m(r22 - (1/2)r22)
ICC' = (1/8)m(r22 - (1/2)r22)
Attached Files:
• Vector Mechanics Dynamics Pg. 719.pdf
File size:
73.1 KB
Views:
491
Last edited: Apr 25, 2012
2. Apr 25, 2012
jaredogden
I've calculated I(bar)AA' = (1/4)[(1/4)(mr22 - (1/2)mr22)] and simplified this to:
I(bar)AA' = (1/32)mr22
Then adding that to md2 where d = (1/2)r1 → (1/4)r2 from the parallel axis theorem to get
(1/32)mr22 + m(1/4)r22 = (9/32)mr22
The answer given by my professor is (5/16)mr22
Did I make a mistake somewhere or is the provided answer wrong? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8723976016044617, "perplexity": 1724.3545782850551}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676593586.54/warc/CC-MAIN-20180722194125-20180722214125-00060.warc.gz"} |
https://jgaa.info/issues.jsp?volume=20&issue=77 | JGAA Volumes Volume 20, no. 1, 2016 Special Issue on Selected Papers from the Ninth International Workshop on Algorithms and Computation (WALCOM 2015) Guest Editor(s): M. Sohel Rahman and Etsuj Tomita Guest Editors' Foreword Vol. 20, no. 1, pp. 1-2, 2016. Dichotomy Theorems for Homomorphism Polynomials of Graph Classes Vol. 20, no. 1, pp. 3-22, 2016. Regular paper. An Improved Algorithm for Parameterized Edge Dominating Set Problem Ken Iwaide and Hiroshi Nagamochi Vol. 20, no. 1, pp. 23-58, 2016. Regular paper. An Efficient Silent Self-Stabilizing 1-Maximal Matching Algorithm in Anonymous Networks Vol. 20, no. 1, pp. 59-78, 2016. Regular paper. The Impact of Communication Patterns on Distributed Self-Adjusting Binary Search Tree Vol. 20, no. 1, pp. 79-100, 2016. Regular paper. Common Unfolding of Regular Tetrahedron and Johnson-Zalgaller Solid Yoshiaki Araki, Takashi Horiyama, and Ryuhei Uehara Vol. 20, no. 1, pp. 101-114, 2016. Regular paper. Threshold Circuits Detecting Global Patterns in Two-dimensional Maps Kei Uchizawa, Daiki Yashima, and Xiao Zhou Vol. 20, no. 1, pp. 115-131, 2016. Regular paper. Simultaneous Drawing of Planar Graphs with Right-Angle Crossings and Few Bends Michael A. Bekos, Thomas C. van Dijk, Philipp Kindermann, and Alexander Wolff Vol. 20, no. 1, pp. 133-158, 2016. Regular paper. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8028514981269836, "perplexity": 2452.1194717072954}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585305.53/warc/CC-MAIN-20211020090145-20211020120145-00405.warc.gz"} |
http://math.stackexchange.com/questions/108155/unbiased-estimator | # Unbiased estimator
Let $X_1, X_2, \dots,X_n$ be the random samples from $f(x,\theta)$ = $\frac{2x}{\theta^2}$, $0 < x < \theta$, $\theta > 0$, then find the Uniformly minimum variance unbiased estimator of $\theta^2$.
-
Do you what you have to find? – Davide Giraudo Feb 11 '12 at 16:09
Do you know that there is a website like this one for stats questions? I think the name is "cross-validated". – Gerry Myerson Feb 12 '12 at 1:11
@GerryMyerson And the site link is Cross Validated(stats.stackexchange.com) – Sasha Feb 12 '12 at 3:14
The likelihood on the sample is $$\mathcal{L}(x_1,\ldots,x_n) = \frac{2^n}{\theta^{2n}} \left( \prod_{k=1}^n x_k \right)\left[ \min(x_1,\ldots,x_n) > 0\right] \cdot \left[ \max(x_1,\ldots,x_n) < \theta \right]$$ where $\left[ \bullet \right]$ is the Iverson bracket.
The function $\theta^{-2 n}$ is monotonically decreasing function, hence the maximum of the likelihood occurs at $\theta = \max(x_1,x_2, \ldots, x_n)$. Thus $T=\max(X_1,X_2, \ldots, X_n) = X_{n:n}$ is the complete sufficient statistics, as a maximum likelihood estimator.
It is easy to see that $\delta(x_1,\ldots,x_n) = \frac{2}{n} \left( x_1^2 + x_2^2 + \cdots + x_n^2 \right)$ is an unbiased estimator for $\theta^2$ as $$\mathbb{E}(\delta(X_1,X_2,\ldots,X_n)) = \mathbb{E}(2 X^2) = \theta^2$$ The MVUE for $\theta^2$ is thus $$\begin{eqnarray} \eta\left( X_1,\ldots,X_n\right) &=& \mathbb{E}\left( \delta(X_1,\ldots,X_n) | T \right) \\ &=& \mathbb{E}\left( \frac{2}{n} \sum_{k=1}^n X_{k:n}^2 | T \right) \\&=& \frac{2}{n} \sum_{k=1}^{n-1} X_{k:n}^2 + \frac{2}{n} T^2 \\ &=& \delta(X_1,\ldots,X_n) \end{eqnarray}$$ Here $X_{k:n}$ denotes $k$-th out of $n$ order statistics.
How can I use the the relation with the Cramer-Rao Lower Bond condition? That is if $\hatT$ is unbaised for $\T(\theta)$ and $var(\hatT)$ = CRLB, then $\hatT$ is the UMVUE of \$T(\theta). – David Feb 11 '12 at 21:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9846771955490112, "perplexity": 356.9640218530525}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657137046.16/warc/CC-MAIN-20140914011217-00064-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/131646/algorithm2e-command-algorithm-already-defined | # algorithm2e Command \algorithm already defined
I want to draw vertical line by using \SetAlgoLined, but LaTeX told me Undefined control sequence.
Then I add \usepackage{algorithm2e} to the end of my preamble. Then it said Command\algorithmalready defined.
\usepackage{graphicx}
\usepackage{xthesis}
\usepackage{xtocinc}
\usepackage{mystyle}
\usepackage{url}
\usepackage{subfigure}
\usepackage{booktabs}
\usepackage{multirow}
\usepackage[printonlyused]{acronym}
\usepackage{algorithm}
\usepackage{algorithmic}
\usepackage{float}
\usepackage{epstopdf}
\usepackage{amssymb,amsmath}
\usepackage{graphicx,epsfig}
\usepackage{multicol}
\usepackage{ifthen}
\usepackage{algorithm2e}
I'm using Window 7 + TexWorks. I have another paper works perfect with algorithm2e and \SetAlgoLined:
\usepackage{cite}
\usepackage{url}
\usepackage{ifthen}
\usepackage{multicol}
\usepackage[utf8]{inputenc}
\usepackage{graphicx}
\usepackage{graphicx,epsfig}
\usepackage{amssymb,amsmath}
\usepackage{subfigure}
\usepackage{epstopdf}
\usepackage{float}
\usepackage{algorithm2e}
\usepackage{multirow}
• – Werner Sep 5 '13 at 1:48
Since the algorithm and algorithm2e packages both define an algorithm environment, simply loading both with no extra precaution will cause a name clash like you experienced.
However, you can still use both packages (if you really need both), but you need to pass the option algo2e to algorithm:
\usepackage[algo2e]{algorithm2e}
This option changes the name of the environment algorithm from the algorithm2e package into algorithm2e and so avoids the conflict with the package which already define an algorithm environment; the option also changes the command name for the list of algorithms to \listofalgorithmes.
A complete example:
\documentclass{article}
\usepackage{algorithm}
\usepackage{algorithmic}
\usepackage[algo2e]{algorithm2e}
\begin{document}
\begin{algorithm}%>- from algorithm package
test
\end{algorithm}
\begin{algorithm2e}%>- from algorithm2e package
test
\end{algorithm2e}
\end{document}
The algorithms bundle (which provides the algorithm and algorithmic packages) is not compatible with algorithm2e. The latter is self-contained, providing both the floating environment algorithm (also provided by algorithms' algorithm package) as well as programming constructs (like If, Else, While, etc).
In terms of your algorithm layout, you have to choose - either construct your algorithms using elements from the algorithms bundle, or algorithm2e. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9827022552490234, "perplexity": 4551.443825212031}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00069.warc.gz"} |
https://academic.oup.com/brain/article/58/2/311/319876/A-NOTE-ON-THE-DEFINITION-OF-THE-MOTOR-AND-PREMOTOR | ## Abstract
Since the anterior boundary of the excitable precentral cortex cannot be accurately defined in anatomical terms, and since the functions of the anterior part of the excitable cortex obviously differ from those of the posterior, a more precise terminology is needed to designate these functionally discrete regions. The term “premotor area” has been used in the past synonymously with the “intermediate precentral” region, but adequate anatomical definition has not hitherto been given. Since these regions correspond roughly with well-recognized cytoarchitectural fields, it is suggested that the following conventions be adopted (fig. 1). That:
1. Motor area be restricted to the area giganto-pyramidalis, i.e. Area 4 of Brodmann, the Vogts and Foerster.
2. Premotor area, which is also motor in function, include Area 6a (upper part) of the Vogts, the posterior part being designated 6a α and the anterior part 6a β.
3. Frontal association areas include Areas 9, 10, 11 and 12 of Brodmann.
The term “excitable area” is applicable to both the motor and premotor regions. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.842149019241333, "perplexity": 4110.552140171197}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501174163.72/warc/CC-MAIN-20170219104614-00624-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://mathhelpforum.com/pre-calculus/18551-need-help-something-fuction.html | # Math Help - Need help with something in a fuction
1. ## Need help with something in a fuction
(Note: When I say delta, I mean the little triangle thing usually meaning change, I can't find the HTML code for it.)
Alright, I have a the function f(x) = x^2+3x-1
I then need to evaluate for f(x+deltax). I know that then should look something like x^2+3x-1 + deltax^2+3x-1.
Please verify that is essentially what should happen, and then help me understand what exactly I am supposed to do with the delta in front of the "x" value.
2. Originally Posted by Lithiux
Alright, I have a the function f(x) = x^2+3x-1
I then need to evaluate for f(x+deltax). I know that then should look something like x^2+3x-1 + deltax^2+3x-1.
$f(x+\Delta x)=(x+\Delta x)^2+3(x+\Delta x)-1$
Can you take it from there?
3. Sorry, I r stoopid.
I'd gotten that far in my head as far as setting it up, but I'm not sure what to do with the delta. I'm not sure what is supposed to change.
4. Originally Posted by Lithiux
Sorry, I r stoopid.
I'd gotten that far in my head as far as setting it up, but I'm not sure what to do with the delta. I'm not sure what is supposed to change.
You are not stupid. Stop that! You simply don't have a lot of experience.
Try thinking of $\Delta x$ as a single variable, not two symbols.
So for example
$(x + \Delta x)^2$
is the same kind of thing as
$(a + b)^2 = a^2 + 2ab + b^2$
where
$a = x \text{ and }b = \Delta x$
So
$(x + \Delta x)^2 = x^2 + 2x(\Delta x) + ( \Delta x)^2$
-Dan
5. Ooooh, so to evaluate the delta, it would need to be an equation of some sort, and as such I just treat it as if it were an x or y or something like that. Cool, thanks.
6. Originally Posted by Lithiux
Ooooh, so to evaluate the delta, it would need to be an equation of some sort, and as such I just treat it as if it were an x or y or something like that. Cool, thanks.
A $\Delta$ usually (not always) indicates the difference in some value. For example
$\Delta x = x_{final} - x_{initial}$
in Physics. But the point here is that $\Delta x$ is just a number.
-Dan
7. Yeah, that was my problem. I was thinking of it in a Physics frame-of-mind. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8273734450340271, "perplexity": 363.0951490487849}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257823072.2/warc/CC-MAIN-20160723071023-00233-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/2-unknowns.54394/ | # Homework Help: 2+ unknowns
1. Nov 28, 2004
### Warwick
Well, my physics teacher loves to make his problems hard, so he makes the physics portion of the problem med difficulty but he makes it so you have to solve with more than 1 unknown present. With big problems on forces I get lost in the algebra so.. i'm trying to improve my algebra skills. If anyone could give a couple algebra problems with 2 unknowns and solve for both, I would be grateful. Give me some nasty ones :)
I was making one up my self and couldn't figure it out.
8=2(2^2)
8=x(k^2)
I tired solving that but I couldn't get the answer, :yuck:
thanks
Last edited: Nov 28, 2004
2. Nov 28, 2004
### Nylex
If you have more than 1 unknown, you need more than 1 equation, unless you want one unknown in terms of the others.
8 = xk^2 => k = (8/x)^1/2
It is impossible to get k (or x) with such little info.
3. Nov 28, 2004
### marlon
http://www.ping.be/~ping1339/index.html#Main-Purpose-=-MATH- [Broken]
Just look at systems of linear equations and look also under matrices and determinants...
regards
marlon
Last edited by a moderator: May 1, 2017
4. Nov 28, 2004
### Warwick
Hmm, yes I remember doing these(not how, heh), never understood the use of them then. So this will work when I have 3 sum of the forces equations in physics when there is more than one unknown?
Can you show me how you would go about solving this one?
u=.065,m=2.5,a=0.12,g=9.8
Tcos20-ukn-mgsin25=ma
n-mgcos25+Tsin20=0
Well this one has only 1 unknown but can you use that matrice technique on it?
5. Nov 28, 2004
### marlon
not necessary, you only need the first equation.
ukn = -ma -mdsin25 + Tcos20
that's all
regards
marlon | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.836955726146698, "perplexity": 1448.3342109001576}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00314.warc.gz"} |
https://www.physicsforums.com/threads/unerstanding-an-integration-question.754151/ | # Unerstanding an Integration question
1. May 17, 2014
### gingermom
1. The problem statement, all variables and given/known data
for -1≤x≤1, F(x) =∫sqrt(1-t^2) from -1 to x ( sorry don't know how to put the limits on the sign
a. What does F(1) represent geometrically?
b. Evaluate F(1)
c. Find F'(x)
2. Relevant equations
3. The attempt at a solution
Since my teacher never seems to give simple questions I am wondering if I am missing something in what is being asked.
a. I know this is a semicircle with radius of 1
b. Evaluate - F(1) - I would think this is just plugging in for x=1 which would be ∏/2
c. It seems like F'(x) would just be the integrand so F' (x) = sqrt(1-t^2)
I feel like maybe I am missing something or am I trying to make this harder than it is?
2. May 17, 2014
### SammyS
Staff Emeritus
a.
F(1) is not a semicircle in and of itself. F(1) is just some number. What does that number represent geometrically? Yes, it's related to that semi-circle.
b.
That's right.
c.
You said: F' (x) = sqrt(1-t^2). That's not right. There is a different independent variable on the left compared to the right.
3. May 17, 2014
### gingermom
Oh, so F(1) would be the area of the semicircle - for C I will have to think on that - Would I find the antiderivative using substitution and then find the derivative of that?
Will go back and review taking the integral with variable in the limits - thanks
4. May 17, 2014
### haruspex
It's simpler than that - use the fundamental theorem of calculus.
For writing limits in forum posts, you could simply use sup and sub: ∫x=01. But it looks much better with LaTeX: $\int_{x=0}^{1}$.
If anyone posts LaTeX you can see how they did it (and copy it) by right-clicking on the text and selecting Show Math As->TeX commands. It doesn't show the controls which bracket the LaTeX. There are, to my knowledge, four ways of doing those. You can use TEX and /TEX, each inside square brackets [], which will put the LaTeX on a line by itself, or use ITEX and /ITEX if you just want it to be part of a longer line. There's a shorthand form for each of these. The first can be done with just a double dollar sign at each end ("", no square brackets); the second with a double hash symbol ("##", # being called a "pound sign" in US).
5. May 17, 2014
### Calu
You may either use a substitution to find F'(x) or use the fundamental theorem of calculus.
Finding a suitable substitution would be faster in an exam situation. Can you spot one?
(I was taught this using substitution 2 years before I was taught the fundamental theorem of calculus).
6. May 17, 2014
### gingermom
so since the upper limit is x it would F '(x) =sqrt(1-x^2) * d/dx X which would be 1 so the answer would be F'(x) = sqrt(1-x^2)
Is that right?
7. May 17, 2014
### SammyS
Staff Emeritus
Part of what you wrote is wrong or mis-stated, part is right.
When you wrote " ... which is 1 ... ", to what does which refer?
8. May 17, 2014
### gingermom
Okay, I think I was making this way harder than it needed to be - since the integral is from -1 to x and the upper limit is not something like x^2, by the Fundamental Rule of Calculus I should just be able to substitute the x for the t. If the upper limit been a limit that involved a function like x^2, then I would have had to use the chain rule. Is that correct?
9. May 17, 2014
### SammyS
Staff Emeritus
That's pretty much it.
F' (x) = sqrt(1-t^2)
but it should have said
F' (x) = sqrt(1-x^2) .
That's all I was getting at for part c . | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.933843731880188, "perplexity": 939.9257605103295}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647545.84/warc/CC-MAIN-20180320224824-20180321004824-00510.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-6-section-6-2-the-law-of-cosines-exercise-set-page-731/49 | ## Precalculus (6th Edition) Blitzer
$63.7\ ft$.
Using the figure as shown in triangle ABC, we have the angle $B=45^\circ$. Using the Law of Cosines, we have $b^2=90^2+60.5^2-2(90)(60.5)cos(45^\circ)\approx4060$ which gives $b\approx63.7\ ft$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9953243732452393, "perplexity": 207.53847325818407}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00287.warc.gz"} |
http://math.stackexchange.com/questions/338563/what-is-the-best-strategy-and-with-that-strategy-what-is-the-probability-of-wi | # What is the best strategy , and with that strategy what is the probability of winning?
In a dice game, the player rolls three dice simultaneously, and then he may roll two more times any number of his dice(0, 1, 2 or 3). The player wins the game if all three dice have the same number on top after the last roll. What is the best strategy , and with that strategy what is the probability of winning?
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Do the two additional rolls have to include the same dice or the same number of dice, or can the player choose independently for the first and second additional roll which of the dice to reroll? – joriki Mar 23 '13 at 7:13
Assuming that the dice to reroll can be independently chosen in the two additional rolls, there is no single optimal strategy. An optimal strategy is one that leaves $2$ or $3$ equal dice unchanged and rerolls the remaining die in case of $2$ equal dice, and rerolls either all dice or all but one die if there are no equal dice.
After the first roll, the probabilities that $2$ or $3$ numbers are equal are $3\cdot6\cdot5/6^3=5/12$ and $6/6^3=1/36$, respectively, so the probability that none are equal is $1-5/12-1/36=5/9$.
If $3$ numbers are equal, we win. If $2$ numbers are equal, we have a $1-(5/6)^2=11/36$ chance of winning by rerolling the remaining die. If no numbers are equal, we might as well reroll completely, and then the probabilities are the same as on the first roll, but now we only have a $1/6$ chance of winning if two dice are equal on the second roll.
$$\frac1{36}+\frac5{12}\cdot\frac{11}{36}+\frac59\left(\frac1{36}+\frac5{12}\cdot\frac16+\frac59\cdot\frac1{36}\right)=\frac{2539}{11664}\approx0.2177\;.$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.804996907711029, "perplexity": 152.5031764805459}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999679512/warc/CC-MAIN-20140305060759-00065-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/113982/finding-a-more-efficient-editor-for-collaboration-of-latex-people-and-word-peopl | # Finding a more efficient editor for collaboration of LaTeX people and Word people
Background
My supervisor often wants to edit my papers, so he frequently asks for word version of the paper. I tried latex2rtf, but it is very limited in terms of packages and customized commands. I asked a similar question and found a related answer. However, every time I look to my beautiful LaTeX PDF, before conversion to Word, I think that why we cannot just edit the title from the output???
Currently LaTeX editors (like TeXStudio, TeXMaker,etc) provide an integrated PDF viewer which provide preview of the compiled document. I always think that if we have an interactive document viewer that identifies different fields like title, paragraph,... and when user double clicks on a field, a new small window opens with the text in it. After user completes his/her revision, LaTeX runs to do typesetting of the new document.
Question
Is there any PDF professional viewer which can be linked and used with LaTeX to improve the experience of the word people in using LaTeX??
For example, many professional pdf viewers like Acrobat Pro, Nitro, Phantom PDF can edit the text in PDF files. I think If we could link the PDF viewer and the LaTeX editor together in the way that edits in a paragraph of a PDF file can be sent backward to tex file. Then we can compile the the TeX file to apply the changes.
Edit1: Current versions of TeXstudio and TeXmaker can highlight the current position of cursor in yellow for few seconds. There is a link between in the TeX source, and Preview. I wish that we can have the the reversed link.
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This isn't really a question on TeX format or typesetting system - while I find it to be a decent question, it's not a good fit for our Q/A format and is likely to be closed. Perhaps send out feelers in chat instead? Otherwise, edit your question to potentially ask for the (hypothetical) name of such software. – Sean Allred May 13 '13 at 2:23
Possible duplicate of Seeking review on a document with people unfamiliar with TeX and Linked Q's. Please refer to Charles Stewart's answer. Suggestion: Two options either convince others or convince yourself in collaboration( part of compromise with word users). Atleast latex typography will be far better than others. Edit :Interactive PDF viewer may have more practical problems than solutions i suppose for colloboration. – texenthusiast May 13 '13 at 2:28
Try to get review on the file only when you are really ready. Before then, talk(!) to your supervisor, and get him to write on a printout of the paper. Then, when you are both ready, invest time in converting with Latex2rtf or pandoc, which do a very good job on simple tex documents. Don't forget you can produce two PDFs from one tex document by using the `\input` command to insert your content into the journal's tex file, and your own simpler \article class. The version produced using `\article` should work with latex2rtf or pandoc. You can specify replacement commands with `\iflatex2rtf`. – Andy Clifton May 13 '13 at 4:01
PDF with annotations. – Nicholas Hamilton May 13 '13 at 4:27
@antmw1361 you can set up forward and backward searching with synctex. This way you can right click the area in the pdf and it opens to that portion of the tex document for editing. Just save and compile. – dustin May 15 '13 at 2:26
If your supervisor does absolutely insist on Word as Editor there is little you can do. However, if it is more the "Word-like user experience" (as opposed to "LaTeX source code user experience"), LyX might be a considerable compromise for the both of you!
LyX is a "WYSIWYW" (What You See is What You Want) text processing system that uses LaTeX as back-end. LyX has a user interface that is close enough to "normal applications" so that "normal users" are able to use it effectively. You can also insert LaTeX commands directly for quick math editing or when a certain feature is not available in LyX.
Regarding interoperability with LaTeX: LyX uses LaTeX as back end, so you can always get from LyX -> LaTeX. However, the internal document format is different. For the LaTeX -> LyX route the converter scripts work pretty well with the standard classes, but may require some manual overwork if you use many own macros or "fancy stuff". So LyX should not be considered as a generic round-trip LaTeX Editor like TeXShop or vim. However, all this works a lot better than any LaTeX -> RTF/OO/DOC/HTML -> LaTeX route.
The main point, which in my experience is the "killer feature" that makes supervisors prefer Word, is that LyX has a built-in change tracking system. I used it quite a bit when sending my thesis to people for proof-reading and it was a pleasure for them to do edits and for me to integrate (or reject) their suggestions.
It's also possible to have mark-ups for the changes in the PDF output:
A subtle side point is that, by using LyX, your supervisor or other coworker immediately gets the LaTeX typesetting experience. For me, this has been a pretty successful path towards the long-term conversion of TeX-illiterate coworkers to LaTeX.
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Are you sure that is a real LyX feature not just calling another command line utility? That looks like latexdiff Perl script? – Predrag Punosevac May 15 '13 at 12:02
@PredragPunosevac: Yes I am. The highlighting is already in the LaTeX code generated by LyX. – Daniel May 15 '13 at 18:58
@LostBrit: I have written about a dozen journal papers (Springer, Elsevier, IEEE, ACM, ...) with LyX and never experienced any problems with the exported LaTeX code. Also LyX does nothing to your TeX source – it does not touch it at all! Recall that LyX should not be considered as a LaTeX editor. LyX is just able to import LaTeX code in many cases, but it manages and stores its own document format and forgets about your TeX code immediately. When you export from LyX to LaTeX, everything is generated. – Daniel May 18 '13 at 20:58
@LostBrit Have you seen what journal offices do to your manuscript under the hood? :) You would never submit anything if you would know. – percusse May 21 '13 at 18:55
@percusse At least a few of the proofs I've had look like they printed out my manuscript, kicked it around on the floor, shuffled the pages and then faxed it to the other side of the world where it was put through OCR into word, before being converted back into 'tex. I've done this enough now that I at least have a template for the "Dear typesetting folks, what did you do to my paper?" letter – Andy Clifton May 22 '13 at 15:40
Is there any PDF professional viewer which can be linked and used with LaTeX to improve the experience of the word people in using LaTeX?
You may like to have a look at BaKoMa TeX Word; a Visual (True WYSIWYG) LaTeX Editor. This stand-alone software is not free though.
Another option, is the online LaTeX editor ShareLaTeX, a truly collaborative cloud solution, now with version control. Using this, there is a chance your supervisor will quickly learn LaTeX himself!
Similar to the above is Authorea, a spin-off of Harvard University. It also supports a subset of Markdown. (Full Pandoc Markdown support would be preferable though.) Cross-editing between LaTeX and Markdown is not possible; one must at document creation.
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I tried BaKoMa and liked it, but it's not free. – antmw1361 May 26 '13 at 0:24
use TeX2Word and Word2TeX. There is a 30-day Evaluation
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I don't think this answers the OP's request for a way to work directly with the PDF from the original files. But, if conversion to word is a way that the OP decides to pursue, he should read the much more detailed answers about converting to word in tex.stackexchange.com/questions/4145/…. – Andy Clifton May 18 '13 at 20:52
You can theoretically edit pdf document with PDFedit the way you described in your question. Before you get too excited let me point out that PDF is a very complex format designed for publishing output, not for any further modifications. PDF unlike its close cousin PostScript programming language is not human readable nor programmable (although PostScript files are mostly machine produced it is theoretically (I have done that practically without help of PSTricks) possible to program pictures and even animations, moreover the language is Turing complete). Acroread (free version) can be used to fill in PDF forms. That is a different thing than editing PDF files. If I recall correctly Acrobat has a paid version which can be used to create those PDF forms. I think we paid one of those full versions of Acroread at the University of Arizona six-seven years ago over \$1000 (that is with all university discounts). The full version can not edit pdf file in a way your described.
So what a hack is PDFedit? It is a low-level tool for technical users that provides structured access to the internal structure of the PDF file. It claims to fame is GUI. It works as described and I tried it but I have no use for it. I am much more familiar with similar non GUI tool PDFtk. Neither of these tools when used to "edit" PDF files are very user friendly. Unless somebody is paying you to repair broken PDF files I would not be waisting my time with them.
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Absolutely: 'writing in PDF' is working at the wrong level. – jon May 15 '13 at 3:05
To be honest, this is not what I'm looking for. – antmw1361 May 15 '13 at 7:43
@antmw1361 I am not really sure than what are you looking for. If you are looking for the tool which will take you back from the PDF viewer into the source file and vise verse that is trivial. That is called inverse and direct search. Any editor which works in server mode can do that. – Predrag Punosevac May 15 '13 at 12:05 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.902906596660614, "perplexity": 1892.998159893012}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464051165777.94/warc/CC-MAIN-20160524005245-00194-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://scholarship.rice.edu/browse?value=Issa%2C+Leila&type=author | Now showing items 1-1 of 1
• #### Source localization in cluttered acoustic waveguides
(2010)
Mode coupling due to scattering by weak random inhomogeneities leads to the loss of coherence in the wave field measured a long distances of propagation. This in turn leads to the deterioration of coherent source localization ... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9151749014854431, "perplexity": 1330.8965127378622}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549425352.73/warc/CC-MAIN-20170725182354-20170725202354-00225.warc.gz"} |
https://community.jmp.com/t5/Discussions/GLM-platform-Nominal-Logistic-does-not-take-random-effects/m-p/4953/highlight/true | Highlighted
Community Trekker
Joined:
Mar 2, 2012
## GLM platform/Nominal Logistic does not take random effects
Hi all,
I want to fit a model with both fixed and random effects but has a binary DV. JMP gives me options for Nominal Logistic / GLM but announces an error when I try to fit the model. Any suggestions are welcome? I figure that JMP must be able to do something at least equivalent to this if it can do do choice models (which this is, but Im having problems getting that function to work with my design).
Cheers,
4 REPLIES
Community Trekker
Joined:
Nov 19, 2011
## Re: GLM platform/Nominal Logistic does not take random effects
I have used JMP 7.0 and know that one cannot use Random effects with logistic regression in that program.
1. Is this still the case for JMP 10.0?? Just want to know since I would consider upgrading (purchasing) if it does this, but definitely not if it does not.
2. Also, will JMP give model selection critieria such as AIC values?
Super User
Joined:
Jun 23, 2011
## Re: GLM platform/Nominal Logistic does not take random effects
hi guys,
i also didn't like the fact that the logistic regression platform can't take a random effect.
as for deer&dog's question, the logistic regression platform give AIC and BIC by default but in the linear regression you will need to ask for it as you can see in the attached output. i produced it from the "big class" data file.
Community Trekker
Joined:
Nov 19, 2011
## Re: GLM platform/Nominal Logistic does not take random effects
Thanks Ron! I'll check my output (just starting an analysis) to see if Jmp 7 produces AIC and BIC. I am amazed that JMP 10 doesn't handle random effects in the log regression. That has been lacking for years now. Bummer.
Community Trekker
Joined:
Dec 25, 2016
## Re: GLM platform/Nominal Logistic does not take random effects
Can JMP Pro 13 fit a model with both fixed and random effects but has a binary dependent variable? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9089669585227966, "perplexity": 3281.4846807385325}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247479838.37/warc/CC-MAIN-20190216024809-20190216050809-00364.warc.gz"} |
http://vergil.chemistry.gatech.edu/notes/grpthy-vib/node5.html | As a last example, we will see that we can create SALC's of atomic orbitals in exactly the same way. Consider the case of ethylene (which is still point-group , see Figure 4). The 4 C-H bonds are symmetry-equivalent, so we can make 4 C-H bonding SALC's.
The four non-zero SALC's are:
These are pictured in Figure 4. Note that four C-H bonds go in to the symmetry-adaptation, and four C-H SALC's come out. We could use similar procedures to construct the remaining four occupied orbitals, which are the symmetric () combination of the two C 1s core orbitals, the antisymmetric () combination of the two C 1s core orbitals, the C-C bonding orbital, and the C-C bonding orbital (see Fig. 4).
Finally, it is worth commenting that we can often come up with the SALC's by intuition more easily than we can work out the projection operators (especially with a little experience and practice). This is certainly true for the ethylene symmetry-adapted orbitals.
Acknowledgments. The author thanks Dr. Sahan Thanthiriwatte for providing the figures. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9122388958930969, "perplexity": 1341.681702030581}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931012025.85/warc/CC-MAIN-20141125155652-00246-ip-10-235-23-156.ec2.internal.warc.gz"} |
http://www.exampleproblems.com/wiki/index.php/Four-vector | Four-vector
In relativity, a four-vector is a vector in a four-dimensional real vector space, called Minkowski space. The usage of the four-vector name tacitly assumes that its components refer to a standard basis. The components transform between these bases as the space and time coordinate differences, ${\displaystyle (\Delta t,\Delta x,\Delta y,\Delta z)}$ under spatial translations, rotations, and boosts (a change by a constant velocity to another inertial reference frame). The set of all such translations, rotations, and boosts (called Poincaré transformations) forms the Poincaré group. The set of rotations and boosts (Lorentz transformations, described by 4×4 matrices) forms the Lorentz group.
Mathematics of four-vectors
A point in Minkowski space is called an "event" and is described in a standard basis by a set of four coordinates such as
${\displaystyle \mathbf {x} :=\left(ct,x,y,z\right)}$
for ${\displaystyle \mu }$ = 0, 1, 2, 3, where c is the speed of light. These coordinates are the components of the position four-vector for the event.
The displacement four-vector is defined to be an "arrow" linking two events:
${\displaystyle \Delta \mathbf {x} :=\left(\Delta ct,\Delta x,\Delta y,\Delta z\right)}$
(Note that the position vector is the displacement vector when one of the two events is the origin of the coordinate system. Position vectors are relatively trivial; the general theory of four-vectors is concerned with displacement vectors.) The inner product of two four-vectors ${\displaystyle u}$ and ${\displaystyle v}$ is defined (using Einstein notation) as
${\displaystyle u\cdot v=u^{\mu }\eta _{\mu \nu }v^{\nu }=\left({\begin{matrix}u^{0}&u^{1}&u^{2}&u^{3}\end{matrix}}\right)\left({\begin{matrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{matrix}}\right)\left({\begin{matrix}v^{0}\\v^{1}\\v^{2}\\v^{3}\end{matrix}}\right)=u^{0}v^{0}-u^{1}v^{1}-u^{2}v^{2}-u^{3}v^{3}}$
where η is the Minkowski metric. Sometimes this inner product is called the Minkowski inner product.
The inner product is often expressed as the effect of the dual vector of one vector on the other:
${\displaystyle u\cdot v=u^{*}(v)=u{_{\nu }}v^{\nu }}$
Here the ${\displaystyle u_{\nu }}$-s are the components of the dual vector ${\displaystyle u^{*}}$ of ${\displaystyle u}$ in the dual basis and called the covariant coordinates of ${\displaystyle u}$, while the original ${\displaystyle u^{\nu }}$ components are called the contravariant coordinates. Lower and upper indices indicate always covariant and contravariant coordinates, respectively.
The relation between the covariant and contravariant coordinates is:
${\displaystyle u_{\mu }=u^{\nu }\eta _{\mu \nu }\,}$.
The four-vectors are arrows on the spacetime diagram or Minkowski diagram.
Four-vectors may be classified as either spacelike, timelike or null. In this article, four-vectors will be referred to simply as vectors. Spacelike, timelike, and null vectors are ones whose inner product with themselves is greater than, less than, and equal to zero respectively.
Examples of four-vectors in dynamics
When considering physical phenomena, differential equations arise naturally; however, when considering space and time derivatives of functions, it is unclear which reference frame these derivatives are taken with respect to. It is agreed that time derivatives are taken with respect to the proper time (τ) in the given reference frame. It is then important to find a relation between this time derivative and another time derivative (taken in another inertial reference frame). This relation is provided by the time transformation in the Lorentz transformations and is:
${\displaystyle {\frac {d\tau }{dt}}={\frac {1}{\gamma }}}$
where γ is the Lorentz factor. Important four-vectors in relativity theory can now be defined, such as the four-velocity of an ${\displaystyle \mathbf {x} (\tau )}$ world line is defined by:
${\displaystyle \mathbf {U} :={\frac {d\mathbf {x} }{d\tau }}={\frac {d\mathbf {x} }{dt}}{\frac {dt}{d\tau }}=\left(\gamma c,\gamma \mathbf {u} \right)}$
where
${\displaystyle u^{i}={\frac {dx^{i}}{dt}}}$
for i = 1, 2, 3. Notice that
${\displaystyle ||\mathbf {U} ||^{2}=U^{\mu }U_{\mu }=c^{2}\,}$
The four-acceleration is given by:
${\displaystyle A={\frac {dU}{d\tau }}=\left(\gamma {\dot {\gamma }}c,\gamma {\dot {\gamma }}\mathbf {u} +\gamma ^{2}\mathbf {\dot {u}} \right)}$
Since the magnitude ${\displaystyle {\sqrt {|U_{\mu }U^{\mu }|}}}$ of ${\displaystyle \mathbf {U} }$ is a constant, the four acceleration is (pseudo-)orthogonal to the four velocity, i.e. the Minkowski inner product of the four-acceleration and the four-velocity is zero:
${\displaystyle A_{\mu }U^{\mu }={\frac {1}{2}}{\frac {\partial (U^{\mu }U_{\mu })}{\partial \tau }}=0\,}$
which is true for all world lines.
The four-momentum is for a massive particle is given by:
${\displaystyle p:=mU=\left(\gamma mc,\mathbf {p} \right)}$
where m is the invariant mass of the particle.
The four-force is defined by:
${\displaystyle F:=mA=\left(\gamma {\dot {\gamma }}mc,\gamma \mathbf {f} \right)}$
where
${\displaystyle \mathbf {f} ={\frac {d\mathbf {p} }{dt}}=m{\dot {\gamma }}\mathbf {u} +m\gamma \mathbf {\dot {u}} }$.
Physics of four-vectors
The power and elegance of the four-vector formalism may be demonstrated by deriving some important relations between the physical quantities energy, mass and momentum.
Deriving E = mc2
Here, an expression for the total energy of a particle will be derived. The kinetic energy (K) of a particle is defined analogously to the classical definition, namely as
${\displaystyle {\frac {dK}{dt}}=\mathbf {f} \cdot \mathbf {u} }$
with f as above. Noticing that ${\displaystyle F^{\mu }U_{\mu }=0}$ and expanding this out we get
${\displaystyle \gamma ^{2}\left(\mathbf {f} \cdot \mathbf {u} -{\dot {\gamma }}mc^{2}\right)=0}$
Hence
${\displaystyle {\frac {dK}{dt}}=c^{2}{\frac {d\gamma m}{dt}}}$
which yields
${\displaystyle K=\gamma mc^{2}+S\,}$
for some constant S. When the particle is at rest (u = 0), we take its kinetic energy to be zero (K = 0). This gives
${\displaystyle S=-mc^{2}\,}$
Thus, we interpret the total energy E of the particle as composed of its kinetic energy K and its rest energy m c2. Thus, we have
${\displaystyle E=\gamma mc^{2}\,}$
Deriving E2 = p2c2 + m2c4
Using the relation ${\displaystyle E=\gamma mc^{2}}$, we can write the four-momentum as
${\displaystyle p=\left({\frac {E}{c}},\mathbf {p} \right)}$.
Taking the inner product of the four-momentum with itself in two different ways, we obtain the relation
${\displaystyle {\frac {E^{2}}{c^{2}}}-p^{2}=P^{\mu }P_{\mu }=m^{2}U^{\mu }U_{\mu }=m^{2}c^{2}}$
i.e.
${\displaystyle {\frac {E^{2}}{c^{2}}}-p^{2}=m^{2}c^{2}}$
Hence
${\displaystyle E^{2}=p^{2}c^{2}+m^{2}c^{4}\,}$
This last relation is useful in many areas of physics.
Examples of four-vectors in electromagnetism
Examples of four-vectors in electromagnetism include the four-current defined by
${\displaystyle J:=\left(\rho c,\mathbf {j} \right)}$
formed from the current density j and charge density ρ, and the electromagnetic four-potential defined by
${\displaystyle \phi :=\left(\psi ,\mathbf {A} c\right)}$
formed from the vector potential A and the scalar potential φ.
A plane electromagnetic wave can be described by the four-frequency defined as
${\displaystyle N:=\left(\nu ,\nu \mathbf {n} \right)}$
where ${\displaystyle \nu }$ is the frequency of the wave and n is a unit vector in the travel direction of the wave. Notice that
${\displaystyle N^{\mu }N_{\mu }=\nu ^{2}\left(n^{2}-1\right)=0}$
so that the four-frequency is always a null vector. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 43, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9777423143386841, "perplexity": 339.0643000659914}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703547333.68/warc/CC-MAIN-20210124044618-20210124074618-00350.warc.gz"} |
https://www.physicsforums.com/threads/finding-dhvap-of-water-from-graph-of-ln-p-of-h2o-in-atm-versus-1-t.559564/ | # Finding ΔHvap of water from graph of ln (p of H2O in atm) versus 1/T
1. Dec 12, 2011
### icecream1023
1. From a lab experiment, I measured the volume of trapped air at various temperatures. I know I did calculations correctly up to the point that I drew the linear relationship between ln (p of H2O in atm) to 1/T (in Kelvin). My textbook says the slope of this graph is supposed to equal -ΔHvap/R, but the first thing that's confusing me is that I don't know what the units are and the second thing that's confusing me is just that the numbers don't work out. The slope of the line in this graph is -5158.73, but I don't understand how units fit into this. As I've said, the y-axis is ln (p of H2O in atm) and the x-axis is 1/T (in Kelvin).
2. I don't...know what to put in this section.
3. If the slope is equal to -ΔHvap/R, then...
-5158.73 = -ΔHvap/R
ΔHvap = 5158.73 * 0.08206 (Again, I have no idea where my units are!!)
ΔHvap = 423.3 *something*
I appreciate any help; thank you!
2. Dec 13, 2011
### cheme101
You're on the right track, you just need to get your units right. Your slope (-ΔH/R) has to be in K because your x-coordinates are in 1/K; their product needs to be dimensionless to agree with ln(P).
With consistent units:
lnP = -(H/R)*T
[unitless]= -([J/mol] / [J/mol K]) * (1/K)
So if you use R=8.314 J / mol K,
you should get -ΔHvap = 8.314 * -5158.73 J/mol
Similar Discussions: Finding ΔHvap of water from graph of ln (p of H2O in atm) versus 1/T | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8687458634376526, "perplexity": 1981.6824412394737}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818685698.18/warc/CC-MAIN-20170919131102-20170919151102-00085.warc.gz"} |
https://crazyproject.wordpress.com/2011/06/11/every-ideal-in-zzi-is-principal/ | ## Every ideal in ZZ[i] is principal
Prove that every ideal in $\mathbb{Z}[i]$ is principal.
We showed here that $\mathbb{Z}[i]$ is a Euclidean domain. Thus it is a principal ideal domain. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 2, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9965435266494751, "perplexity": 324.31722623539974}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189031.88/warc/CC-MAIN-20170322212949-00067-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.helsinki.fi/en/news/science-news/citizen-science-discovers-a-new-form-of-the-northern-lights?fbclid=IwAR1CVmj6ZBttYc9LosS_QZdohgIHVnyFFVRkzEnFtXs1dNtzqXoxJULDyK0 | # Citizen science discovers a new form of the Northern Lights
29.1.2020
Working together with space researchers, Finnish amateur photographers have discovered a new auroral form. Named 'dunes' by the hobbyists, the phenomenon is believed to be caused by waves of oxygen atoms glowing due to a stream of particles released from the Sun.
Working together with space researchers, Finnish amateur photographers have discovered a new auroral form. Named 'dunes' by the hobbyists, the phenomenon is believed to be caused by waves of oxygen atoms glowing due to a stream of particles released from the Sun. Time lapse video of a new auroral form called Dunes. Credit: Kari Saari
In the recently published study, the origins of the dunes were tracked to a wave guide formed within the mesosphere and its boundary, the mesopause. The study also posits that this new auroral form provides researchers with a novel way to investigate conditions in the upper atmosphere.
The study was published in the first issue of the high-impact journal AGU Advances.
## An unknown fingerprint appears in the sky
Minna Palmroth, Professor of Computational Space Physics at the University of Helsinki, heads a research group developing the world's most accurate simulation of the near-Earth space and space weather that cause auroral emissions.
The sun releases a steady flow of charged particles, known as the solar wind. Reaching the Earth’s ionised upper atmosphere, the ionosphere, they create auroral emissions by exciting atmospheric oxygen and nitrogen atoms. The excitation state is released as auroral light.
In late 2018, Palmroth published a book entitled 'Revontulibongarin opas' (‘A guide for aurora borealis watchers’). The book was born out of Palmroth's cooperation with Northern Lights enthusiasts and the answers she provided to questions about the physics of the phenomenon in the hobbyists' Facebook group.
Thousands of magnificent photographs of the Northern Lights taken by hobbyists were surveyed and categorised for the book. Each auroral form is like a fingerprint, typical only of a certain phenomenon in the auroral zone. During the classification, hobbyists pointed out that a certain auroral form did not fit into any of the pre-existing categories. Palmroth set aside these unusual forms for later consideration.
By an almost unbelievable coincidence, just days after the book was published, the hobbyists saw this unusual form again and immediately informed Palmroth. The form appeared as a green-tinged and even pattern of waves resembling a striped veil of clouds or dunes on a sandy beach.
"One of the most memorable moments of our research collaboration was when the phenomenon appeared at that specific time and we were able to examine it in real time", says Northern Lights and astronomy hobbyist Matti Helin.
## Waves newly revealed by the aurora
Investigations into the phenomenon were launched, with hobbyist observations and scientific methods coming together to explain the waves.
"It was like piecing together a puzzle or conducting detective work," says Helin. "Every day we found new images and came up with new ideas. Eventually, we got to the bottom of it…"
The phenomenon was photographed at the same time in both Laitila and Ruovesi, southwest Finland, with the same detail observed in the auroral emission in both images. Maxime Grandin, a postdoctoral researcher in Palmroth’s team, identified stars behind the emission and determined the azimuths and elevations of the stars with the help of the astronomy software program Stellarium. This made it possible to use the stars as points of reference when calculating the altitude and extent of the auroral phenomenon.
Grandin found that the auroral dunes occur at a relatively low altitude of 100 kilometres, in the upper parts of the mesosphere. The wavelength of the wave field was measured to be 45 kilometres.
A total of seven similar events – where a camera had recorded the same even pattern of waves – were further identified from the 'Taivaanvahti' ('Sky Watch') service maintained by Ursa Astronomical Association.
## Unexplored region
The part of the auroral zone where the Earth's electrically-neutral atmosphere meets the edge of space is an extremely challenging environment for satellites and other space-borne instruments. Palmroth says this is why it is one of the least studied places on our planet.
"Due to the difficulties in measuring the atmospheric phenomena occurring between 80 and 120 kilometres in altitude, we sometimes call this area 'the ignorosphere'," she says.
The dunes were observed precisely in that particular region of the auroral zone. The observed phenomenon guided the researchers towards a middle ground between atmospheric research and space research, as the usual methodology of space physics could not explain it alone.
"The differences in brightness within the dune waves could be due to either waves in the precipitating particles coming from space, or in the underlying atmospheric oxygen atoms," says Palmroth. "We ended up proposing that the dunes are a result of increased oxygen atom density."
Next, the team had to determine how the variability in the density of the oxygen atoms caused by gravity waves in the atmosphere results in such an even and widespread field of waves. Normally at the altitude of study there are many different kinds of gravity waves travelling in different directions at different wavelengths, which is why they do not easily form the even wavefields exhibited by the dunes.
## The Northern Lights illuminate a tidal bore
The study suggests that the phenomenon in question is a mesospheric bore, a rare and little-studied phenomenon that takes place in the mesosphere. The tidal bore phenomenon is a wave common to many rivers, where the tide travels up the river channel.
Various types of gravity wave are born in the atmosphere and then rise. In very rare cases, gravity waves can get filtered as they rise between the mesopause and an inversion layer that is intermittently formed below the mesopause. The inversion layer makes the filtered waves bend and enables them to travel long distances through the channel without attenuation.
Very rarely, a gravity wave rising up in the atmosphere can be filtered and bent to travel between the mesopause and an inversion layer intermittently formed below the mesopause. The mesopause and the inversion layer are colder than the other layers of the atmosphere. In the wave channel established between these two layers, gravity waves coming from below can travel long distances without subsiding. Dune-shaped auroral emissions are created when solar wind charges the oxygen atoms surging through the channel. (Graphic credit: Jani Närhi)
When the oxygen atoms in the bore collide with the electrons precipitating down upon the atmosphere, they become excited. When releasing this excitation, they create the auroral light. This is why mesospheric bores – a phenomenon thus far considered a very challenging subject of research – can occasionally be seen with the naked eye.
## Space researchers focus on the atmosphere
Prior to this discovery, mesospheric bores were not observed in the auroral zone, nor have they been investigated via auroral emissions.
"The auroral zone as a whole is usually discounted in studies focused on the bore, as auroral emissions impair the technique used to identify mesospheric bores," says Palmroth.
Traditionally, researchers specialising in the atmosphere and space have largely investigated their topics of interest separate from each other. This is because there are only a handful of known mechanisms of interaction between the ionosphere bathing in the precipitating electrons, and the neutral atmosphere.
With the help of measuring devices operated by the Finnish Meteorological Institute, the dunes were found to occur simultaneously and in the same region where the electromagnetic energy originating in space is transferred to the ignorosphere.
"This could mean that the energy transmitted from space to the ionosphere may be linked with the creation of the inversion layer in the mesosphere," says Palmroth. "In terms of physics, this would be an astounding discovery, as it would represent a new and previously unobserved mechanism of interaction between the ionosphere and the atmosphere."
Reference:
M. Palmroth, M. Grandin, M. Helin, P. Koski, A. Oksanen, M. A. Glad, R. Valonen, K. Saari, E. Bruus, J. Norberg, A. Viljanen, K.4 Kauristie, and P. T Verronen. Citizen scientists discover a new auroral form: Dunes provide insight into the upper atmosphere. AGU Advances 1, 1-12, 28.1.2020, https://doi.org/10.1029/2019AV000133
Contact details
Minna Palmroth, professor, University of Helsinki and Finnish Meteorological Institute
+358 50 311 1950 (CET+1 ja UTC+2)
[email protected]
@MinnaPalmroth
Maxime Grandin, postdoctoral researcher, University of Helsinki
[email protected],
@Maxime_Grandin
Matti Helin, astronomy hobbyist
+358 44 359 0866 (CET+1 ja UTC+2)
[email protected]
Johanna Pellinen, Head of Communications, +358 43 824 5394, [email protected] | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8018917441368103, "perplexity": 2820.6056269389064}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107884755.46/warc/CC-MAIN-20201024194049-20201024224049-00117.warc.gz"} |
https://www.physicsforums.com/threads/missing-sumthing-really-dumb-please-help.179705/ | 1. Aug 8, 2007
1. The problem statement, all variables and given/known data
A block of mass M = 3 kg is released from rest and slides down an incline that makes an angle q = 32° with the horizontal. The coefficient of kinetic friction between the block and the incline is µk = 0.15.
What is the acceleration of the block down the inclined plane?
2. Relevant equations
this is the easiste part of 4 part question and i cant get it.
3. The attempt at a solution
OK heres what i know
Normal = 3kg * (9.8sin32)
so friction is
fs = .15*15.58 = 2.34N
so i'm taking x axis as the incline plane, (pos x = down the plane)
friction opposes the motion, so force runs along neg axis = -2.34N
forces in pos x direction are
3 * 9.8cos32 = 24.93N
24.93N - 2.34N = 22.59N (this is the net force)
Finally F/m = a
22.69/3 = 7.53m/s/s
= wrong.
??
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Last edited: Aug 8, 2007
2. Aug 8, 2007
HallsofIvy
Staff Emeritus
Looks to me like you have sine and cosine reversed. You draw the inclined plane as a right triangle having angle 32 degrees at the bottom. The force of gravity is straight down, the two legs of that triangle are perpendicular and parallel to the inclined plane- the 32 degree angle is the bottom of that triangle. The normal force is given by cos(32), the force along the incline by sin(32).
3. Aug 8, 2007
yeah yer right
thanks | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9324337840080261, "perplexity": 1168.0545049758075}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948587496.62/warc/CC-MAIN-20171216084601-20171216110601-00356.warc.gz"} |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-1-section-1-7-absolute-value-equations-and-inequalities-1-7-exercises-page-119/62 | ## Intermediate Algebra (12th Edition)
$\left[ 0,6 \right]$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|2x-6| \le 6 ,$ use the definition of absolute value inequalities. Use the properties of inequalities to isolate the variable. For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$ For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} -6 \le 2x-6 \le 6 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -6+6 \le 2x-6+6 \le 6+6 \\\\ 0 \le 2x \le 12 \\\\ \dfrac{0}{2} \le \dfrac{2x}{2} \le \dfrac{12}{2} \\\\ 0 \le x \le 6 .\end{array} In interval notation, the solution set is $\left[ 0,6 \right] .$ The colored graph is the graph of the solution set. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9994725584983826, "perplexity": 549.933228278706}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125944479.27/warc/CC-MAIN-20180420155332-20180420175332-00414.warc.gz"} |
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1366629/?tool=pubmed | Biophys J. 2005 Aug; 89(2): 782–795.
Published online 2005 May 6.
PMCID: PMC1366629
# The Physics of Filopodial Protrusion
## Abstract
Filopodium, a spike-like actin protrusion at the leading edge of migrating cells, functions as a sensor of the local environment and has a mechanical role in protrusion. We use modeling to examine mechanics and spatial-temporal dynamics of filopodia. We find that >10 actin filaments have to be bundled to overcome the membrane resistance and that the filopodial length is limited by buckling for 10–30 filaments and by G-actin diffusion for >30 filaments. There is an optimal number of bundled filaments, ∼30, at which the filopodial length can reach a few microns. The model explains characteristic interfilopodial distance of a few microns as a balance of initiation, lateral drift, and merging of the filopodia. The theory suggests that F-actin barbed ends have to be focused and protected from capping (the capping rate has to decrease one order of magnitude) once every hundred seconds per micron of the leading edge to initiate the observed number of filopodia. The model generates testable predictions about how filopodial length, rate of growth, and interfilopodial distance should depend on the number of bundled filaments, membrane resistance, lamellipodial protrusion rate, and G-actin diffusion coefficient.
## INTRODUCTION
The crawling motion of animal cells over a substrate has been described as the succession of protrusion, attachment, and retraction (1). The first step in this sequence, protrusion, is driven by actin polymerization at the leading edge of the cell (2). A common type of protrusive specialization of the leading edge of the cell is the lamellipodium—a flat, leaf-like extension filled with a dense branched network of short (tenths of micron long) actin filaments (Fig. 1). According to the dendritic nucleation model (3), nascent filaments branch from the sides or tips of existing filaments in a sterically precise way. Filaments' barbed ends are oriented forward at roughly ±35° to the direction of protrusion (4). The barbed ends elongate at tenths of micron per second and are capped within seconds. After capping, the filaments lag behind the leading edge and are replaced by the next generation of filaments.
Organization and characteristic scales of filopodia and lamellipodia.
The lamellipodial leading edge is interspersed with filopodia—bundles of actin filaments that are packed tightly together and protrude forward (5,6) (Fig. 1). Similar to the lamellipodial filaments, filopodial filaments are polarized with their barbed ends in the direction of protrusion, but in contrast, they are parallel, long, and turn over very slowly (7) (Fig. 1). Filopodial and lamellipodial protrusions rely on different mechanisms: filament treadmilling (8) and array (9) treadmilling, respectively. They are regulated by different signaling pathways (10,11), yet they are intimately connected, because the filopodial bundles emerge from the lamellipodial network (12,13).
Filopodial protrusions can be “guiding” devices probing space ahead of the lamellipodium. They can also be mechanical devices “penetrating” the environment and serving as a robust scaffold for the lamellipodial protrusion. The role of filopodia as the sensors of the local environment and as sites for adhesion and signaling is well documented (14). In some cells, filopodia are essential for navigation: when filopodia are suppressed, the nerve growth cones can advance but cannot navigate (15). However, fish keratocytes, for example, migrate without filopodia at all (16). It is also worth mentioning that three-dimensional (3D) cell migration through extracellular matrices or engineered scaffolds seems to rely more on the filopodial protrusions in contrast with two-dimensional (2D) cell crawling on flat surfaces (17,18).
There are two major questions about the filopodial protrusion: how is it maintained and how is it initiated? One possibility is that the filopodial filaments are initiated from the cell leading edge by specialized structures (19). Recent evidence, however, points out that the lamellipodial filaments themselves can bend together and “zipper” into parallel bundles of actin filaments (12,20). First, these bundles do not protrude much from the leading edge (such bundles are called Λ-precursors (12) because of their shape). Then, they either mature into the filopodia, or merge with other bundles. To initiate such bundles, the barbed ends have to be locally associated with each other and protected from capping. Protein VASP, elevated in the region of frequent filopodial emergence (12), inhibits capping (21). VASP also transiently binds the barbed ends (22), so association of VASP molecules with each other or some protein cluster could be sufficient to create the tip of the actin bundle. Bundling protein fascin, which is enriched near the tip of the bundle (12), assists VASP and likely other proteins in filopodial initiation. Emerged filopodia have to “outrun” the lamellipodial protrusion, so they have to overcome the membrane resistance, and the G-actin has to be delivered to the filopodial tips.
Available quantitative data (12,23,24) allow theoretical examination of the filopodial mechanics. In the next section, we investigate the restrictions that buckling, membrane resistance, and G-actin diffusion impose on the filopodial dynamics. Then, we find the connection between the spacing between adjacent filopodia and the rates of the filopodial initiation and lateral drift. Finally, we discuss the modeling implications for the biology of filopodial protrusions.
## MECHANICS AND MAINTENANCE OF FILOPODIA
Filopodial protrusions are a few tenths of micron in diameter (25), a few microns in length (7,12,13) (see Discussion for exceptions), and contain 10–30 bundled filaments (12,25) (Fig. 1). The distances between neighboring filopodia are in the micron range. In this article, we explain how these characteristic scales emerge from the physics of the actin bundle. Models' parameters and variables are listed in Tables 1 and and2,2, respectively.
Model variables
Model parameters
### More than 10 bundled filaments are required for filopodia not to buckle
Membrane bending and tension result in the resistance force at the filopodial tip estimated theoretically as F ∼10–20 pN for a membrane cylinder of radius 50–100 nm (26). More detailed recent modeling results in a similar estimate (S. Sun, Johns Hopkins University, personal communication). Experiments with pulling membrane tethers the width and length of which are similar to characteristic filopodial dimensions also give the forces in similar range of 10–50 pN (27,28). Note that in these experiments both membrane bending force and breaking of membrane-cortex links contribute to the resistance, which is likely similar to the resistance to the filopodial protrusion.
Mechanically, the cross-linked filopodial bundle is an effective elastic rod, to which tip the membrane resistance force is applied. The critical force that buckles such rod is equal to:
(1)
where kBT ≈ 4.1 pN nm is the thermal energy (26), Lp ∼ 10 μm is the F-actin persistence length (29,30), and L is the length of the filopodial protrusion. Here π2kBTLp/4L2 is the buckling force for one filament (31), and I(N) is the nondimensional factor, which is responsible for the dependence of the bundle stiffness on the number of the bundled filaments, N. There are two limiting cases: if the filaments are bundled weakly (for example, the distance between the cross-links along the filaments is large and bundling protein is very flexible), then the filaments buckle independently, and I(N) = N. If, in the opposite limit, the bundling is so frequent and tight that the filaments are effectively “glued” to each other, then the bundle can be considered as a single thick rod. The cross-section area of such rod is equal to the number of the filaments times one filament's cross-section area, and the rod's effective radius is proportional to the square root of the number of the filaments. The stiffness is proportional to the radius in power four (31), so in this case I(N) ∼ N2. Numerical simulations described below suggest that I(N) ≈ 0.5 × N2.
Using Eq. 1, we can estimate the critical length at which the membrane resistance force buckles the filopodial bundle:
(2)
We plotted as function of N in two limiting cases (I(N) = N and ) in Fig. 2. The plot demonstrates that the weakly cross-linked bundle of 10–30 filaments would buckle at length below 0.5 μm. Strong bundling can support the length in micron range, in agreement with numerous observations.
Dependence of the critical length, at which the filopodium would buckle, on the number of cross-linked and not cross-linked filaments (solid curves), as predicted by Eq. 2. The dotted lines show the predicted length range for the characteristic numbers ...
We used computer simulations to derive the function I(N) in biologically relevant situation. In the filopodial bundle, the filaments are not packed densely (the cross-section area of the protrusion is ∼0.01 μm2, whereas the total cross-section area of the 25 filopodial filaments is ∼0.001 μm2), so it is likely that each filament is cross-linked with only a few neighbors. We considered variable number (315) of elastic rods, 2 μm in length, with the same mechanical characteristics as those of F-actin. The rods were arranged in a parallel stack, and each rod was connected by elastic cross-links to 2–4 nearest neighbors. The bundling protein, fascin, has length between 10 and 15 nm (32), and its stiffness is likely to be similar to that of F-actin (33). We varied both the length of the cross-links from 10 to 50 nm, and their stiffness from 10 times less to 10 times more than that of F-actin. The electron microscopy (EM) data (12) indicate that the interfascin distance along a filament in the filopodia is of the order of tens of nanometers, so we varied the average corresponding distance from 20 to 500 nm. (In the Appendix, we consider a model of fascin distribution along the bundle.)
We used FEMLAB to solve the buckling problem of elasticity theory (the “Structural Mechanics” module solves an eigenvalue problem, such that the lowest eigenvalue corresponds to the buckling force, whereas the corresponding eigenfunction gives the shape of the buckled bundle). The shape of the buckled actin bundle in a sample simulation is shown in Fig. 3 A. (Filaments were arranged in parallel in 2D; general principles of the linear elasticity theory suggest that the 3D bundle would buckle at similar forces.) The simulation results showed that if the average distance between the neighboring cross-links along an actin filament in the bundle is of the order of 1 μm, then the stiffness of the bundle scaled as ≈N (Fig. 3, B and C), in agreement with simple physical arguments above. Also in agreement with these qualitative arguments, at small average distance between the neighboring cross-links ≈0.1 μm, the stiffness of the bundle scaled as ≈0.5 × N2 (Fig. 3, B and C). We conclude that the observed bundling is tight and in the biologically relevant regime the stiffness of the filopodial bundle increases with the number of the bundled filaments approximately as ≈0.5 × N2. The observed number of the filaments can support the bundle of 1–1.5-μm long against buckling (Fig. 2).
(A) Computed shape of the actin filaments bundled by short elastic links. (B) The computed buckling force (scaled by the buckling force for one 2-μm-long filament) is plotted as the function of the number of bundled filaments N for the average ...
D. Mullins (University of California, San Francisco, personal communication) observed recently that a filopodial bundle of 25 filaments has effective persistence length of 14 mm. This agrees with our simulations, according to which In this observation, bundle grew to be 40-μm long, an order of magnitude greater than predicted by the model. We discuss the difference in the Discussion.
### G-actin diffusion limits the length of thick bundles; membrane resistance limits the length of thin bundles
G-actin diffusion, in addition to the membrane resistance, limits the length of the filopodium. In this article, by G-actin we mean a part of the G-actin pool that can assemble onto the barbed ends (GTP-G-actin not sequestered by thymosin; see Mogilner and Edelstein-Keshet (34)). Let L(t) be the time-dependent length of the N-filament bundle, and a(x, t) be the concentration of the G-actin along the filopodial length (Fig. 4 A). The x axis is oriented forward, and its origin is at the base of the filopodium at the leading edge (Fig. 4 A). In the filopodium, G-actin diffuses and drifts with the cytoplasmic fluid. This drift rate is roughly equal to the rate of the filopodial protrusion dL/dt, because on the relevant timescale the membrane is impermeable to water (35), and due to incompressibility, the cytoplasm has to fill the filopodium at the rate of protrusion. Equation for the G-actin concentration has the form:
(3)
(A) Results of computer simulations of the 2-D G-actin distribution (a(x, t)) in the filopodium and the small adjacent part of the lamellipodium (distance is in microns; G-actin concentration illustrated with shading is in nondimensional units, see Appendix ...
Here D is the effective G-actin diffusion coefficient. The boundary condition at the filopodial base (x = 0) is that the G-actin concentration there is equal to that at the leading lamellipodial edge, a0. (In the Appendix, we examine this assumption by simulating a two-dimensional G-actin distribution (shown in Fig. 4 A) in the filopodium and adjacent part of the lamellipodium.) The boundary condition at the filopodial tip (x = L(t)), which is similar to that at the lamellipodial leading edge (34), is that the G-actin diffusive flux there, −D(∂a/∂x)(L), is equal to the number of monomers assembling per second onto the tips of N filaments. This number is equal to the number of the filaments times the rate of elongation of the filopodial filaments Vf and divided by the half-monomer size δ; η is the geometric coefficient converting the number of monomers into micromolar units (see the Appendix). The drift part of the G-actin flux is not included into the boundary condition, because the filopodial tip and the cytoplasm move together.
The rate of the filopodial extension, dL/dt, is equal to the difference between the rates of the filopodial filaments' elongation Vf and of the lamellipodial filaments' extension in the direction of protrusion, Vl: dL/dt = VfVl. Neglecting small rate of filaments' disassembly, Vfkonδa(L) exp(−/kBTN) (34). Here V0 = konδa(L) is the free polymerization rate proportional to the G-actin concentration at the filopodial tip (kon is the assembly rate); F/N is the membrane resistance load force per filament. The exponential factor is responsible for slowing the protrusion rate by the membrane resistance (34). It is convenient to introduce the effective number of filaments that can support the filopodial protrusion, N0 = /kBT ≈ 13, then Vfkonδa(L) exp(−N0/N).
In the lamellipodium, the leading edge filaments are distributed over a wide range of angles (4). Those elongating at acute angles to the direction of protrusion grow slower, than those elongating at greater angles, simply because the filaments' elongation rate is the increasing function of the angle, Velong = Vl/ cos θ, where Vl is the same for all filaments (Fig. 4 A). This means that slower growing filaments at small angles are generating disproportionately large force, whereas the filaments at greater angles are “free-loaders” growing faster against smaller force. This also means that a large “critical angle” θc exists, such that filaments growing at this angle elongate against zero force at free polymerization rate V0 = konδa0, whereas filaments growing at even greater angles simply cannot keep up with the leading edge and lag behind it (36,24). Then, Vl = konδa0 cos θc, and
(4)
Equations 3 and 4 together represent a difficult free boundary problem. Fortunately, the timescale separation (G-actin diffusion is much faster than the filopodial growth and cytoplasmic drift) allows approximate analytical solution of the problem, which is sketched in the Appendix. The result is that the G-actin concentration decreases linearly from the base to the tip of the filopodium (Fig. 4 A), so that the concentration gradient is the function of the filopodial length:
(5)
This gradient induces the G-actin flux, which is “consumed” at the tip and makes it grow, but the longer the filopodium becomes, the smaller is the G-actin concentration at the tip, and the slower is the rate of growth. Solution of Eqs. 3 and 4 (see the Appendix) is plotted in Fig. 4 B for N = 20, θc = 80°. It has the following asymptotics:
(6)
(7)
Thus, for the first few seconds, when the filopodium is short, the G-actin concentration at its tip is almost equal to that at the lamellipodial leading edge, and the filopodial filaments grow in the direction of protrusion much faster than tilted lamellipodial filaments. Then, the G-actin concentration decreases, and the filopodial growth slows down exponentially at great (over micron) lengths. The maximal length of the filopodium, when its elongation slows down to match the lamellipodial expansion rate, is given by the formula:
(8)
In Fig. 4 C, we plotted the stationary filopodial length Lmax as a function of N at three different values of the critical angle, θc = 60°, 75°, 80°. The data (4) indicate that the critical angle in rapidly moving cells is likely to be not <60°.
The greater the critical angle is, the slower the lamellipodial protrusion, and the farther the filopodium can extend (Fig. 4 C). This is in agreement with the observation that in the rapidly and steadily moving keratocyte cells the filopodia are absent (16). Another prediction is that the filopodial length is linearly proportional to the G-actin diffusion coefficient.
Dependence of the filopodial length on the number of the bundled filaments is biphasic. At great N, the factor exp(N0/N) ≈ 1 (many filaments easily overcome the membrane resistance), and the length is inversely proportional to the filament number, because many growing filament tips deplete the actin monomeric pool. When N is small, the exponential factor exp(N0/N) increases rapidly, and the filopodial length dramatically decreases. In fact, if the filament number is less than:
(9)
the bundle cannot protrude at all. These results are similar to effects of the membrane resistance and G-actin diffusion in lamellipodial protrusion (35,34).
Note that we made the estimates assuming that the filaments are not moving relative to the substratum. In fact, the model is also valid in the presence of retrograde flow of actin, which is almost always the case (37). Indeed, let Ve be the elongation rate of the lamellipodial filaments in the direction of protrusion, and Vr be the rate of lamellipodial network's retrograde movement. Then, the lamellipodial extension rate would be Ve = VlVr. The filopodial bundles are embedded into the lamellipodial network and move rearward with the same rate (7,24), so the rate of the filopodial extension is VfVr, where Vf is the rate of growth of the filopodial filaments. Then, the filopodial length changes with the rate (VlVr) − (VfVr), same as in Eq. 4. The G-actin diffusion and drift are unaffected by the retrograde flow. Note also, that our theory is not applicable to the acrosomal protrusion of Thyone (38,39), where the physics and biology is different, and length, rate of extension, and actin concentration are many times greater. Finally, there is a possibility that filopodial tip complex proteins, such as formins, change the polymerization kinetics, in which case the estimates would change.
### Length of 10–30-filament bundle is limited by buckling; length of >30-filament bundle is limited by G-actin diffusion
Equations 2 and 8 give the maximal attainable filopodial lengths limited by buckling and diffusion, respectively, as functions of the number of the bundled filaments. The resulting observed length is the minimum of these two lengths, if at given N the filopodium buckles at shorter length than that allowed by diffusion, then the growth would be stopped by buckling, and vice versa. We plotted the function Lmax(N) in Fig. 5 A. Our model predicts that >7–8 bundled filaments can maintain the filopodial protrusion. When the filament number is <10, the membrane resistance limits the filopodial length to submicron range. The length of the bundle of 10–30 filaments is limited by buckling, and is proportional to the filament number. The length of the optimal, 30-filament bundle, reaches 1.5 μm. The length of the thicker bundle decreases inversely proportionally to the filament number, because more filament tips deplete G-actin.
(A) Predicted filopodial length limited by the membrane resistance, buckling, and G-actin diffusion as a function of the number of bundled filaments (θc = 80°). (B) Length distribution for 26 filopodial protrusions gleaned from ...
Quantitative observations reported in Argiro et al. (23) partially corroborate our theoretical findings. The maximal length of the observed filopodia was 2–10 μm, which is greater, but the same order of magnitude as predicted. In the Discussion, we speculate on the factors that can explain the difference. The observed rate of the filopodial extension, ≈0.12 μm/s, in agreement with Eq. 6, was maximal just after filopodial initiation and declined thereafter, similar to the predicted time series in Fig. 4 B. The growth did not end with asymptotic slowing down, rather, the filopodia collapsed pivoting or buckling when the maximal length was reached. The initial rate of extension directly correlated with the eventual length of the filopodium (23). This is explained by Eq. 6: greater N means faster initial extension rate (more filaments are less affected by the membrane resistance), and also greater final length when the bundle buckles. Interesting model prediction is that for thicker bundles, length of which is limited by the diffusion, the initial extension rate should be correlated negatively with the final filopodial length. Also, we measured the lengths of 26 adjacent filopodial protrusions in Fig. 2 of Oldenbourg et al. (24) (plotted in Fig. 5 B). Most of the filopodia observed have lengths of 2 μm, in agreement with our estimates.
## BALANCE OF LATERAL DRIFT AND EMERGENCE RATE OF Λ-PRECURSORS REGULATES INTERFILOPODIAL SPACING
The data on molecular mechanisms are too sketchy to attempt detailed quantitative modeling of the filopodial initiation. Here we address the easier question of spacing of the filopodial protrusions along the lamellipodial leading edge. In the next section we also discuss the implications of the estimates that we derive below for the “convergence-elongation” model (12) of the actin bundle initiation. In the Appendix, we consider a simple model that explains fascin-mediated bundling near the tip of the bundle.
Growing lamellipodial barbed end tilted at angle θ relative to the direction of protrusion drifts with velocity Vl tan θ relative to the leading edge protruding with the rate Vl (8) (Figs. 4 A and 6 A). Convergence, “zippering”, and elongation of a few such lamellipodial filaments would produce an actin bundle, either remaining embedded into the lamellipodium, or making filopodial protrusion (Fig. 6 A). Such bundle would be also tilted at some smaller angle and undergo the lateral drift (24). When two such bundles “collide” at the leading edge, their filaments align with each other and the bundles merge (12). As a result, the number of the bundles decreases. Here we show that the interfilopodial spacing can be explained by the balance between the bundle initiation and merging caused by the lateral drift. We neglect simple disappearance of filipodia, because filaments in a filopodium are stable for >1000 s (11).
(A) Illustration of the lateral drift. Dashed lines represent the lamellipodial leading edge at four consecutive moments of time. Barbed ends of the individual filaments and the Λ-precursor change their position along the leading edge as the edge ...
We investigate the spacing between filopodia using first a continuous deterministic model that reveals important biological scales, and then performing realistic stochastic simulations. In the continuous model, we introduce densities (numbers per micron) of Λ-precursors, λ(x, t), and of filopodial protrusions, f(x, t), along the lamellipodial leading edge. These densities change according to the following dynamics:
(Fig. 6 A). Here b is the rate of initiation of Λ-precursors (bundling), and m is the rate of “maturation” of the Λ-precursors into the filopodial protrusions; r1 is the effective rate of “collision” of the Λ-precursors, as a result of which two colliding Λ-precursors merge into one filopodial protrusion due to the increase of the number of filaments in the merged bundle; r2 is the effective rate of “collision” of a Λ-precursor with a filopodial protrusion, as a result of which the Λ-precursor disappears merging with the filopodial protrusion. Finally, r3 is the effective rate of “collision” of two filopodia that merge into one. This simple model can be made more sophisticated by making the number of filaments in actin bundles the independent model variable and assuming some rules of when merging of Λ-precursors results in a thicker Λ-precursor, and when it results in a filopodial protrusion depending on the numbers of bundled filaments. However, this does not change qualitatively simple results derived below.
Order of magnitude of the rate r1 can be estimated as average inverse time before collision of two Λ-precursors, which is equal to the average distance between the precursors, 1/λ, divided by the average rate of the lateral drift, vd: r1vdλ. Similarly, r2vdf, r3vdf. Thus, we obtain the system of equations (in the mean field approximation neglecting correlations between filopodia) for the actin bundle densities:
(10)
These equations can be nondimensionalized by scaling the densities using the balance between the bundling rate and merging rate: vdλ2b, so the density scale is Timescale is equal to the characteristic life time of an individual λ-precursor before it merges with another, Equations for nondimensional variables have the form:
(11)
Here is the dimensionless ratio of the maturation rate to the effective rate of merging of actin bundles.
Phase plane analysis of nonlinear Eq. 11 shows that there is a unique stable biologically relevant stationary solution. It can be found analytically in two limits. First, when the maturation rate is very slow, the average precursor and filopodial densities are almost equal: On the other hand, when the λ-precursors mature fast, the λ-precursors density is low, λ′ ≈ 1/ε, while f′ ≈ 1. For intermediate values of ε, f′ ∼ 1. The important conclusion is that the order of magnitude of the stationary filopodial density is (in dimensional variables).
We measured the distances between 26 adjacent filopodia in Fig. 2 of Oldenbourg et al. (24) and plotted the results in Fig. 6 B. The average interfilopodial distance is ∼2 μm, and f ∼ 0.5/μm. The average angle at which the actin bundles are tilted relative to the direction of protrusion is <30°, and the average lateral drift rate is a few fold less than the rate of protrusion, vd ∼ 0.01μm/s. We can estimate the rate of emergence of the Λ-precursors as bvdf2 ∼ 0.001–0.01 μm−1s−1. Thus, a new Λ-precursor has to appear once every few hundreds of seconds per micron of the leading edge. There are no measurements of this rate available, but this estimate seems to compare well with observations reported in Svitkina et al. (12).
Also, the micrographs in Svitkina et al. (12) indicate that the densities of Λ-precursors and filopodial protrusions are comparable, so, according to our analysis, the rate of precursors' maturation cannot be faster than s. In other words, on the average, individual Λ-precursors can be observed for ∼100 s before merging or maturing into a filopodium.
This analysis is supported by the following stochastic simulations, which are essential because of the dispersal of the actin bundles' orientations, correlations between the bundles and large fluctuations of the bundles' number. Each actin bundle (either Λ-precursor, or filopodium) is characterized by its position along the lamellipodial leading edge, xi(t), rate of lateral movement, vi(t), and maturity index, mi, equal to zero for a precursor and to unity for a filopodium. We consider a 30-μm-long segment of the leading edge (Fig. 6 C) and generate new precursors on it at random location with the rate b = 0.01 μm−1s−1. Each nascent precursor is tilted to the protrusion direction at a random angle uniformly distributed in the interval −30° < θi < 30°, and vi(t) = v × tan(θi), where v = 0.05 μm/s is the protrusion rate. The precursors mature (mi switches from 0 to 1) into the filopodia with the constant rate m = 0.01/s. The trajectories of the precursors (light gray) and filopodia (dark gray) from a sample simulation are shown in Fig. 6 C, where time in seconds is shown on the y axis. In the simulations, we update the positions of the tips of the precursors and bases of the filopodia along the leading edge each time step (5 s). Actin bundles that run into the edges of the segment “disappear”.
We consider “collision events” of pairs of the actin bundle, when the distance between them is smaller than 30 nm. Each collision results in the merger of the pair. When two precursors collide, the resulting actin bundle becomes another precursor, or changes into a filopodium with equal probability. (Numerical experiments show that assigning weighed probabilities does not change the results qualitatively.) When either a precursor collides with a filopodium, or two filopodia collide, a single filopodium results. The lateral movement rate of the resulting bundle is equal to that of one of the colliding pair having minimal absolute value, if either two precursors, or two filopodia collide. If a precursor and a filopodium collide, then the lateral movement rates of “mother” and “daughter” filopodia are the same.
Repeating simulations like those shown in Fig. 6 C, we plotted the histogram of the interfilopodial distances (Fig. 6 D). We chose the rate of initiation of Λ-precursors so that the observed and calculated mean interfilopodial distances are the same order of magnitude. Simulations demonstrate that the observed and calculated variances of these distances are also similar (Fig. 6, B and D). We also tested numerically the predicted dependence of the analytical model. The stochastic simulations confirm that the density of filopodia along the leading edge is proportional to the square root of the rate of initiation of Λ-precursors and inversely proportional to the square root of the drift rate (Fig. 6, E and F).
## DISCUSSION
### Filopodial length
Estimates in this article show that to overcome the membrane resistance, >10 actin filaments have to be bundled in filopodia. The length of the filopodial bundle of 10–25 filaments is limited to 1–2 μm due to buckling of the bundle by the membrane resistance force. Thicker bundles are stronger, but growth of >30 filaments bundled together is limited by G-actin diffusion: more barbed ends consume so many monomers that diffusion cannot maintain bundles longer than 2 μm. This analysis explains the observed number (tens) of actin filaments in filopodia and their length (microns): in fibroblasts, macrophages, and nerve growth cones the filopodial length rarely exceeds 10 μm. Our findings can also explain the experimental observations (23) of the rate of the filopodial growth of the order of 0.1 μm/s and its correlation with the final filopodial length.
However, filopodia sometimes grow longer: in sea urchin embryo, where filopodia were first seen live in 1961 (40), they were 5–35-μm long. R. D. Mullins (University of California, San Francisco, personal communication cited above) observed recently the filopodial bundle 40-μm long. Also, some observations indicate that the rate of filopodial elongation does not slow down with filopodial length as fast as predicted by our theory (7). These discrepancies point out that simple G-actin diffusion and linear elastic stability of the cross-linked filament bundle cannot fully explain the observed filopodial behavior. Thus, perhaps the most valuable lesson from our model is that additional mechanisms have to be at work in filopodia.
There are few possible explanations for these discrepancies between theory and experiment. First, decreasing membrane resistance by two orders of magnitude increases the filopodial length limited by buckling by one order of magnitude, from a few microns to a few tens of microns (buckling length is proportional to the square root of the force). This can be accomplished by regulation of the membrane tension (41). Second, adhesion of the filopodia to the substratum, which we did not consider, can strengthen the filopodia significantly: long filopodia adheres to the surface, whereas filopodia without adhesions bends laterally (13). Third, as far as the diffusion-limited growth is concerned, our estimates were made for the steadily protruding lamellipodial leading edge. In fact, this protrusion in most cells consists of irregular cycles of protrusion and retraction (21). Filopodial growth can continue past the micron range, with slowing speed, if the lamellipodial leading edge is stalled. Finally, other means of transport not considered here, for example those mediated by unconventional myosin motors (42), can contribute to filopodial elongation. Indeed, there are indications that unconventional myosin motors are responsible for transport of adhesion molecules (14) and of Mena/VASP proteins (43).
### Note about protrusion force generation
The polymerization ratchet mechanism of force generation (44) requires frequent bending of either filament tips, or membrane, or both, so that the transient gap between the filaments' tips and membrane is >δ ≈ 2.7 nm. Unlike the tilted lamellipodial filaments, the filopodial filaments are perpendicular to the resisting membrane, and the transient gap due to their thermal bending is smaller. Its magnitude can be estimated as the shortening of the end-to-end distance for elastic rod of length lc and persistence length Lp due to the thermal bending: (45). For actin, Lp ∼ 10 μm, and lc ∼ 20–30 nm is of the order of the average distance between the fascin cross-links. The value of δ1 is <1 nm at these parameters, so filament bending is not sufficient. However, the membrane bending is sufficient: in Mogilner and Oster (44) we derived the formula: for the corresponding gap. Substituting the values of the membrane bending modulus, B ∼ 50 kBT, the membrane resistance force, F ∼ 20 pN, and the area of the filopodial tip, A ∼ 0.01 μm2 we estimate the corresponding gap as ∼10 nm.
More thorough stochastic simulations taking into account detailed membrane dynamics and polymerization kinetics confirm this conclusion (S. Sun, Johns Hopkins University, personal communication). However, future modeling is needed because the filopodial tip is loaded with proteins and its mechanical properties are unknown. Also, abundance of VASP at the tip could lead to frequent attachment of the filaments to the membrane (46). It is possible that other models of force generation based on complex mechanochemical cycles of barbed ends associated with auxiliary proteins are relevant for the filopodial protrusion (47,48). If this is the case, then the exact values for the generated polymerization force and G-actin kinetics rates at the filopodial tip would change, but their orders of magnitude would not, so the order of magnitude estimates in this article would remain valid.
### Model implications for molecular mechanisms of filopodia initiation
The model explains the characteristic distance between adjacent filopodia in micron range as the balance of initiation and lateral drift and merging of the actin bundles. The theory suggests that F-actin barbed ends have to be locally focused and protected from capping approximately once every hundred seconds per micron of the lamellipodial leading edge to initiate the observed number of filopodia.
From the EM data reported in Svitkina et al. (12) we can glean ∼100 barbed ends per micron of the leading edge (this estimate compares well with 250 ends per micron reported in (49)). At protrusion rate v = 0.05 μm/s and average angle between filament growth and leading edge protrusion θ = 35°, the average lateral drift rate is v × tan(θ) ≈ 0.035 μm/s, so barbed ends tilted to the right/left converge at the rate 0.07 μm/s. Total height of the lamellipod is ∼0.2 μm (49), so each filament (which is ∼0.005 μm in diameter) would, on the average, “collide” due to the lateral drift with another filament at the rate (∼50 filaments/μm) × (0.07 μm/s) × (0.005 / 0.2 μm) ∼ 0.1/s. Filaments are tenths of microns long, so the capping rate at the leading edge is of the order of 0.1/s, so there is a significant probability that any growing filament would “collide” with an oppositely tilted filament. If the barbed ends of such pair of filaments are kept together either by a dynamic cross-linker that stays close to the growing filament tips, or by a “processive capper”, such as formin (50), which in turn is associated with a nascent “filopodial tip complex” (reviewed in Small et al. (51)), then the filaments would bend into parallel configuration and start to grow almost in the direction of protrusion. The corresponding bending force is in subpiconewton range and can be easily generated by the polymerization ratchet mechanism (44,50).
Other filament tips would collide with the pair and could be trapped in the growing bundle creating a nascent Λ-precursor. In order for this precursor to assemble a bundle of ∼10 filaments, the effective capping rate in the vicinity of the precursor tip has to decrease to ∼0.01/s. Indeed, the average stationary number of the growing tips in the bundle can be estimated as the ratio of the rate of collisions of the bundle with lamellipodial barbed ends, ∼0.1/s, to the capping rate, so the latter can be estimated as ∼(0.1/s) / 10 = 0.01/s. It would take ∼10 / (0.1/s) = 100 s to assemble the actin bundle.
Such a low capping rate cannot be maintained along the whole leading edge, because the filaments would grow a few microns long and buckle (21). Therefore, our estimates suggest that once every hundred seconds per micron of the lamellipodial leading edge, a nascent filopodial tip complex (or part of it) self-assembles, such that its components both associate with the filament tips physically, and protect them from capping. Then, in hundred seconds, a Λ-precursor develops and matures into a filopodium or merges with other actin bundles. Likely, some positive feedbacks are involved in this process. For example, transient changes in membrane curvature have been shown to cause filopodia perhaps by activating pathways that trigger actin polymerization (52), and in turn concentration of filaments in actin bundles curves the membrane locally. It is premature to speculate about specific pathways of the filopodial initiation. The value of our model is that it predicts the rate of filopodial precursors initiation and the local capping rate posing quantitative constraints for future models.
### Model predictions
The model generates the following testable predictions:
• There is an “optimal” filament number at which the maximal filopodial length is achieved.
• Decreasing membrane stiffness would lead to increasing of the filopodial lengths for actin bundles with small filament numbers (the length of which is limited by the membrane resistance), whereas the length of thicker filament bundles (the length of which is limited by the G-actin diffusion) would not change.
• Faster lamellipodial protrusion correlates with shorter average filopodial lengths.
• Faster lamellipodial protrusion correlates with greater average distance between adjacent filopodia.
• Initial growth rate of thin (thick) filopodial bundles is an increasing (decreasing) function of the filament number and correlates positively (negatively) with the final filopodial length.
From the physical point of view, it is tempting to compare the filopodial and lamellipodial protrusions. In terms of G-actin “consumption”, the lamellipodial filaments (hundreds per micron of the leading edge (49)) deplete G-actin pool equally with the filopodial bundle (tens of filaments per one-tenth of micron of the leading edge). On smooth surfaces, lamellipodial organization of actin filaments is optimal for the elastic polymerization ratchet mechanism of protrusion force generation, because in the lamellipodium the filaments are cross-linked neither too heavily, nor too lightly, so they do not buckle, yet are flexible enough (46). However, filopodial protrusions would be more efficient for crawling through extracellular matrix and on surfaces of other cells. Another possible role of relatively rigid actin bundles embedded into the lamellipodial actin sheet is to strengthen the lamellipodium against buckling, by analogy with engineered macroscopic structures (53). Future modeling efforts can help to elucidate other filopodial important functions, such as being guides for microtubules (54).
## Acknowledgments
We are grateful to G. G. Borisy, T. Schaus, R. Cheney, T. Switkina, and K. Tosney for fruitful discussions, and to R. D. Mullins, G. G. Borisy, and S. Sun for sharing unpublished data.
The work is supported by National Institutes of Health GLUE grant “Cell Migration Consortium” (NIGMS U54 GM64346) and National Science Foundation grant DMS-0315782.
## APPENDIX I: ANALYSIS OF THE G-ACTIN DIFFUSION AND THE GROWTH OF THE FILOPODIUM
The factor η [μM−1μm−1] converts μM concentration units into the number of molecules per unit length of the filopodium, given that the filopodial radius is ∼0.1 μm. As noted above, most of the volume inside the filopodium is free for the monomers to diffuse in. A concentration of 1 μM corresponds to molecules per μm3, and this figure corresponds to π × (0.1 μm)2 × 600/μm molecules per 1 μm of the filopodium. Thus, .
The following scales are characteristic for the filopodial protrusion: a0 ≈ 10 μM for the G-actin concentration, for the lamellipodial length, and s for time. Rescaling Eqs. 3 and 4, we obtain the nondimensionalized equations for variables (we keep the same notations for the rescaled variables):
(12)
(13)
Here On the relevant scale, the G-actin diffusion is much faster than the cytoplasmic drift and filopodial growth, and the left-hand side and the second term on the right-hand side in Eq. 12 can be neglected: over seconds, the diffusion establishes a quasistationary gradient of the G-actin concentration, which slowly follows changes of the filopodial length over tens of seconds. So, and using the boundary conditions, we obtain: a(x, t) ≈ 1 − α(L(t))x, where α ≈ 1/(1 + L(t)). Corresponding dimensional formula is Eq. 5. Substituting these expressions into Eq. 13 gives:
Integrating this first order ordinary differential equation (L(0) = 0), we find the solution implicitly:
This formula can be used to plot the solution numerically (Fig. 4 B) and to find the asymptotic behavior of the filopodial length:
Corresponding dimensional formulas are Eqs. 6 and 7.
## APPENDIX II: G-ACTIN GRADIENT IN THE FILOPODIUM “SEAMLESSLY” MATCHES THAT IN THE LAMELLIPODIUM
The boundary condition a(0) = a0 for the G-actin concentration at the base of the filopodium (Fig. 4 A), where a0 is the concentration at the lamellipodial leading edge, is nontrivial, because the “consumption” of the G-actin at the filopodial tip and corresponding diffusive flux can, in principle, locally deplete the G-actin concentration at the leading edge. To examine this boundary condition, we used FEMLAB to solve the following 2D (the lamellipodium is flat) diffusion problem. We considered the 0.1-μm-wide and 1-μm-long filopodium and 1-μm-wide and 0.8-μm-long adjacent part of the lamellipodium (Fig. 4 A). We solved the G-actin diffusion equation on this combined domain using the parameters described above and the following boundary conditions: i), the G-actin concentration at the “back” of the lamellipodial part of the domain is 1.2 (in the units of a0); ii), the G-actin concentration at the “front” of the lamellipodial part of the domain is 1; iii), the G-actin flux at the tip of the filopodial part of the domain is given by Eq. 12; iv), the G-actin flux at the sides of both lamellipodial and filopodial parts of the domain is zero. Conditions i and ii are chosen so that the G-actin flux at the lamellipodial leading edge matches the “consumption” of the G-actin at the edge at characteristic protrusion rate and F-actin density at the edge (34). The stationary solution of this diffusion problem illustrated with shading in Fig. 4 A shows that the G-actin gradients at the lamellipodium and filopodium “seamlessly” match each other, and that the G-actin concentration at the base of the filopodium is, indeed, a0. This is the consequence of the fact that the characteristic number of the filopodial filaments per filopodial size, ∼20/0.1 μm, is the same as the characteristic number of the lamellipodial filaments per leading edge length, ∼200/1 μm (49), so the filopodium “consumes” proportional share of G-actin and does not deplete the lamellipodial G-actin pool.
## APPENDIX III: DYNAMIC MODEL OF FASCIN DISTRIBUTION IN THE FILOPODIUM
The bundling protein fascin is turned on and off by regulated phosphorylation (32). It is likely that this regulation takes place at the filopodial tip complex (G. G. Borisy, Northwestern University, personal communication). This suggests the following model that explains the observed increased fascin presence near the tips of actin bundles. Let us consider the (L = 2 μm)-long filopodium, place the x axis directed backward with the origin at the tip, and consider the linear densities of fascin bound to the actin filaments, fb(x, t), diffusing “inactive” fascin dissociated from the actin filaments, fi(x, t), and diffusing “active” fascin dissociated from the actin filaments, fa(x, t). The dynamics of fascin is described by the following system of equations:
(14)
(15)
(16)
(17)
The first term in the right-hand side of Eq. 14 is responsible for the kinematic drift of the bound fascin with the protrusion rate v, due to treadmilling together with F-actin, away from the filopodial tip. The first terms in the right-hand side of Eqs. 15 and 16 describes the fascin diffusion. We use the value of the diffusion coefficient Df = 2 μm2/s, scaled proportionally to length from the known value of the G-actin diffusivity. The second terms in the right-hand side of Eqs. 14 and 15 describe the inactivation and dissociation of fascin from F-actin with the rate k1. The second term in the right-hand side of Eq. 16 and the third term in the right-hand side of Eq. 14 are responsible for association of activated fascin with F-actin with the rate k2. We use the values k1 = k2 = 1/s, which are characteristic for the kinetics of actin-binding proteins (33). Equation 17 gives the boundary conditions: due to the drift, there is no bound fascin at the tip, fb(0) = 0. All inactivated fascin is activated at the tip, so fi(0) = 0, and the fluxes of the inactivated and activated fascin balance (last formula in Eq. 17). We assume that at the base of the filopodium there is no activated fascin (like in the lamellipodium), and the concentration of the inactivated fascin is equal to that in the lamellipodium, f0.
We used FEMLAB to solve Eqs. 1417. The solutions are plotted in Fig. 7. The model predicts that the bundling fascin concentration is maximal at approximately 300 nm from the filopodial tip, much closer to the tip, than to the base of the filopodium. Near the base, the cross-linking density decreases. This is unlikely to affect the buckling force. Finally, we repeated the simulations for the Λ-precursor (allowing the diffusion to be two-dimensional). The corresponding bound fascin density is shown with the dotted curve in Fig. 7. Again, the fascin density is maximal near the bundle's tip.
### FIGURE 7
Nondimensionalized linear densities of fascin along the filopodial actin bundle. The horizontal axis shows distance in microns; the origin corresponds to the filopodial tip. (Solid line) Fascin associated with F-actin; (dashed line) “activated” (decreasing) and “inactivated” (increasing) fascin; (dotted line) fascin associated with F-actin along the Λ-precursor's bundle.
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http://physics.stackexchange.com/questions/18277/solar-wind-and-the-earths-magnetic-field | # Solar wind and the Earth's magnetic field
I have again an old question from a comprehensive exam I took a couple of months ago. Lucky for me one could pick 5 out of 8 questions, because on some of the problems I didn't even know how to start. Now that classes are over I've now the time to revisit those problems I was dumbfounded by, such as this one: (Abriged version)
Life on earth would be impossible if we were constantly exposed to charged solar particles. Luckily, earth's magnetic field protects us from them. The solar particles have a typical energy spectrum of $d\Phi / dE \propto E^{-3}$ particles/$m^2/s/J$. What is the minimal field strength of the earth's magnetic field based on the anthropic principle, i.e., it couldn't be weaker or else we wouldn't live to observe it.
Well, this quantity $d\Phi/dE$ looks like a flux, so I guess the general setting is that of a scattering problem. But first, $d\Phi / dE$ isn't given completely, only a rough form of its energy dependence. And second, I'm not sure what a reasonably simple model for this entire process would be. Easiest in terms of calculation would be to assume some sort of homogeneous magnetic field aligned with the earth's magnetic axis, because I guess it's a pain to calculate the path of a particle in a dipole field...
Maybe the idea of this is to calculate the total cross section of earth's magnetic field and then demand that it should "cover" the earth? Or they want me to solve an equation of motion for incoming solar particles and show that all of them are deflected? My problem right now is that I don't even know how to interpret the $d\Phi/dE$ quantity whose energy dependence I'm given. I guess it makes more sense to someone with a background in elementary particle physics?
Right now I'm trying to write a vector potential $\vec{A} = \mu_o/(4\pi r^2) \vec{m} \cdot \vec{e}_r$ where $\vec{e}_r$ is the unit vector in $r$-direction in spherical coordinates, and then try to get equations of motion from the Hamiltonian $$H = \frac{(\vec{p} + q\vec{A})^2}{2m}$$ but I am not sure if I'll be able to solve whatever comes out of that, or if I'm completely on the wrong track with this.
EDIT
Image taken from here
Maybe it helps trying to understand this schematic, but I cannot easily see how the Lorentz force would create such a trajectory.
ANOTHER EDIT
From further searching, I know suspect that this has something to do with how a plasma current (the charged particles) interact with a magnetic field. That would mean that I have to calculate the radius of the ensuing magnetosphere and then demand, via the anthropic principle, that it should be at least of the same size (or larger) as the radius of earth. So the Lorentz force would probably not directly have anything to do with it. But I also have no training in plasma physics. (Some of the problems in the exam were specifically geared towards Astronomy students, so I guess they'd find it a breeze).
-
Any chance you could get more specific than "Any hints..."? After all, we don't let the newbie posters ask that kind of question so it wouldn't quite be fair to let it go in this case ;-) Also, are you expected to be able to do this without external resources? Do they want an exact answer or a rough (perhaps order-of-magnitude) estimate? – David Z Dec 14 '11 at 17:43
I'll try to elaborate. No external resources are allowed. I guess an order of magnitude estimate is okay. – Lagerbaer Dec 14 '11 at 17:46
If an order of estimate then could we not assume $Bqv=mv^2/r$ and then make an assumption about $v\sim 10^8 m/s$. The other quantities we can assume to be for say an electron. Of course, this means the flux relationship given is a red herring! – Omar Dec 14 '11 at 18:32
@Omar Maybe one can combine this? If I use your reasoning, there exists a critical particle energy $E_c$ below which a particle gets deflected and above which it hits the earth. The total number of particles hitting the earth then scales like $\int_{E_c}^\infty \frac{1}{E^3}$ ~ $1/E_c^2$. Maybe then one can make some smart argument for what $E_c$ should be... – Lagerbaer Dec 14 '11 at 18:43
@Lagerbaer That could be an interesting approach. The problem is that it is a powerlaw flux relationship. I guess you could make some assumptions about the underlying particle powerlaw distribution and then assume they have a characteristic temperature (a little dubious for a powerlaw distribution). All that seems really complicated for a no-external-resources question! – Omar Dec 14 '11 at 19:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9297122955322266, "perplexity": 262.0489553409849}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931004237.15/warc/CC-MAIN-20141125155644-00121-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/have-a-cone-and-divide-it-into-infinately-small-slices.26195/ | # Have a cone and divide it into infinately small slices
• Start date
• #1
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0
If I have a cone and divide it into infinately small slices. Wouldn't both sides of one slice have the same area and wouldn't the next slice (and so on) have the same area as the slice before. So wouldn't your cone actually be a cylinder?
My answer is no, because the reasoning is wrong. If I had infinately small slices I would never complete the cone/cylinder in the first place. And the assumption of both sides having the same area is an assumption to be able to integrate, but is not reality. If we're talking about the perfect cone then both sides of the slices should have different areas even if the slices were infinately small.
• #2
AKG
Homework Helper
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Lorentz said:
If I have a cone and divide it into infinately small slices. Wouldn't both sides of one slice have the same area and wouldn't the next slice (and so on) have the same area as the slice before. So wouldn't your cone actually be a cylinder?
It would something more like, as the number of slices approaches infinity, the width of each slice approaches zero, and the areas on the two faces of the slice approach each other.
If I had infinately small slices I would never complete the cone/cylinder in the first place.
Here's something to think about. You should know that a line is made up of infinite points. Each point has zero size. So, given an infinite number of zero-sized points put together, how is it that you get a line with non-zero-size?
And the assumption of both sides having the same area is an assumption to be able to integrate, but is not reality.
Well, be careful here, because you can't really make any arguments "from reality" when dealing with math. Math is a useful tool in modelling reality, but that doesn't mean that it is based on reality (it is based on its own axioms, some of which seem rather unnatural), nor does it mean that it is an accurate tool in modelling reality.
If we're talking about the perfect cone then both sides of the slices should have different areas even if the slices were infinately small.
What does it mean to be infinitely small? Can something be smaller than infinitely small? What would the difference in area be between the two faces?
• #3
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AKG said:
Can something be smaller than infinitely small? What would the difference in area be between the two faces?
erm... the difference would be infinitely small? But that would still be a difference which makes it possible to glue the slices together and get the cone back again rather then a cylinder. If the difference would be zero we would get the cilinder.
• #4
58
0
This question just popped into my mind: Is there a difference between infinitely small and zero?
• #5
HallsofIvy
Homework Helper
41,847
966
Lorentz said:
This question just popped into my mind: Is there a difference between infinitely small and zero?
Yes, IF you are working in "non-standard analysis" and, by "infinitely small", you mean "infinitesmal". Otherwise "infinitely small" is just a (misleading) shorthand for "in the limit".
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5K | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8375326991081238, "perplexity": 682.4883855520472}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057039.7/warc/CC-MAIN-20210920131052-20210920161052-00601.warc.gz"} |
http://en.wikipedia.org/wiki/Hypoexponential_distribution | # Hypoexponential distribution
Parameters $\lambda_{1},\dots,\lambda_{k} > 0\,$ rates (real) $x \in [0; \infty)\!$ Expressed as a phase-type distribution $-\boldsymbol{\alpha}e^{x\Theta}\Theta\boldsymbol{1}$ Has no other simple form; see article for details Expressed as a phase-type distribution $1-\boldsymbol{\alpha}e^{x\Theta}\boldsymbol{1}$ $\sum^{k}_{i=1}1/\lambda_{i}\,$ $\ln(2)\sum^{k}_{i=1}1/\lambda_{i}$ $(k-1)/\lambda$ if $\lambda_{k} = \lambda$, for all k $\sum^{k}_{i=1}1/\lambda^2_{i}$ $2(\sum^{k}_{i=1}1/\lambda_{i}^3)/(\sum^{k}_{i=1}1/\lambda_{i}^2)^{3/2}$ no simple closed form $\boldsymbol{\alpha}(tI-\Theta)^{-1}\Theta\mathbf{1}$ $\boldsymbol{\alpha}(itI-\Theta)^{-1}\Theta\mathbf{1}$
In probability theory the hypoexponential distribution or the generalized Erlang distribution is a continuous distribution, that has found use in the same fields as the Erlang distribution, such as queueing theory, teletraffic engineering and more generally in stochastic processes. It is called the hypoexponetial distribution as it has a coefficient of variation less than one, compared to the hyper-exponential distribution which has coefficient of variation greater than one and the exponential distribution which has coefficient of variation of one.
## Overview
The Erlang distribution is a series of k exponential distributions all with rate $\lambda$. The hypoexponential is a series of k exponential distributions each with their own rate $\lambda_{i}$, the rate of the $i^{th}$ exponential distribution. If we have k independently distributed exponential random variables $\boldsymbol{X}_{i}$, then the random variable,
$\boldsymbol{X}=\sum^{k}_{i=1}\boldsymbol{X}_{i}$
is hypoexponentially distributed. The hypoexponential has a minimum coefficient of variation of $1/k$.
### Relation to the phase-type distribution
As a result of the definition it is easier to consider this distribution as a special case of the phase-type distribution. The phase-type distribution is the time to absorption of a finite state Markov process. If we have a k+1 state process, where the first k states are transient and the state k+1 is an absorbing state, then the distribution of time from the start of the process until the absorbing state is reached is phase-type distributed. This becomes the hypoexponential if we start in the first 1 and move skip-free from state i to i+1 with rate $\lambda_{i}$ until state k transitions with rate $\lambda_{k}$ to the absorbing state k+1. This can be written in the form of a subgenerator matrix,
$\left[\begin{matrix}-\lambda_{1}&\lambda_{1}&0&\dots&0&0\\ 0&-\lambda_{2}&\lambda_{2}&\ddots&0&0\\ \vdots&\ddots&\ddots&\ddots&\ddots&\vdots\\ 0&0&\ddots&-\lambda_{k-2}&\lambda_{k-2}&0\\ 0&0&\dots&0&-\lambda_{k-1}&\lambda_{k-1}\\ 0&0&\dots&0&0&-\lambda_{k} \end{matrix}\right]\; .$
For simplicity denote the above matrix $\Theta\equiv\Theta(\lambda_{1},\dots,\lambda_{k})$. If the probability of starting in each of the k states is
$\boldsymbol{\alpha}=(1,0,\dots,0)$
then $Hypo(\lambda_{1},\dots,\lambda_{k})=PH(\boldsymbol{\alpha},\Theta).$
## Two parameter case
Where the distribution has two parameters ($\mu_1 \neq \mu_2$) the explicit forms of the probability functions and the associated statistics are[1]
CDF: $F(x) = 1 - ( \mu_1 e^{-x \mu_2} - \mu_2 e^{-x \mu_1}) / ( \mu_1 - \mu_2 )$
PDF: $f(x) = \mu_1\mu_2( e^{-x \mu_2} - e^{-x \mu_1} ) / ( \mu_1 - \mu_2 )$
Mean: $1/\mu_1+1/\mu_2$
Variance: $1/\mu_1^2+1/\mu_2^2$
Coefficient of variation: $( \mu_1 + \mu_2 )^{0.5} / ( \mu_1 + \mu_2 )$
The coefficient of variation is always < 1.
Given the sample mean ($\bar{x}$) and sample coefficient of variation ($c$) the parameters $\mu_1$ and $\mu_2$ can be estimated:
$\mu_1= ( 2 / \bar{x} ) ( 1 + ( 1 + 2 ( c^2 - 1 ) )^{(0.5)} )^{-1}$
$\mu_2 = ( 2 / \bar{x} ) ( 1 - ( 1 + 2 ( c^2 - 1 ) )^{(0.5)} )^{-1}$
## Characterization
A random variable $\boldsymbol{X}\sim Hypo(\lambda_{1},\dots,\lambda_{k})$ has cumulative distribution function given by,
$F(x)=1-\boldsymbol{\alpha}e^{x\Theta}\boldsymbol{1}$
and density function,
$f(x)=-\boldsymbol{\alpha}e^{x\Theta}\Theta\boldsymbol{1}\; ,$
where $\boldsymbol{1}$ is a column vector of ones of the size k and $e^{A}$ is the matrix exponential of A. When $\lambda_{i} \ne \lambda_{j}$ for all $i \ne j$, the density function can be written as
$f(x) = \sum_{i=1}^k \lambda_i e^{-x \lambda_i} \left(\prod_{j=1, j \ne i}^k \frac{\lambda_j}{\lambda_j - \lambda_i}\right) = \sum_{i=1}^k \ell_i(0) \lambda_i e^{-x \lambda_i}$
where $\ell_1(x), \dots, \ell_k(x)$ are the Lagrange basis polynomials associated with the points $\lambda_1,\dots,\lambda_k$.
The distribution has Laplace transform of
$\mathcal{L}\{f(x)\}=-\boldsymbol{\alpha}(sI-\Theta)^{-1}\Theta\boldsymbol{1}$
Which can be used to find moments,
$E[X^{n}]=(-1)^{n}n!\boldsymbol{\alpha}\Theta^{-n}\boldsymbol{1}\; .$
## General case
In the general case where there are $a$ distinct sums of exponential distributions with rates $\lambda_1,\lambda_2,\cdots,\lambda_a$ and a number of terms in each sum equals to $r_1,r_2,\cdots,r_a$ respectively. The cumulative distribution function for $t\geq0$ is given by
$F(t) = 1 - \left(\prod_{j=1}^a \lambda_j^{r_j} \right) \sum_{k=1}^a \sum_{l=1}^{r_k} \frac{\Psi_{k,l}(-\lambda_k) t^{r_k-l} \exp(-\lambda_k t)} {(r_k-l)!(l-1)!} ,$
with
$\Psi_{k,l}(x) = -\frac{\partial^{l-1}}{\partial x^{l-1}} \left(\prod_{j=0,j\neq k}^a \left(\lambda_j+x\right)^{-r_j} \right) .$
with the additional convention $\lambda_0 = 0, r_0 = 1$.
## Uses
This distribution has been used in population genetics[2] and queuing theory[3][4]
## References
1. ^ Bolch, Gunter; Greiner, Stefan; de Meer, Hermann; Trivedi, Kishor Shridharbhai (2006). Queueing Networks and Markov Chains: Modeling and Performance Evaluation with Computer Science Applications (2nd ed.). Wiley-Blackwell. ISBN 978-0-471-56525-3.
2. ^ Strimmer K, Pybus OG (2001) "Exploring the demographic history of DNA sequences using the generalized skyline plot", Mol Biol Evol 18(12):2298-305
3. ^ http://www.few.vu.nl/en/Images/stageverslag-calinescu_tcm39-105827.pdf
4. ^ Bekker R, Koeleman PM (2011) "Scheduling admissions and reducing variability in bed demand". Health Care Manag Sci, 14(3):237-249 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 55, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9813053607940674, "perplexity": 685.6410113876618}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368697420704/warc/CC-MAIN-20130516094340-00007-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://ipsc.ksp.sk/2015/real/solutions/a.html | # Internet Problem Solving Contest
## Solution to Problem A – A+B
Our task is to rearrange the string of digits into two integers a and b such that a + b is as large as possible. Let’s see how we can construct such a and b, and thus find the value of a + b.
In the easy subproblem, we had exactly three digits. The answer will certainly be of the form $\overline{ab}+\overline{\vphantom{b}c}$, where a, b and c are the three given digits (in some order), and the overline denotes a number that consists of the given digits.
The value of the above number is 10a + b + c. From this it is obvious that a should be the largest of the three given digits and that the order of b and c does not matter.
The above observation can easily be generalized to solve the hard subproblem as well. The optimal solution is to read the string of digits, sort them in non-ascending order, and then take the first n − 1 digits as one of the numbers and the last digit as the other number. For example, the optimal solution for the input 12345 is 5432+1.
### Formal proof
It is easy to verify that the solution described above will never run into troubles with unnecessary leading zeros – if there are zeros in the input, one of them will be b (which is valid) and all others will be at the end of a (which is also valid).
Lemma: In the optimal solution one of the two numbers will consist of just a single digit.
Proof: Consider an arbitrary arrangement of digits. Label the two numbers a and b so that a > b. Suppose that b has more than one digit. Let’s now take the last digit of b and append it to a instead. It should be obvious that we didn’t create any new leading zeros anywhere, so the solution remains valid.
How will the sum change? Let’s write the original b as 10b′ + d. The sum before we moved the digit was a + 10b′ + d. After the move the new sum is 10a + d + b. The change is 9(a − b′) > 0, therefore the new arrangement is better than the old one, so the old arrangement cannot be optimal.
Theorem: The arrangement described in our solution above is optimal.
Proof: From the lemma we know that each optimal solution has the form $\overline{d_1\dots d_{n-1}} + \overline{d_n}$. The value of this number is 10n − 2d1 + 10n − 3d2 + ⋯ + 10dn − 2 + dn − 1 + dn.
Hence, it is clearly optimal to assign the largest digit to d1, the second largest to d2, and so on. The order of the last two digits does not matter. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8703495860099792, "perplexity": 143.1666629557465}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780055684.76/warc/CC-MAIN-20210917151054-20210917181054-00139.warc.gz"} |
http://math.stackexchange.com/questions/778334/multiply-by-t-to-account-for-et-in-the-homogeneous-solution | # Multiply by $t$ to account for $e^t$ in the homogeneous solution
$$y'' - y' = e^t \sin{t}$$
So far I have $\lambda(\lambda - 1) = 0 \quad\implies\quad \lambda_1 = 0 \quad\land\quad \lambda_2 = 1 \qquad\implies y_h(t) = c_1+c_2e^t$
This line brings me my question: $$y_p(t) = t\left(c_1e^t\sin{t} + c_2e^t\cos{t} \right)$$
Why do I not have to multiply by $t$? I noticed WolframAlpha disagreed.
The general method I was under the impression of, was that if anything in the form of a solution I am currently looking for was in the homogeneous solution, I had to expand by $t$. Clearly that is not correct(?).. Would a kind soul be able to shed some light on this matter?
Thank you so much for any help. :-)
-
If you multiply by $e^{-t}$ at the beginning you then have an integrating factor and it becomes easier to solve. – user88595 May 2 '14 at 13:51
@user88595, I'm aware - but I need practice with undetermined coefficients. :-) – Erlend May 2 '14 at 14:20
$e^t$ is a solution to the homogeneous equation, but neither $e^t\cos(t)$ nor $e^t\sin(t)$ are solutions to the homogeneous equation, so it suffices to try a particular solution of the form $Ae^t\sin(t) + Be^t\cos(t)$. This might be easier to see using annihilators. Your equation is $(D^2-D)y = e^t\sin(t)$, and $e^t\sin(t)$ is annihilated by $(D-1)^2 + 1$, so $y$ satisfies $(D^2-D)((D-1)^2+1)y = 0$, whence $y = c_1 + c_2 e^t + Ae^t\cos(t) + Be^t\sin(t)$, and noting that the first two terms are the complementary solution, the last two must be the particular solution. – Nicholas Stull May 2 '14 at 14:34
You should make that an answer, because it did answer my question! Thank you, @NicholasStull – Erlend May 2 '14 at 14:50
Note that while $e^t$ is a solution to the homogeneous equation, neither $e^t\cos(t)$ nor $e^t\sin(t)$ are solutions to the homogeneous equation, so it suffices to try a particular solution of the form $$Ae^t\sin(t) + Be^t\cos(t)$$
If instead you had an equation such as $y'' - 2y' + 2y = e^t\sin(t)$, then because the complementary solution is $y_c = c_1 e^t\cos(t) + c_2 e^t\sin(t)$, here, you do need to multiply by $t$, so that you would have a trial solution for the particular solution of the form $$y_p = t(Ae^t\cos(t) + Be^t\sin(t))$$
One thing I might add is (as I said in my comment), this might be easier to see in terms of annihilators. Looking at your equation, which was (with $D$ the derivative) $$y'' - y' = (D^2-D)y = e^t\sin(t)$$ we notice that $D^2 - D = D(D-1)$ exactly annihilates all functions of the form $c_1 + c_2e^t$, which is our complementary solution, and then we notice that $(D-1)^2+1$ annihilates $e^t\sin(t)$ (this is an easy thing to check, and is just a computation). So our equation could be written $$((D-1)^2+1)D(D-1)y = 0$$ which has solution precisely $$y = c_1 + c_2 e^t + Ae^t\cos(t) + Be^t\sin(t)$$ and since the first two terms are exactly our complementary solution, the rest of the terms must be the trial solution which will give the particular solution after we use the method of undetermined coefficients.
Also, returning to the example above, in terms of annihilators, if we had the equation $$y'' - 2y' + 2y = e^t\sin(t)$$ then this can be rewritten (as above) in terms of annihilators as $$((D-1)^2+1)^2y = 0$$ which has solution of the form $$y = c_1 e^t \cos(t) + c_2 e^t\sin(t) + t\left( A e^t \cos(t) + B e^t\sin(t)\right)$$ and since the first two terms are exactly the complementary solution, the rest of the terms give the form of the trial solution we use the method of undetermined coefficients on to find the particular solution. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.948567807674408, "perplexity": 140.14471540763034}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430453759945.83/warc/CC-MAIN-20150501041559-00034-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://cs.stackexchange.com/questions/106680/how-do-we-prove-the-time-complexity-of-this-simple-problem-in-probabilistic-infe | # How do we prove the time complexity of this simple problem in probabilistic inference on a Bayesian network?
Suppose we have a simple Bayesian network with two rows of nodes: $$x_1, x_2, \ldots, x_n$$ and $$y_1, y_2, \ldots, y_n$$. Each node $$x_k$$ takes a state of either 0 or 1 with equal probability. Each node $$y_k$$ takes state 1 with probability $$p_{k,0}$$ if $$x_k$$ is state 0 and probability $$p_{k,1}$$ if $$x_k$$ is state 1.
Is exponential time required to compute the probability that all $$y_k$$ are 1? Please provide a proof either way.
• It's probably best if you first tried to solve it on your own. – Yuval Filmus Apr 8 at 21:28
• @SapereAude instead of deleting the question you could answer the question yourself, expanding on the order of the factoring and marginalizing. This way if anyone else comes across this problem or a similar one, they can learn from your example and about factoring orders in general. – ryan Apr 9 at 0:35
• @ryan A meaningful suggestion. – SapereAude Apr 9 at 4:14
First let's define some additional notation. Let $$X = (x_1, x_2, \ldots, x_n)$$ and $$Y = (y_1, y_2, \ldots, y_n)$$ -- that is, two $$n$$-tuples of our variables. Let $$\mathbf{1}$$ denote an $$n$$-tuple of 1s, $$(1, 1, \ldots, 1)$$. And let $$S_n$$ denote the set of all possible $$n$$-tuples of $$0$$ and $$1$$. In CS literature, the elements of $$S_n$$ are sometimes called "strings". If $$\sigma$$ is one such element, we write $$\sigma(k)$$ for its $$k$$th component.
In this new notation, the task is to compute $$\Pr(Y = \mathbf{1})$$. We can marginalize out the configuration of 0s and 1s on $$X$$ as follows: $$\Pr(Y = \mathbf{1}) = \sum_{\sigma \in S_n} \Pr(Y = \mathbf{1} | X = \sigma) \Pr (X = \sigma).$$
From the prompt we know that each $$x_k$$ is 0 or 1 with equal probability, so $$\Pr (X = \sigma) = 2^{-n}$$ for every $$\sigma \in S_n$$. Additionally, for any given $$\sigma \in S_n$$, we know that $$\Pr(Y = \mathbf{1} | X = \sigma) = \prod_{k=1}^n p_{k,\sigma(k)}$$, since the probabilities $$p_{k,\sigma(k)}$$ are independent. Combining these observations, we obtain $$\Pr(Y = \mathbf{1}) = 2^{-n} \sum_{\sigma \in S_n} \prod_{k=1}^n p_{k,\sigma(k)}.$$
Our next step is to show that we can factor the left-hand side of this equation as follows: $$\sum_{\sigma \in S_n} \prod_{k=1}^n p_{k,\sigma(k)} = \prod_{k=1}^n (p_{k,0}+p_{k,1}).$$ We prove this inductively. When $$n = 1$$, both sides are $$p_{1,0}+p_{1,1}$$, so the base case holds. For the inductive case, we assume the $$(n-1)$$th case and write the $$n$$th case as $$(p_{n,0}+p_{n,1}) \sum_{\sigma \in S_{n-1}} \prod_{k=1}^{n-1} p_{k,\sigma(k)} = (p_{n,0}+p_{n,1}) \prod_{k=1}^{n-1} (p_{k,0}+p_{k,1}).$$ Both sides simplify to those of the above identity, respectively, so the identity is proved.
This leaves us with the result that $$\Pr(Y = \mathbf{1}) = 2^{-n} \prod_{k=1}^n (p_{k,0}+p_{k,1}).$$ For a hash table representation, look-up times are $$O(1)$$ for each $$p_{k,0}$$ and $$p_{k,1}$$, so we can compute this product with a simple for-loop in a runtime of $$O(n)$$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 44, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9694162011146545, "perplexity": 149.47160479370885}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670743.44/warc/CC-MAIN-20191121074016-20191121102016-00299.warc.gz"} |
http://mathhelpforum.com/algebra/1522-algebra.html | # Math Help - Algebra
1. ## Algebra
x power 4 -(x-z)whole power 4.
Factorisation
2. Originally Posted by shaurya
x power 4 -(x-z)whole power 4.
Factorisation
I guess that you are expected to use the difference of two squares result
here:
$u^2-v^2=(u-v)(u+v)$.
Here you have:
$x^4-(x-z)^4$.
Now use $x^2$ for $u$ and $(x-z)^2$ for $v$, this will give you the first stage
of factorisation. The difference of squares can then be applied again to further.
factorise the expression.
RonL | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9126706719398499, "perplexity": 1672.7188285516002}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657137046.16/warc/CC-MAIN-20140914011217-00071-ip-10-234-18-248.ec2.internal.warc.gz"} |
http://www.reddit.com/r/cheatatmathhomework/comments/1jejbc/linear_algebra_eigenspaces/ | [–] 1 point2 points (1 child)
sorry, this has been archived and can no longer be voted on
First, a basis for a subspace isn't unique, so it's slightly misleading to say 'don't change' even in the case when Ak and A have identical eigenspaces.
Second, raising to a power can collapse eigenspaces into the same. For example, if A has eigenvalues 1 and -1, then in A2 those spaces are subspaces of the eigenspace for 1.
Third, not every matrix diagonalizes. When it doesn't then the eigenspace for lambdak in Ak can be larger for a different reason. Say A=[0,1;0,0] with a one dimension eigenspace for lambda=0, then A2=[0,0;0,0] has a two dimensional eigenspace for lambda2=0 that includes the eigenspace of A.
So without any extra information really all you have is that the eigenspace for lamdba in A is a subspace to the eigenspace for lambdak in Ak.
Adding a multiple of the identity is simpler. Every eigenspace stays the same, the eigenvalues simply shift. A+2I has eigenvalues 2 greater than A.
The number of eigenspaces is the number of distinct linear roots of the characteristic polynomial. In your example 3, with an eigenspace for 1,3,4.
In one direction Av=lambda*v with v!=0 implies (A-lambda*I)v=0 with v!=0 implies det(A-lambda*I)=0. In the other, det(A-lambda*I)=0 implies exists v!=0 with (A-lambda*I)v=0, etc.
[–][S] 0 points1 point (0 children)
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thanks!
[–] 0 points1 point (2 children)
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Yeah, basically.
It makes sense when you consider that if k is an eigenvalue, and x is a corresponding eigenvector then using the commutativity of constant multipliers:
AAx = Akx = Axk = kxk = kkx = k2 x
the number of eigenspaces in your example would be 3, one with dimension 1, one with dimension 2, and one with dimension 3. The dimensionality of each eigenspace is equal to the exponent.
This might make it clear:
If you have two eigenvectors x,y with the same eigenvalue k, then
Ax=kx
Ay=ky
Now doing some distributing,
A(x+y)=Ax+Ay=kx+ky=k(y+x)
So y+x is also an eigenvector, and more generally, any linear combination ay+bx is an eigenvector with the same eigenvalue, meaning x and y is a basis for the eigenspace, and this naturally expands to whatever multiplicity of eigenvalues you have.
[–][S] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
thanks!
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
That characteristic polynomial clearly splits, but you would need to know that A is diagonalizable in order to say the dimension of each eigenspace equals the multiplicity of the corresponding eigenvalue.
At best you know that 1 <= dim(eigenspace of lambda) <= m where m is the multiplicity of the eigenvalue seen from the characteristic polynomial. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8500614166259766, "perplexity": 1150.458765368914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928562.33/warc/CC-MAIN-20150521113208-00307-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://kullabs.com/class-9/science-9/machine | ## Machine
Subject: Science
### Lesson Info
• Notes 3
• Videos 10
• Exercises 65
• Practice Test 21
• Skill Level Medium
#### Overview
After completion of this lesson, student must be able to:
• Describe mechanical advantage, velocity ratio and efficiency of a simple machine (lever, pulley, wheel and axle and inclined plane).
• Solve numerical problems related to mechanical advantage, velocity ratio and efficiency of the simple machines mentioned above.
• Describe the law of moment in lever with an example.
#### Machine
Simple machine helps us by magnifying force, accelerating work and by changing the direction of force. This notes gives us the information about machine, its importance, mechanical advantage, velocity ratio. Learn More
#### Types of Simple Machine
A lever is a rigid bar may be straight or bent which is capable of rotating fixed point called fulcrum. A pulley is a metallic or wooden disc with a grooved rim. The rim rotates about a horizontal axis passing through its centre. This note gives us information about types of simple machine. Learn More
#### Moment
The law of moment states that “In the equilibrium condition of lever, the sum of the anticlockwise moment is equal to the sum of the clockwise moment”. This note gives us further information about moment of force. Learn More | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.806626558303833, "perplexity": 3319.1248571576994}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057119.85/warc/CC-MAIN-20210920221430-20210921011430-00425.warc.gz"} |
https://math.stackexchange.com/questions/775434/the-point-open-game-and-omega-covers | # The point-open game and $\omega$-covers
Let $X$ be a topological space. The point-open game $G_{po}(X)$ is defined as folows. It is played by two players ONE and TWO. In the n'th step $(n \in \omega)$, ONE choose a finite subset $F$ of $X$, and TWO selects an open $G_n$ in $X$, $F_n \subset G_n$. ONE wins if $\bigcup \{ G_n : n \in \omega \} = X$, otherwise TWO wins.
Also:
If $\langle A_n : n \in \omega \rangle$ is a sequence of subsets of a set $X$, $$\underline{Lim} A_n = \{ x \in X : \exists n_0 \in \omega \forall n \geq n_0, x \in A_n \}$$ If $\mathcal A$ is a family of subsets of a set $X$, then, $L(\mathcal A)$ denotes the smallest family of subsets of $X$ containing $\mathcal A$ and closed under $\underline{Lim}$.
I am trying to prove that (a)$\Rightarrow$(b) where:
(a) If $\mathcal I$ is an open $\omega$-cover of $X$, then, there is a sequence $G_n \in \mathcal I$, with $\underline{Lim} G_n = X$.
(b) If $\mathcal I$ is an open $\omega$-cover of $X$, then $X \in L(\mathcal I)$.
A family $\mathcal A$ of subsets of a set $A$ is said to be an $\omega$-cover of $X$, if for any finite subset $F$ of $X$,, there is an $A \in \mathcal A$ with $F \subset A$.
• A couple questions. (1) What does your question have to do with the point-open game? (Your question seems to only involve $\underline{\mathrm{Lim}}$, $L(\mathcal{I})$, and open $\omega$-covers.) (2) Isn't the implication trivial? If (a) holds, then given any open $\omega$-cover $\mathcal{I}$ of $X$ there is a sequence $\langle G_n \rangle_n$ in $\mathcal{I}$ such that $\underline{\mathrm{Lim}}_n G_n = X$. Since $G_n \in L(\mathcal{I})$ for each $n$ and $L(\mathcal{I})$ is closed under the $\underline{\mathrm{Lim}}$ operation, it must be that $X \in L(\mathcal{I})$. – user642796 Apr 30 '14 at 9:06
• Also, I think what you're calling the "point-open" game should be called the "finite-open" game; it would be the "point-open" game if the finite sets $F_n$ were required to be singletons. (Although the two games do seem to be more or less equivalent.) – bof Apr 30 '14 at 9:14
• The Claim I stated is a part of more general proposition which incudes also point-open game.. It is in the bottom of page 153 of this article: ac.els-cdn.com/0166864182900657/… Also, the definition for point open game is from that article. Anyhow, I see now, that it is trivial. I was missing the part of "Closed under Lim". Thank you both!! – topsi Apr 30 '14 at 10:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9389917254447937, "perplexity": 160.7000123644605}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999740.32/warc/CC-MAIN-20190624211359-20190624233359-00181.warc.gz"} |
http://swmath.org/software/15428 | hglm
R package hglm: Hierarchical Generalized Linear Models. Procedures for fitting hierarchical generalized linear models (HGLM). It can be used for linear mixed models and generalized linear mixed models with random effects for a variety of links and a variety of distributions for both the outcomes and the random effects. Fixed effects can also be fitted in the dispersion part of the mean model.
Keywords for this software
Anything in here will be replaced on browsers that support the canvas element
References in zbMATH (referenced in 6 articles , 1 standard article )
Showing results 1 to 6 of 6.
Sorted by year (citations) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.812217652797699, "perplexity": 1761.8636784543237}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364008.55/warc/CC-MAIN-20210302125936-20210302155936-00490.warc.gz"} |
https://math.libretexts.org/TextMaps/Analysis_TextMaps/Map%3A_Partial_Differential_Equations_(Miersemann)/3%3A_Classification/3.3.0%3A_Systems_of_First_Order | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 3.3: Systems of First Order
Consider the quasilinear system
\label{syst1}
\sum_{k=1}^nA^k(x,u)u_{u_k}+b(x,u)=0,
where $$A^k$$ are $$m\times m$$-matrices, sufficiently regular with respect to their arguments, and
$$u=\left(\begin{array}{c} u_1\\ \vdots\\u_m \end{array}\right),\ \ u_{x_k}=\left(\begin{array}{c} u_{1,x_k}\\ \vdots\\u_{m,x_k} \end{array}\right),\ \ b=\left(\begin{array}{c} b_1\\ \vdots\\b_m \end{array}\right).$$
We ask the same question as above: can we calculate all derivatives of $$u$$ in a neighborhood of a given hypersurface $$\mathcal{S}$$ in $$\mathbb{R}$$ defined by $$\chi(x)=0$$, $$\nabla\chi\not=0$$, provided $$u(x)$$ is given on $$\mathcal{S}$$?
For an answer we map $$\mathcal{S}$$ onto a flat surface $$\mathcal{S}_0$$ by using the mapping $$\lambda=\lambda(x)$$ of Section 3.1 and write equation (\ref{syst1}) in new coordinates. Set $$v(\lambda)=u(x(\lambda))$$, then
$$\sum_{k=1}^nA^k(x,u)\chi_{x_k}v_{\lambda_n}=\mbox{terms known on}\ \mathcal{S}_0.$$
We can solve this system with respect to $$v_{\lambda_n}$$, provided that
$$\det\left(\sum_{k=1}^nA^k(x,u)\chi_{x_k}\right)\not=0$$
on $$\mathcal{S}$$.
Definition. Equation
$$\det\left(\sum_{k=1}^nA^k(x,u)\chi_{x_k}\right)=0$$
is called characteristic equation associated to equation (\ref{syst1}) and a surface $${\mathcal{S}}$$: $$\chi(x)=0$$, defined by a solution $$\chi$$, $$\nabla\chi\not=0$$, of this characteristic equation is said to be characteristic surface.
Set
$$C(x,u,\zeta)=\det\left(\sum_{k=1}^nA^k(x,u)\zeta_k\right)$$
for $$\zeta_k\in\mathbb{R}$$.
Definition.
1. The system (\ref{syst1}) is hyperbolic at $$(x,u(x))$$ if there is a regular linear mapping $$\zeta=Q\eta$$, where $$\eta=(\eta_1,\ldots,\eta_{n-1},\kappa)$$, such that there exists $$m$$ {\it real} roots $$\kappa_k=\kappa_k(x,u(x),\eta_1,\ldots,\eta_{n-1})$$, $$k=1,\ldots,m$$, of $$D(x,u(x),\eta_1,\ldots,\eta_{n-1},\kappa)=0$$ for all $$(\eta_1,\ldots,\eta_{n-1})$$, where $$D(x,u(x),\eta_1,\ldots,\eta_{n-1},\kappa)=C(x,u(x),x,Q\eta).$$
2. System (\ref{syst1}) is parabolic if there exists a regular linear mapping $$\zeta=Q\eta$$ such that $$D$$ is independent of $$\kappa$$, that is, $$D$$ depends on less than $$n$$ parameters.
3. System (\ref{syst1}) is elliptic if $$C(x,u,\zeta)=0$$ only if $$\zeta=0$$.
Remark. In the elliptic case all derivatives of the solution can be calculated from the given data and the given equation.
### Contributors
• Integrated by Justin Marshall. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.993510901927948, "perplexity": 305.4818074682495}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818694719.89/warc/CC-MAIN-20170926032806-20170926052806-00709.warc.gz"} |
https://chitowntutoring.com/secant-sec-cosecant-csc-and-cotangent-cot/ | Secant (sec), Cosecant (csc), and Cotangent (cot)
This section will show you how the functions of trigonometry like cotangent, secant, and cosecant are related to the other trigonometric functions of sine, cosine, and tangent.
Trigonometric functions
Functions of trigonometric, angles’ functions, are common in the real world and in mathematics. The sound which comes out of the speakers of computers is generated by the waves of trigonometric, or the sound waves transmitted out of the speakers, which are seen in the sine waveform.
At this point, you are already aware of the sine, cosine and tangent functions. These are just three basic functions of trigonometry. All other trigonometric functions are totally based on these three functions, which you will see.
Do you know the functions of sine, cosine, and tangent? If not, then recall it with their three sides of opposite, hypotenuse and adjacent sides. Look at the right-angle triangle, do you remember how we defined those three functions?
The sine function is defined as the opposite side/hypotenuse and the cosine is defined as the adjacent side/ hypotenuse, and the tangent is the ratio of the opposite/adjacent side. Now we have reviewed these three functions of trigonometric, let us see the other three functions of trigonometry of cotangent, secant, and cosecant.
Cotangent
First of all, we have the function of cotangent, which is defined as the reciprocal of the inverse of the function of a tangent. In mathematics, we write it as cot θ = 1 / tan θ. All our trigonometric functions are shortened to three letters when written the function while evaluating.
Now, because the cotangent is the reciprocal of the tangent function, we also define it as the inverse function of the tangent. If the tangent is the opposite/adjacent side, then the cotangent function is the reciprocal of it – adjacent/opposite. We can write all such information like this:
Cotangent function is evaluated as:
Cot θ = 1 / tan θ = adjacent side / opposite side
Secant function
Now we have the function of secant. We can define it as the reciprocal of the cosine function. The three letters, or say a short form of secant is sec. Do you know how cosine is defined? Yes, the cosine is written as – adjacent side/hypotenuse. Then the secant is the reciprocal of the cosine function which is the reciprocal of the cosine function –
Sec θ = 1 / cos θ = hypotenuse / adjacent side.
Cosecant function
Now we have the function of the cosecant. This function is the reciprocal of the sine function which has the short form cosec or CSC. As this is the reciprocal of the sine function and the sine function is defined as the opposite side/hypotenuse, we can alternately define cosecant function as:
Cosecant θ = 1 / sin θ = hypotenuse / opposite side
Thus, here you have a clear idea of the concept of all the three trigonometric functions. Examine the following diagram:
Example 1: What is the csc of x?
A is the opposite side, B is the adjacent side, and C is the hypotenuse.
Since csc = hypotenuse /opposite,
csc x = C/A
Example 2: What is the sec of x?
Again, A is the opposite side, B is the adjacent side, and C is the hypotenuse.
Since sec = hypotenuse / adjacent,
sec x = C/B
Example 3: What is the cot of x?
Since cot = adjacent / opposite,
cot x = B/A | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9823336601257324, "perplexity": 628.026659571416}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703499999.6/warc/CC-MAIN-20210116014637-20210116044637-00781.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-1-equations-and-inequalities-1-3-solve-linear-equations-1-3-exercises-skill-practice-page-21/18 | ## Algebra 2 (1st Edition)
$-7/2$
Subtracting 3 from each side gives, $8/7*d=-4$ Then multiplying both sides by $7/8$ gives $d=-7/2$ Checking gives: $3-4=-1$ which is true. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9376088976860046, "perplexity": 520.0596527006745}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144708.87/warc/CC-MAIN-20200220070221-20200220100221-00276.warc.gz"} |
http://math.stackexchange.com/questions/234857/help-here-derivative-question | # Help here derivative question?
Prove that for every $x$, we have $\Delta[f(x)+g(x)]=\Delta f(x)+ \Delta g(x)$. Thanks in advance.
-
What is $\Delta(f(x))$ for you? – Sigur Nov 11 '12 at 13:43
Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Julian Kuelshammer Nov 11 '12 at 13:44
The laplacian $\Delta$ is a linear differential operator. Don't you understand my answer? Well, try to explain what you are asking, please. – Siminore Nov 11 '12 at 14:26
Since Notyathing said "derivative question", it's likely $\Delta$ is the Laplacian (or perhaps some other differential operator). It does not represent general differences as in your answer. – Mark S. Nov 11 '12 at 15:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9024894833564758, "perplexity": 667.738299000597}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510275111.45/warc/CC-MAIN-20140728011755-00183-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/118534/what-can-be-said-about-zeros-of-zetas-sharing-the-largest-real-part?answertab=active | What can be said about zeros of $\zeta(s)$ sharing the largest real part?
Specifically, if $\rho$ is such that $\zeta(\rho)=0$ and $\max_{\rho}\Re(\rho)= \Theta$, can anything interesting be said about the number/distribution of zeros on the vertical line $\sigma=\Theta$?
Clearly this question is almost as hypothetical as they get, so I welcome conditional answers (though not on RH please), consequences of the Bohr-Landau theorem, consequences of the known behavior of $\zeta(s)$ in the critical strip, etc.
Maybe you know something along the lines of If there are finitely/ infinitely many, then...''?
I am also interested in why your answer may simply be No.''
-
Is there any reason (short of RH) to believe that the maximum exists at all? In fact, since there are no zeros with Re=1, this would imply that all zeros have $\Re(\rho)\le1-\epsilon$ for some constant $\epsilon>0$, and this is an open problem. – Emil Jeřábek Jan 10 '13 at 13:52
Good point. I think so: roughly speaking, either there is a maximum (whose value is not known), or there is at least one rogue zero $r$ with $1-\epsilon<\Re(r)<1$, for every $\epsilon > 0$. But then, owing to the known zero-free region, $\Im(r)>T_{\epsilon}\rightarrow\infty$ as $\epsilon\rightarrow 0$. – Kevin Smith Jan 10 '13 at 15:53
That’s not necessarily true either. If there is no maximum, i.e., the supremum is not attained, there is no telling whether the supremum is 1 or smaller. Of course, if the maximum does not exist and $\rho_n$ is any sequence of zeros whose real parts tend to the supremum, then $\lim_n|\Im(\rho_n)|=\infty$, since the roots of any meromorphic function are isolated; this does not need any sophisticated information on zero-free regions. – Emil Jeřábek Jan 10 '13 at 16:08
Obviously - clearly I had implicitly assumed you were arguing about zeros near the line $\sigma=1$, and whether I was discussing an open problem (which is off-topic). Isolation of zeros is not strong enough to prove the statement I made in the above comment. Your original objection is valid - indeed I have not given a good reason to believe a maximum does exist if the supremum is $\leq 1-C$, but I am remaining on topic.. Clearly I need to be more precise here. – Kevin Smith Jan 10 '13 at 17:38
This is getting off topic, but I must say the statement you made above is not very clear to me. Is $T_\epsilon$ some specific function? If not, the following lemma follows easily from the fact that roots are isolated. Let $f$ be a meromorphic function such that $s:=\sup\{\Re(\rho):f(\rho)=0\}$ is finite, and not attained. Then there exists an unbounded nonincreasing function $T\colon(0,+\infty)\to[0,+\infty)$ with the property that $|\Im(\rho)|\ge T(\epsilon)$ whenever $\rho$ is such that $f(\rho)=0$ and $\Re(\rho)\ge s-\epsilon$. – Emil Jeřábek Jan 11 '13 at 11:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.91813725233078, "perplexity": 227.6772538858993}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701150206.8/warc/CC-MAIN-20160205193910-00211-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://math.stackexchange.com/users/23599/deftfyodor?tab=activity | # deftfyodor
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Aug7 comment Is there any function that never gives an answer other than 0/0 when applying L'Hôpital's rule? I provided the answer before the clarification. Others have now given better examples. Aug5 answered Is there any function that never gives an answer other than 0/0 when applying L'Hôpital's rule? Aug5 comment Adjustable Sigmoid Curve (S-Curve) from (0,0) to (1,1) I'm not sure if it has the analytic properties that you want, but the CDF of a beta distribution is pretty versatile and meets your explicit requirements. Jul16 answered Generating Bifurcation Animations Jul11 awarded Scholar Jul11 accepted Graph of continuous function from compact space is compact. Jul10 awarded Student Jul10 comment Graph of continuous function from compact space is compact. All right, I cannot say I'm terribly satisfied (I've been trying to find some way to make this work for almost a day now), but that does ease my mind. Thank you. Jul10 asked Graph of continuous function from compact space is compact. May4 comment What explains this bizarre behavior? It is somewhat reminiscent of the logistic map, which is a typical example of chaos in fairly simple systems. May2 awarded Commentator May2 comment How did Euler and Bernoulli prove this limit? There is a pretty nice proof for why the above expression is equal to $\sum \frac{1}{n!}$, but that is really just a conversion between two definitions of $e$. Apr23 comment genetic algorithm binary encoding I think that the GA should converge to giving low fitness values to elements with the wrong sign bit. As you suggested, the problem is somewhat less with two's compliment, as flipping a bit leads to a change of at most $2^{n-1}$ (n is the number of bits), whereas using the sign bit would allow the value to change by $2^n$ if the sign bit was flipped. Apr23 awarded Yearling Apr23 answered genetic algorithm binary encoding Apr23 answered Please tell me what I am doing wrong for this multivariable Calculus Problem Apr3 comment Stupid question - How do I calculate $\Phi(1.5)$? If by $\Phi$ is all you have, you mean that you are limited in your use of technology or tables, you could calculate the Taylor Expansion to a few terms and integrate that- however that sounds like a miserable waste of time. Mar22 comment A question comparing $\pi^e$ to $e^\pi$ He's working from a college algebra textbook, there is no reason to suspect that he even knows what differentiation is. Mar22 awarded Critic Feb8 awarded Editor | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8511759638786316, "perplexity": 734.22127697267}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345758904/warc/CC-MAIN-20131218054918-00096-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://science.sciencemag.org/content/311/5768/1754 | Report
# Measurements of Time-Variable Gravity Show Mass Loss in Antarctica
See allHide authors and affiliations
Science 24 Mar 2006:
Vol. 311, Issue 5768, pp. 1754-1756
DOI: 10.1126/science.1123785
## Abstract
Using measurements of time-variable gravity from the Gravity Recovery and Climate Experiment satellites, we determined mass variations of the Antarctic ice sheet during 2002–2005. We found that the mass of the ice sheet decreased significantly, at a rate of 152 ± 80 cubic kilometers of ice per year, which is equivalent to 0.4 ± 0.2 millimeters of global sea-level rise per year. Most of this mass loss came from the West Antarctic Ice Sheet.
View Full Text | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9642820954322815, "perplexity": 2875.263544303608}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934804965.9/warc/CC-MAIN-20171118132741-20171118152741-00463.warc.gz"} |
https://proofwiki.org/wiki/Open_Real_Intervals_are_Homeomorphic | # Open Real Intervals are Homeomorphic
## Theorem
Consider the real numbers $\R$ as a metric space under the Euclidean metric.
Let $I_1 := \left({a \,.\,.\, b}\right)$ and $I_2 := \left({c \,.\,.\, d}\right)$ be non-empty open real intervals.
Then $I_1$ and $I_2$ are homeomorphic.
## Proof
By definition of open real interval, for $I_1$ and $I_2$ to be non-empty it must be the case that $a < b$ and $c < d$.
In particular it is noted that $a \ne b$ and $c \ne d$.
Thus $a - b \ne 0$ and $c - d \ne 0$.
Consider the real function $f: I_1 \to I_2$ defined as:
$\forall x \in I_1: f \left({x}\right) = c + \dfrac {\left({d - c}\right) \left({x - a}\right)} {b - a}$
Then after some algebra:
$\forall x \in I_2: f^{-1} \left({x}\right) = a + \dfrac {\left({b - a}\right) \left({x - c}\right)} {d - c}$
Both of these are defined as $a - b \ne 0$ and $c - d \ne 0$.
By the Combination Theorem for Continuous Functions, both $f$ and $f^{-1}$ are continuous on the open real intervals on which they are defined.
Hence the result by definition of homeomorphic.
$\blacksquare$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9542945623397827, "perplexity": 150.55084197327997}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737204.32/warc/CC-MAIN-20200807143225-20200807173225-00444.warc.gz"} |
https://math.libretexts.org/Bookshelves/Linear_Algebra/Book%3A_Linear_Algebra_(Schilling%2C_Nachtergaele_and_Lankham)/13%3A_Appendices/13.02%3A_Summary_of_Algebraic_Structures | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 13.2: Summary of Algebraic Structures
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
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Loosely speaking, an algebraic structure is any set upon which "arithmetic-like'' operations have been defined. The importance of such structures in abstract mathematics cannot be overstated. By recognized a given set $$S$$ as an instance of a well-known algebraic structure, every result that is known about that abstract algebraic structure is then automatically also known to hold for $$S$$. This utility is, in large part, the main motivation behind abstraction.
Before reviewing the algebraic structures that are most important to the study of Linear Algebra, we first carefully define what it means for an operation to be "arithmetic-like''.
## C.1 Binary operations and scaling operations
When discussing an arbitrary nonempty set $$S$$, you should never assume that $$S$$ has any type of "structure'' (algebraic or otherwise) unless the context suggests differently. Put another way, the elements in $$S$$ can only every really be related to each other in a subjective manner. E.g., if we take $$S = \{\text{Alice},\,\text{Bob},\,\text{Carol}\}$$, then there is nothing intrinsic in the definition of $$S$$ that suggests how these names should objectively be related to one another.
If, on the other hand, we take $$S = \mathbb{R}$$, then you have no doubt been conditioned to expect that a great deal of "structure'' already exists within $$S$$. E.g., given any two real numbers $$r_{1}, r_{2} \in \mathbb{R}$$, one can form the sum $$r_{1} + r_{2}$$, the difference $$r_{1} - r_{2}$$, the product $$r_{1}r_{2}$$, the quotient $$r_{1} / r_{2}$$ (assuming $$r_{2} \neq 0$$), the maximum $$\max\{r_{1}, r_{2}\}$$, the minimum $$\min\{r_{1}, r_{2}\}$$, the average $$(r_{1} + r_{2})/2$$, and so on. Each of these operations follows the same pattern: take two real numbers and "combine'' (or "compare'') them in order to form a new real number.
Moreover, each of these operations imposes a sense of "structure'' within $$\mathbb{R}$$ by relating real numbers to each other. We can abstract this to an arbitrary nonempty set as follows:
Definition C.1.1. A binary operation on a nonempty set $$S$$ is any function that has as its domain $$S \times S$$ and as its codomain $$S$$.
In other words, a binary operation on $$S$$ is any rule $$f : S \times S \to S$$ that assigns exactly one element $$f(s_{1}, s_{2}) \in S$$ to each pair of elements $$s_{1}, s_{2} \in S$$. We illustrate this definition in the following examples.
Example C.1.2.
1. Addition, subtraction, and multiplication are all examples of familiar binary operations on $$\mathbb{R}$$. Formally, one would denote these by something like
$+ : \mathbb{R} \times \mathbb{R} \to \mathbb{R}, \ - : \mathbb{R} \times \mathbb{R} \to \mathbb{R}, \ \text{and} \ * : \mathbb{R} \times \mathbb{R} \to \mathbb{R}, \ \text{respectively}.$
Then, given two real numbers $$r_{1}, r_{2} \in \mathbb{R}$$, we would denote their sum by $$+(r_{1}, r_{2})$$, their difference by $$-(r_{1}, r_{2})$$, and their product by $$*(r_{1}, r_{2})$$. (E.g., $$+(17, 32) = 49$$, $$-(17, 32) = -15$$, and $$*(17, 32) = 544$$.) However, this level of notational formality can be rather inconvenient, and so we often resort to writing $$+(r_{1}, r_{2})$$ as the more familiar expression $$r_{1} + r_{2}$$, $$-(r_{1}, r_{2})$$ as $$r_{1} - r_{2}$$, and $$*(r_{1}, r_{2})$$ as either $$r_{1} * r_{2}$$ or $$r_{1}r_{2}$$.
2. The division function $$\div : \mathbb{R} \times \left( \mathbb{R}\setminus\{0\} \right) \to \mathbb{R}$$ is not a binary operation on $$\mathbb{R}$$ since it does not have the proper domain. However, division is a binary operation on $$\mathbb{R}\setminus\{0\}$$.
3. Other binary operations on $$\mathbb{R}$$ include the maximum function $$\max:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$$, the minimum function $$\min:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$$, and the average function $$(\cdot + \cdot)/2:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$$.
4. An example of a binary operation $$f$$ on the set $$S = \{\text{Alice},\,\text{Bob},\,\text{Carol}\}$$ is given by
$f(s_{1}, s_{2}) = \begin{cases} s_{1} & {\rm{if~}} s_{1} {\rm{~alphabetically ~precedes~}} s_{2}, \\ \text{Bob} & \text{otherwise}. \end{cases}$
This is because the only requirement for a binary operation is that exactly one element of $$S$$ is assigned to every ordered pair of elements $$(s_{1}, s_{2}) \in S \times S$$.
Even though one could define any number of binary operations upon a given nonempty set, we are generally only interested in operations that satisfy additional "arithmetic-like'' conditions. In other words, the most interesting binary operations are those that, in some sense, abstract the salient properties of common binary operations like addition and multiplication on $$\mathbb{R}$$. We make this precise with the definition of a so-called "group'' in Section C.2.
At the same time, though, binary operations can only be used to impose "structure'' within a set. In many settings, it is equally useful to additional impose "structure'' upon a set. Specifically, one can define relationships between elements in an arbitrary set as follows:
Definition C.1.3. A scaling operation (a.k.a. external binary operation) on a nonempty set $$S$$ is any function that has as its domain $$\mathbb{F} \times S$$ and as its codomain $$S$$, where $$\mathbb{F}$$ denotes an arbitrary field. (As usual, you should just think of $$\mathbb{F}$$ as being either $$\mathbb{R}$$ or $$\mathbb{C}$$).
In other words, a scaling operation on $$S$$ is any rule $$f : \mathbb{F} \times S \to S$$ that assigns exactly one element $$f(\alpha, s) \in S$$ to each pair of elements $$\alpha \in \mathbb{F}$$ and $$s \in S$$. This abstracts the concept of "scaling'' an object in $$S$$ without changing what "type'' of object it already is. As such, $$f(\alpha, s)$$ is often written simply as $$\alpha s$$. We illustrate this definition in the following examples.
Example C.1.4.
1. Scalar multiplication of $$n$$-tuples in $$\mathbb{R}^{n}$$ is probably the most familiar scaling operation to you. Formally, scalar multiplication on $$\mathbb{R}^{n}$$ is defined as the following function:
$\left( \alpha, (x_{1}, \ldots, x_{n}) \right) \longmapsto \alpha (x_{1}, \ldots, x_{n}) = (\alpha x_{1}, \ldots, \alpha x_{n}), \ \forall \, \alpha \in \mathbb{R}, \ \forall \, (x_{1}, \ldots, x_{n}) \in \mathbb{R}^n.$
In other words, given any $$\alpha \in \mathbb{R}$$ and any $$n$$-tuple $$(x_{1}, \ldots,x_{n}) \in \mathbb{R}^n$$, their scalar multiplication results in a new $$n$$-tupledenoted by $$\alpha (x_{1}, \ldots, x_{n})$$. This new $$n$$-tuple is virtually identical to the original, each component having just been "rescaled'' by $$\alpha$$.
2. Scalar multiplication of continuous functions is another familiar scaling operation. Given any real number $$\alpha \in \mathbb{R}$$ and any function $$f \in \mathcal{C}(\mathbb{R})$$, their scalar multiplication results in a new function that is denoted by $$\alpha f$$, where $$\alpha f$$ is defined by the rule $(\alpha f)(r) = \alpha (f(r)), \forall \, r \in \mathbb{R}.$ In other words, this new continuous function $$\alpha f \in \mathcal{C}(\mathbb{R})$$ is virtually identical to the original function $$f$$; it just rescales'' the image of each $$r \in \mathbb{R}$$ under $$f$$ by $$\alpha$$.
3. The division function $$\div : \mathbb{R} \times \left( \mathbb{R}\setminus\{0\} \right) \to \mathbb{R}$$ is a scaling operation on $$\mathbb{R}\setminus\{0\}$$. In particular, given two real number $$r_{1}, r_{2} \in \mathbb{R}$$ and any non-zero real number $$s \in \mathbb{R}\setminus\{0\}$$, we have that $$\div(r_{1}, s) = r_{1}(1/s)$$ and $$\div(r_{2}, s) = r_{2}(1/s)$$, and so $$\div(r_{1}, s)$$ and $$\div(r_{2}, s)$$ can be viewed as different scalings'' of the multiplicative inverse $$1/s$$ of $$s$$.
This is actually a special case of the previous example. In particular, we can define a function $$f \in \mathcal{C}(\mathbb{R}\setminus\{0\})$$ by $$f(s) = 1/s$$, for each $$s \in \mathbb{R}\setminus\{0\}$$. Then, given any two real numbers $$r_{1}, r_{2} \in \mathbb{R}$$, the functions $$r_{1}f$$ and $$r_{2}f$$ can be defined by
$r_{1}f(\cdot) = \div(r_{1}, \cdot) \ \ \text{and} \ \ r_{2}f(\cdot) = \div(r_{2}, \cdot), \ \text{respectively}.$
4. Strictly speaking, there is nothing in thedefinition that precludes $$S$$ from equaling $$\mathbb{F}$$. Consequently, addition, subtraction,and multiplication can all be seen as examples ofscaling operations on $$\mathbb{R}$$.
As with binary operations, it is easy to define any number of scaling operations upon a given nonempty set $$S$$. However, we are generally only interested in operations that are essentially like scalar multiplication on $$\mathbb{R}^{n}$$, and it is also quite common to additionally impose conditions for how scaling operations should interact with any binary operations that might also be defined upon $$S$$. We make this precise when we present an alternate formulation of the definition for a vector space in Section C.2.
Put another way, the definitions for binary operation and scaling operation are not particularly useful when taken as is. Since these operations are allowed to be any functions having the proper domains, there is no immediate sense of meaningful abstraction. Instead, binary and scaling operations become useful when additionally conditions are placed upon them so that they can be used to abstract "arithmetic-like'' properties. In other words, we are usually only interested in operations that abstract the salient properties of familiar operations for combining things like numbers, $$n$$-tuples, and functions.
## C.2 Groups, fields, and vector spaces
We begin this section with the following definition, which is unequivocably one of the most fundamental and ubiquitous notions in all of abstract mathematics.
Definition C.2.1.
Let $$G$$ be a nonempty set, and let $$*$$ be a binary operation on $$G$$. (In other words, $$*:G \times G \to G$$ is a function with $$*(a, b)$$ denoted by $$a*b$$, for each $$a, b \in G$$.) Then $$G$$ is said to form a group under $$*$$ if the following three conditions are satisfied:
1. (associativity) Given any three elements $$a, b, c \in G$$, $(a * b) * c = a * (b * c).$
2. (existence of an identity element) There is an element $$e \in G$$ such that, given any element $$a \in G$$, $a * e = e * a = a.$
3. (existence of inverse elements) Given any element $$a \in G$$, there is an element $$b \in G$$ such that $a * b = b * a = e.$
You should recognize these three conditions (which are sometimes collectively referred to as the group axioms) as properties that are satisfied by the operation of addition on $$\mathbb{R}$$. This is not an accident. In particular, given real numbers $$\alpha, \beta \in \mathbb{R}$$, the group axioms form the minimal set of assumptions needed in order to solve the equation $$x + \alpha = \beta$$ for the variable $$x$$, and it is in this sense that the group axioms are an abstraction of the most fundamental properties of addition of real numbers.
A similar remark holds regarding multiplication on $$\mathbb{R}\setminus\{0\}$$ and solving the equation $$\alpha x = \beta$$ for the variable $$x$$. Note, however, that this cannot be extended to all of $$\mathbb{R}$$.
Because the group axioms are so general, they are particularly useful in building more complicated algebraic structures. This is done by adding any number of additional axioms, the most fundamental of which is as follows.
Definition C.2.2. Let $$G$$ be a group under binary operation $$*$$. Then $$G$$ is called an abelian group (a.k.a. commutative group) if, given any two elements $$a, b \in G$$, $$a * b = b * a$$.
Examples of groups are everywhere in abstract mathematics. We now give some of the more important examples that occur in Linear Algebra. Please note, though, that these examples are primarily aimed at motivating the definitions of more complicated algebraic structures. (In general, groups can be much "stranger'' than those below.)
Example C.2.3.
1. If $$G \in \left\{ \mathbb{Z}, \,\mathbb{Q}, \,\mathbb{R}, \,\mathbb{C} \right\}$$, then $$G$$ forms an abelian group under the usual definition of addition.
Note, though, that the set $$\mathbb{Z}_{+}$$ of positive integers does not form a group under addition since, e.g., it does not contain an additive identity element.
1. Similarly, if $$G \in \left\{ \,\mathbb{Q}\setminus\{0\}, \,\mathbb{R}\setminus\{0\}, \,\mathbb{C}\setminus\{0\} \right\}$$, then $$G$$ forms an abelian group under the usual definition of multiplication.
Note, though, that $$\mathbb{Z}\setminus\{0\}$$ does not form a group under multiplication since only $$\pm 1$$ have multiplicative inverses.
1. If $$m, n \in \mathbb{Z}_{+}$$ are positive integers and $$\mathbb{F}$$ denotes either $$\mathbb{R}$$ or $$\mathbb{C}$$, then the set $$\mathbb{F}^{m \times n}$$ of all $$m \times n$$ matrices forms an abelian group under matrix addition.
Note, though, that $$\mathbb{F}^{m \times n}$$ does not form a group under matrix multiplication unless $$m = n = 1$$, in which case $$\mathbb{F}^{1 \times 1} = \mathbb{F}$$.
1. Similarly, if $$n \in \mathbb{Z}_{+}$$ is a positive integer and $$\mathbb{F}$$ denotes either $$\mathbb{R}$$ or $$\mathbb{C}$$, then the set $$GL(n, \mathbb{F})$$ of invertible $$n \times n$$ matrices forms a group under matrix multiplications. This group, which is often called the general linear group, is non-abelian when $$n \geq 2$$.
Note, though, that $$GL(n, \mathbb{F})$$ does not form agroup under matrix addition for any choice of $$n$$ since, e.g., the zero matrix $$0_{n \times n} \notin GL(n, \mathbb{F})$$.
In the above examples, you should notice two things. First of all, it is important to specify the operation under which a set might or might not be a group. Second, and perhaps more importantly, all but one example is an abelian group. Most of the important sets in Linear Algebra possess some type of algebraic structure, and abelian groups are the principal building block of virtually every one of these algebraic structures. In particular, fields and vector spaces (as defined below) and rings and algebra (as defined in Section C.3) can all be described as "abelian groups plus additional structure''.
Given an abelian group $$G$$, adding "additional structure'' amounts to imposing one or more additional operation on $$G$$ such that each new operations is "compatible'' with the preexisting binary operation on $$G$$. As our first example of this, we add another binary operation to $$G$$ in order to obtain the definition of a field:
Definition C.2.4. Let $$F$$ be a nonempty set, and let $$+$$ and $$*$$ be binary operations on $$F$$. Then $$F$$ forms a field under $$+$$ and $$*$$ if the following three conditions are satisfied:
1. $$F$$ forms an abelian group under $$+$$.
2. Denoting the identity element for $$+$$ by $$0$$, $$F\setminus\{0\}$$ forms an abelian group under $$*$$.
3. ($$*$$ distributes over $$+$$) Given any three elements $$a, b, c \in F$$, $a * (b + c) = a * b + a * c.$
You should recognize these three conditions (which are sometimes collectively referred to as the field axioms) as properties that are satisfied when the operations of addition and multiplication are taken together on $$\mathbb{R}$$. This is not an accident. As with the group axioms, the field axioms form the minimal set of assumptions needed in order to abstract fundamental properties of these familiar arithmetic operations. Specifically, the field axioms guarantee that, given any field $$F$$, three conditions are always satisfied:
1. Given any $$a, b \in F$$, the equation $$x + a = b$$ can be solved for the variable $$x$$.
2. Given any $$a \in F\setminus\{0\}$$ and $$b \in F$$, the equation $$a * x = b$$ can be solved for $$x$$.
3. The binary operation $$*$$ (which is like multiplication on $$\mathbb{R}$$) can be distributed over (i.e., is "compatible'' with) the binary operation $$+$$ (which is like addition on $$\mathbb{R}$$).
Example C.2.5.
It should be clear that, if $$F \in \left\{\mathbb{Q}, \,\mathbb{R}, \,\mathbb{C} \right\}$$, then $$F$$ forms a field under the usual definitions of addition and multiplication.
Note, though, that the set $$\mathbb{Z}$$ of integers does not form a field under these operations since $$\mathbb{Z} \setminus \{0\}$$ fails to form a group under multiplication. Similarly, none of the other sets from Example C.2.3 can be made into a field.
In some sense $$\mathbb{Q}$$, $$\mathbb{R}$$, and $$\mathbb{C}$$ are the only easily describable fields. While there are many other interesting and useful examples of fields, none of them can be described using entirely familiar sets and operations. This is because the field axioms are extremely specific in describing algebraic structure. As we will see in the next section, though, we can build a much more general algebraic structure called a "ring'' by still requiring that $$F$$ form an abelian group under $$+$$ but simultaneously relaxing the requirement that $$F$$ simultaneously form an abelian group under $$*$$.
For now, though, we close this section by taking a completely different point of view. Rather than place an additional (and multiplication-like) binary operation on an abelian group, we instead impose a special type of scaling operation called scalar multiplication. In essence, scalar multiplication imparts useful algebraic structure on an arbitrary nonempty set $$S$$ by indirectly imposing the algebraic structure of $$\mathbb{F}$$ as an abelian group under multiplication. (Recall that $$\mathbb{F}$$ can be replaced with either $$\mathbb{R}$$ or $$\mathbb{C}$$.)
Definition C.2.6. Let $$S$$ be a nonempty set, and let $$*$$ be a scaling operation on $$S$$. (In other words, $$* : \mathbb{F} \times S \to S$$ is a function with $$*(\alpha, s)$$ denoted by $$\alpha*s$$ or even just $$\alpha s$$, for every $$\alpha \in \mathbb{F}$$ and $$s \in S$$.) Then $$*$$ is called scalar multiplication if it satisfies the following two conditions:
1. (existence of a multiplicative identity element for $$*$$) Denote by $$1$$ the multiplicative identity element for $$\mathbb{F}$$. Then, given any $$s \in S$$, $$1 * s = s$$.
2. (multiplication in $$\mathbb{F}$$ is quasi-associative with respect to $$*$$) Given any $$\alpha, \beta \in \mathbb{F}$$ and any $$s \in S$$, $(\alpha \beta) * s = \alpha * (\beta * s).$
Note that we choose to have the multiplicative part of $$\mathbb{F}$$ "act'' upon $$S$$ because we are abstracting scalar multiplication as it is intuitively defined in Example C.1.4 on both $$\mathbb{R}^{n}$$ and $$\mathcal{C}(\mathbb{R})$$. This is because, by also requiring a "compatible'' additive structure (called vector addition), we obtain the following alternate formulation for the definition of a vector space.
Definition C.2.7.
Let $$V$$ be an abelian group under the binary operation $$+$$, and let $$*$$ be a scalar multiplication operation on $$V$$ with respect to $$\mathbb{F}$$. Then $$V$$ forms a vector space over $$\mathbb{F}$$ with respect to $$+$$ and $$*$$ if the following two conditions are satisfied:
1. ($$*$$ distributes over $$+$$) Given any $$\alpha \in \mathbb{F}$$ and any $$u, v \in V$$, $\alpha * (u + v) = \alpha * u + \alpha * v.$
2. ($$*$$ distributes over addition in $$\mathbb{F}$$) Given any $$\alpha, \beta \in \mathbb{F}$$ and any $$v \in V$$, $(\alpha + \beta) * v = \alpha * v + \beta * v.$
## C.3 Rings and algebras
In this section, we briefly mention two other common algebraic structures. Specifically, we first "relax'' the definition of a field in order to define a ring, and we then combine the definitions of ring and vector space in order to define an algebra. In some sense, groups, rings, and fields are the most fundamental algebraic structures, with vector spaces and algebras being particularly important variants within the study of Linear Algebra and its applications.
Definition C.3.1.
Let $$R$$ be a nonempty set, and let $$+$$ and $$*$$ be binary operations on $$R$$. Then $$R$$ forms an (associative) ring under $$+$$ and $$*$$ if the following three conditions are satisfied:
1. $$R$$ forms an abelian group under $$+$$.
2. ($$*$$ is associative) Given any three elements $$a, b, c \in R$$, $$a * (b * c) = (a * b) * c$$.
3. ($$*$$ distributes over $$+$$) Given any three elements $$a, b, c \in R$$,
$a * (b + c) = a * b + a * c \ \ \text{and} \ \ (a + b) * c = a * c + b * c.$
As with the definition of group, there are many additional properties that can be added to a ring; here, each additional property makes a ring more field-like in some way.
Definition C.3.2. Let $$R$$ be a ring under the binary operations $$+$$ and $$*$$.
Then we call $$R$$
1. commutative if $$*$$ is a commutative operation; i.e., given any $$a, b \in R$$, $$a * b = b * a$$.
2. unital if there is an identity element for $$*$$; i.e., if there exists an element $$i \in R$$ such that, given any $$a \in R$$, $$a * i = i * a = a$$.
3. a commutative ring with identity (a.k.a. CRI) if it's both commutative and unital.
In particular, note that a commutative ring with identity is almost a field; the only thing missing is the assumption that every element has a multiplicative inverse. It is this one difference that results in many familiar sets being CRIs (or at least unital rings) but not fields. E.g., $$\mathbb{Z}$$ is a CRI under the usual operations of addition and multiplication, yet, because of the lack of multiplicative inverses for all elements except $$\pm 1$$, $$\mathbb{Z}$$ is not a field.
In some sense, $$\mathbb{Z}$$ is the prototypical example of a ring, but there are many other familiar examples. E.g., if $$F$$ is any field, then the set of polynomials $$F[z]$$ with coefficients from $$F$$ is a CRI under the usual operations of polynomial addition and multiplication, but again, because of the lack of multiplicative inverses for every element, $$F[z]$$ is itself not a field. Another important example of a ring comes from Linear Algebra. Given any vector space $$V$$, the set $$\mathcal{L}(V)$$ of all linear maps from $$V$$ into $$V$$ is a unital ring under the operations of function addition and composition. However, $$\mathcal{L}(V)$$ is not a CRI unless $$\dim(V) \in \{0, 1\}$$.
Alternatively, if a ring $$R$$ forms a group under $$*$$ (but not necessarily an abelian group), then $$R$$ is sometimes called a skew field (a.k.a. division ring). Note that a skew field is also almost a field; the only thing missing is the assumption that multiplication is commutative. Unlike CRIs, though, there are no simple examples of skew fields that are not also fields.
As you can probably imagine, many other properties that can be appended to the definition of a ring, some of which are more useful than others. We close this section by defining the concept of an algebra over a field. In essence, an algebra is a vector space together with a "compatible'' ring structure. Consequently, anything that can be done with either a ring or a vector space can also be done with an algebra.
Definition C.3.3.
Let $$A$$ be a nonempty set, let $$+$$ and $$\times$$ be binary operations on $$A$$, and let $$*$$ be scalar multiplication on $$A$$ with respect to $$\mathbb{F}$$. Then $$A$$ forms an (associative) algebra over $$\mathbb{F}$$ with respect to $$+$$, $$\times$$, and $$*$$ if the following three conditions are satisfied:
1. $$A$$ forms an (associative) ring under $$+$$ and $$\times$$.
2. $$A$$ forms a vector space over $$\mathbb{F}$$ with respect to $$+$$ and $$*$$.
3. ($$*$$ is quasi-associative and homogeneous with respect to $$\times$$) Given any element $$\alpha \in \mathbb{F}$$ and any two elements $$a, b \in R$$, $\alpha * (a \times b) = (\alpha * a) \times b {\rm{~and~}} \alpha * (a \times b) = a \times (\alpha * b).$
Two particularly important examples of algebras were already defined above: $$F[z]$$ (which is unital and commutative) and $$\mathcal{L}(V)$$ (which is, in general, just unital). On the other hand, there are also many important sets in Linear Algebra that are not algebras. E.g., $$\mathbb{Z}$$ is a ring that cannot easily be made into an algebra, and $$\mathbb{R}^{3}$$ is a vector space but cannot easily be made into a ring (since the cross product operation from Vector Calculus is not associative).
## Contributors
Both hardbound and softbound versions of this textbook are available online at WorldScientific.com. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9798421859741211, "perplexity": 141.21232332557983}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541301598.62/warc/CC-MAIN-20191215042926-20191215070926-00531.warc.gz"} |
https://planetmath.org/SubnormalSeries | # subnormal series
Let $G$ be a group with a subgroup $H$, and let
$G=G_{0}\rhd G_{1}\rhd\cdots\rhd G_{n}=H$ (1)
be a series of subgroups with each $G_{i}$ a normal subgroup of $G_{i-1}$. Such a series is called a subnormal series or a subinvariant series.
If in addition, each $G_{i}$ is a normal subgroup of $G$, then the series is called a normal series.
A subnormal series in which each $G_{i}$ is a maximal normal subgroup of $G_{i-1}$ is called a composition series.
A normal series in which $G_{i}$ is a maximal normal subgroup of $G$ contained in $G_{i-1}$ is called a principal series or a chief series.
Note that a composition series need not end in the trivial group $1$. One speaks of a composition series (1) as a composition series from $G$ to $H$. But the term composition series for $G$ generally means a composition series from $G$ to $1$.
Similar remarks apply to principal series.
Some authors use normal series as a synonym for subnormal series. This usage is, of course, not compatible with the stronger definition of normal series given above.
Title subnormal series Canonical name SubnormalSeries Date of creation 2013-03-22 13:58:42 Last modified on 2013-03-22 13:58:42 Owner mclase (549) Last modified by mclase (549) Numerical id 8 Author mclase (549) Entry type Definition Classification msc 20D30 Synonym subinvariant series Related topic SubnormalSubgroup Related topic JordanHolderDecompositionTheorem Related topic Solvable Related topic DescendingSeries Related topic AscendingSeries Defines composition series Defines normal series Defines principal series Defines chief series | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 18, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.918563723564148, "perplexity": 641.0413473068048}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703515075.32/warc/CC-MAIN-20210118154332-20210118184332-00693.warc.gz"} |
http://math.stackexchange.com/questions/214646/recover-n-integers-using-m-more-integers?answertab=oldest | # recover N integers using M more integers
Suppose we have $N$ integers $a_1,a_2,\dots,a_N$, Given $M$ more integers $b_1,b_2,\dots,b_M$($b_i$ is calculated from $a_1\dots a_n$ by some ways)
Now remove any $M$ numbers from $a_1,a_2,\dots,a_N, b_1,b_2,\dots,b_M$, I want to recover $a_1,a_2,\dots,a_N$
My question is, Can I find a way to calculate such $b_1,b_2,\dots,b_m$?
For example, suppose $M=1$, we can calculate $b_1$ as $$b_1=a_1\oplus a_2\oplus\dots\oplus a_N$$ so if $a_i$ is missing ,we just need to XOR $b_1$ and left $a_i$.
For any $M$, my idea is to make $b_i$ as a linear combinations of $a_i$, that is $b_i = \sum_{j=1}^{N}k_{ij}a_j, 1\le i\le M$
Define A as a $(M+N)\times N$ matrix $$A = \left[ \begin{array}{cccc} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \vdots& \vdots& & \vdots\\ 0 & 0 & \dots & 1\\ k_{11}& k_{12} &\dots & k_{1N}\\ \vdots& \vdots & & \vdots\\ k_{M1}& k_{M2} &\dots& k_{MN} \\ \end{array} \right]$$ The first $N$ rows form an identity matrix $I_N$
The problem is to find $k_{ij}$, such that remove any M rows of $A$, the left $N\times N$ matrix is still full rank.
I'm not sure whether we define $k_{ij}=i^{j-1}$ will work .
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You are looking for a matrix whose square submatrices are all nonsingular. These have been studied in coding theory --- in a "Maximum Distance Separable" (or, MDS) code, the generator matrix has this property. For example, the problem is discussed in Lacan and Fimes, Systematic MDS Erasure Codes Based on Vandermonde Matrices, IEEE Communications Letters 8 (2004) 570-572.
In any event, I think your choice of $k_{ij}$ is fine; I think it leads to a Vandermonde matrix, and there are formulas for the determinant of Vandermonde matrices which show that every square submatrix of a Vandermonde matrix with positive entries is nonsingular.
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Since $\mathbb Z^n$ is countable and it's straightforward to construct an enumeration (e.g. in a similar spirit as the diagonal enumeration of $\mathbb Z^2$), you can encode all $N$ integers in a single value $b$. If you take all $b_i=b$, then either they all get removed and you still have all the $a_i$, or at least one of them remains and you can reconstruct all the $a_i$ from that one.
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Thank you, joriki, though this answer seems a bit tricky – Benson Oct 16 '12 at 5:47
@Benson: I don't think it's tricky in the sense of difficult (to implement). Perhaps you mean it seems like a trick? – joriki Oct 16 '12 at 5:52
Yes, I mean in actual use, we need a lot of space to store such $b_i$. If each $a_i$ takes 4Bytes to store, than each $b_i$ needs $4*N$ Bytes to store – Benson Oct 16 '12 at 6:30
Somewhere in between the two previous answers of joriki and Gerry Myerson, let me point out that there exists an entire theory devoted to this question, known as the theory of error-correcting codes or coding theory: how to encode information (a bunch of numbers) such that even with limited information (fewer numbers, or some numbers incorrect) we can recover the original information.
The scheme you propose in your question (and Gerry Myerson in his answer) is a particular specific error-correcting code, and the one in joriki's answer (pick an injection $\mathbb{Z}^n \to \mathbb{Z}$ and use it in your encoding — BTW, on such polynomial functions, rejecting exponential solutions like $2^{a_1}3^{a_2}\dots$, see the nice article "Bert and Ernie" by Zachary Abel) is also an error-correcting code. The theory in general includes analysis of the tradeoffs between size of the encoding, efficiency of encoding/decoding, the extent to which loss can happen while still leaving recovery possible, etc. Here is a good free book that touches on it.
For instance, here is an approach that answers your question in the sense of "Given $N$ numbers, generate $N+M$ numbers such that even if any $M$ numbers are removed, the original $N$ numbers can be recovered". Given the $N$ numbers $a_1, \dots, a_N$, construct a polynomial of degree $N-1$ e.g. $p(x) = a_1 + a_2x + \dots + a_Nx^{N-1}$ in some field, and let the $N+M$ values $b_i$ be the values $p(x_i)$ of this polynomial at some pre-chosen values $x_1, \dots, x_{N+M}$. Then given any $N$ of these values (and knowing which ones), we can reconstruct the polynomial and hence the $a_i$s, through polynomial interpolation. This is the idea behind Reed-Solomon codes, used in CDs and DVDs.
If you insist that the $N+M$ values must be the original $N$ values and $M$ others, then with this constraint too there are many error-correcting codes (and joriki points out below that the previous idea can also be made to work), for instance in the class known as cyclic redundancy checks. Your $M=1$ example of using a parity bit is precisely one such check (a variant is used in ISBN and UPC numbers; see check digit).
Those involve polynomials, in general. If you further insist that the $N+M$ values must be given by a linear transformation with a matrix of the form $A$ as you wrote in the question, then see Gerry Myerson's answer, I guess.
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Thank you, ShreevatsaR – Benson Oct 16 '12 at 6:26
The interpolation approach can also be used with $N$ original values and $M$ additional values by letting $p$ be the unique polynomial interpolating between the $a_i$ considered as function values at $x_1,\dotsc,x_N$. – joriki Oct 16 '12 at 6:27
@joriki: Oh indeed! That's a nice idea; thanks for the observation. – ShreevatsaR Oct 16 '12 at 6:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9301508069038391, "perplexity": 275.90251151869256}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999669442/warc/CC-MAIN-20140305060749-00016-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://socratic.org/questions/eight-less-than-the-product-of-6-and-a-number-equals-5-how-do-i-write-this-as-an | Algebra
Topics
# Eight less than the product of 6 and a number equals 5. How do I write this as an equation?
Mar 15, 2018
$6 x - 8 = 5$
#### Explanation:
"Less than" implies that that number will come after what is next.
$\text{something} \setminus - \setminus 8$
"The product of" implies multiplication, and in this case, you would replace "a number" with a value, such as $x$.
$\left(\text{6 times} \setminus x\right) - 8$
$\left(6 x\right) - 8$
The final part, "equals", is hopefully self-explanatory.
$\left(6 x\right) - 8 = 5$
##### Impact of this question
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http://tex.stackexchange.com/questions/37168/when-to-use-math-mode/37210 | # When to use math mode?
In my document, the numbers inside math mode appear differently than those numbers out. Sometimes, I have numbers like "10 squared" within a paragraph, so it seems useful to use $10^2$. However, maybe on the same line, I have "10 km". The style of the 10 is different. In this case, should I also use $10$ km?
• Is there a general rule for when it is best to use math mode within a document?
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Another vote for siunitx, as suggested by uli. It properly spaces units from magnitudes. If you do have to do this manually for simple stuff, use 10\,km outside of math mode. The \, adds a half-space which is the neatest looking gap. – qubyte Dec 5 '11 at 11:11
Don Knuth touched on this topic in his article for TUGboat -- "Typesetting Concrete Mathematics". His examples don't include units (for that, the siunitx package is a good choice, as already mentioned), but the method for determining what is math and what isn't is well illustrated otherwise.
(The article is set in Knuth's Concrete fonts, and shows some of the special techniques used in setting that book. Irrelevant for this question, but interesting nonetheless.)
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I find particularly interesting the last paragraph on page 31 (ending on p. 32). – egreg Dec 5 '11 at 15:07
As the math font and the main text font are likely to have different looking numbers you should aim for consistency. Whenever you refer to a part of a document, e.g. chapter 4, theorem 3.4, bullet point 2, figure 9.3, table 12.1 or similar elements, stick with the same font, which is likely to be your main text font.
Whenever your talk about parts of a mathematical expression, e.g. the leading coefficient, then be consistent and use the same font as used for typesetting the formular.
In case of physical quantities I would recommend the use of siunitx that allows for a consistent application of the SI system throughout the document.
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As uli said, siuntix is recommended. I use it to typeset all numbers.
Example
\documentclass{minimal}
\usepackage{siunitx}
\sisetup{locale=UK}
\begin{document}
Lorem Ipsum is simply \SI{10.5}{\kilo\meter} dummy text of the printing.
Lorem Ipsum has been the \num{2e-19} industry's standard dummy text ever since the
1500s, when an unknown printer took a galley of type and scrambled it to
make a type \SI{2,6}{\volt\per\meter} specimen book. It has survived not only five centuries, but
also the leap into electronic typesetting, remaining essentially unchanged.
It was popularised in the 1960s with the release of Letraset sheets
containing Lorem Ipsum passages, and more recently with desktop publishing
software like Aldus PageMaker including versions of Lorem Ipsum.
\end{document}
Note the handling of 10.5 and 2,6 (both with . in output) and of 2e-9. The behavior of \per (in \volt\per\meter) is customizable.
I didn’t found a solution to write soemthing like \num{2^3}. Does anybody know if this is possible?
As said in the comments it is possible to use \num[parse-numbers=false]{2^3}. But this affects an e12 part too.
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2^3 is a formula, rather than a number, so $$2^3$$ is the answer. Or maybe $$\num{2.6}^3$$ if you want to be fussy. – egreg Dec 5 '11 at 15:24
You could say \num[exponent-base=2]{e3}. – Torbjørn T. Dec 5 '11 at 15:24
Thanks. @egreg: That could work but I can imagine a case like \SI{2,6^2e9}{\volt} where this solution would be inconvenient and inflexible (immune to \sisetup): \num{2,6}^2 \times 10^9\,\si{\volt} – Tobi Dec 5 '11 at 15:44
I can't imagine why somebody would write something like that. :) – egreg Dec 5 '11 at 15:46
@Aditya I suspect the old 'example code not actually used for the example output' issue. siunitx certainly knows the difference between a volt and a volt per metre. – Joseph Wright Dec 5 '11 at 16:59
My simple rule, which is sort of like what uli and barbara beeton wrote:
Write numerals in plain text and numbers in math.
More casually, you could make the distinction that it's a numeral if it could belong (in context) in, say, an essay on literary criticism, and a number if a scientist might write it. Or with an eye towards utility, put it in math if you can imagine it being next to a plus sign.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8758812546730042, "perplexity": 2168.247932622803}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416405292026.28/warc/CC-MAIN-20141119135452-00106-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-evaluate-2x-2-y-if-x-2-frac-1-2-y-3-frac-3-5 | Algebra
Topics
# How do you evaluate 2x ^ { 2} y if x = 2\frac { 1} { 2} ,y = - 3\frac { 3} { 5}?
Oct 4, 2017
$2 {x}^{2} y = - 45$
#### Explanation:
You can evaluate $2 {x}^{2} y$ by subbing in the given points, $x = 2 \frac{1}{2}$ and $y = - 3 \frac{3}{5}$.
First, it's best to turn these fractions into improper fractions:
$x = \frac{5}{2}$
$y = - \frac{18}{5}$
Subbing these into the expression we get:
$2 {x}^{2} y$
$= 2 {\left(\frac{5}{2}\right)}^{2} \left(- \frac{18}{5}\right)$
$= 2 \left(\frac{25}{4}\right) \left(- \frac{18}{5}\right)$
These fractions can be further simplified before you multiply them together:
$= 1 \left(\frac{25}{2}\right) \left(- \frac{18}{5}\right)$
$= \frac{5}{2} \left(- \frac{18}{1}\right)$
$= \frac{5}{1} \left(- 9\right)$
$= - 45$
So in short, $2 {x}^{2} y = - 45$
Oct 4, 2017
$- 45$
#### Explanation:
$\text{change the mixed numbers into "color(blue)"improper fractions}$
$\Rightarrow 2 \frac{1}{2} = \frac{5}{2} \text{ and } 3 \frac{3}{5} = \frac{18}{5}$
$\Rightarrow 2 {x}^{2} y$
$= 2 \times {\left(\frac{5}{2}\right)}^{2} \times - \frac{18}{5}$
$= 2 \times \frac{25}{4} \times - \frac{18}{5}$
$\textcolor{b l u e}{\text{cancel common factors }}$ on numerators/denominators.
$= {\cancel{2}}^{1} \times {\cancel{25}}^{5} / {\cancel{4}}^{2} \times - \frac{18}{\cancel{5}} ^ 1$
$= 1 \times \frac{5}{\cancel{2}} ^ 1 \times - {\cancel{18}}^{9} / 1$
$= 1 \times 5 \times - 9 = - 45$
##### Impact of this question
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http://boris-belousov.net/2017/09/10/gaussian-process-linear-regression/ | # Gaussian process vs kernel ridge regression
Consider the prediction problem: given a dataset $\mathcal{D} = \{ (\mathbf{x}_i, y_i) \}_{i=1}^{N}$ of pairs of inputs $\mathbf{x}_i \in \mathbb{R}^n$ and outputs $y_i \in \mathbb{R}$, one wishes to predict the output $y$ given an input $\mathbf{x}$. We first solve this problem using plain ridge regression, also known as weight decay or Tikhonov regularization. After that, we transform the solution by means of the matrix inversion lemma to a form amenable to kernelization—at this point, inner products of feature vectors can be replaced by a kernel function, leading to the kernel ridge regression. Direct comparison of the kernel ridge regression solution with the mean of the predictive distribution returned by a Gaussian process with the same kernel establishes their equivalence. Thus, if the uncertainty of a prediction is irrelevant, Gaussian process regression and kernel ridge regression can be used interchangeably.
## Ridge regression
Consider the model $p(y|\mathbf{w}) = \mathcal{N}\left(y | \mathbf{w}^\mathrm{T} \boldsymbol{\phi}(\mathbf{x}), \beta^{-1}\right)$ linear in features $\boldsymbol{\phi}(\mathbf{x}) \in \mathbb{R}^M$ with the output $y$ corrupted by zero-mean Gaussian noise with precision $\beta$. Assembling $N$ outputs into a vector $\mathbf{y} \in \mathbb{R}^N$ and assuming that data points are drawn independently, the likelihood splits into a product of independent terms (see Bishop, Formula 3.10)
Given a Gaussian prior over the parameters $p(\mathbf{w}) = \mathcal{N}(\mathbf{w}|\mathbf{0}, \alpha^{-1}\mathbf{I})$, the mean of the posterior distribution $p(\mathbf{w}|\mathbf{y})$ is a linear function of the targets $\mathbf{y}$ (see Bishop, Formula 3.53)
where $\boldsymbol{\Phi} \in \mathbb{R}^{N \times M}$ is the design matrix with rows $\boldsymbol{\phi}(\mathbf{x}_i)^\mathrm{T}$ and $\lambda = \alpha / \beta$ is a regularization parameter. When a test input $\mathbf{x}$ arrives, the maximum a posteriori estimate of the output is given by the mean of the predictive distribution (see Bishop, Formula 3.58)
Ridge regression $$\label{RR} \mu_{y|\mathbf{y}} = \boldsymbol{\phi}^\mathrm{T} \boldsymbol{\mu}_{\mathbf{w}|\mathbf{y}} = \boldsymbol{\phi}^\mathrm{T} \left( \lambda \mathbf{I} + \boldsymbol{\Phi}^\mathrm{T} \boldsymbol{\Phi} \right)^{-1} \boldsymbol{\Phi}^\mathrm{T} \mathbf{y}$$
where $\boldsymbol{\phi} = \boldsymbol{\phi}(\mathbf{x})$ is the feature vector of the test input $\mathbf{x}$. Formula \eqref{RR} reveals an interesting property of the linear regression: the predictive mean $\mu_{y|\mathbf{y}}$ is a linear combination of the targets $y_i$ from the dataset with weights
That is, $\mu_{y | \mathbf{y}} = \sum_{i=1}^N \omega_i y_i$. Regression functions, such as this, which make predictions by taking linear combinations of the training set target values are known as linear smoothers.
## Kernel ridge regression
In the classical regime, the dimensionality of the feature space $M$ is smaller than the number of data points $N$. Thus, matrix $\boldsymbol{\Phi}$ is skinny and it is advisable to use Formula \eqref{RR} because it involves inversion of a small matrix $\boldsymbol{\Phi}^\mathrm{T}\boldsymbol{\Phi}$. However, one may wish to use more expressive high-dimensional representations for which there may be no hope to obtain enough data to get to the classical regime. In this case, $M > N$ and matrix $\boldsymbol{\Phi}$ is fat. One can still apply Formula \eqref{RR} but it may be extremely impractical. Luckily, the matrix inversion lemma allows one to shift the transposition sign and invert $\boldsymbol{\Phi}\boldsymbol{\Phi}^\mathrm{T}$ instead, which may be significantly simpler in the considered circumstances. With the help of the identity
the predictive mean \eqref{RR} can be computed as
Ridge regression after applying the matrix inversion lemma $$\label{RR_MIL} \mu_{y|\mathbf{y}} = \boldsymbol{\phi}^\mathrm{T} \boldsymbol{\Phi}^\mathrm{T} \left(\boldsymbol{\Phi} \boldsymbol{\Phi}^\mathrm{T} + \lambda \mathbf{I} \right)^{-1} \mathbf{y}.$$
Formula \eqref{RR_MIL} is remarkable not only because it allows us to invert a smaller matrix but also because it can be entirely expressed in terms of inner products of feature vectors without requiring access to the feature vectors themselves. This is the basis of the so-called kernel trick. By introducing the kernel function
we can rewrite \eqref{RR_MIL} in the kernelized form as
Kernel ridge regression $$\label{KRR} \mu_{y|\mathbf{y}} = \mathbf{k}^\mathrm{T} \left( \mathbf{K} + \lambda \mathbf{I} \right)^{-1} \mathbf{y}$$
where $\mathbf{k} = \mathbf{k}(\mathbf{x}) \in \mathbb{R}^N$ is a vector with elements $k_i(\mathbf{x}) = k(\mathbf{x}_i, \mathbf{x})$ and $\mathbf{K} \in \mathbb{R}^{N \times N}$ is the Gram matrix of the set of vectors $\boldsymbol{\phi}_i = \boldsymbol{\phi}(\mathbf{x}_i)$ with elements $K_{ij} = \boldsymbol{\phi}_i^\mathrm{T} \boldsymbol{\phi}_j$. Formula \eqref{KRR}, referred to as kernel ridge regression, has a wider scope of applicability than the ridge regression formulas \eqref{RR} and \eqref{RR_MIL} we started with. Indeed, Formula \eqref{KRR} does not restrict one to the use of finite-dimensional feature vectors but allows for utilization of infinite-dimensional ones, opening the door into the beautiful reproducing kernel Hilbert space.
## Gaussian process regression
The final piece of the puzzle is to derive the formula for the predictive mean in the Gaussian process model and convince ourselves that it coincides with the prediction \eqref{KRR} given by the kernel ridge regression. Starting with the likelihood
where $\mathbf{f} = (f_1,\dots,f_N)^\mathrm{T}$ is a vector of evaluations $f_i = f(\mathbf{x}_i)$ of the function $f$ at every point in the dataset, and combining it with a Gaussian process prior over functions $f$ concisely expressed as $p(\mathbf{f}) = \mathcal{N}(\mathbf{f}|\mathbf{0}, \mathbf{K})$, we find the marginal distribution $p(\mathbf{y}) = \mathcal{N}\left( \mathbf{y} | \mathbf{0}, \mathbf{K} + \lambda \mathbf{I} \right)$, which is crucial for making predictions.
Prediction in the Gaussian process model is done via conditioning. In order to find the predictive distribution $p(y|\mathbf{y})$, one conditions the joint distribution
on the observed targets $\mathbf{y}$, which yields (see Bishop, Formula 6.66)
Gaussian process regression $$\label{GPR} \mu_{y|\mathbf{y}} = \mathbf{k}^\mathrm{T} \left( \mathbf{K} + \lambda \mathbf{I} \right)^{-1} \mathbf{y}$$
for the predictive mean. Needless to say, Formula \eqref{GPR} for the Gaussian process regression is exactly the same as Formula \eqref{KRR} for the kernel ridge regression.
We conclude that Gaussian process conditioning results in kernel ridge regression for the conditional mean in the same way as plain Gaussian conditioning results in linear regression. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 4, "x-ck12": 0, "texerror": 0, "math_score": 0.9590234756469727, "perplexity": 286.69764732848085}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370510287.30/warc/CC-MAIN-20200403030659-20200403060659-00151.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-7-section-7-2-rational-exponents-7-2-exercises-page-448/59 | ## Intermediate Algebra (12th Edition)
$t^{8/15}$
$\bf{\text{Solution Outline:}}$ Use the definition of rational exponents and the laws of exponents to convert the given expression, $\dfrac{\sqrt[3]{t^4}}{\sqrt[5]{t^4}} ,$ to exponential form. $\bf{\text{Solution Details:}}$ Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{t^{\frac{4}{3}}}{t^{\frac{4}{5}}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} t^{\frac{4}{3}-\frac{4}{5}} .\end{array} To simplify the expression, $\dfrac{4}{3}-\dfrac{4}{5} ,$ find the $LCD$ of the denominators $\{ 3,5 \}.$ The $LCD$ is $15$ since it is the lowest number that can be divided by both denominators. Multiplying both the numerator and the denominator of each term by the constant that will make the denominators equal to the $LCD$ results to \begin{array}{l}\require{cancel} t^{\frac{4}{3}\cdot\frac{5}{5}-\frac{4}{5}\cdot\frac{3}{3}} \\\\= t^{\frac{20}{15}-\frac{12}{15}} \\\\= t^{\frac{8}{15}} \\\\= t^{8/15} .\end{array} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9983229637145996, "perplexity": 387.048791558436}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583510754.1/warc/CC-MAIN-20181016134654-20181016160154-00114.warc.gz"} |
https://myaptitude.in/cat/quant/the-milk-and-water-in-two-vessels-a-and-b-are-in-the-ratio-4-3-and-2-3-respectively | The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel C consisting half milk and half water?
1. 8 : 3
2. 4 : 3
3. 2 : 3
4. 7 : 5
Milk in Vessel A = 4/7 (Dearer Value)
Milk in Vessel B = 2/5 (Cheaper Value)
Milk in Vessel C = 1/2 (Mean Value)
Dearer : Cheaper = 1/2 - 2/5 : 4/7 - 1/2 = 1/10 : 1/14
Required ratio is 14 : 10 = 7 : 5
The correct option is D. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8921045660972595, "perplexity": 1177.730197623459}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986647517.11/warc/CC-MAIN-20191013195541-20191013222541-00051.warc.gz"} |
https://math.eretrandre.org/tetrationforum/showthread.php?tid=676&pid=6100 | • 0 Vote(s) - 0 Average
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Tetration and imaginary numbers. robo37 Junior Fellow Posts: 15 Threads: 6 Joined: Jun 2009 07/12/2011, 03:22 PM Thanks for the help I got with my last question, now here's something else. i^i = 0.207879576..., which is interesting, so I wounder if there is any way to find out what i^^i is? Furthermore, what is i sroot i, i itteratedroot i, and the ith exponential factorial? Thanks. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 07/13/2011, 01:03 PM (This post was last modified: 07/13/2011, 02:51 PM by sheldonison.) (07/12/2011, 03:22 PM)robo37 Wrote: Thanks for the help I got with my last question, now here's something else. i^i = 0.207879576..., which is interesting, so I wounder if there is any way to find out what i^^i is? Furthermore, what is i sroot i, i itteratedroot i, and the ith exponential factorial? Thanks. There is an attracting fixed point ($\approx 0.438282936727032 + 0.360592471871385i$), which can be used to develop a superfunction for base i. When I used the attracting fixed point the result I got was, $^i i \approx 0.500129061733810 + 0.324266941212720i$. The equation I used was $\text{superf}(\text{superf}^{-1}(1)+i)$, where superf is developed from the attracting fixed point for base i. edit, I made a correction here I forget how to figure out the nth sroot.... so you'll have to report back the results for your other questions. Is the "ith sroot" equation perhaps $\text{superf}(\text{superf}^{-1}(1)-i)$? If it is, than the result is $^{-i} i \approx -1.13983245176083 + 0.702048300301002i$ - Sheldon robo37 Junior Fellow Posts: 15 Threads: 6 Joined: Jun 2009 07/13/2011, 03:25 PM (This post was last modified: 07/13/2011, 06:05 PM by bo198214.) (07/13/2011, 01:03 PM)sheldonison Wrote: (07/12/2011, 03:22 PM)robo37 Wrote: Thanks for the help I got with my last question, now here's something else. i^i = 0.207879576..., which is interesting, so I wounder if there is any way to find out what i^^i is? Furthermore, what is i sroot i, i itteratedroot i, and the ith exponential factorial? Thanks. There is an attracting fixed point ($\approx 0.438282936727032 + 0.360592471871385i$), which can be used to develop a superfunction for base i. When I used the attracting fixed point the result I got was, $^i i \approx 0.424801328697548 + 0.424973603314731i$. The equation I used was $\text{superf}(\text{superf}^{-1}(i)+i)$, where superf is developed from the attracting fixed point for base i. I forget how to figure out the nth sroot.... so you'll have to report back the results for your other questions. Is the "ith sroot" equation perhaps $\text{superf}(\text{superf}^{-1}(i)-i)$? If it is, than the result is $\approx -0.0723270995404099 - 0.323973330391954i$ - Sheldon Wow, thanks for that. It's interesting that the imaginary part is almost as big as the real part with the first resault, but I'm sure that's just coincidence. I'm rather interested with the imaginary and complex plain; I've already found out, with a little help from Google Calculator, that $i root i = 4.81047738$ $i! = 0.498015668 - 0.154949828 i$, $F (i) = 0.379294534 + 0.215939518 i$ and the ith square triangular number is -$0.120450647 - 1.87314977*10[-16i]$, at the moment I'm on the ith partition number, but I'm having difficulty as there seems to be no closed finite function to use. I don't suppose anyone could herlp me out here? $p(i) = \frac1i\cdot\sum_{k=0}^{i-1}\sigma(i-k)\cdot p(k) = ?$ $p(i) = 1+\sum_{k=1}^{\lfloor \frac{1}{2}i \rfloor} p(k,i-k) = ?$ « Next Oldest | Next Newest »
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https://infoscience.epfl.ch/record/231554 | ## Revisiting the quest for a universal log-law and the role of pressure gradient in "canonical" wall-bounded turbulent flows
The trinity of so-called "canonical" wall-bounded turbulent flows, comprising the zero pressure gradient turbulent boundary layer, abbreviated ZPG TBL, turbulent pipe flow, and channel/duct flows has continued to receive intense attention as new and more reliable experimental data have become available. Nevertheless, the debate on whether the logarithmic part of the mean velocity profile, in particular the Karman constant kappa, is identical for these three canonical flows or flow-dependent is still ongoing. In this paper, the asymptotic matching requirement of equal. in the logarithmic overlap layer, which links the inner and outer flow regions, and in the expression for the centerline/free-stream velocity is reiterated and shown to preclude a universal logarithmic overlap layer in the three canonical flows. However, the majority of pipe and channel flowstudies at friction Reynolds numbers Re-tau below approximate to 10(4) extract from near-wall profiles the same kappa of 0.38-0.39 as in the ZPG TBL. This apparent contradiction is resolved by a careful reanalysis of high-quality mean velocity profiles in the Princeton "Superpipe" and other pipes, channels, and ducts, which shows that the mean velocity in a near-wall region extending to around 700 "+" units in channels and ducts and 500 "+" units in pipes is the same as in the ZPG TBL. In other words, all the "canonical" flow profiles contain the lower end of the ZPG TBL log-region, which starts at a wall distance of 150-200 "+" units with a universal kappa of kappa(ZPG) approximate to 0.384. This interior log-region is followed by a second logarithmic region with a flow specific. > kappa(ZPG), which increases monotonically with pressure gradient. This second, exterior log-layer is the actual overlap layer matching up to the outer expansion, which implies equality of the exterior. and kappa(CL) obtained from the evolution of the respective centerline velocity with Reynolds number. The location of the switch-over point implies furthermore that this second log-layer only becomes clearly identifiable, i.e., separated from the wake region, for Re-tau well beyond 10(4) (see Fig. 1). This explains the discrepancies between the Karman constants of 0.38-0.39, extracted from near-wall pipe profiles below Re-tau approximate to 10(4) and the kappa's obtained from the evolution of the centerline velocity with Reynolds number. The same analysis is successfully applied to velocity profiles in channels and ducts even though experiments and numerical simulations have not yet reached Reynolds numbers where the different layers have even started to clearly separate.
Published in:
Physical Review Fluids, 2, 9, 094602
Year:
2017
Publisher:
College Pk, Amer Physical Soc
ISSN:
2469-990X
Laboratories: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8813235759735107, "perplexity": 2491.3723164945336}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221215222.74/warc/CC-MAIN-20180819145405-20180819165405-00685.warc.gz"} |
http://mathhelpforum.com/pre-calculus/158427-logarithm.html | 1. ## Logarithm
Hi,
Are there alternative ways of solving this equation? Could logarithms be used?
$a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$
2. Originally Posted by Hellbent
Hi,
Are there alternative ways of solving this equation? Could logarithms be used?
$a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$
logs could be used ... but the way you did it more direct.
3. I need assistance with the logs part, please.
Tried using logs prior to posting question, but the answers are inconsistent.
4. Originally Posted by Hellbent
Hi,
Are there alternative ways of solving this equation? Could logarithms be used?
$a^{60} = 2.044 \implies a = 2.044^{1/60} = 1.012$
$60\log{a} = \log{2.044}$
$\log{a} = \frac{log{2.044}}{60}
\RightArrow
= .00517
$
so $10^{(.00517)} = 1.01199$ or $1.012$
as mentioned your first method is better..
5. $a^{60} = k$ , where $k = 2.044$
$60\log{a} = \log{k}
$
$\displaystyle \log{a} = \frac{\log{k}}{60}$
$\displaystyle a = e^{\frac{\log{k}}{60}}$
$a \approx 1.012$
note that $\displaystyle a = e^{\frac{\log{k}}{60}} = \left(e^{\log{k}}\right)^{\frac{1}{60}} = k^{\frac{1}{60}} = 2.044^{\frac{1}{60}}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9787487387657166, "perplexity": 1427.7279235884234}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560279368.44/warc/CC-MAIN-20170116095119-00060-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://math.msu.edu/seminars/TalkView.aspx?talk=4065 | ## Student Algebra
• Nicholas Ovenhouse, MSU
• Log-Canonical Poisson Brackets on the Algebra of Rational Functions
• 10/05/2016
• 4:10 PM - 5:00 PM
• C304 Wells Hall
On a symplectic manifold, there are always canonical coordinates around any point, where the symplectic form looks like the standard one on R^2n. In terms of Poisson geometry, this means the bracket of any two coordinate functions is constant. We ask whether such a thing is possible in the algebraic situation. That is, given a Poisson bracket, is there some change of coordinates, using only rational functions, which makes the bracket between coordinate functions constant?
## Contact
Department of Mathematics
Michigan State University
619 Red Cedar Road
C212 Wells Hall
East Lansing, MI 48824
Phone: (517) 353-0844
Fax: (517) 432-1562
College of Natural Science | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9289822578430176, "perplexity": 2656.867775104035}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039747665.82/warc/CC-MAIN-20181121092625-20181121114625-00181.warc.gz"} |
https://www.arxiv-vanity.com/papers/0706.0652/ | CERN–PH–TH/2007-087 DCPT/07/50, IPPP/07/25
MPP–2007–64 UMN–TH–2606/07, FTPI–MINN–07/19
The Supersymmetric Parameter Space in Light of
-physics Observables and Electroweak Precision Data
J. Ellis, S. Heinemeyer, K.A. Olive, A.M. Weber and G. Weiglein
TH Division, Physics Department, CERN, Geneva, Switzerland
Instituto de Fisica de Cantabria (CSIC-UC), Santander, Spain
William I. Fine Theoretical Physics Institute,
University of Minnesota, Minneapolis, MN 55455, USA
Max-Planck-Institut für Physik, Föhringer Ring 6, D–80805 Munich, Germany
IPPP, University of Durham, Durham DH1 3LE, UK
Abstract
Indirect information about the possible scale of supersymmetry (SUSY) breaking is provided by -physics observables (BPO) as well as electroweak precision observables (EWPO). We combine the constraints imposed by recent measurements of the BPO , , and with those obtained from the experimental measurements of the EWPO , , , and , incorporating the latest theoretical calculations of these observables within the Standard Model and supersymmetric extensions. We perform a fit to the parameters of the constrained minimal supersymmetric extension of the Standard Model (CMSSM), in which the SUSY-breaking parameters are universal at the GUT scale, and the non-universal Higgs model (NUHM), in which this constraint is relaxed for the soft SUSY-breaking contributions to the Higgs masses. Assuming that the lightest supersymmetric particle (LSP) provides the cold dark matter density preferred by WMAP and other cosmological data, we scan over the remaining parameter space. Within the CMSSM, we confirm the preference found previously for a relatively low SUSY-breaking scale, though there is some slight tension between the EWPO and the BPO. In studies of some specific NUHM scenarios compatible with the cold dark matter constraint we investigate () planes and find preferred regions that have values of somewhat lower than in the CMSSM.
December 3, 2020
## 1 Introduction
The dimensionality of the parameter space of the minimal supersymmetric extension of the Standard Model (MSSM) [1, 2] is so high that phenomenological analyses often make simplifying assumptions that reduce drastically the number of parameters. One assumption that is frequently employed is that (at least some of) the soft SUSY-breaking parameters are universal at some high input scale, before renormalization. One model based on this simplification is the constrained MSSM (CMSSM), in which all the soft SUSY-breaking scalar masses are assumed to be universal at the GUT scale, as are the soft SUSY-breaking gaugino masses and trilinear couplings . The assumption that squarks and sleptons with the same gauge quantum numbers have the same masses is motivated by the absence of identified supersymmetric contributions to flavour-changing neutral interactions and rare decays (see Ref. [3] and references therein). Universality between squarks and sleptons with different gauge interactions may be motivated by some GUT scenarios [4]. However, the universality of the soft SUSY-breaking contributions to the Higgs scalar masses is less motivated, and is relaxed in the non-universal Higgs model (NUHM) [5, 6, 7].
There are different possible approaches to analyzing the reduced parameter spaces of the CMSSM and the NUHM. One minimal approach would be to approximate the various theoretical, phenomenological, experimental, astrophysical and cosmological constraints naively by functions, determine the domains of the SUSY parameters allowed by their combination, and not attempt to estimate which values of the parameters might be more or less likely. This approach would perhaps be adequate if one were agnostic about the existence of low-energy SUSY. On the other hand, if one were more positive about its existence, and keen to find which SUSY parameter values were more ‘probable’, one would make a likelihood analysis and take seriously any possible hints that the Standard Model (SM) might not fit perfectly the available data. This is the approach taken in this paper. We perform a combined analysis of electroweak precision observables (EWPO), going beyond previous such analyses [8, 9] (see also Ref. [10]), and of -physics observables (BPO), including some that have not been included before in comprehensive analyses of the SUSY parameter space (see, however, Ref. [11]). In the past, the set of EWPO included in such analyses have been the boson mass , the effective leptonic weak mixing angle , the anomalous magnetic moment of the muon , and the mass of the lightest MSSM Higgs boson mass . Since our previous study, the theoretical link between experimental observables and within the Standard Model has become more precise, changing the distribution for the possible MSSM contribution. We also include in this analysis a new EWPO, namely the total boson width . In addition, we now include four BPO: the branching ratios , and , and the mass mixing parameter . For each observable, we construct the function including both theoretical and experimental systematic uncertainties, as well as statistical errors. The largest theoretical systematic uncertainty is that in , mainly associated with the renormalization-scale ambiguity. Since this is not a Gaussian error, we do not add it in quadrature with the other errors. Instead, in order to be conservative, we prefer to add it linearly.
For our CMSSM analysis, the fact that the cold dark matter density is known from astrophysics and cosmology with an uncertainty smaller than fixes with proportional precision one combination of the SUSY parameters, enabling us to analyze the overall value as a function of for fixed values of and . The value of is fixed by the electroweak vacuum conditions, the value of is fixed with a small error by the dark matter density, and the Higgs mass parameters are fixed by the universality assumption. As in previous analyses, we consider various representative values of for the specific choices . Also as previously, we find a marked preference for relatively small values of for , respectively, driven largely by with some assistance from . This preference would have been more marked if the BPO were not taken into account. Indeed, there is a slight tension between the EWPO and the BPO, with the latter disfavouring smaller , particularly for large . As corollaries of this analysis, we present the distributions for the masses of various MSSM particles, including the lightest Higgs boson mass . This shows a strong preference for , allowing as high as with .
In view of the slight tension between the EWPO and BPO within the CMSSM, we have gone on to explore the NUHM, which effectively has and as additional free parameters as compared to the CMSSM. In particular, we have investigated whether the NUHM reconciles more easily the EWPO and BPO, and specifically whether there exist NUHM points with significantly lower . As pointed out previously, generic NUHM parameter planes in which the other variables are held fixed do not satisfy the cold dark matter density constraint imposed by WMAP et al. In this paper, we introduce ‘WMAP surfaces’, which are () planes across in which the other variables are adjusted continuously so as to maintain the LSP density within the WMAP range. We then examine the values of the EWPO and BPO in the NUHM as functions over these WMAP surfaces 111A more complete characterization of these WMAP surfaces will be given elsewhere [12], as well as a discussion of their possible use as ‘benchmark scenarios’ for evaluating the prospects for MSSM Higgs phenomenology at the Tevatron, the LHC and elsewhere.. In each of the WMAP surfaces we find localized regions preferred by the EWPO and BPO and, in some cases, the minimum value of is significantly lower than along the WMAP strips in the CMSSM, indicating that the NUHM may help resolve the slight tension between the EWPO and the BPO. We explore this possibility further by investigating lines that explore further the NUHM parameter space in neighbourhoods of the low- points in the WMAP surfaces.
In Sect. 2 we review the current status of the EWPO and BPO that we use, our treatment of the available theoretical calculations and their errors, as well as their present experimental values. The analysis within the CMSSM can be found in Sect. 3, while the NUHM investigation is presented in Sect. 4. Sect. 5 summarizes our principal conclusions.
## 2 Current Experimental Data
The relevant data set includes five EWPO: the mass of the boson, , the effective leptonic weak mixing angle, , the total boson width, , the anomalous magnetic moment of the muon, , and the mass of the lightest MSSM Higgs boson, . In addition, we include four BPO: the branching ratios , , and as well as the mass-mixing parameter . A detailed description of the EWPO and can be found in Refs. [8, 13, 9, 14].
In this Section we start our analysis by recalling the current precisions of the experimental results and the theoretical predictions for all these observables. We also display the CMSSM predictions for the EWPO (where new results are available), and also for the BPO. These predictions serve as examples of the expected ranges of the EWPO and BPO values once SUSY corrections are taken into account.
In the following, we refer to the theoretical uncertainties from unknown higher-order corrections as ‘intrinsic’ theoretical uncertainties and to the uncertainties induced by the experimental errors of the SM input parameters as ‘parametric’ theoretical uncertainties. We do not discuss here the theoretical uncertainties in the renormalization-group running between the high-scale input parameters and the weak scale, see Ref. [15] for a recent discussion in the context of calculations of the cold dark matter (CDM) density. At present, these uncertainties are less important than the experimental and theoretical uncertainties in the precision observables.
Assuming that the nine observables listed above are uncorrelated, a fit has been performed with
χ2≡7∑n=1⎡⎣(Rexpn−Rtheonσn)2+2log(σnσminn)⎤⎦+χ2Mh+χ2Bs. (1)
Here denotes the experimental central value of the th observable (, , , and , ), ), is the corresponding MSSM prediction and denotes the combined error, as specified below. Additionally, is the minimum combined error over the parameter space of each data set as explained below, and and denote the contribution coming from the experimental limits on the lightest MSSM Higgs boson mass and on , respectively, which are also described below.
We also list below the parametric uncertainties in the predictions on the observables induced by the experimental uncertainty in the top- and bottom-quark masses. These errors neglect, however, the effects of varying and on the SUSY spectrum that are induced via the RGE running. In order to take the and parametric uncertainties correctly into account, we evaluate the SUSY spectrum and the observables for each data point first for the nominal values [16]222Using the most recent experimental value, [17] would have a minor impact on our analysis. and , then for and , and finally for and . The latter two evaluations are used by appropriate rescaling to estimate the full parametric uncertainties induced by the experimental uncertainties [16]333Using the most recent experimental error of [17] would also have a minor impact on our analysis. and . These parametric uncertainties are then added to the other errors (intrinsic, parametric, and experimental) of the observables as described in the text below.
We preface our discussion by describing our treatment of the cosmological cold dark matter density, which guides our subsequent analysis of the EWPO and BPO within the CMSSM and NUHM.
### 2.1 Cold Dark Matter Density
Throughout this analysis, we focus our attention on parameter points that yield the correct value of the cold dark matter density inferred from WMAP and other data, namely [18]. The fact that the density is relatively well known restricts the SUSY parameter space to a thin, fuzzy ‘WMAP hypersurface’, effectively reducing its dimensionality by one. The variations in the EWPO and BPO across this hypersurface may in general be neglected, so that we may treat the cold dark matter constraint effectively as a function. For example, in the CMSSM we focus our attention on ‘WMAP lines’ in the () planes for discrete values of the other SUSY parameters and [19, 20]. Correspondingly, in the following, for each value of , we present theoretical values for the EWPO and BPO corresponding to the values of on WMAP strips.
We note, however, that for any given value of there may be more than one value of that yields a cold dark matter density within the allowed range, implying that there may be more than one WMAP line traversing the the plane. Specifically, in the CMSSM there is, in general, one WMAP line in the coannihilation/rapid-annihilation funnel region and another in the focus-point region, at higher . Consequently, each EWPO and BPO may have more than one value for any given value of . In the following, we restrict our study of the upper WMAP line to the part with for and for , restricting in turn the range of .
The NUHM, with and , has two more parameters than the CMSSM, which characterize the degrees of non-universality of the two Higgs masses. The WMAP lines therefore should, in principle, be generalized to three-volumes in the higher-dimensional NUHM parameter space where the cold dark matter density remains within the WMAP range. We prefer here to focus our attention on ‘WMAP surfaces’ that are slices through these three-volumes with specific fixed values for (combinations of) the other NUHM parameters. These WMAP surfaces are introduced in more detail in the subsequent section describing our NUHM analysis, and will be discussed in more detail in Ref. [12].
In regions that depend sensitively on the input values of and , such as the focus-point region [21] in the CMSSM, the corresponding parametric uncertainty can become very large. In essence, the ‘WMAP hypersurface’ moves significantly as varies (and to a lesser extent also ), but remains thin. Incorporating this large parametric uncertainty naively in eq. (1) would artificially suppress the overall value for such points. This artificial suppression is avoided by adding the second term in eq. (1), where is the value of the combined error evaluated for parameter choices which minimize over the full data set.
### 2.2 The W Boson Mass
The boson mass can be evaluated from
M2W(1−M2WM2Z)=πα√2GF(1+Δr), (2)
where is the fine structure constant and the Fermi constant. The radiative corrections are summarized in the quantity [22]. The prediction for within the SM or the MSSM is obtained by evaluating in these models and solving eq. (2) for .
We use the most precise available result for in the MSSM [23]. Besides the full SM result, for the MSSM it includes the full set of one-loop contributions [24, 25, 23] as well as the corrections of [26] and of [27, 28] to the quantity ; see Ref. [23] for details.
The remaining intrinsic theoretical uncertainty in the prediction for within the MSSM is still significantly larger than in the SM. For realistic parameters it has been estimated as [28]
ΔMintr,currentW\raisebox−3.0pt$<∼$10MeV , (3)
depending on the mass scale of the supersymmetric particles. The parametric uncertainties are dominated by the experimental error of the top-quark mass and the hadronic contribution to the shift in the fine structure constant. Their current errors induce the following parametric uncertainties [14]
δmcurrentt=2.1GeV ⇒ ΔMpara,mt,currentW≈13MeV, (4) δ(Δαcurrenthad)=35×10−5 ⇒ ΔMpara,Δαhad,currentW≈6.3MeV . (5)
The present experimental value of is [30, 30, 31, 32, 33]
Mexp,currentW=80.398±0.025GeV. (6)
We add the experimental and theoretical errors for in quadrature in our analysis.
The current status of the MSSM prediction and the experimental resolution is shown in Fig. 1. We note that the CMSSM predictions for in the coannihilation and focus-point regions are quite similar, and depend little on . We also see that small values of are slightly preferred, reflecting the familiar fact that the experimental value of is currently somewhat higher than the SM prediction.
### 2.3 The Effective Leptonic Weak Mixing Angle
The effective leptonic weak mixing angle at the boson peak can be written as
sin2θeff=14(1−Reveffaeff) , (7)
where and denote the effective vector and axial couplings of the boson to charged leptons. We use the most precise available result for in the MSSM [14]. The prediction contains the same classes of higher-order corrections as described in Sect. 2.2.
In the MSSM with real parameters, the remaining intrinsic theoretical uncertainty in the prediction for has been estimated as [28]
Δsin2θintr,currenteff\raisebox−3.0pt$<∼$7×10−5, (8)
depending on the SUSY mass scale. The current experimental errors of and induce the following parametric uncertainties [14]
δmcurrentt=2.1GeV ⇒ Δsin2θpara,mt,currenteff≈6.3×10−5, (9) δ(Δαcurrenthad)=35×10−5 ⇒ Δsin2θpara,Δαhad,currenteff≈12×10−5. (10)
The experimental value is [30, 30]
sin2θexp,currenteff=0.23153±0.00016 . (11)
We add the experimental and theoretical errors for in quadrature in our analysis.
As compared with our older analyses [8, 9] we now use a new result for , obtained recently, that differs non-negligibly from that used previously, due to the inclusion of more higher-order corrections (which also result in a smaller intrinsic error). The corresponding new results in the CMSSM are shown in Fig. 2 for (left) and (right) as functions of . Whereas previously the agreement with the experimental result was best for , we now find best agreement for large values. However, taking all uncertainties into account, the deviation for generally stays below the level of one sigma. We note that the predictions for in the coannihilation and focus-point regions are somewhat different.
### 2.4 The Total Z Boson Decay Width
The total boson decay width, , is given by
ΓZ=Γl+Γh+Γ~χ01 , (12)
where are the rates for decays into SM leptons and quarks, respectively, and denotes the decay width to the lightest neutralino. We have checked that, for the parameters analyzed in this paper, always . However, SUSY particles enter via virtual corrections to and . We use the most precise available result for in the MSSM [14]. The prediction contains the same classes of MSSM higher-order corrections as described in Sect. 2.2.
So far no estimate has been made of the intrinsic uncertainty in the prediction for in the MSSM. Following the numerical analysis in Ref. [14], we use a conservative value of
ΔΓintr,currentZ\raisebox−3.0pt$<∼$1.0MeV (13)
The current experimental errors of and induce the following parametric uncertainties [14]
δmcurrentt=2.1GeV ⇒ ΔΓpara,mt,currentZ≈0.51MeV, (14) δ(Δαcurrenthad)=35×10−5 ⇒ ΔΓpara,Δαhad,currentZ≈0.32MeV. (15)
The experimental value is [30, 30]
Γexp,currentZ=2495.2±2.3MeV . (16)
We add the experimental and theoretical errors for in quadrature in our analysis.
A comparison of the MSSM prediction with the experimental value is shown in Fig. 3. We see that the experimental value is within a standard deviation of the CMSSM value at large , which corresponds to the SM value with the same Higgs boson mass. The marginal improvement in the CMSSM prediction at small is not significant. We note that the predictions for in the coannihilation and focus-point regions are somewhat different.
### 2.5 The Anomalous Magnetic Moment of the Muon
The SM prediction for the anomalous magnetic moment of the muon (see Refs. [34, 35, 36, 37, 38] for reviews) depends on the evaluation of QED contributions (see Refs. [39, 40] for recent updates), the hadronic vacuum polarization and light-by-light (LBL) contributions. The former have been evaluated in Refs. [41, 42, 43, 44, 38, 45, 46] and the latter in Refs. [47, 48, 49, 50, 51]. The evaluations of the hadronic vacuum polarization contributions using and decay data give somewhat different results. In view of the fact that recent measurements tend to confirm earlier results, whereas the correspondence between previous data and preliminary data from BELLE is not so clear, and also in view of the additional uncertainties associated with the isospin transformation from decay, we use here the latest estimate based on data [46]:
atheoμ=(11659180.5±4.4had±3.5LBL±0.2QED+EW)×10−10, (17)
where the source of each error is labeled. We note that the new data sets that have recently been published in Refs. [52, 53, 54] have been partially included in the updated estimate of .
The SM prediction is to be compared with the final result of the Brookhaven experiment E821 [55, 56], namely:
aexpμ=(11659208.0±6.3)×10−10, (18)
leading to an estimated discrepancy [46, 57]
aexpμ−atheoμ=(27.5±8.4)×10−10, (19)
equivalent to a 3.3- effect444Three other recent evaluations yield slightly different numbers [38, 43, 37], but similar discrepancies with the SM prediction.. While it would be premature to regard this deviation as a firm evidence for new physics, within the context of SUSY, it does indicate a preference for a non-zero contribution.
Concerning the MSSM contribution, the complete one-loop result was evaluated a decade ago [58]. In view of the correlation between the signs of and of [59], variants of the MSSM with are already severely challenged by the present data on , whether one uses either the or decay data, so we restrict our attention in this paper to models with . In addition to the full one-loop contributions, the leading QED two-loop corrections have also been evaluated [60]. Further corrections at the two-loop level have been obtained recently [61, 62], leading to corrections to the one-loop result that are . These corrections are taken into account in our analysis according to the approximate formulae given in Refs. [61, 62].
The current status of the CMSSM prediction and the experimental resolution is shown in Fig. 4, where the 1- and 2- bands are shown. We note that the coannihilation and focus-point region predictions for are quite different. For , the focus-point prediction agrees less well with the data, whereas for the focus-point prediction does agree well in a limited range of .
### 2.6 The Mass of the Lightest MSSM Higgs Boson
The mass of the lightest -even MSSM Higgs boson can be predicted in terms of the other MSSM parameters. At the tree level, the two -even Higgs boson masses are obtained as functions of , the -odd Higgs boson mass , and , whereas other parameters enter into the loop corrections. We employ the Feynman-diagrammatic method for the theoretical prediction of , using the code FeynHiggs [63, 64, 65], which includes all numerically relevant known higher-order corrections. The status of these results can be summarized as follows. For the one-loop part, the complete result within the MSSM is known [66, 67, 68]. Computation of the two-loop effects is quite advanced: see Ref. [69] and references therein. These include the strong corrections at and Yukawa corrections at to the dominant one-loop term, and the strong corrections from the bottom/sbottom sector at . In the case of the sector corrections, an all-order resummation of the -enhanced terms, , is also known [70, 71]. Most recently, the and corrections have been derived [72] 555 A two-loop effective potential calculation has been presented in Ref. [73], including now even the leading three-loop corrections [74], but no public code based on this result is currently available. . The current intrinsic error of due to unknown higher-order corrections has been estimated to be [69, 75, 13, 76]
ΔMintr,currenth=3GeV , (20)
which we interpret effectively as a confidence level limit: see below.
It should be noted that, for the unconstrained MSSM with small values of and values of which are not too small, a significant suppression of the coupling can occur compared to the SM value, in which case the experimental lower bound on may be more than 20 GeV below the SM value [77]. However, we have checked that within the CMSSM and the other models studied in this paper, the coupling is always very close to the SM value. Accordingly, the bounds from the SM Higgs search at LEP [78] can be taken over directly (see e.g. Refs. [79, 80]).
Concerning the analysis, we use the complete likelihood information available from LEP. Accordingly, we evaluate as follows the contribution to the overall function 666 We thank P. Bechtle and K. Desch for detailed discussions and explanations. . Our starting points are the values provided by the final LEP results on the SM Higgs boson search, see Fig. 9 in Ref. [78] 777 We thank A. Read for providing us with the values. . We obtain by inversion from the corresponding value of determined from Ref. [81]
12erfc(√12~χ2(Mh))≡CLs(Mh) , (21)
and note the fact that implies that as is appropriate for a one-sided limit. Correspondingly we set . The theoretical uncertainty is included by convolving the likelihood function associated with and a Gaussian function, , normalized to unity and centered around , whose width is :
χ2(Mh)=−2log(∫∞−∞e−~χ2(x)/2~Φ1.5(Mh−x)dx) . (22)
In this way, a theoretical uncertainty of up to is assigned for of all values corresponding to one parameter point. The final is then obtained as
χ2Mh=χ2(Mh)−χ2(116.4GeV) for Mh≤116.4GeV , (23) χ2Mh=0 for Mh>116.4GeV , (24)
and is then combined with the corresponding quantities for the other observables we consider, see eq. (1).
We show in Fig. 5 the predictions for in the CMSSM for (left) and (right). The predicted values of are similar in the coannihilation and focus-point regions. They depend significantly on , particularly in the coannihilation region, where negative values of tend to predict very low values of that are disfavoured by the LEP direct search. Also shown in Fig. 5 is the present nominal 95 % C.L. exclusion limit for a SM-like Higgs boson, namely [78], and a hypothetical LHC measurement of . We recall that we use the numerical value of the LEP Higgs likelihood function in our combined analysis.
### 2.7 The decay b→sγ
Since this decay occurs at the loop level in the SM, the MSSM contribution might a priori be of similar magnitude. A recent theoretical estimate of the SM contribution to the branching ratio at the NNLO QCD level is [82]
BR(b→sγ)=(3.15±0.23)×10−4 . (25)
We record that the error estimate for is still under debate, and that other SM contributions to have been calculated Refs. [83, 84], but these corrections are small compared with the theoretical uncertainty quoted in (25).
For comparison, the present experimental value estimated by the Heavy Flavour Averaging Group (HFAG) is [85, 3]
BR(b→sγ)=(3.55±0.24+0.09−0.10±0.03)×10−4, (26)
where the first error is the combined statistical and uncorrelated systematic uncertainty, the latter two errors are correlated systematic theoretical uncertainties and corrections respectively.
Our numerical results have been derived with the evaluation provided in Refs. [88, 86, 87], incorporating also the latest SM corrections provided in Ref. [82]. The calculation has been checked against other approaches [90, 91, 89]. For the current theoretical intrinsic uncertainty of the MSSM prediction for we use the SM uncertainty given in eq. (25) and add linearly the intrinsic MSSM corrections [91, 89] and the last two errors given by HFAG of [3]. The full intrinsic error is then added linearly to the sum in quadrature of the experimental error given by HFAG as and the parametric error.
In Fig. 6 we show the predictions in the CMSSM for for as functions of , compared with the 1- experimental error (full line) and the full error (dashed line, but assuming a negligible parametric error). For , we see that positive values of are disfavoured at small , and that small values of are disfavoured for all the studied values of if .
### 2.8 The Branching Ratio for Bs→μ+μ−
The SM prediction for this branching ratio is [92], and the present experimental upper limit from the Fermilab Tevatron collider is at the C.L. [93], providing ample room for the MSSM to dominate the SM contribution. The current Tevatron sensitivity is based on an integrated luminosity of about 780 collected at CDF. The exclusion bounds can be translated into a function for each value of 888 We thank C.-J. Stephen and M. Herndon for providing the numbers. A slightly more stringent upper limit of at the C.L. has been announced more recently by the D0 Collaboration [94]. However, the corresponding function is not available to us. Since the difference to the result employed here is small, we expect only a minor impact on our analysis. :
~χ2(Bs)≡χ2(BR(Bs→μ+μ−)) , (27)
with . The theory uncertainty is included by convolving the likelihood function associated with and a Gaussian function, , normalized to unity and centered around , whose width is given by the theory uncertainty, see below. Consequently,
χ2(Bs)=−2log(∫∞−∞e−~χ2(x)/2~Φth(BR(Bs→μ+μ−)−x)dx) . (28)
The final is then obtained as
χ2Bs=χ2(Bs)−χ2(0.266×10−7) for BR(Bs→μ+μ−)≥0.266×10−7 , (29) χ2Bs=0 for BR(Bs→μ+μ−)<0.266×10−7 . (30)
The Tevatron sensitivity is expected to improve significantly in the future. The limit that could be reached at the end of Run II is assuming 8 collected with each detector [95]. A sensitivity even down to the SM value can be expected at the LHC. Assuming the SM value, i.e. , it has been estimated [96] that LHCb can observe 33 signal events over 10 background events within 3 years of low-luminosity running. Therefore this process offers good prospects for probing the MSSM.
For the theoretical prediction we use results from Ref. [97], which are in good agreement with Ref. [98]. This calculation includes the full one-loop evaluation and the leading two-loop QCD corrections.
The theory error is estimated as follows. We take into account the parametric uncertainty induced by [99]
fBs=230±30MeV . (31)
The most important SUSY contribution to scales as
BR(Bs→μ+μ−)∼f2BsM4A . (32)
In the models that predict the value of at the low-energy scale, i.e. in our case the CMSSM, we additionally include the parametric uncertainty due to the shift in in eq. (32) that is induced by the experimental errors of and in the RGE running [98]. These errors are added in quadrature. The intrinsic error is estimated to be negligible as compared to the parametric error. Thus the parametric error constitutes our theory error entering in eq. (28).
In Fig. 7 the CMSSM predictions for for as functions of are compared with the present Tevatron limit. For (left plot) the CMSSM prediction is significantly below the present and future Tevatron sensitivity. However, already with the current sensitivity, the Tevatron starts to probe the CMSSM coannihilation region for and , whereas the CMSSM prediction in the focus-point region is significantly below the current sensitivity.
### 2.9 The Branching Ratio for Bu→τντ
The decay has recently been observed by BELLE [100], and the experimental world average is given by [100, 101, 11]
BR(Bu→τντ)exp=(1.31±0.49)×10−4 . (33)
We follow Ref. [102] for the theoretical evaluation of this decay. The main new contribution within the MSSM comes from the direct-exchange of a virtual charged Higgs boson decaying into . Taking into account the resummation of the leading enhanced corrections, within scenarios with minimal flavor violation such as the CMSSM and the NUHM, the ratio of the MSSM result over the SM result can be written as
BR(Bu→τντ)MSSMBR(Bu→τντ)SM=[1−(m2BuM2H±)tan2β1+ε0tanβ]2 . (34)
Here denotes the effective coupling of the charged Higgs boson to up- and down-type quarks, see Ref. [102] for details. The deviation of the experimental result from the SM prediction can be expressed as
BR(Bu→τντ)epxBR(Bu→τντ)SM=0.93±0.41 , (35)
where the error includes the experimental error as well as the parametric errors from the various SM inputs. We use eq. (34) for our theory evaluation, which can then be compared with eq. (35), provided that the value of agrees sufficiently well in the SM and in the MSSM (which we assume here). As an error estimate we use the combined experimental and parametric error from eq. (35), an estimated intrinsic error of , and in the CMSSM, as for , an additional parametric error from , evaluated from RGE running. These errors have been added in quadrature.
We show in Fig. 8 the theoretical results for the ratio of CMSSM/SM for as functions of for . These results are also compared with the present experimental result. The central (solid) line indicates the current experimental central value, and the other solid (dotted) lines show the current - ranges from eq. (35). For the SM result is reproduced over most of the parameter space. Only very small values give a ratio visibly smaller than 1. For the result varies strongly between 0 and 1, and the CMSSM could easily account for the small deviation of the central value of the experimental result from the SM prediction, should that become necessary. The prediction in the focus-point region is somewhat closer to the SM value.
### 2.10 The Bs–¯Bs Mass Difference ΔMBs
The oscillation frequency and consequently the the mass difference has recently been measured by the CDF Collaboration [103],
(ΔMBs)exp=17.77±0.12 ps−1 , (36)
which is compatible with the broader range of the result from D0 [104].
We follow Ref. [102] for the theory evaluation. The main MSSM contribution to the oscillation comes from the exchange of neutral Higgs bosons, but we use here the full result given in Ref. [102] (taken from Ref. [105]), where the leading dependence is given as
1−(ΔMBs)MSSM(ΔMBs)SM∼mb(mb)ms(mb)M2A . (37)
The SM value, obtained from a global fit, is given by [106]
(ΔMBs)SM=19.0±2.4 ps−1 , (38)
resulting in
(ΔMBs)exp(ΔMBs)SM=0.93±0.13 . (39)
The error in eq. (39) is supplemented by the parametric errors in eq. (37) from and, in the case of the CMSSM, as for , an additional parametric error from . These errors are added in quadrature. The intrinsic error, in comparison, is assumed to be negligible.
In Fig. 9 we show the results for the ratio of CMSSM/SM for as functions of for . These are also compared with the present experimental result. The central (solid) line indicates the current experimental central value, and the other solid (dotted) lines show the current - ranges from eq. (39). For the SM result is reproduced over the whole parameter space. Only for and in the coannihilation region can the CMSSM prediction be significantly lower than 1. Here the CMSSM could account for the small deviation of the experimental result from the central value SM prediction, should that be necessary.
## 3 CMSSM Analysis Including EWPO and BPO
We now use the analyses of the previous Section to estimate the combined function for the CMSSM as a function of , using the master formula (1). As a first step, Fig. 10 displays the distribution for the EWPO alone.
In the case (left panel of Fig. 10), we see a well-defined minimum of for when , which disappears for large negative and is not present in the focus-point region. The rise at small is due both to the lower limit on coming from the direct search at LEP and to , whilst the rise at large is mainly due to (see Fig. 4). The measurement of (see Fig. 1) leads to a slightly lower minimal value of , but there are no substantial contributions from any of the other EWPO. The preference for in the coannihilation region is due to (see Fig. 5), and the relative disfavour for the focus-point regions is due to its mismatch with (see Fig. 4).
In the case (right panel of Fig. 10), we again see a well-defined minimum of , this time for to 500 GeV, which is similar for all the studied values of . In this case, there is also a similar minimum of for the focus-point region at . The increase in at small is due to as well as , whereas the increase at large is essentially due to . We note that the overall minimum of is similar for both values of , and represents an excellent fit in each case.
Fig. 11 shows the corresponding combined for the BPO alone. For both values of , these prefer large values of , reflecting the fact that there is no hint of any deviation from the SM, and the overall quality of the fit is good. Small values of are disfavoured, particularly in the coannihilation region with , mainly due to . The focus-point region is generally in very good agreement with the BPO data, except at very low for .
Finally, we show in Fig. 12 the combined values for the EWPO and BPO, computed in accordance with eq. (1). We see that the global minimum of for both values of . This is quite a good fit for the number of experimental observables being fitted, and the is similar to the one for the EWPO alone. This increase in the total reflects the fact that the BPO exhibit no tendency to reinforce the preference of the EWPO for small : rather the reverse, in fact. For both values of , the focus-point region is disfavoured by comparison with the coannihilation region, though this effect is less important for . For , and are preferred, whereas, for , and are preferred. This change-over is largely due to the impact of the LEP constraint for and the constraint for .
We display in Fig. 13 the functions for various SUSY masses in the CMSSM for , including (a) , (b) and (which are very similar), (c) , (d) , (e) and (f) . We see two distinct populations of points, corresponding to the coannihilation (which is favoured) and focus-point regions (which is disfavoured). In the latter region, very low values of are preferred, as can be seen in panels (a) and (f), relatively small values of , as can be seen in panel (b), large values of , as can be seen in panels (c) and (e), and large values of , as can (not) be seen in panel (d). Compared to the analysis in Ref. [9], where was the only BPO included, and where a top quark mass of was used, there is no significant shift of the values of the masses where has its minimum, which is in the coannihilation region. As before, the present analysis gives hope for seeing squarks and gluinos in the early days of the LHC (panels (e) and (f)), and also hope for seeing charginos, neutralinos and staus at the ILC (panels (a), (b) and (c)), whereas observing the heavier Higgs bosons would be more challenging (panel (d)).
In Fig. 14 we show the analogous functions for various SUSY masses in the CMSSM for : (a) , (b) and (which are very similar), (c) , (d) , (e) and (f) . We again see the clear separation between the focus-point and coannihilation regions, interpolated by a light-Higgs pole strip, and that the coannihilation region is somewhat preferred. As for lower , small values of and larger values of are preferred, and also small values of and larger values of . Again as for , compared to the analysis in Ref. [9], where was the only BPO included and where a top quark mass of was used, we do not find a significant shift in the values of the masses with lowest . The sparticle masses are generally higher than for : finding squarks and gluinos should still be ‘easy’ at the LHC, but seeing charginos, neutralinos and staus at the ILC would be more challenging, depending on its center-of-mass energy.
Analogously to the sparticle masses in Figs. 13 and 14, we display in Fig. 15 the total functions for , as calculated in the CMSSM for (left panel) and (right panel). We recall that this theoretical prediction has an intrinsic uncertainty of , which should be combined with the experimental error in . It is a clear prediction of this analysis that should be very close to the LEP lower limit, and probably , though a value as large as is possible (but is disfavoured), particularly if .
In the case of the SM, it is well known that tension between the lower limit on from the LEP direct search and the relatively low value of preferred by the EWPO has recently been increasing [30, 31]. This tension is strongly reduced within the CMSSM, particularly for . We display in Fig. 16 the global functions for the EWPO and BPO, but this time omitting the contribution for the LEP Higgs search. This corresponds to the fitted value of in the CMSSM. Comparing Fig. 16 and Fig. 15, we see that all data (excluding ) favour a value of if and if . On the other hand, the currently best-fit value of | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9488250017166138, "perplexity": 1017.9198350551906}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00194.warc.gz"} |
https://www.physicsforums.com/search/4753358/ | # Search results
1. ### Light refracting through a plastic sphere
Homework Statement A small light bulb is placed 10.0 cm from the center of a plastic sphere of radius 1.0 cm and refractive index 1.40. Where is the image of the bulb? Homework Equations 1) Thin lens equation 2) Thick lens equation The Attempt at a Solution I realize that I'm...
2. ### Area of triangle given 3 vectors pointing to vertices
Yep, you're right. One of those was supposed to be an A cross B. Thanks.
3. ### Area of triangle given 3 vectors pointing to vertices
So, the area of the triangle between two vectors (let's say A and B) is 0.5|Axb| right? I still don't see how I can use that to solve this. I can find the area of every triangle but the one I need. EDIT: Alright, I was just being a dummy. I redrew my picture so that each of the vectors point...
4. ### Area of triangle given 3 vectors pointing to vertices
Homework Statement Three vectors A, B, C point from the origin O to the three corners of a triangle. Show that the area of the triangle is given by area = \frac{1}{2}|(B\timesC) + (C\timesA) + (A\timesC)| Homework Equations area of triangle with sides a, b, c = \frac{1}{2}|a\timesc|...
5. ### Metal block sliding horizontally
Homework Statement A metal block of mass m slides on a horizontal surface that has been lubricated with a heavy oil so that the block suffers a viscous resistance that varies as the 3/2 power of the speed: F(v) = -cv3/2 If the initial speed of the block is vo at x = 0, show that the block...
6. ### Trajectory of a ball
Homework Statement A cannon shoots a ball at an angle \theta above the horizontal ground a) Neglecting air resistence, find the ball's position (x(t) and y(t)) as a function of time. b) Take the above answer and find an equation for the ball's trajectory y(x). Homework Equations... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8278947472572327, "perplexity": 539.4297722718019}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057225.57/warc/CC-MAIN-20210921161350-20210921191350-00343.warc.gz"} |
http://sites.maths.cf.ac.uk/swngsn/category/abstracts/ | # Category Archives: Abstracts
Abstracts
## Weak Rigidity/Compactness Problems in Nonlinear Partial Differential Equations
Two of the fundamental issues in the analysis of generalised solutions for nonlinear PDEs are the weak rigidity/continuity of nonlinear PDEs and the compactness/convergence of approximate/multiscale solutions. In this talk, we will discuss some recent developments on these issues for several important classes of nonlinear PDEs
## Carlo Mercuri – A compactness result for Schrödinger-Poisson systems.
I will present a compactness result for certain sequences of approximated critical points of functionals (Palais-Smale sequences) related to a class of Schrödinger-Poisson systems. Applications will be discussed in relation to the minimax approach for finding positive solutions to these systems. This is a joint work with Megan Tyler (PhD student, Swansea University).
## Elaine Crooks – Invasion speeds in a competition-diffusion model with mutation.
We consider a reaction-diffusion system modelling the growth, dispersal and mutation of two phenotypes. This model was proposed in by Elliott and Cornell (2012), who presented evidence that for a class of dispersal and growth coefficients and a small mutation rate, the two phenotypes spread into the unstable extinction state at a single speed that… Read More »
## Nicolas Dirr – Existence of solutions and convergence of a finite-element scheme for a stochastic pororus-medium equation with multiplicative noise in divergence form.
We show existence by showing convergence of a suitable finite element scheme. This is joint work with G. Gruen and H. Grillmeyer.
## PDEs and probability – Horatio Boedihardjo
We will discuss the classical relationships between probability and PDEs, as well as some recent developments. In particular, we will explain our ongoing study of an eigenvalue problem associated with a linear matrix-valued elliptic PDE from probability theory. Joint work with Ni Hao (UCL).
## Stochastic homogenisation of high-contrast media – Mikhail Cherdantsev
Using a suitable stochastic version of the compactness argument of V. V. Zhikov, we develop a probabilistic framework for the analysis of heterogeneous media with high contrast. We show that an appropriately defined multiscale limit of the field in the original medium satisfies a system of equations corresponding to the coupled “macroscopic” and “microscopic” components… Read More »
## On the existence and uniqueness of vectorial absolute minimisers in Calculus of Variations in L-infinity – Nikos Katzourakis
Calculus of Variations in the space L-infinity has a relatively short history in Analysis. The scalar-valued theory was pioneered by the Swedish mathematician Gunnar Aronsson in the 1960s and since then has developed enormously. The general vector-valued case delayed a lot to be developed and its systematic development began in the 2010s. One of the… Read More »
## Uniqueness of minimisers of Ginzburg-Landau functionals – Luc Nguyen
We provide necessary and sufficient conditions for the uniqueness of minimisers of the Ginzburg-Landau functional for $\RR^n$-valued maps under suitable convexity assumption on the potential and for $H^{1/2} \cap L^\infty$ boundary data that is non-negative in a fixed direction $e\in \SSphere^{n-1}$. Furthermore, we show that, when minimisers are non-unique, the set of minimisers is invariant… Read More »
TBC | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8174152374267578, "perplexity": 1157.746086866752}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00202.warc.gz"} |
https://www.physicsforums.com/threads/pwm-for-motor-control.268310/ | # PWM for motor control
1. Oct 30, 2008
### cepheid
Staff Emeritus
I recently encountered (not for the first time) a situation in which a PWM output from a microcontroller is sent to a MOSFET bridge that drives a motor (a very high current, inductive load). I started to wonder exactly WHY it is that the duty cycle of the PWM controls the motor speed. I know that the average of the waveform is proportional to the duty cycle, but it seemed strange that the load would behave as though it were being exposed to the average voltage when in fact it was effectively being exposed to a rapidly time-varying one. Then I read something that hinted that the inductive load was somehow providing a smoothing effect. I figured that if I could just solve the system and figure out the current through the motor as a function of time, I'd be set. After all...
...presumably the current through the motor is proportional to its speed (if somebody knows otherwise, please let me know!)...
...so I modelled the system as a series RL circuit with a square wave voltage source v(t) (is this reasonable?)
I set up the typical DE:
$$v(t) = i(t)R + L\frac{di(t)}{dt}$$
After solving using an integrating factor, you get:
$$i(t) = \frac{e^{-\frac{R}{L}t}}{L}\int_0^t e^{\frac{R}{L}\tau} v(\tau)\, d\tau$$
I guess we can define the square wave piecewise, letting it have period T and duty cycle D where D is a number between 0 and 1 telling you for what fraction of the period it's high:
$$v(t) = V_{max}, \ \ \ \ 0 \leq t \leq DT$$
$$v(t) = 0, \ \ \ \ DT \leq t \leq T$$
$$v(t+T) = v(t)$$
So what's i(t)? if t = T, the integral only goes over one square pulse:
$$i(T) = \frac{e^{-\frac{R}{L}T}}{L}\int_0^{DT} V_{max} e^{\frac{R}{L}\tau} \, d\tau$$
Now it took me a while, but I finally figured that at some *arbitrary* time t, the integral will have gone over n square pulses where:
$$n = \left \lceil \frac{t}{T} \right\rceil$$
That's all well and good, but I don't know how to calculate that to get some sensible result for i(t). The integral works out to:
$$i(t) = n\frac{e^{-\frac{R}{L}t}}{R}V_{max} e^{\frac{R}{L}DT}$$
So...what do I do now? Ideally I'd like to get the result that i(t) is constant and equal to
$$D\frac{V_{max}}{R}$$
That would be *awesome*, because it would mean that the current is a fraction of the max that could be drawn, the fraction being determined by the duty cycle, and everything would make sense. But I don't know how to get there.
1. Does anybody know how to solve this math problem?
2. Am I thinking about this in the right way? Because I can't think of any other way that "PWM duty cycle controls motor speed" remotely makes any sort of sense.
Last edited: Oct 30, 2008
2. Oct 31, 2008
### cepheid
Staff Emeritus
Something just struck me when I was considering a simpler example (a PWM light dimmer)
Does this work as a good qualitative explanation for the *motor* as well? Still, it doesn't quite cut it. In the lightbulb, it doesn't matter that the current drops to zero during the off portion of the cycle, because we can't perceive it. Or maybe it doesn't drop to zero, due to back emf...in which case we're back to where we started...trying to solve for i(t) to figure out exactly what happens! Can somebody help me do the math?
3. Oct 31, 2008
### dlgoff
I'm no expert but I think you will need to model using a fourier integral. Since a pulse v(t) is the sum of all frequencies each having their own phase.
4. Oct 31, 2008
### MATLABdude
I won't address the mathematical points of this discussion, but as for the qualitative hand-wavy arguments, well...
In the case of a motor driver driving a motor, the motor (and wheel, and robot / car, etc.) all have a certain (rotational) inertia. If the motor driver frequency were low enough, you would see the motor stop and go in jerky motions. But go above this critical frequency, and only smooth motion results.
In the case of an (incandescent) light, if you assume that the number of photons emitted might be high or zero (in reality, you receive a portion of the sine waveform), but you receive some total number of photons at your eyes every 1/30 of a second or so--roughly the upper limit of your eyes' 'sampling frequency'. As a result, you see the average of the intensity. For all of us, 50/60 Hz AC driven incandescent lights look continuous, despite a zero crossing 120 times a second--this might be due to the incandescent lights' inductance--but someone else would have to confirm that. Fluorescent lights are able to respond fast enough that they do flicker at 100/120 Hz. For most of us, this is imperceptible, but I know people who are driven crazy by fluorescents because they're able to see this constant flickering--or some beat frequency, at least.
If you're familiar with communications and the Nyquist theorem, these are both sort of physical manifestations of that.
5. Nov 5, 2008
### cepheid
Staff Emeritus
Yeah, that's a really good point, and I can't believe I didn't think of it. For this reason, the steadiness of i(t) doesn't matter quite as much, and so I've abandoned the mathematical analysis (I have other things to do).
Yes I am, and we have since discussed this further in the other thread as well. Thank you very much for your remarks!
6. Nov 10, 2008
### famousken
Energy=power X time In a pulse width modulation circuit, you are not varying the power put into a load, that stays the same, but you are varying the time it is applied to it, thus varying the total energy delivered.
7. Nov 10, 2008
### Averagesupernova
You can't really change one without changing the other in this case. Average the power and it certainly changes with duty cycle.
8. Nov 10, 2008
### famousken
average, yes, which is factors time into the equation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8413575887680054, "perplexity": 707.1832587691457}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607862.71/warc/CC-MAIN-20170524211702-20170524231702-00549.warc.gz"} |
http://mathoverflow.net/questions/112536/is-there-something-interesting-in-the-uniqueness-condition-for-a-sheaf | # Is there something interesting in the uniqueness condition for a sheaf?
After digesting the Presheaf definition by the very first time, one feels (at least I felt) a strange sensation noticing the existence and uniqueness conditions to graduate that Presheaf as a sheaf, but although some "natural" examples are given to show that the existence condition is not garanted (bounded functions is the canonical one), all examples that I occur are bizarre and absolutely unnatural, in the text books I've seen I found nothing.
So the question is: Is there some "interesting" and/or "natural" Presheaf (I mean a Presheaf useful for something at least pedagogically) which supports existence and fails only the uniqueness condition?
Thanks
-
The answer, of course, depends on which presheaves you count as natural or non-pathological. If you're only willing to consider presheaves F on X of the type "F(U) = {functions on X satisfying some condition}" then you're always going to have uniqueness. – Tom Leinster Nov 16 '12 at 3:52
Well, all "natural" presheafs are presheafs of functions, for which uniqueness is automatic. However, the presheaf quotient of a sheaf by a subpresheaf need not satisfy uniqueness. For example, consider at the presheaf quotient of the sheaf of locally constant functions on a space by the subpresheaf of constant functions. – anon Nov 16 '12 at 3:54 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.939039945602417, "perplexity": 918.4748969050863}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257831770.41/warc/CC-MAIN-20160723071031-00324-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://www.maths.kisogo.com/index.php?title=Characteristic_property_of_the_quotient_topology | # Characteristic property of the quotient topology
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.
AKA: the Characteristic property of the quotient topology which redirects here
## Statement
In this commutative diagram[ilmath]f[/ilmath] is continuous[ilmath]\iff[/ilmath][ilmath]f\circ q[/ilmath] is continuous [ilmath]\xymatrix{ X \ar[d]_{q} \ar[dr]^{f\circ q} & \\ Y \ar[r]_f & Z }[/ilmath]
Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]q:X\rightarrow Y[/ilmath] be a quotient map. Then[1]:
• For any topological space, [ilmath](Z,\mathcal{ H })[/ilmath] a map, [ilmath]f:Y\rightarrow Z[/ilmath] is continuous if and only if the composite map, [ilmath]f\circ q[/ilmath], is continuous
## Proof
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9775208830833435, "perplexity": 2913.588399244041}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572833.95/warc/CC-MAIN-20220817032054-20220817062054-00298.warc.gz"} |
http://forrestbao.blogspot.com/2014/06/what-determinant-really-determines.html?utm_source=feedburner&utm_medium=feed&utm_campaign=Feed%3A+ThinkForrestThink+%28Think%2C+Forrest%21+Think%21%29 | ### What a determinant really determines
The determinant is an important concept in linear algebra. Since a determinant is defined for a square matrix, many people would think it as a property of a matrix. But what does it really determine? In other words, why did mathematicians invent it? And why is it defined so? For example, why $\ \det\left ( \left [ \array{ a & b \cr c & d} \right ] \right) = ad - bc$?
Answer: It determines whether a system of linear equations has a unique solution.
The concept of matrix does not come from nowhere. It is strongly related to linear equations. Let's consider a system of linear equations: \begin{align} ax + by & = C_1\\ cx + dy & = C_2 \label{eq:two_eqs} \end{align}
If $C_1=C_2=0$, eliminating $x$ will result in this: $(ad-bc)y=0$. Pay attention to the coefficient: $ad-bc$. Does it look like $\ \det\left ( \left [ \array{ a & b \cr c & d} \right ] \right)$?
Now if $ad-bc=0$, then $y$ can take whatever value to be a solution of the equations. Namely, the equation has unlimited amount of solutions.
If $C_1 \not = 0$ and $C_2 \not = 0$, the result of eliminating $x$ will have a non-zero constant on the right-hand side: $(ad-bc)y=C_3$. Now if $ad-bc = 0$, there is no way for the equations to have a solution.
Therefore, the determinant actually determines whether a system of linear equations has a unique solution. A system of linear equations can be represented as a matrix. So the determinant of the matrix defines the property of the linear system that the set of equations defines.
References:
1. System of Linear Equations, http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/system/system.html
2. Determinant, Wolfram MathWorld, http://mathworld.wolfram.com/Determinant.html
3. Determinant, Wikipedia, http://en.wikipedia.org/wiki/Determinant | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9605260491371155, "perplexity": 162.82116937812}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218188926.39/warc/CC-MAIN-20170322212948-00197-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/weather-station-question.380355/ | # Homework Help: Weather Station question
1. Feb 21, 2010
### Claire_01
A weather station at the airport measured a station pressure of 1014 mb. The density of dry air is 1.3 kg/m^3. The gas constant for dry air, R, is 287 J/kg-K. Calculate the temperature of the dry air at the airport.
1 mb = 100 J/m ^3
AND
The surface pressure at the airport then decreased to 1010 mb but the air temperature reamined the same (as the answer in #1). Calculate new density of air.
*I'm really confused and I would greatly appreciate it if someone can walk me through this problem
This is what I have so far, I know you have to use the Ideal Gas law
P= p R T
1014 mb = 1.3 kg/m^3 x 287 j/kg-k x T
Last edited: Feb 21, 2010
2. Feb 21, 2010
### Mindscrape
Hey, that looks really good so far! Surely you know enough algebra to take it from there. Just as a comment though, make sure you watch your units! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8045719861984253, "perplexity": 1483.3628541809878}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376825349.51/warc/CC-MAIN-20181214022947-20181214044447-00225.warc.gz"} |
http://vkxq.jacobsweb.it/graphs-of-sine-and-cosine-functions-answer-key.html | The graph is a smooth curve. Trigonometric Functions and Their Graphs Notes. Hint: Sine is odd and cosine is. graph 20) Identify the inte on [0, 2m] for which the cosine function is increasing. Transformations of Sine and Cosine Functions A sinusoid is a transformation of the graph of the sine function. Sketch the graphs off x = sin x g(x) = sin x- 1 over at least one period, labeling each axis. 12/17 T or 12/18 W. Some of the worksheets for this concept are Graphing trig functions, Graphs of trig functions, Amplitude and period for sine and cosine functions work, Graphing sine and cosine functions, Work 15 key, Honors algebra 2 name, Sinusoidal functions work, Of the sine and cosine functions. 00 -900 900 1 00 -2700 -1 00 o 6300. The _____ represents half the distance between the max and min of a sine or cosine graph. Day 2 - Listing Discontinuities Homework Answer Key. y 5 sin x, 2p units right 16. 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Use the unit circle to help evaluate the sine function y = sin(x) for values of x that are multiples of 4 S between 2S and 2S. The second derivative test 89 39. Graphing Sine and Cosine Fill in the blanks and graph. 2 The Unit Circle and Circular Functions - 6. Notice that both the sine and cosine have a maximum value of 1 and a minimum value of -1. In these examples we will graph a sine and cosine function using a table of values. The international standard pitch has been set at a frequency of 440 cycles/second. On the same grid, sketch the graph of f x. Show Answer. 6 Introduction to Trig Identities Answers 1. sin 7! 2! 1 1 1 Find the values of •for which each equation is true. Determining the equation of a circle by completing the square. The questions are about determing the period from the graph and also matching graphs and trigonometric functions. Practice Quiz - Graphing Sine and Cosine Use transformations to graph each of the following functions. View the graph and select. 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Then the amplitude of f is the number 2 M m Example 1: Specify the period and amplitude of the given function Now let’s talk about the graphs of the sine and cosine functions. 4 Investigation: Sketching the graphs of: f(x) = sin x f(x) = cos x f(x) = tan x 5. 2 - Graphs of Rational Functions; Assign 3. Trigonometric functions repeat every 2π radians. 00 -900 900 1 00 -2700 -1 00 o 6300. Precalculus Chapter 6 Worksheet Graphing Sinusoidal Functions in Degree Mode Find the amplitude, period, phase (horizontal) displacement and translation (vertical displacement). The remaining trigonometric functions can be most easily defined in terms of the sine and cosine, as usual: tanx = sinx cosx cotx = cosx sinx secx = 1 cosx cscx = 1 sinx and they can also be defined as the corresponding ratios of coordinates. 7153 8)cos-1 -0. (Check your answer with your graphing calculator!) f x x( ). The graphs of all sine and cosine functions are related to the graphs of y = sin x and y = cos x which are shown below. When you write a sine or cosine function for a sinusoid, you need to find the values of a, b>0, h, and kfor y= a sin b(x º h) + k or y = a cos b(x º h) + k. Thus, key points in graphing sine functions are obtained by dividing the period into four equal parts. Graphing A Trig Function You. Then find. Give the amplitude and period of each function. Learn how to graph trigonometric functions and how to interpret those graphs. The cosine function of an angle. Note also that the cosine has a maximum value of 1, and a minimum value of −1. Using degrees, find the amplitude and period of each function. This note explains the following topics: Foundations of Trigonometry, Angles and their Measure, The Unit Circle: Cosine and Sine, Trigonometric Identities, Graphs of the Trigonometric Functions, The Inverse Trigonometric Functions, Applications of Trigonometry, Applications of Sinusoids, The Law of Sines and cosines, Polar Form of Complex Numbers. Videos, practice questions, ask and answer questions. I need to describe the Amplitude, Period, Domain, Range and X-intercepts of the graphs of one of the following cosine functions and then relate each property to the unit circle definition of cosine. Each question is a chance to learn. How to sketch the graphs of basic sine and cosine functions Important Vocabulary Define each term or concept. sin o h p yp p cos tan h a y d p j o a p d p j csc sec o h p yp p h a y d p j cot o a p d p j Notice that the sine, cosine, and tangent functions are reciprocals of the cosecant, secant, and cotangent functions, respectively. Using Key Points to Sketch a Sine Curve Sketch the graph of on the interval. Students should discuss the related heights on the unit. Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Label x-axis in terms of π. It explains how to identify the amplitude, period, phase shift, vertical shift, and midline of a sine or cosine function. Note also that the cosine has a maximum value of 1, and a minimum value of −1. Graph the reciprocal function of. The graph of y=sin(x) is like a wave that forever oscillates between -1 and 1, in a shape that repeats itself every 2π units. DO NOT GRAPH!! 1. They are both expressed according to the triangle on the right, where each letter represents one side-length (lower-case) and the angle opposite to it (upper-case). Answer Keys Answers for math worksheets, quiz, homework, and lessons. Lesson 6 Basic Graphs of Sine and Cosine. Give the amplitude and period of each function. Determine the amplitude, period, and vertical shift of each function. You've already learned the basic trig graphs. Lakeland Community College Lorain County Community College. To account for a phase shift of , subtract from the x-values of each of the key points for the graph of y = 2 sin 5x. Corrective Assignment. • Use amplitude and period to help sketch graphs. y: 3 sin x y: 3 cos 4500 —x 19. A sine function has the following key features: Period = π Amplitude = 2 Midline: y= −2 y-intercept: (0, -2) The function is a reflection of its parent function over the x-axis. ) Graph f(x) = 8cos(2x) + 1 along the domain –π < x < π. y = -2 sin(-2x). 5) 6)y = cos-1 - 2 2 Give the value of the function in radians. Answer: The amplitude is 0. In other words, instead of the graph's midline being the x-axis, it's going to be the line y = -1. Solution: Cosine Function. Displaying top 8 worksheets found for - Graphs Of Sine And Cosine Functions. y = -2 sin(-2x). Students will then identify the amplitude and period of other sine and cosine functions, although they will not be required to graph. For the sine graph the key points are these points 0, 0 pi of a 2, 1, pi 0 3pi over 2 negative 1 and 2 pi, 0. 9) 10) Domain: Range: Domain: Range: Amplitude: 2 Period: Amplitude: 1 Period: π. Answer: We are given the tangent function. Graphs of the Sine and Cosine Functions Divide the interval into four equal parts to obtain the values for which sin bx or cos bx equal -1, 0, or 1. 7 Inverse Trigonometric Functions p. y = sin 4x 2. Geometry Diagnostic Test Answer Section MULTIPLE CHOICE 1. The students have looked at the graphs of sine and cosine over the last several days and are beginning to remember how the look. The period of any sine or cosine function is 2π, dividing one complete revolution into quarters, simply the period/4. Graphs provided. This is how I like to introduce sine and cosine graphs this unit (after spending time with the unit circle and rotations it is a great way to see how we get the sinusoidal graph from a circle, see my blog post here for details). Graphing Sine And Cosine Functions Worksheet Answers - The easiest way of implying a worksheet is that it is a mono spreadsheet that is present into the package supplied by Microsoft. Sine and cosine functions are periodic functions. Chapter 8: Sinusoidal Functions 510 Getting Started: Sine and Cosine Patterns 512 8. Finding the equation of a parabola using focus and directrix. If f is sine or cosine, then −1 ≤ a ≤ 1 and, if f is tangent, then a ∈ R. 1 Practice — Graphing Sine and Cosine Pre-Calculus Name: For 1-3, identify the amplitude, period, frequency and vertical shift of each function. 7a)-- 1 point. 6 Graphs of the Sine and Cosine Function Graph each function using degrees. Students will then identify the amplitude and period of other sine and cosine functions, although they will not be required to graph. The value of h indicates a translation left (h < 0) or right (h > 0). Identify the phase for a sine or cosine function. y = cos 5x 3. Extreme Values. 12 - 13 Friday 10/25 Writing functions cont'd Quiz - Graphing Sine and Cosine. • Learn how to look at a graph of a transformed sine or cosine function and to write a function to represent that graph explore several real-world settings and represent the situation with a trigonometric function that can be used to answer questions about the situation. Tues 4/21: More Work with Graphing Cosine and Sine Functions (Unwrapping the Unit Circle) Complete the worksheet for today that builds on yesterday’s lesson with Cosine and Sine graphs. Give your answer correct to 3 significant figures Diagram NOT accurately 1500 60 m Angle 1500. Determining the equation of a circle by completing the square. The general sine and cosine graphs will be illustrated and applied. You can make copies of the Answer Keys to hand out to your class, but. DO NOT GRAPH!! 1. 3 The Tangent and Cotangent Functions Sec 5. Find the midpoint of the interval by adding the x-values of the endpoints and dividing by 2. This article will teach you how to graph the sine and cosine functions by hand, and how each variable in the standard equations transform the shape, size, and direction of the graphs. y = 7 cos - 1 5. But just as you could make the basic quadratic, y = x 2, more complicated, such as y = -(x + 5) 2 - 3, so also trig graphs can be made more complicated. 6 graphing other 4 trig functions worksheet practice test review worksheet (answers part 1 and. The graph of y= sin 1 xlooks like:. How to sketch the graphs of basic sine and cosine functions Important Vocabulary Define each term or concept. Vertical Shifting of Sinusoidal Graphs. 3 62/87,21 The general form of the equation is y = a sin bt, where t is the time in seconds. A sine graph is a graph of the function =y sin θ. I can graph sine function and its translations. Free trigonometric equation calculator - solve trigonometric equations step-by-step This website uses cookies to ensure you get the best experience. Free worksheet(pdf) and answer key on graphing sine , cosine ,tangent with phase shifts. Because the. It is mandatory to procure user consent prior to running these cookies on your website. The motion of the toy starts at its highest position of 5 inches above its rest point, bounces down to its lowest position of 5 inches below its rest point, and then bounces back to its highest position in a total of 4 seconds. Graphing Trig Functions Practice Worksheet With Answers Students will practice graphing sine and cosine curves : a) identify period and amplitude based on equation or on the graph b) write equation from graph c) write. Precalculus (6th Edition) answers to Chapter 6 - The Circular Functions and Their Graphs - 6. Practice B Graphs of Sine and Cosine Using f x sinx or g x cosx as a guide, graph each function. Graphs of Sine and Cosine Functions : Questions like Determine the amplitude and period of each function, …. Unit 7: Graphing Trigonometric Functions. Different sounds create different waves. 5 and b = 4 is y = 1. Then the amplitude of f is the number 2 M m Example 1: Specify the period and amplitude of the given function Now let’s talk about the graphs of the sine and cosine functions. 22 scaffolded questions on equation, graph involving amplitude and period. Then its graph is:-6 (The hash marks on the x-axis are in increments of ˇ=2. functions using different representations. Students will match 10 graphs to 10 sine or cosine equations by finding the amplitude and period of each function. 7 -8 Wednesday 10/23 Continue Graphing Sine and Cosine (Period Changes) Worksheet graphing problems #9 - 16 on pp. What is the range of f(x) = sin(x)? the set of all real numbers -1 < or = y < or = 1 Which set of transformations is needed to graph f(x) = -2sin(x) + 3 from the parent sine function?. ANS: B PTS: 1 DIF: Easy OBJ: Section 5. y = –4 sin 3x + 2 5. Both graphs are shown below to emphasize the difference in the final results (but we can see that the above functions are different without graphing the functions). The student will submit a synopsis at the beginning of the semester for approval from the departmental committee in a specified format. Graphing Trigonometric Functions Scavenger Hunt This walk around activity will help students practice identifying the key characteristics of the sine, cosine, and tangent functions and matching them to their graphs. , the the function is one-to-one and so it does have an inverse. y 5 sin (x 2 p) 2 1 Write an equation for each of the following translations. six trig functions. Vocab: Midline/Sinusoidal Axis. Solution: Cosine Function. Mathematics 5 SN SINUSOIDAL GRAPHS AND WORD PROBLEMS The tuning fork is a device used to verify the standard pitch of musical instruments. Sine and Cosine Graphs: Translations. The problem is as follows: A buoy in the harbor of San Juan, Puerto Rico, bobs up and down. What is the graph of each translation in the interval 0 Q2 è? a. In this topic, we’re going to focus on three trigonometric functions that specifically concern right-angled triangles. Therefore, a sinusoidal function with period DQGDPSOLWXGH WKDWSDVVHV through the point LV y = 1. The trigonometry equation that represents this relationship is. In comparing the graphs of the cosecant and secant functions with those of the sine and cosine functions, respectively, note that the "hills" and "_____" are interchanged. Some of the worksheets for this concept are Graphing trig functions, Graphs of trig functions, Amplitude and period for sine and cosine functions work, Graphing sine and cosine functions, Work 15 key, Honors algebra 2 name, Sinusoidal functions work, Of the sine and cosine functions. Graphing Sine Functions. Find the value of the coordinates of the points A, B, and C. The cosecant, secant, and cotangent ratios can be expressed in terms of sine, cosine. Amplitude = | a | Let b be a real number. The Period goes from one peak to the next (or from any point to the next matching point): The Amplitude is the height from the center line to the peak (or to the trough). 6 3 Graphing Sine and Cosine Functions Objective Use the graphs from Graphing Sine And Cosine Functions Worksheet, source:slideplayer. Student needs to show proof. 22 matching sine/cosine graphs (excluding horizontal shift) with 4 extra graphs thrown into the answer bank. Pa Functions And Their Graphs Solutions Examples S. 4 Writing the equation of Sine and Cosine: 11. • Sketch translations of these functions. The graphs overlap. Sample Test Answer Key Trigonometric Functions and Their Graphs. Trigonometry Name Pd Date Graphing Sine and Cosine Practice Worksheet Graph the following functions over two periods, one in the positive direction and one in the negative direction. Key included. WORD ANSWER KEY. 3-20 Domain and range. The motion of the toy starts at its highest position of 5 inches above its rest point, bounces down to its lowest position of 5 inches below its rest point, and then bounces back to its highest position in a total of 4 seconds. The trigonometric functions are also known as the circular functions. SWBAT : Identify and draw a sine and cosine graph. sin = o_pp hyp cos = _adj hyp tan = o_pp adj The domain of each of these trigonometric functions is the set of all acute angles of a right triangle. 57 = 90^@). Lesson-- Graphing Sine and Cosine Functions Assignment 1--- Graphing the basic sine and cosine functions: This will be handed out on a worksheet No Answer Key for this Mrs. Corrective Assignment. Understand key features of graphs of trig functions o Graph of the sine function o Graph of the cosine function o Key features of the sine and cosine function o Graph of the tangent function o Key features of the tangent function o Practice Solutions Back to Table of Contents. circle and the nature of the curve of the function graph. On a sheet of graph paper, predict what the following graphs would look like. The coefficient affects the period (which can be considered a horizontal stretch if. Add and Subtract Complex Numbers. Therefore the sine and cosine of an acute angle are always positive numbers less than 1. Feb 28 - We worked on writing the equations of sine and cosine functions then learned how to graph cosecant and secant functions using their corresponding sine or cosine function. When the period of a sine function doubles the frequency (1) doubles. Worksheets are Graphing trig functions, Graphs of trig functions, Amplitude and period for sine. Write two different equations for the same graph below. To sketch the graphs of the basic sine and cosine functions by hand, it helps to note five key points in one period of each graph: the intercepts, maximum points, and minimum points (see Figure 4. 7: Slope Fields ; Chapter 2 Test; Chapter 2 Answer Key ; Chapter 3: Derivatives and Graphs [ Chapter 3 pdf ]. They are both expressed according to the triangle on the right, where each letter represents one side-length (lower-case) and the angle opposite to it (upper-case). All graphs were computer generated and adjusted to be easy to read for students. On a sheet of graph paper, predict what the following graphs would look like. Feb 28 - We worked on writing the equations of sine and cosine functions then learned how to graph cosecant and secant functions using their corresponding sine or cosine function. How long is the hypotenuse? 8. Create AccountorSign In. 4 Graphing Sine and Cosine Functions. Notice that both the sine and cosine have a maximum value of 1 and a minimum value of -1. Day 1 - Parent Graphs and Transformations Worksheet 1 - Answer Key. Worksheet by Kuta Software LLC MAC 1114 - Trigonometry Name_____ 7. The amplitude is a=2 and the period is. 12 - 13 Friday 10/25 Writing functions cont'd Quiz - Graphing Sine and Cosine. Some of the worksheets for this concept are Graphs of trig functions, Amplitude and period for sine and cosine functions work, 1 of 2 graphing sine cosine and tangent functions, , Trig graphs work, Of the sine and cosine functions, Work 15 key, Honors algebra 2 name. 1_solutions. 8 Sketching Trig Functions. Worksheet graphing problems # 1- 8 on pp. Evaluate the function for The function passes through. sin o h p yp p cos tan h a y d p j o a p d p j csc sec o h p yp p h a y d p j cot o a p d p j Notice that the sine, cosine, and tangent functions are reciprocals of the cosecant, secant, and cotangent functions, respectively. 2n for which the sine function is increasing. The graphs of y = a sin (bx + c) + d and y = a cos (bx + c) + d have the following characteristics. 2 Exploring Graphs of Periodic Functions 521 History Connection: Not as Easy as ! 526 8. sin 7! 2! 1 1 1 Find the values of •for which each equation is true. Sketch the graph of the function over the interval -2( ≤ x ≤ 2(. (a) Using your calculator, sketch the graph on the grid to the right. Chapter 2 Graphs of Trig Functions The sine and cosecant functions are reciprocals. What is the equation for the sine function graphed here? Gimme a Hint. So, tangent function crosses x-axis at , n is the set of integers. Graphing Trig Functions - 1 - www. However, there are yet many people who afterward don't as soon as reading. y = 4 sin x 12. Practice Quiz – Graphing Sine and Cosine Use transformations to graph each of the following functions. This Graphs of Other Trigonometric Functions Presentation is suitable for 10th - 12th Grade. 8 Sketching Trig Functions. It explains how to identify the amplitude, period, phase shift, vertical shift, and midline of a sine or cosine function. the graph has an extreme point, (0, 0). 3-21 graphing inverse functions. College Math MCQs: Multiple Choice Questions and Answers (Quiz & Tests with Answer Keys) provides mock tests for competitive exams to solve 803 MCQs. Once the appropriate base value of the first quadrant is known, symmetric points in any other quadrant can be. Solution : Factor the expression on the left and set each factor to zero. a) y = sin x Domain_____. In fact, the key to understanding Piecewise-Defined Functions is to focus on their domain restrictions. y = F(x + 1) 7. Inverse Sine Function (Arcsine) Each of the trigonometric functions sine, cosine, tangent, secant, cosecant and cotangent has an inverse (with a restricted domain). Displaying all worksheets related to - Graphing Sine Functions. to save your graphs! + New Blank Graph. Therefore, the sum of the zeros of the function is equal to –b. Graphs of the Sine and Cosine Functions Divide the interval into four equal parts to obtain the values for which sin bx or cos bx equal -1, 0, or 1. Use the sine tool to graph the function. I attached an image but just in case it doesn't show up properly, the prompt is to write \frac{\csc(x)\cot(x)}{\sec(x)} in terms of sine and cosine. 12 Graph translations of sine and cosine functions S. y = sin 2 θ 21. Extreme Values. Practice Quiz – Graphing Sine and Cosine Use transformations to graph each of the following functions. What is the range of f(x) = sin(x)? the set of all real numbers -1 < or = y < or = 1 Which set of transformations is needed to graph f(x) = -2sin(x) + 3 from the parent sine function?. Solution: Cosine Function. Express your answer as a fraction in lowest terms. identities that it knows about to simplify your expression. Graph each translation of y 5 sin x in the interval from 0 to 2π. Sample Test Answer Key Trigonometric Functions and Their Graphs. 2 The Unit Circle and Circular Functions - 6. Verify your answer with graphing software or a graphing calculator. cosθ=x and sinθ=y We could use the sine and cosine graphs, however the unit circle is more useful for these problems. The international standard pitch has been set at a frequency of 440 cycles/second. Plotting the points from the table and continuing along the x-axis gives the shape of the sine function. LESSON 2: GRAPHING QUADRATIC FUNCTIONS Study: Putting the Pieces Together Use key components such as vertex, axis of symmetry, and x- and y-intercepts to sketch the graphs of quadratic functions and solve quadratic inequalities. 6 Angles of Elevation and Depression R 19 MAY 2016 - 8. Learn how to construct trigonometric functions from their graphs or other features. Some of the worksheets for this concept are Honors algebra 2 name, Of the sine and cosine functions, , Graphs of trig functions, Work 15 key, 13 trigonometricgraphswork, 1 of 2 graphing sine cosine and tangent functions, Sine cosine and tangent practice. 5, Graphs of Sine and Cosine Functions Homework: 4. y = 3 sin x + 1. The graph. Experiment with the graph of a sine or cosine function. Graphing Trig Functions Worksheet With Answers Quiz How Graph The from Graphing Sine And Cosine Worksheet, source:deargraham. These basic waves have the property that they deviate from the t-axis by no more than one unit. Lesson Notes. Day 2 - Graphing Rational Functions - Notes. 4 Part 2 : Applications of Trigonometric Functions. Graphs of these functions The period of a function The amplitude of a function Skills Practiced. Lesson 6 Basic Graphs of Sine and Cosine. The input to the sine and cosine functions is the rotation from the positive x-axis, and that may. Trigonometric Functions, you will begin by learning about the inverses of quadratics and other functions. 5 ~ Graphs of Sine and Cosine Functions In this lesson you will: • Sketch the graphs of basic sine and cosine functions. SWBAT: Use Sine and Cosine to define given functions. Example: We know that the derivative of the sine function is the cosine function. Graphs Of Sine. They also apply two basic transformations, one vertical translation and one horizontal translation, to the sine graph as well as determine any changes that may have occurred to the domain and range. View answers. y 2 cos x y=3sin2x cos(—3x) 14. Displaying all worksheets related to - Graphing Sine. 8_writing_tan. Sorry but it won't allow me to copy and paste the graph so I hope you guys know how this graph looks like from the function. You can use these points to sketch the graphs of y = a sin bx and y = a cos bx. 4 Part 1 : Solving Trigonometric Equations Sec 5. So: tan à L 1 cot à and cot à L 1 tan à. See Example. 6 Introduction to Trig Identities Answers 1. y: 3 sin x y: 3 cos 4500 —x 19. 1 Graphing Sine and Cosine Functions Sec 5. The absolute value of a is the amplitude of the function y = a sin x. Graphing Trig Functions Practice Worksheet With Answers Students will practice graphing sine and cosine curves : a) identify period and amplitude based on equation or on the graph b) write equation from graph c) write. c Use trigonometric (sine, cosine) functions to model and solve problems; justify results: Develop and use the law of sines and the law of cosines. 1 Graphing Sine and Cosine. We summarize these facts in the following theorem. Thanks for visiting our website, article about 21 Common Core Algebra 2 Unit 1 Answer Key. The Definition of the Sine and Cosine Functions. 9 - 10 Thursday 10/24 Writing Equations of sine and cosine functions (Notes p. 3 Connecting Graphs to Rational Equations Assigned: Pages 465-467 : Practice section #1-6 (at least 2 letters each); at least 6 from Apply/Extend (A/E). Article objectives; To learn about the properties of graphs of trigonometric functions. Watch the two videos on APEX {8. y = –4 sin 3x + 2 5. Then the amplitude of f is the number 2 M m Example 1: Specify the period and amplitude of the given function Now let’s talk about the graphs of the sine and cosine functions. y = - cos 2x 15. Explore how changing the values in the equation can translate or scale the graph of the function. Here we will do the opposite, take the side lengths and find the angle. The graphs of tan x, cot x, sec x and csc x are not as common as the sine and cosine curves that we met earlier in this chapter. From the graph, you. of the Answer Keys to hand out to your class, but. 4 Part 2 : Applications of Trigonometric Functions. Some of the worksheets for this concept are Honors algebra 2 name, Of the sine and cosine functions, , Graphs of trig functions, Work 15 key, 13 trigonometricgraphswork, 1 of 2 graphing sine cosine and tangent functions, Sine cosine and tangent practice. A set of questions, with their answers, on identifying the graphs of trigonometric functions sin (x), cos (x), tan(x), cot (x), sec (x) and csc (x) are presented in this page. Some of the worksheets for this concept are Graphing trig functions, Graphs of trig functions, Amplitude and period for sine and cosine functions work, Graphing sine and cosine functions, Work 15 key, Honors algebra 2 name, Sinusoidal functions work, Of the sine and cosine functions. When you click the button, this page will try to apply 25 different trig. In general, the graph of y = f(x) + k is the graph of y = f(x) translated k units vertically. In these examples we will graph a sine and cosine function using a table of values. 1 Graphing Sine, Cosine, and Tangent Functions 831. The value of k indicates a translation up (k > 0) or down (k < 0). Extra Practice - Combined Transformations Note: Answer key is provided on the backside of the sheet. Graphs of the Sine and Cosine Functions Divide the interval into four equal parts to obtain the values for which sin bx or cos bx equal -1, 0, or 1. Our mission is to provide a free, world-class education to anyone, anywhere. 4 Graphing Sine and Cosine Functions. 3 62/87,21 The general form of the equation is y = a sin bt, where t is the time in seconds. Since -1<=cosx<=1 AA x in R , the cosine function is bounded. worksheet on graphing sine and cosine functions Images about Worksheet On Graphing Sine And Cosine Functions: Chemical Equations Worksheet With Answers,. when , where n is the set of integers. 1) y = sin (θ − 135) 90 ° 180. The student correctly compares transformations of a function, and then graphs the function over a different transformation. asked by Anononymous on January 14, 2020; Mathematics. Student needs to show proof. Dugopolski’s Precalculus: Functions and Graphs, Fourth Edition gives students the essential strategies they need to make the transition to calculus. 6: Powers of Trig Functions: Secant and Tangent ; 2. Determining the equation of a circle by completing the square. U Lsin T E4 b. Let f be a periodic function and let m and M denote, respectively, the minimum and maximum values of the function. Unit 3: Trigonometry of General Triangles and Trigonometric Functions 3. −≤ ≤ππx Is the cosine function even, odd, or neither? Communicate Your Answer 3. The Cosine Graph a. to save your graphs! + New Blank Graph. 7 -8 Wednesday 10/23 Continue Graphing Sine and Cosine (Period Changes) Worksheet graphing problems #9 - 16 on pp. 1 Linear Functions; 2. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval \([ −1,1 ]$$. m) and the axle height is thus 40 m (the mean of 10 m and 70 m). 4 y 3csc 2x 1 3 GRAPHING INVERSE TRIG FUNCTIONS Find the domain, range, and sketch a complete graph of each function. Graphs of tan, cot, sec and csc. When the period of a sine function doubles the frequency (1) doubles. Their behavior will only be explored in this lesson. Edmonds will check it after you graph it. Evaluate tan ˇ 3 and sec ˇ 4. 5, then check. Sine Rule - The sine rule is given by; a/sinA =b/sinB =c/sinC There are two conditions when you can use the sine rule; When two angles and one side is given. In other words, instead of the graph's midline being the x-axis, it's going to be the line y = -1. I like to share this Solving Trigonometric Equations with you all through my article. 2 Graphing Sinusoidal Functions using 5 Points Method Sec 5. The Unit Circle. ) The point (pi/2, -4) is on the graph. answer choices. Just as a quick review, the polar coordinate system is very similar to that of the rectangular coordinate system. 1) y = sin (θ − 135) 90 ° 180. 1 - Parent Functions and transformations: p. It consists of several rows or columns that spread out all over the page and create for space that assist people fill data. trigonometric graphs use mini whiteboards to answer. • Use amplitude and period to help sketch graphs. 5 and b = 4 is y = 1. x to create a table Answer: Use the unit circle and of values. sin 7! 2! 1 1 1 Find the values of •for which each equation is true. Graph the functions applying transformations using this information. Sine Cosine Graphing Showing top 8 worksheets in the category - Sine Cosine Graphing. Evaluate tan ˇ 3 and sec ˇ 4. Understanding how to create and draw these functions is essential to these classes, and to nearly anyone working in a scientific field. Chapter(14(-(TrigonometricFunctions(andIdentities(Answer'Key(CK912Algebra(II(with(Trigonometry(Concepts( 16! 14. Since the cosine function has an extreme point for x = 0,. Note how the sine and cosecant curves are reciprocals of each other as are the cosine and secant curves. Pa Functions And Their Graphs Solutions Examples S. , identify the given information and graph the trig function. Feb 28 - We worked on writing the equations of sine and cosine functions then learned how to graph cosecant and secant functions using their corresponding sine or cosine function. The angle of elevation from the boat to the top of the lighthouse is 26 degrees. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval[latex]\,\left[-1,1\right]. If I subtract the portion on the right, which is 1/4 and add it to the left, the length of the green graph is still 21T, and you only have 1 cycle between 0 & 2 IT. These will be key points on the graphs of y = sin x and y = cos x. Displaying top 8 worksheets found for - Graphs Of Sine. 5 Graphing Sine and Cosine Imagine taking the circumference of the unit circle and 'peeling' it off the circle and straightening it out so that the radian measures from 0 to 2π lie on the x‐axis. The graphs of. The second derivative test 89 39. Hint: Sine is odd and cosine is. ; Hornsby, John; Schneider, David I. Graphing calculators will NOT be permitted on the quiz. 1_solutions. Click here. 18 Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases… d) Graph trigonometric functions, showing period, midline, and amplitude. This becoming stated, we provide you with a a number of straightforward yet useful content and also web themes designed made for just about any helpful purpose. In general, the graph of y = f(x) + k is the graph of y = f(x) translated k units vertically. 8 Applications and Models p. Mar 12/13 9. This is a problem. High School Geometry High School Statistics Algebra 1 Algebra 2. Give the amplitude and period of each function. The range of sine and tangent is in Quadrants I and IV, while the range of cosine is Quadrants I and II. Precalculus (6th Edition) answers to Chapter 6 - The Circular Functions and Their Graphs - 6. Gimme a Hint. But this graph is shifted down by one unit. 4 Trigonometric Functions of Any Angle p. Since most of the issue is been already printed for him. Then graph. Sine and Cosine of A ± B. The period is π. 2 Graphs of the Sine and Cosine Functions A Periodic Function and Its Period A nonconstant function f is said to be periodic if there is a number p > 0 such that f(x + p ) = f(x) for all x in the domain of f. Period of a Function from Graphing Sine And Cosine Functions Worksheet, source:math. Sine and Cosine Functions. Graph each translation of y 5 sin x in the interval from 0 to 2π. There are at and The maximum and minimum points are indicated by the voice balloons. High School Geometry High School Statistics Algebra 1 Algebra 2. Amplitude and Period of Sine and Cosine Functions The amplitude of y = a sin ( x ) and y = a cos ( x ) represents half the distance between the maximum and minimum values of the function. 1that geometrically this means the graphs of the cosine and sine functions have no jumps, gaps, holes in the graph, asymptotes, 1See section1. Like all functions, the sine function has an input and an output. Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period - Domain & Range - Duration: 18:35. Trigonometric Equations. View answers. y = 2 cos x. PDF LESSON. What is the equation for the cosine function graphed here? Gimme a Hint. com Graphs Sine And Cosine Worksheet Free Worksheets Library from Graphing Sine And Cosine Functions Worksheet, source:comprar-en-internet. Write a sine function with the given characteristics. Introduction: In this lesson, the period and frequency of basic graphs of sine and cosine will be discussed and illustrated as well as vertical shift. 2 Trigonometric Ratios of any angle 5. Notice that both the sine and cosine have a maximum value of 1 and a minimum value of -1. Figure $$\PageIndex{2}$$: The sine function Notice how the sine values are positive between $$0$$ and $$\pi$$, which correspond to the values of the sine function in quadrants I and II on the unit circle, and the sine values are negative between $$\pi$$ and $$2. [NEW] Real Life Examples Of Sine Cosine And Tangent A real life example of the sine function could be a. x 0 π 6 π 4 π 3 π 2 3 4 π π 3 2 π 2π yx=sin 0 0. Free worksheetpdf and answer key on grpahing sine and cosine curves. You write (sin(t))^2 for the square of sin(t), and never sin^2t. 7 Inverse Trigonometric Functions p. Answer Key to Quarter 3 Exam Review. • Complete the review and practice test exercises from the textbook. They are asked to find the domain and range of the sine graph. Learn how to graph trigonometric functions and how to interpret those graphs. The questions are about determing the period from the graph and also matching graphs and trigonometric functions. [Math] Algebra 2 - Graphing Sine and Cosine (A's) Which function describes the graph shown below? Wondering if anyone knew of a better answer key for this god. See Figure \(\PageIndex{2}$$. What are the period and midline of ? Gimme a Hint. 5, Graphs of Sine and Cosine Functions Homework: 4. Displaying all worksheets related to - Graphing Sine Functions. Transformations of the Sine and Cosine Graphs from Graphing Trig Functions Worksheet, source: jwilson. So: tan à L 1 cot à and cot à L 1 tan à. Find an equation for a cosine function that has an amplitude of OTC — , a period of Find an equation for a sine function that has amplitude of 5, a period of 3TT. Powered by Create your own unique website with customizable templates. The function 5ysin x is called a periodic functionwith a periodof 2p because for every x in the domain of the sine function, sin x5 sin (x1 2p). 5-1 1 0 π_ 2 3__π 2 5__π 2-π_ 2 Period Period One Cycle 3__π 2 5__π - 2-y = sin θ θ. Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. Answer Key 14. Write two different equations for the same graph below. 1) Answers to Graphing Sine and Cosine 1) p 2 p3p 2 2p-6-4-2 2 4 6. 8 Sketching Trig Functions. Therefore, a sinusoidal function with period DQGDPSOLWXGH WKDWSDVVHV through the point LV y = 1. Write a sine function that can be used to model the initial behavior of a sound wave with the frequency and amplitude given. The following sheets list the key concepts which are taught in the specified math course. 64 Key points in graphing the sine function Graph variations of y = sin x. 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https://socratic.org/questions/if-log-2-x-1-log-4-x-and-log-8-4x-are-consecutive-terms-of-a-geometric-sequence- | Precalculus
Topics
# If log_2 x, 1 + log_4 x and log_8 4x are consecutive terms of a geometric sequence, what are all possible values of x?
Feb 18, 2017
$x = \frac{1}{4} , 64$
#### Explanation:
We use the property of a geometric sequence that $r = {t}_{2} / {t}_{1} = {t}_{3} / {t}_{2}$.
$\frac{1 + {\log}_{4} x}{{\log}_{2} x} = \frac{{\log}_{8} 4 x}{1 + {\log}_{4} x}$
Convert everything to base $2$ using the rule ${\log}_{a} n = \log \frac{n}{\log} a$.
$\frac{1 + \log \frac{x}{\log} 4}{\log \frac{x}{\log} 2} = \frac{\frac{\log 4 x}{\log} 8}{1 + \log \frac{x}{\log} 4}$
Apply the rule $\log {a}^{n} = n \log a$ now.
(1 + logx/(2log2))/(logx/log2) = ((log4x)/(3log2))/(1 + logx/(2log2)
$\frac{1 + \frac{1}{2} {\log}_{2} x}{{\log}_{2} x} = \frac{\frac{1}{3} {\log}_{2} \left(4 x\right)}{1 + \frac{1}{2} {\log}_{2} x}$
Apply ${\log}_{a} \left(n m\right) = {\log}_{a} n + {\log}_{a} m$.
$\frac{1 + \frac{1}{2} {\log}_{2} x}{{\log}_{2} x} = \frac{\frac{1}{3} {\log}_{2} 4 + \frac{1}{3} {\log}_{2} x}{1 + \frac{1}{2} {\log}_{2} x}$
$\frac{1 + \frac{1}{2} {\log}_{2} x}{{\log}_{2} x} = \frac{\frac{2}{3} + \frac{1}{3} {\log}_{2} x}{1 + \frac{1}{2} {\log}_{2} x}$
Now let $u = {\log}_{2} x$.
$\frac{1 + \frac{1}{2} u}{u} = \frac{\frac{2}{3} + \frac{1}{3} u}{1 + \frac{1}{2} u}$
$\left(1 + \frac{1}{2} u\right) \left(1 + \frac{1}{2} u\right) = u \left(\frac{2}{3} + \frac{1}{3} u\right)$
$1 + u + \frac{1}{4} {u}^{2} = \frac{2}{3} u + \frac{1}{3} {u}^{2}$
$0 = \frac{1}{12} {u}^{2} - \frac{1}{3} u - 1$
Now multiply both sides by $12$.
$12 \left(0\right) = 12 \left(\frac{1}{12} {u}^{2} - \frac{1}{3} u - 1\right)$
$0 = {u}^{2} - 4 u - 12$
$0 = \left(u - 6\right) \left(u + 2\right)$
$u = 6 \mathmr{and} u = - 2$
Revert to the original variable, $x$.
Since $u = {\log}_{2} x$:
$6 = {\log}_{2} x , - 2 = {\log}_{2} x$
$x = 64 , x = \frac{1}{4}$
Hopefully this helps!
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https://riemann.unizar.es/~mmarco/DME/DME.html | # DME cryptosystem
## Intro: Multivariate cryptosystems
DME is a cryptosystem of the multivariate family. That is, it consists in a polynomial map $\begin{array}{ccc} g:K^n & \to & K^m \\ \left( \begin{array}{c} x_1 \\ \vdots \\ x_n \end{array} \right) & \to & \left( \begin{array}{c} g_1(x_1,\ldots,x_n) \\ \vdots \\ g_m(x_1,\ldots,x_n) \end {array} \right) \end{array}$
where the $$g_i$$ are polynomials with coefficients in the field $$K$$.
The idea is quite symple: your public key is just the polynomials $$g_1,\ldots,g_m$$, and when someone wants to encrypt some message, just encode it as the values $$(x_1,\ldots,x_n)$$, evaluate the polynomial map in those values, and send the correspondent values to you. Then you need some method (which essentially will be your private key) to find the original values from the encrypted message. That is, decrypting is esentially the same as inverting the map $$f$$.
If these polynomials have degree $$1$$, we are in the realm of linear algebra, so we know easily to invert the map just by computing the inverse of a matrix.
However, if the degrees of the polynomials are higher, we get into a completely different scenario. Here it is not even easy to determine if the map is bijective or not. It could be bijective, it could map seven different values to a single one, it could map infinetly many to one, or it could even be the case that for some values it is injective, and for others it is not.
Given a value in $$K^m$$, finding the possible preimages is essentally solving a system of polynomial equations. There is a general method for that, based on elimination theory; so one might think that this family of cryptosystems are doomed. But the caveat is that the general method makes use of Gröbner basis, which are really expensive to compute (and by really expensive I mean something like double exponential in the worst case, which in practice means that toy examples are ok, but as soon as the size grows, it becomes unfeasible).
Ok, so now that we know that in general it is difficult to decrypt the messages in this kind of cryptosystems, we have another problem: how does the legitimate recipient of the message decrypt it? Or in other words: how can we construct the map $$f$$ in such a way that someone that knows some secret information (the private key) can invert it, but it remains hard to do so without that secret information?
The usual way to solve that question is to make $$f$$ by composing several maps that are easy to invert. The list of those easy to invert maps will be the public key, whereas the final composition, would mix them in a way that is hard to recover. As a very easy example, consider the following maps:
• $$(x,y)\to (2x+y,x+y)$$
• $$(x,y)\to (x+y^2,y)$$
• $$(x,y)\to (x,y+x^3)$$
• $$(x,y)\to (x+3y,2x+7y)$$
As you can easiyly see, the first and last ones are just linear maps, and the rest consist on adding a function of one variable to the other one; so their inverses can be easily computed:
• $$(x,y)\to (2x+y,x+y)$$
• $$(x,y)\to (x-y^2,y)$$
• $$(x,y)\to (x,y-x^3)$$
• $$(x,y)\to (7x-3y,-2x+y)$$
Now, if we compose all the three of them, we get that $$(x,y)$$ maps to
$\left( 2 x^{6} + 36 x^{5} y + 270 x^{4} y^{2} + 1080 x^{3} y^{3} + 2430 x^{2} y^{4} + 2916 x y^{5} + 1458 y^{6} - 8 x^{4} - 100 x^{3} y - 468 x^{2} y^{2} - 972 x y^{3} - 756 y^{4} - x^{3} - 9 x^{2} y - 27 x y^{2} - 27 y^{3} + 8 x^{2} + 56 x y + 98 y^{2} + 4 x + 13 y , x^{6} + 18 x^{5} y + 135 x^{4} y^{2} + 540 x^{3} y^{3} + 1215 x^{2} y^{4} + 1458 x y^{5} + 729 y^{6} - 4 x^{4} - 50 x^{3} y - 234 x^{2} y^{2} - 486 x y^{3} - 378 y^{4} - x^{3} - 9 x^{2} y - 27 x y^{2} - 27 y^{3} + 4 x^{2} + 28 x y + 49 y^{2} + 3 x + 10 y \right)$
which doesn't look that easy to invert at all (note, this case is actually small enough to be inverted with the elimination techniques mentioned before, just take it as a toy example to showcase how the complex systems can be obtained by composing simple ones). In this example we also observe a phenomenon that we have to be careful with: the complexity of the polynomial system (which will translate into the size of the public key) can grow a lot if we don't choose the simple pieces to compose carefully.
So, summarizing, the challenge of creating a multivariate cryptosystem consists on choosing some elementary maps such that:
• Each one of the elementary maps is easy to invert
• The composition of all of them is a polynomial map that:
• is hard to invert
• is not too long to write
## The DME cryptosystem
DME stands for double matrix exponentiation, because two of the elementary pieces mentioned before are matrix exponentiations (we will see later what this means). Besides these two matrix exponentiations, there are some linear maps and some auxilary transformations that basically consist on choosing the right way to represent the objects. Don't worry if this paragraph makes no sense to you now, we will explain it all carefully now.
### Step by step
We will see how the map is constructed step by step. For the purpose of this section, we will assume we have fixed some parameters. Later we will explain what different choices could have been made. For starters, assume we work in some binary field. In the example implementation, we use $$\mathbb{F}_q$$ with $$q=2^{48}$$. This will be our basic way to represent data. That is, all data will be seen as a list of elements in this field. Since it is a binary field, it is easy to represent a list of 48 bits as an element of this field. Just represent them as a degree $$48$$ polynomial in $$t$$ with coefficients in $$F_2$$, where the coefficients are the values of the bits. This polynomial should be considered as a representative of its class modulo some irreducible polynomial. The choice of this irreducible polynomial doesn't really matter from the mathematical point of view (all possible choices are isomorphic), but maybe some choices can allow more efficient implementations, so we choose $$f=t^{48}+t^{28}+t^{27}+t+1$$.
So now that we have a way to represent series of bits as elements of our field, we will consider vectors with $$6$$ entries in this field. That is, we will see a series of $$288$$ bits as a vector
$(x_0,x_1,x_2,x_3,x_4,x_5)$
where the $$x_i$$ are polynomials in $$t$$ modulo $$f$$ as before. That will be our plaintext. For reasons that you will later see, if too many of these vectors are zero, our system will fail, so we have to force them to be nonzero. This is done by adding some padding: that is, instead of considering $$288$$ bits, we will consider a few less, and insert some $$1$$'s between them. In particular it is enough to introduce 3 such $$1$$'s, but in order not to break bytes, we actually add $$3$$ bytes of padding; that is, we encode $$33$$ bytes and interleave $$3$$ bytes with the value $$1$$ in the last positions of $$x_1$$, $$x_3$$ and $$x_5$$.
#### First linear map $$L_1$$
Ok, so now that we have our plaintext correctly padded and expressed as a vector, the first step we take is to apply a linear transformation to it; that is, multiply it by a matrix with entries in $$\mathbb{F}_q$$.
But remember that we mentioned that we have to be careful to prevent the complexity of the total map to explode; so the matrix that we use here must have a specific shape. In particular, it must look as follows:
$L_1=\left( \begin{array}{cccccc} a_{1,1} & a_{1,2} & 0 & 0 & 0 & 0 \\ a_{2,1} & a_{2,2} & 0 & 0 & 0 & 0 \\ 0 & 0 & a_{3,3} & a_{3,4} & 0 & 0 \\ 0 & 0 & a_{4,3} & a_{4,4} & 0 & 0 \\ 0 & 0 & 0 & 0 & a_{5,5} & a_{5,6} \\ 0 & 0 & 0 & 0 & a_{6,5} & a_{6,6} \end{array} \right)$
That is, we have divided our vector of 6 entries in three parts of two entries each, and aplied a linear transformation to each one of those three pairs. The entries of this matrix $$a_{i,j}$$ will be the first part of our private key (to be precise, what we use for decryption is the inverse of this matrix, but to keep things simple just imagine that we keep this matrix, and apply its inverse when decrypting).
#### First exponentiation, $$F_1$$
Ok, so now that we have aplied a linear map, we have a new vector
$(y_0,y_1,y_2,y_3,y_4,y_5)$
It's time to apply one of those misterious matrix exponentiations. But first, we have to change the representation of our vector. The reason for that is that the matrix exponentiation does not act over $$\mathbb{F}_q$$, but over $$\mathbb{F}_{q^2}$$. That is, we have to see our vector of six entries in the field of $$2^{48}$$ elements, as three elements in the field of $$2^{96}$$ elements. That is actually quite easy, since we can just see $$\mathbb{F}_{q^2}$$ as the set of degree 1 polynomials over $$\mathbb{F}_q$$, modulo some irreducible polynomial. The actual choice of this polynomial is not important from the mathematical point of view, so just assume that we have picked some irreducible polynomial $$f_2\in \mathbb{F}_q[T]$$ of degree 2, and consider our vector as
$(Y_0,Y_1,Y_2)=(y_0+y_1 T, y_2+y_3 T, y_4+y_5 T)$
Ok, so now we have our data expressed as elements on a bigger field. What do we need now to apply a matrix exponentiation? Well, a matrix of course! But in this case, the entries of the matrix are not elements of any finite field, just integers. Again, to make sure that the complexity of the total map does not explode, we must be careful with the choice of the matrix. Take a matrix of the form
$A=\left( \begin{array}{ccc} 2^{E_{1,1}} & 2^{E_{1,2}} & 0 \\ 2^{E_{2,1}} & 0 & 2^{E_{2,3}} \\ 0 & 2^{E_{3,2}} & 2^{E_{3,3}} \end{array} \right)$
and apply it to our vector in $$\mathbb{F}_{q^2}$$. But wait a second! this is not a linear transformation, this is a matrix exponentiation. This means that the matrix is not aplied as a multiplicative operator: it is aplied as an exponent. That is, mimic the process you follow yo multiply by a matrix, but change the products by exponentiations, and the sums as products.
Confusing? Ok, let's just put it down explicitely. Formally we are aplying an operation defined as follows:
$(Y_0,Y_1,Y_2)^{ \left( \begin{array}{ccc} 2^{E_{1,1}} & 2^{E_{1,2}} & 0 \\ 2^{E_{2,1}} & 0 & 2^{E_{2,3}} \\ 0 & 2^{E_{3,2}} & 2^{E_{3,3}} \end{array} \right) } = (Y_0^{2^{E_{1,1}}}Y_1^{2^{E_{1,2}}}, Y_0^{2^{E_{2,1}}}Y_2^{2^{E_{2,3}}}, Y_1^{2^{E_{3,2}}}Y_2^{2^{E_{3,3}}})$
Ok, so we know how to apply this matrix exponentiations. But wait a sec, how can we apply the inverse transformation? Don't worry, there is a way: just make sure that you matrix $$A$$ is invertible over the integers modulo $$2^{96}-1$$ and take its inverse there. Now, by using the finite field version of Fermat's little theorem it can be seen that the matrix exponentiation corresponding to the inverse matrix takes us back to the starting point (well, not exactly, if there are zeros involved, things behave differently, that is why we needed to add some padding to make sure that we got not zeros at this point).
One would be tempted to consider the polynomial $$f_2$$ and the values $$E_{i,j}$$ as part of the private key, but it happens that it would add no extra security (they are either irrelvant or can be easily recovered), so in order to keep keys more compact, they will be agreed on in advance, as part of the standard setup.
Now that we have applied the matrix exponentiation, we have a new vector of three entries in $$\mathbb{F}_{q^2}$$. Just like we did before, that is the same as having a vector of six entries in $$\mathbb{F}_q$$:
$(Z_0,Z_1,Z_2)=(z_0+z_1T,z_2+z_3T,z_4+z_5T) \leftrightarrow (z_0,z_1,z_2,z_3,z_4,z_5)$
#### Second linear map $$L_2$$
Next step is easy: just another linear transformation. In this case, the matrix will have a different form. In particular we will multiply our vector by the matrix
$L_2=\left( \begin{array}{cccccc} b_{1,1} & b_{1,2} & b_{2,3} & 0 & 0 & 0 \\ b_{2,1} & b_{2,2} & b_{3,3} & 0 & 0 & 0 \\ b_{3,1} & b_{3,2} & b_{3,3} & 0 & 0 & 0 \\ 0 & 0 & 0 & b_{4,4} & b_{4,5} & b_{4,6} \\ 0 & 0 & 0 & b_{5,4} & b_{5,5} & b_{5,6} \\ 0 & 0 & 0 & b_{6,4} & b_{6,5} & b_{6,6} \end{array} \right)$
The $$b_i$$ are just values in in $$\mathbb{F}_q$$, and the only thing we have to make sure is that the matrix is invertible. As before, this matrix (or its inverse) will be part of the private key.
After multiplying by it, we get a new vector $$(s_0,s_1,s_2,s_3,s_4,s_5)\in{\mathbb{F}_q}^6$$.
#### Second exponentiation
The next step in our encryption process is a new matrix exponentiation; but this time it will happen in a bigger field. Now we will see our vector in $${\mathbb{F}_q}^6$$ as a vector in $${\mathbb{F}_{q^3}}^2$$, that is
$(s_0,s_1,s_2,s_3,s_4,s_5) \leftrightarrow (s_0+s_1S+s_2S^2,s_3+s_4S+s_5S^2)=(S_1,S_2)$ where the polynomials in $$S$$ are considered modulo some irreducible one $$f_3\in \mathbb{F}_q[S]$$ of degree three.
Now, just like we did before, we fix some matrix of the form
$\left( \begin{array}{cc} 2^{F_{1,1}} & 2 ^{F_{1,2}} \\ 2^{F_{2,1}} & 2^{F_{2,2}} \end{array} \right)$
that is invertible modulo $$2^{144}-1$$. Aplying the corresponding matrix exponentiation we obtain
$(S_1,S_2)^{\left( \begin{array}{cc} 2^{F_{1,1}} & 2 ^{F_{1,2}} \\ 2^{F_{2,1}} & 2^{F_{2,2}} \end{array} \right)}= (S_1^{2^{F_{1,1}}}S_2^{2^{F_{1,2}}}, S_1^{2^{F_{2,1}}}S_2^{2 ^{F_{2,2}}})= (R_1,R_2)$
and again, we see it as a vector in $${\mathbb{F}_q}^6$$
$(R_1,R_2)=(r_0+r_1S+r_2S^2,r_3+r_4S+r_5S^2) \leftrightarrow (r_0,r_1,r_2,r_3,r_4,r_5)$
Just as before, the choice of this matrix and the minimal polynomial will be fixed in the setup.
#### Final linear map
Now we apply a new linear map and we are done. The final part of our private key is another matrix of the form
$L3=\left( \begin{array}{cccccc} c_{1,1} & c_{1,2} & c_{1,3} & 0 & 0 & 0 \\ c_{2,1} & c_{2,2} & c_{2,3} & 0 & 0 & 0 \\ c_{3,1} & c_{3,2} & c_{3,3} & 0 & 0 & 0 \\ 0 & 0 & 0 & c_{4,4} & c_{4,5} & c_{4,6} \\ 0 & 0 & 0 & c_{5,4} & c_{5,5} & c_{5,6} \\ 0 & 0 & 0 & c_{6,4} & c_{6,5} & c_{6,6} \end{array} \right)$
which will be multiplied by our vector to obtain the final cyphertext.
### The public key
Summarizing, we have three elementary transformations, each of one easy to invert, that we compose to get our encryption map. But wait a second, those exponential transformations are not expressed as polynomial maps! Well, it turns out they are. If you try to follow the track of what happens when you compute it, you can verify it. So we can express the total composition of maps as a polynomial map. And the six polynomials that will express it will be the public key of our system.
You can also check that the strange choices we did (the shapes of the matrices, and the facts that the exponents are powers of two) will ensure that the polynomials that appear in this expression will have exactly 64 monomials. Moreover, since we have fixed the exponentiation matrices, each monomial will be the product of four powers of variables, and the exponents that appear will always be the same. So in order to encode the public key we only need to write the coefficients that appear. That is, the public key will consist on a list of $$64\cdot 6$$ elements of $$\mathbb{F}_q$$. The private key will be the coefficients of the linear maps, that is, $$48$$ elements of $$\mathbb{F}_q$$.
Now a little problem needs to be addressed: how do we compute the public key? One way would be to rewrite all elementary maps as explicit polynomial maps, and compose them all. But that is not specially efficient. There is a faster way. Since we know exactly what exponents will appear in the polynomials, for a given plain text we can compute the values of those powers. Then the final result will be a linear map applied to the vector of those products. The matrix that determines this map can be determined by linear algebra if we know enough pairs preimage-image. So we can choose a lot of random plaintexts, compute their corresponding monomials, and the corresponding cyphertext step by step (that is, applying the five elementary transformations). With enough of them, we can obtain the matrix with the coefficients of the polynomial transformation.
## Generalization
We have seen how the DME cryptosystem works with a specific choice of the setup. In particular, we have chosen a prime $$p=2$$, a finite field $$F_{2^48}$$ of characteristic $$p$$, two positive integers $$n=2, m=3$$, two field extensions of order $$n$$ and $$m$$ over our original field, and two special matrices $$A$$ and $$B$$. These parameters can be changed, and then we would obtain different incarnations of the cryptosystem.
In general, a setup would consist of the following choices:
• A prime number $$p$$
• A finite field $$F$$ of size $$q=p^d$$
• Two positive integers $$n<m$$
• Two field extensions of $$F$$ of degrees $$n$$ and $$m$$
• One $$m\times m$$ matrix with integer entries such that:
• its entries are either zeros or powers of $$p$$
• there are exactly two nonzero entries in each row and each column
• is invertible modulo $$p^{dm}-1$$
• rows and columns cannot be arranged such that the matrix is a diagonal sum of boxes
• One $$n\times n$$ matrix like before, but using $$p^{dn}-1$$ instead of $$p^{dm}-1$$
With this choices, we have a cryptosystem that can encrypt vectors of $$n\cdot m$$ entries in $$F$$ (minus the padding). The choice $$n=2,m=3$$ that we showed here is the one implemented in the submission for the NIST, but right after writing it, we found that there are ways to reduce the security of the private key, so we recommend other choices. $$n=2,m=6$$ might be a better one.
It is still not clear what choices of matrices are the most secure. An intuitive guess is that the inverses having a big Hamming weight could be better than smaller ones, but it is just an intuition. Further research would be necessary to understand how these matrices should be chosen. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9213575124740601, "perplexity": 207.33453434258163}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575751.84/warc/CC-MAIN-20190922221623-20190923003623-00318.warc.gz"} |
https://www.physicsforums.com/threads/maximum-height-of-a-projectile-thrown-from-a-rooftop.735334/ | # Homework Help: Maximum height of a projectile thrown from a rooftop
1. Jan 28, 2014
### s.dyseman
1. The problem statement, all variables and given/known data
A man stands on the roof of a building of height 14.6m and throws a rock with a velocity of magnitude 30.8m/s at an angle of 33.2∘ above the horizontal. You can ignore air resistance.
Calculate the maximum height above the roof reached by the rock.
2. Relevant equations
Velocity and position equations
Basic trigonometry
3. The attempt at a solution
Initially, I solved for the y-component of the velocity vector given: V=30.8*Sin(33.2)=16.86m/s
Then, I solved for the amount of time it would take for the rock to reach maximum height, where the velocity of the y-component vector is equal to 0: Vy=Voy+g*t=16.86-9.8t=1.72s
I plug this time into the position equation of Y=Yo+Voy*t+g*t^2=14.6+16.86(1.72)-4.9(1.72)^2=29.1m
So, the maximum height should be equal to 29.1m. Not sure why this is incorrect... Perhaps I calculated the vector incorrectly?
2. Jan 28, 2014
### Staff: Mentor
You calculated the maximum height above the ground (and I can confirm this value).
3. Jan 28, 2014
### s.dyseman
Ah, so I just needed to subtract the height of the roof... Simple detail I missed... Thank you! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9042815566062927, "perplexity": 976.8972948312337}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794865809.59/warc/CC-MAIN-20180523200115-20180523220115-00402.warc.gz"} |
http://cstheory.stackexchange.com/questions/14675/is-there-some-mathematical-closed-form-or-somewhat-tight-asymptotic-one-for-g | Is there some mathematical closed form (or somewhat tight asymptotic one) for “Google Eggs Puzzle”?
The following brief description of the known "Google Eggs Puzzle" comes mainly from the web site Google Eggs:
Google Eggs Puzzle: Given n floors and m eggs, what is the approach to find the highest floor from which eggs can be thrown safely, while minimizing the throws (not broken eggs).
The so called "highest floor" in the above problem deserves more formal definition:
"highest:" there must be a floor f (in any sufficiently tall building) such that an egg dropped from the f th floor breaks, but one dropped from the (f-1)st floor will not. Then, f-1 here is the highest floor.
Actually, the description of "highest" is an excerpt from the book "The Algorithm Design Manual (Second Edition)" by Steven S. Skiena. Being an exercise in Chapter 8 "Dynamic Programming", there are plenty of resources in Web devoted to solving the puzzle by the means of dynamic programming, like Google Eggs and The Two Egg Problem.
However, there is a question from the above book:
Show that $E(n, m) = \Theta(n^{\frac{1}{m}})$, where $E(\cdot)$ is the minimum number of throws. (Note: I have changed the notations used in book for consistency.)
It is the question that motivates my problem:
My Problem: Is there some mathematical closed form for general "Google Eggs Puzzle" with n floors and m eggs, instead of dynamic programming recurrence, and of course tighter than the $E(n, m) = \Theta(n^{\frac{1}{m}})$ one?
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I don't think the asymptotic bound is tight. It works when $m$ is a constant, but if you take $m = \log n$, your bound gives you a constant, which is false. I think a tight bound is $\Theta(\min_{k\leq m} kn^{1/k})$, which reproduces your bound for constant $m$, but also gives $\log n$ as the number of throws when $m$ is large enough to support the naive binary search based strategy. – Robin Kothari Dec 9 '12 at 15:40
@RobinKothari I agree with you. The numerical experiments in the material Joy of Egg-Dropping support your observation. However, I don't catch the meaning of $\Theta(\min_{k \le m} kn^{\frac{1}{k}})$. As my guess, the parameter $k$ is the actual number of eggs in use. Then, what does it mean as a factor in $kn^{\frac{1}{k}}$ ? Thanks a lot. – hengxin Dec 12 '12 at 1:23
I can try to explain the meaning, but it's a bit long so I'll post it as an answer. – Robin Kothari Dec 12 '12 at 3:36
With m eggs and k measurements the most floors that can be checked is exactly $$n(m,k)={k \choose 0} + {k \choose 1} + \ldots + {k \choose m},$$ (maybe $\pm 1$ depending on the exact def). Proof is trivial by induction. This expression has no closed form inverse but gives good asymptotic.
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Just sketching the dropping strategy a little would make the answer more complete. Maybe it's not appropriate, since I guess it's not research-level. Anyway, with 2 eggs, you can skip $k$ floors on your first drop, and if it doesn't break, skip $k - 1$, and if it doesn't break skip $k - 2$, etc. Which gives $k(k+1) / 2$ as the highest floor you could reach using this strategy. – Joe Dec 10 '12 at 23:58
@domotorp It seems constructive to examine the puzzle from the perspective you have just shown. And the equation about $n(m,k)$ can be proved by induction on $m$ and $k$. Although there is no clear closed form for the right hand side of this equation, can it give the asymptotic expression $k(n,m) = \Theta(n^{\frac{1}{m}})$? – hengxin Dec 11 '12 at 13:45
@hengxin, yesish, because $\binom{k}{m}$ is a polynomial in $k$ of degree $m$, so this shows that holding $m$ constant gives $n(m, k) = \Theta(k^m)$. But see Robin's comment on the question. The more interesting question is whether this exact expression allows a more precise bound by approximating the binomial tail e.g. with erf. – Peter Taylor Dec 11 '12 at 13:51
In my comment above I said perhaps $\Theta(\min_{k \le m} kn^{\frac{1}{k}})$ is a tight bound. I'm not sure about the lower bound, but since you just want an explanation for what $k$ means, I can explain the intuition using the upper bound.
As you guessed, $k$ is the number of eggs actually used. That explains the $\min$ on the outside. Now once we've decided to use $k$ eggs, here's a strategy that works: Think of the number $n$ as being written out in base $n^{1/k}$. So $n$'s representation will have $k$ "digits" (the word "digit" is usually reserved for base 10, but I'll use it here), and each digit holds a value from 0 to $n^{1/k}-1$. With our $k$ eggs, we're trying to extract the digits of $n$ one by one. First we start with the most significant digit. This can be determined by throwing an egg from the floor numbered $100..00$, $200..00$, and so on. After at most $n^{1/k}-1$ throws, we've learnt what the most significant bit is, and in the worst case we've broken only 1 egg. Now we do this for all the other digits. Since there are $k$ digits, we'll need $O(kn^{1/k})$ throws.
As a sanity check, observe that when $k=1$, this strategy boils down to dropping eggs from each floor one by one starting from floor 1. When $k = \log n$, we're just working in base 2. So this yields the binary search algorithm.
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An interesting feature of domotorp's solution, unlike yours, is that it does not require knowing n in advance! – JɛffE Dec 12 '12 at 4:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9419028759002686, "perplexity": 280.4480018351266}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375093899.18/warc/CC-MAIN-20150627031813-00133-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://www.reference.com/browse/natural-number | Definitions
# Natural number
In mathematics, a natural number (also called counting number) can mean either an element of the set {1, 2, 3, ...} (the positive integers) or an element of the set {0, 1, 2, 3, ...} (the non-negative integers). The former is generally used in number theory, while the latter is preferred in mathematical logic, set theory, and computer science. A more formal definition will follow.
Natural numbers have two main purposes: they can be used for counting ("there are 3 apples on the table"), and they can be used for ordering ("this is the 3rd largest city in the country").
Properties of the natural numbers related to divisibility, such as the distribution of prime numbers, are studied in number theory. Problems concerning counting, such as Ramsey theory, are studied in combinatorics.
## History of natural numbers and the status of zero
The natural numbers had their origins in the words used to count things, beginning with the number one.
The first major advance in abstraction was the use of numerals to represent numbers. This allowed systems to be developed for recording large numbers. For example, the Babylonians developed a powerful place-value system based essentially on the numerals for 1 and 10. The ancient Egyptians had a system of numerals with distinct hieroglyphs for 1, 10, and all the powers of 10 up to one million. A stone carving from Karnak, dating from around 1500 BC and now at the Louvre in Paris, depicts 276 as 2 hundreds, 7 tens, and 6 ones; and similarly for the number 4,622.
A much later advance in abstraction was the development of the idea of zero as a number with its own numeral. A zero digit had been used in place-value notation as early as 700 BC by the Babylonians, but, they omitted it when it would have been the last symbol in the number. The Olmec and Maya civilization used zero as a separate number as early as 1st century BC, developed independently, but this usage did not spread beyond Mesoamerica. The concept as used in modern times originated with the Indian mathematician Brahmagupta in 628. Nevertheless, medieval computists (calculators of Easter), beginning with Dionysius Exiguus in 525, used zero as a number without using a Roman numeral to write it. Instead nullus, the Latin word for "nothing", was employed. The first systematic study of numbers as abstractions (that is, as abstract entities) is usually credited to the Greek philosophers Pythagoras and Archimedes. However, independent studies also occurred at around the same time in India, China, and Mesoamerica.
In the nineteenth century, a set-theoretical definition of natural numbers was developed. With this definition, it was convenient to include zero (corresponding to the empty set) as a natural number. Including zero in the natural numbers is now the common convention among set theorists, logicians and computer scientists. Other mathematicians, such as number theorists, have kept the older tradition and take 1 to be the first natural number.
## Notation
Mathematicians use N or $mathbb\left\{N\right\}$ (an N in blackboard bold, displayed as ℕ in Unicode) to refer to the set of all natural numbers. This set is countably infinite: it is infinite but countable by definition. This is also expressed by saying that the cardinal number of the set is ($aleph_0$).
To be unambiguous about whether zero is included or not, sometimes an index "0" is added in the former case, and a superscript "*" is added in the latter case:
$mathbb\left\{N\right\}$0 = { 0, 1, 2, ... } ; $mathbb\left\{N\right\}$* = { 1, 2, ... }.
(Sometimes, an index or superscript "+" is added to signify "positive". However, this is often used for "nonnegative" in other cases, as ℝ+ = [0,∞) and ℤ+ = { 0, 1, 2,... }, at least in European literature. The notation "*", however, is standard for nonzero or rather invertible elements.)
Some authors who exclude zero from the naturals use the term whole numbers, denoted $mathbb\left\{W\right\}$, for the set of nonnegative integers. Others use the notation $mathbb\left\{P\right\}$ for the positive integers.
Set theorists often denote the set of all natural numbers by a lower-case Greek letter omega: ω. This stems from the identification of an ordinal number with the set of ordinals that are smaller. When this notation is used, zero is explicitly included as a natural number.
## Algebraic properties
addition multiplication closure: a + b is a natural number a × b is a natural number associativity: a + (b + c) = (a + b) + c a × (b × c) = (a × b) × c commutativity: a + b = b + a a × b = b × a existence of an identity element: a + 0 = a a × 1 = a distributivity: a × (b + c) = (a × b) + (a × c) No zero divisors: if ab = 0, then either a = 0 or b = 0 (or both)
## Formal definitions
Historically, the precise mathematical definition of the natural numbers developed with some difficulty. The Peano postulates state conditions that any successful definition must satisfy. Certain constructions show that, given set theory, models of the Peano postulates must exist.
### Peano axioms
• There is a natural number 0.
• Every natural number a has a natural number successor, denoted by S(a).
• There is no natural number whose successor is 0.
• Distinct natural numbers have distinct successors: if ab, then S(a) ≠ S(b).
• If a property is possessed by 0 and also by the successor of every natural number which possesses it, then it is possessed by all natural numbers. (This postulate ensures that the proof technique of mathematical induction is valid.)
It should be noted that the "0" in the above definition need not correspond to what we normally consider to be the number zero. "0" simply means some object that when combined with an appropriate successor function, satisfies the Peano axioms. All systems that satisfy these axioms are isomorphic, the name "0" is used here for the first element, which is the only element that is not a successor. For example, the natural numbers starting with one also satisfy the axioms.
### Constructions based on set theory
#### A standard construction
A standard construction in set theory, a special case of the von Neumann ordinal construction, is to define the natural numbers as follows:
We set 0 := { }, the empty set,
and define S(a) = a ∪ {a} for every set a. S(a) is the successor of a, and S is called the successor function.
If the axiom of infinity holds, then the set of all natural numbers exists and is the intersection of all sets containing 0 which are closed under this successor function.
If the set of all natural numbers exists, then it satisfies the Peano axioms.
Each natural number is then equal to the set of natural numbers less than it, so that
*0 = { }
*1 = {0} = {{ }}
*2 = {0,1} = {0, {0}} = {{ }, {{ }}}
*3 = {0,1,2} = {0, {0}, {0, {0}}} = {{ }, {{ }}, {{ }, {{ }}}}
*n = {0,1,2,...,n−2,n−1} = {0,1,2,...,n−2} ∪ {n−1} = (n−1) ∪ {n−1}
and so on. When a natural number is used as a set, this is typically what is meant. Under this definition, there are exactly n elements (in the naïve sense) in the set n and nm (in the naïve sense) if and only if n is a subset of m.
Also, with this definition, different possible interpretations of notations like Rn (n-tuples versus mappings of n into R) coincide.
Even if the axiom of infinity fails and the set of all natural numbers does not exist, it is possible to define what it means to be one of these sets. A set n is a natural number means that it is either 0 (empty) or a successor, and each of its elements is either 0 or the successor of another of its elements.
#### Other constructions
Although the standard construction is useful, it is not the only possible construction. For example:
one could define 0 = { }
and S(a) = {a},
producing
0 = { }
1 = {0} = {{ }}
2 = {1} = {{{ }}}, etc.
Or we could even define 0 = {{ }}
and S(a) = a U {a}
producing
0 = {{ }}
1 = {{ }, 0} = {{ }, {{ }}}
2 = {{ }, 0, 1}, etc.
Arguably the oldest set-theoretic definition of the natural numbers is the definition commonly ascribed to Frege and Russell under which each concrete natural number n is defined as the set of all sets with n elements. This may appear circular, but can be made rigorous with care. Define 0 as $\left\{\left\{\right\}\right\}$ (clearly the set of all sets with 0 elements) and define $sigma\left(A\right)$ (for any set A) as $\left\{x cup \left\{y\right\} mid x in A wedge y notin x\right\}$(see set-builder notation). Then 0 will be the set of all sets with 0 elements, $1=sigma\left(0\right)$ will be the set of all sets with 1 element, $2=sigma\left(1\right)$ will be the set of all sets with 2 elements, and so forth. The set of all natural numbers can be defined as the intersection of all sets containing 0 as an element and closed under $sigma$ (that is, if the set contains an element n, it also contains $sigma\left(n\right)$). This definition does not work in the usual systems of axiomatic set theory because the collections involved are too large (it will not work in any set theory with the axiom of separation); but it does work in New Foundations (and in related systems known to be consistent) and in some systems of type theory.
## Properties
One can recursively define an addition on the natural numbers by setting a + 0 = a and a + S(b) = S(a + b) for all a, b. This turns the natural numbers (N, +) into a commutative monoid with identity element 0, the so-called free monoid with one generator. This monoid satisfies the cancellation property and can be embedded in a group. The smallest group containing the natural numbers is the integers.
If we define 1 := S(0), then b + 1 = b + S(0) = S(b + 0) = S(b). That is, b + 1 is simply the successor of b.
Analogously, given that addition has been defined, a multiplication × can be defined via a × 0 = 0 and a × S(b) = (a × b) + a. This turns (N*, ×) into a free commutative monoid with identity element 1; a generator set for this monoid is the set of prime numbers. Addition and multiplication are compatible, which is expressed in the distribution law: a × (b + c) = (a × b) + (a × c). These properties of addition and multiplication make the natural numbers an instance of a commutative semiring. Semirings are an algebraic generalization of the natural numbers where multiplication is not necessarily commutative.
If we interpret the natural numbers as "excluding 0", and "starting at 1", the definitions of + and × are as above, except that we start with a + 1 = S(a) and a × 1 = a.
For the remainder of the article, we write ab to indicate the product a × b, and we also assume the standard order of operations.
Furthermore, one defines a total order on the natural numbers by writing a b if and only if there exists another natural number c with a + c = b. This order is compatible with the arithmetical operations in the following sense: if a, b and c are natural numbers and ab, then a + cb + c and acbc. An important property of the natural numbers is that they are well-ordered: every non-empty set of natural numbers has a least element. The rank among well-ordered sets is expressed by an ordinal number; for the natural numbers this is expressed as "$omega$".
While it is in general not possible to divide one natural number by another and get a natural number as result, the procedure of division with remainder is available as a substitute: for any two natural numbers a and b with b ≠ 0 we can find natural numbers q and r such that
a = bq + r and r < b.
The number q is called the quotient and r is called the remainder of division of a by b. The numbers q and r are uniquely determined by a and b. This, the Division algorithm, is key to several other properties (divisibility), algorithms (such as the Euclidean algorithm), and ideas in number theory.
The natural numbers including zero form a commutative monoid under addition (with identity element zero), and under multiplication (with identity element one).
## Generalizations
Two generalizations of natural numbers arise from the two uses:
• A natural number can be used to express the size of a finite set; more generally a cardinal number is a measure for the size of a set also suitable for infinite sets; this refers to a concept of "size" such that if there is a bijection between two sets they have the same size. The set of natural numbers itself and any other countably infinite set has cardinality ($aleph_0$).
• Ordinal numbers "first", "second", "third" can be assigned to the elements of a totally ordered finite set, and also to the elements of well-ordered countably infinite sets like the set of natural numbers itself. This can be generalized to ordinal numbers which describe the position of an element in a well-order set in general. An ordinal number is also used to describe the "size" of a well-ordered set, in a sense different from cardinality: if there is an order isomorphism between two well-ordered sets they have the same ordinal number. The first ordinal number that is not a natural number is expressed as $omega$; this is also the ordinal number of the set of natural numbers itself.
$aleph_0$ and $omega$ have to be distinguished because many well-ordered sets with cardinal number $aleph_0$ have a higher ordinal number than $omega$, for example, $omega^\left\{omega^\left\{omega6+42\right\}cdot1729+omega^9+88\right\}cdot3+omega^\left\{omega^omega\right\}cdot5+65537$; $omega$ is the lowest possible value (the initial ordinal).
For finite well-ordered sets there is one-to-one correspondence between ordinal and cardinal number; therefore they can both be expressed by the same natural number, the number of elements of the set. This number can also be used to describe the position of an element in a larger finite, or an infinite, sequence.
Other generalizations are discussed in the article on numbers.
## References
• Edmund Landau, Foundations of Analysis, Chelsea Pub Co. ISBN 0-8218-2693-X.
• Richard Dedekind, Essays on the theory of numbers, Dover, 1963, ISBN 0486210103 / Kessinger Publishing, LLC , 2007, ISBN 054808985X | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 22, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9124787449836731, "perplexity": 329.89391282116725}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936462426.4/warc/CC-MAIN-20150226074102-00019-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://gingkoapp.com/test2 | • General
so what is this has to add to ginko
• Introduction
• This document hold all the project details for the ExpressPack Quality 2014 (EPQ 2014) version, included all the analyses, design and development details. If a details exist on another document or web page a references will be included. It is the intention to cover all the necessary information and data to build and deploy EPQ 2014. Included but not limited to: requirements, analyses, UML diagrams, ER diagrams, test plan, principals.
This document is a living document and may evolve through updates, be expanded as needed. The changes will happen through revision and is under version control. The changes are done by the document owner(s) or there secondary.
The original with the latest update and the prior versions can be found on neomatics portal add link
It is the duty of every team member to:
• To read and understand the whole document.
• To respect and to stick to the definitions, designs, guidelines and principals described in the document.
• Report any inconsistent, wrong or missing information.
In any doubt, question or obstacle contact the document owner.
While the document describes the building of EPQ 2014 some deliverables will be part of ExpressPack (EP) Framework. This will have impacts on the chosen architecture, namespaces and artifacts.
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http://ldkoffice.com/standard-error/sampling-average-error.html | Home > Standard Error > Sampling Average Error
# Sampling Average Error
## Contents
In that case, the mean you estimate is the parameter. As will be shown, the standard error is the standard deviation of the sampling distribution. Suppose the population standard deviation is 0.6 ounces. The data set is ageAtMar, also from the R package openintro from the textbook by Dietz et al.[4] For the purpose of this example, the 5,534 women are the entire population his comment is here
## Standard Error Formula
Consider a sample of n=16 runners selected at random from the 9,732. So if this up here has a variance of-- let's say this up here has a variance of 20. For the purpose of this example, the 9,732 runners who completed the 2012 run are the entire population of interest.
So here, just visually, you can tell just when n was larger, the standard deviation here is smaller. This approximate formula is for moderate to large sample sizes; the reference gives the exact formulas for any sample size, and can be applied to heavily autocorrelated time series like Wall Instead of weighing every single cone made, you ask each of your new employees to randomly spot check the weights of a random sample of the large cones they make and Standard Error Of The Mean Definition Well, we're still in the ballpark.
For an upcoming national election, 2000 voters are chosen at random and asked if they will vote for candidate A or candidate B. Standard Error Mean Now, this guy's standard deviation or the standard deviation of the sampling distribution of the sample mean, or the standard error of the mean, is going to the square root of The effect of the FPC is that the error becomes zero when the sample size n is equal to the population size N. Secondly, the standard error of the mean can refer to an estimate of that standard deviation, computed from the sample of data being analyzed at the time.
However, different samples drawn from that same population would in general have different values of the sample mean, so there is a distribution of sampled means (with its own mean and Sampling Error Example A low sampling error means that we had relatively less variability or range in the sampling distribution. T-distributions are slightly different from Gaussian, and vary depending on the size of the sample. But anyway, hopefully this makes everything clear.
## Standard Error Mean
Search this site: Leave this field blank: Home Overview ResearchMethods Experiments Design Statistics FoundationsReasoning Philosophy Ethics History AcademicPsychology Biology Physics Medicine Anthropology Self-HelpSelf-Esteem Worry Social Anxiety Sleep Anxiety Write Paper Assisted So we take 10 instances of this random variable, average them out, and then plot our average. Standard Error Formula They report that, in a sample of 400 patients, the new drug lowers cholesterol by an average of 20 units (mg/dL). Standard Error Calculator The proportion or the mean is calculated using the sample.
If σ is known, the standard error is calculated using the formula σ x ¯ = σ n {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt {n}}}} where σ is the this content The standard error of the mean now refers to the change in mean with different experiments conducted each time. If σ is not known, the standard error is estimated using the formula s x ¯ = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} where s is the sample So 1 over the square root of 5. Standard Error Vs Standard Deviation
The margin of error and the confidence interval are based on a quantitative measure of uncertainty: the standard error. Blackwell Publishing. 81 (1): 75–81. The standard error is the standard deviation of the Student t-distribution. http://ldkoffice.com/standard-error/sampling-error-of-the-mean.html Louis, MO: Saunders Elsevier.
Well, Sal, you just gave a formula. Standard Error Regression What's the margin of error? (Assume you want a 95% level of confidence.) It's calculated this way: So to report these results, you say that based on the sample of 50 Moreover, this formula works for positive and negative ρ alike.[10] See also unbiased estimation of standard deviation for more discussion.
## Standard errors provide simple measures of uncertainty in a value and are often used because: If the standard error of several individual quantities is known then the standard error of some
Here are the steps for calculating the margin of error for a sample mean: Find the population standard deviation and the sample size, n. They report that, in a sample of 400 patients, the new drug lowers cholesterol by an average of 20 units (mg/dL). The following expressions can be used to calculate the upper and lower 95% confidence limits, where x ¯ {\displaystyle {\bar {x}}} is equal to the sample mean, S E {\displaystyle SE} Sampling Error Formula Consider a sample of n=16 runners selected at random from the 9,732.
A practical result: Decreasing the uncertainty in a mean value estimate by a factor of two requires acquiring four times as many observations in the sample. References Sarndal, Swenson, and Wretman (1992), Model Assisted Survey Sampling, Springer-Verlag, ISBN 0-387-40620-4 Fritz Scheuren (2005). "What is a Margin of Error?", Chapter 10, in "What is a Survey?", American Statistical It will be shown that the standard deviation of all possible sample means of size n=16 is equal to the population standard deviation, σ, divided by the square root of the check over here A crucial midway concept you need to understand is the sampling distribution. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9811792969703674, "perplexity": 792.5375876880403}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812306.16/warc/CC-MAIN-20180219012716-20180219032716-00110.warc.gz"} |
http://mathhelpforum.com/calculus/49955-indefinite-integral.html | # Math Help - Indefinite integral
1. ## Indefinite integral
Having no luck with this question. Any help would be great.
Give an expression for the indefinite integral
Integral of 1/(sqrt x) (1+sqrt x)^2 dx (x>0)
2. Originally Posted by offahengaway and chips
Having no luck with this question. Any help would be great.
Give an expression for the indefinite integral
Integral of 1/(sqrt x) (1+sqrt x)^2 dx (x>0)
do a substitution. $u = 1 + \sqrt{x}$
you will end up with $2 \int u^{-2}~du$
which i am sure you can handle
3. I would also use the substitution method, but I did this to make it look different (not really sure if it helps): | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9490521550178528, "perplexity": 1575.8074839233543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931009271.61/warc/CC-MAIN-20141125155649-00142-ip-10-235-23-156.ec2.internal.warc.gz"} |
http://en.wikipedia.org/wiki/Magnus_effect | # Magnus effect
The Magnus effect, depicted with a back-spinning cylinder or ball in an air stream. The arrow represents the resulting lifting force. The curly flow lines represent a turbulent wake. The airflow has been deflected in the direction of spin.
The Magnus effect is the commonly observed effect in which a spinning ball (or cylinder) curves away from its principal flight path. It is important in many ball sports. It affects spinning missiles, and has some engineering uses, for instance in the design of rotor ships and Flettner aeroplanes.
In terms of ball games, top spin is defined as spin about a horizontal axis perpendicular to the direction of travel, where the top surface of the ball is moving forward with the spin. Under the Magnus effect, top spin produces a downward swerve of a moving ball, greater than would be produced by gravity alone, and back spin has the opposite effect.[1] Likewise side-spin causes swerve to either side as seen during some baseball pitches.[2]
The overall behaviour is similar to that around an airfoil (see lift force) with a circulation which is generated by the mechanical rotation, rather than by airfoil action.[3]
It is named for Gustav Magnus, the German physicist who investigated it. The force on a rotating cylinder is known as Kutta-Joukowski lift,[4] after Martin Wilhelm Kutta and Nikolai Zhukovsky (or Joukowski) who first analyzed the effect.
## Physics
A valid intuitive understanding of the phenomenon is possible, beginning with the fact that, by conservation of momentum, the deflective force on the body is no more or less than a reaction to the deflection that the body imposes on the air-flow. The body "pushes" the air down, and vice versa. As a particular case, a lifting force is accompanied by a downward deflection of the air-flow. It is an angular deflection in the fluid flow, aft of the body.
In fact there are several ways in which the rotation might cause such a deflection. By far the best way to know what actually happens in typical cases is by wind-tunnel experiments. Lyman Briggs[5] made a definitive wind tunnel study of the Magnus effect on baseballs, and others have produced interesting images of the effect.[5][6][7][8] The studies show a turbulent wake behind the spinning ball. The wake is to be expected and is the cause of aerodynamic drag. However there is a noticeable angular deflection in the wake and the deflection is in the direction of the spin.
The process by which a turbulent wake develops aft of a body in an air-flow is complex but well-studied in aerodynamics. It is found that the thin boundary layer detaches itself ("flow separation") from the body at some point and this is where the wake begins to develop. The boundary layer itself may be turbulent or not; this has a significant effect on the wake formation. Quite small variations in the surface conditions of the body can influence the onset of wake formation and thereby have a marked effect on the downstream flow pattern. The influence of the body's rotation is of this kind.
It is said[citation needed] that Magnus himself wrongly postulated a theoretical effect with laminar flow due to skin friction and viscosity as the cause of the Magnus effect. Such effects are physically possible but slight in comparison to what is produced in the Magnus effect proper.[5] In some circumstances the causes of the Magnus effect can produce a deflection opposite to that of the Magnus effect.[8]
The diagram at the head of this article shows lift being produced on a back-spinning ball. The wake and trailing air-flow have been deflected downwards. The boundary layer motion is more violent at the underside of the ball where the spinning movement of the ball's surface is forward and reinforces the effect of the ball's translational movement. The boundary layer generates wake turbulence after a short interval.
On a cylinder, the force due to rotation is known as Kutta-Joukowski lift. It can be analyzed in terms of the vortex produced by rotation. The lift on the cylinder per unit length, F/L, is the product of the velocity, V, the density of the fluid, $\rho$, and the strength of the vortex that is established by the rotation, G:[4]
$F/L= \rho V G$
## History
German physicist Heinrich Gustav Magnus described the effect in 1852.[9][10] However, in 1672, Isaac Newton had described it and correctly inferred the cause after observing tennis players in his Cambridge college.[11][12]
In 1742, Benjamin Robins, a British mathematician, ballistics researcher, and military engineer, explained deviations in the trajectories of musket balls in terms of the Magnus effect.[13][14][15][16]
## In sport
The Magnus effect explains commonly observed deviations from the typical trajectories or paths of spinning balls in sport, notably association football (soccer), table tennis, tennis,[17] volleyball, golf, baseball, cricket and in paintball marker balls.
The curved path of a golf ball known as slice or hook is due largely to the ball's spinning motion (about its vertical axis) and the Magnus effect, causing a horizontal force that moves the ball from a straight-line in its trajectory.[18] Back-spin (upper surface rotating backwards from the direction of movement) on a golf ball causes a vertical force that counteracts the force of gravity slightly, and enables the ball to remain airborne a little longer than it would were the ball not spinning: this allows the ball to travel farther than a non-spinning (about its horizontal axis) ball.
In table tennis, the Magnus effect is easily observed, because of the small mass and low density of the ball. An experienced player can place a wide variety of spins on the ball. Table tennis rackets usually have a surface made of rubber to give the racket maximum grip on the ball, to impart a spin.
The Magnus effect is not responsible for the movement of the cricket ball seen in swing bowling,[19] although it does contribute to the motion known as drift in spin bowling.
In airsoft, a system known as Hop-Up is used to create a backspin on a fired BB, which will greatly increase its range, using the Magnus effect in a similar manner as in golf.
In paintball, Tippmann's Flatline Barrel System also takes advantage of the Magnus effect by imparting a backspin on the paintballs, which increases their effective range by counteracting gravity.
In baseball, pitchers often impart different spins on the ball, causing it to curve in the desired direction due to the Magnus effect. The PITCHf/x system measures the change in trajectory caused by Magnus in all pitches thrown in Major League Baseball.[20]
The match ball for the 2010 FIFA World Cup has been criticised for the different Magnus effect from previous match balls. The current ball is described as having less Magnus effect and as a result flies farther but with less controllable swerve.[21]
## In external ballistics
The Magnus effect can also be found in advanced external ballistics. Firstly, a spinning bullet in flight is often subject to a crosswind, which can be simplified as blowing either from the left or the right. In addition to this, even in completely calm air a bullet experiences a small sideways wind component due to its yawing motion. This yawing motion along the bullet's flight path means that the nose of the bullet is pointing in a slightly different direction from the direction in which the bullet is travelling. In other words, the bullet is "skidding" sideways at any given moment, and thus it experiences a small sideways wind component in addition to any crosswind component.[22]
The combined sideways wind component of these two effects causes a Magnus force to act on the bullet, which is perpendicular both to the direction the bullet is pointing and the combined sideways wind. In a very simple case where we ignore various complicating factors, the Magnus force from the crosswind would cause an upward or downward force to act on the spinning bullet (depending on the left or right wind and rotation), causing an observable deflection in the bullet's flight path up or down, thus changing the point of impact.
Overall, the effect of the Magnus force on a bullet's flight path itself is usually insignificant compared to other forces such as aerodynamic drag. However, it greatly affects the bullet's stability, which in turn affects the amount of drag, how the bullet behaves upon impact, and many other factors. The stability of the bullet is affected[citation needed] because the Magnus effect acts on the bullet's centre of pressure instead of its centre of gravity. This means that it affects the yaw angle of the bullet: it tends to twist the bullet along its flight path, either towards the axis of flight (decreasing the yaw thus stabilizing the bullet) or away from the axis of flight (increasing the yaw thus destabilizing the bullet). The critical factor is the location of the centre of pressure, which depends on the flowfield structure, which in turn depends mainly on the bullet's speed (supersonic or subsonic), but also the shape, air density and surface features. If the centre of pressure is ahead of the centre of gravity, the effect is destabilizing; if the centre of pressure is behind the centre of gravity, the effect is stabilizing.[citation needed]
## In flying machines
Some flying machines have been built which use the Magnus effect to create lift with a rotating cylinder at the front of a wing, allowing flight at lower horizontal speeds.[4] The earliest attempt to use the Magnus Effect for a heavier than air aircraft was in 1910 by a US member of Congress, Butler Ames of Massachusetts. The next attempt was in the early 1930s by three inventors in New York state.[23]
## Ship stabilization
The effect is used in a special type of ship stabilizer consisting of a rotating cylinder mounted beneath the waterline and emerging laterally. By controlling the direction and speed of rotation, strong lift or downforce can be generated.[24] The largest deployment of the system to date is in the Eclipse yacht.
## References
1. ^ http://math.ucr.edu/home/baez/physics/General/golf.html
2. ^ The Curveball, The Physics of Baseball.
3. ^ Clancy, L.J., Aerodynamics, Section 4.6
4. ^ a b c "Lift on rotating cylinders". NASA Glenn Research Center. 2010-11-09. Retrieved 2013-11-07.
5. ^ a b c Briggs, Lyman (1959). "Effect of Spin and Speed on the Lateral Deflection (Curve) of a Baseball and the Magnus Effect for Smooth Spheres". American Journal of Physics 27 (8): 589. Bibcode:1959AmJPh..27..589B. doi:10.1119/1.1934921.
6. ^ Brown, F (1971). See the Wind Blow. University of Notre Dame.
7. ^ Van Dyke, Milton (1982). An album of Fluid motion. Stanford University.
8. ^ a b Cross, Rod. "Wind Tunnel Photographs". Physics Department, University of Sydney. p. 4. Retrieved 10 February 2013.
9. ^ G. Magnus (1852) "Über die Abweichung der Geschosse," Abhandlungen der Königlichen Akademie der Wissenschaften zu Berlin, pages 1-23.
10. ^ G. Magnus (1853) "Über die Abweichung der Geschosse, und: Über eine abfallende Erscheinung bei rotierenden Körpern" (On the deviation of projectiles, and: On a sinking phenomenon among rotating bodies), Annalen der Physik, vol. 164, no. 1, pages 1-29.
11. ^ Isaac Newton, "A letter of Mr. Isaac Newton, of the University of Cambridge, containing his new theory about light and color," Philosophical Transactions of the Royal Society, vol. 7, pages 3075-3087 (1671-1672). (Note: In this letter, Newton tried to explain the refraction of light by arguing that rotating particles of light curve as they moved through a medium just as a rotating tennis ball curves as it moves through the air.)
12. ^ Gleick, James. 2004. Isaac Newton. London: Harper Fourth Estate.
13. ^ Benjamin Robins, New Principles of Gunnery: Containing the Determinations of the Force of Gun-powder and Investigations of the Difference in the Resisting Power of the Air to Swift and Slow Motions (London: J. Nourse, 1742). (On page 208 of the 1805 edition of Robins' New Principles of Gunnery, Robins describes the experiment in which he observed the Magnus effect: A ball was suspended by a tether consisting of two strings twisted together, and the ball was made to swing. As the strings unwound, the swinging ball rotated, and the plane of its swing also rotated. The direction in which the plane rotated depended on the direction in which the ball rotated.)
14. ^ Tom Holmberg, "Artillery Swings Like a Pendulum..." in "The Napoleon Series"
15. ^ Steele, Brett D. (April 1994) "Muskets and pendulums: Benjamin Robins, Leonhard Euler, and the ballistics revolution," Technology and Culture, vol. 35, no. 2, pages 348-382.
16. ^ Newton's and Robins' observations of the Magnus effect are reproduced in: Peter Guthrie Tait (1893) "On the path of a rotating spherical projectile," Transactions of the Royal Society of Edinburgh, vol. 37, pages 427-440.
17. ^ Lord Rayleigh (1877) "On the irregular flight of a tennis ball," Messenger of Mathematics, vol. 7, pages 14–16.
18. ^ Clancy, L.J., Aerodynamics, Section 4.5
19. ^ Clancy, L.J., Aerodynamics, Figure 4.19
20. ^ Nathan, Alan M. (October 18, 2012). "Determining Pitch Movement from PITCHf/x Data". Retrieved 18 October 2012.
21. ^ SBS 2010 FIFA World Cup Show interview 22 June 2010 10:30pm by Craig Johnston
22. ^ Ruprecht Nennstiel. "Yaw of repose". Nennstiel-ruprecht.de. Retrieved 2013-02-22.
23. ^ Whirling Spools Lift This Plane. Popular Science. Nov 1930. Retrieved 2013-02-22.
24. ^ "Quantum Rotary Stabilizers". Jun 2, 2009. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8184590339660645, "perplexity": 1601.429083366229}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999654052/warc/CC-MAIN-20140305060734-00026-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://findfilo.com/maths-question-answers/if-a-is-the-area-and-2s-the-sum-of-three-sides-of-5eq | If A is the area and 2s the sum of three sides of a triangle, then | Filo
Class 11
Math
Algebra
Sequences and Series
554
150
If A is the area and 2s the sum of three sides of a triangle, then
1. None of these
Solution: We know that, A.MG.M
Ans: A
554
150
Connecting you to a tutor in 60 seconds. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8288694024085999, "perplexity": 875.9418230174789}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00520.warc.gz"} |
https://opus4.kobv.de/opus4-zib/frontdoor/index/index/docId/680 | ## Linear convergence of an interior point method for linear control constrained optimal control problems
Please always quote using this URN: urn:nbn:de:0297-zib-6809
• The paper provides a detailed analysis of a short step interior point algorithm applied to linear control constrained optimal control problems. Using an affine invariant local norm and an inexact Newton corrector, the well-known convergence results from finite dimensional linear programming can be extended to the infinite dimensional setting of optimal control. The present work complements a recent paper of Weiser and Deuflhard, where convergence rates have not been derived. The choice of free parameters, i.e. the corrector accuracy and the number of corrector steps, is discussed.
$Rev: 13581$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8690038919448853, "perplexity": 521.5068382809727}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560279248.16/warc/CC-MAIN-20170116095119-00183-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://dsp.stackexchange.com/questions/4901/circular-time-delay-in-signal-effect-on-phase-spectra-of-dft | # Circular Time Delay in signal, effect on phase spectra of DFT
If you circularly shift a signal x[n]= [-3 -2 -1 0 1 2 3 2 1 0 -1 -2 ] to the right by M=1 compared to M=2, the phase spectra has a lot many zeros when M=2 compared to M=1.
here, I am talking about Circular Time shifting.
I know that if x[n] is even, then DFT is purely real and hence phase is zero. However, why are there way many more zeros when M=2 compared to M=1?
Thanks
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If by phase spectrum you mean the sequence $$\bigr(\angle X[0], \angle X[1], \angle X[2], \ldots, \angle X[N-1]\bigr),$$ then remember that a circular shift by $M$ in the time domain results in each $X[n]$ being multiplied by $\exp(-j2\pi Mn/N)$, and thus causes a change in $\angle X[n]$; in particular, $\angle X[n]$ decreases by $2\pi Mn/N$. For the case $M=1$, the multipliers are distinct $N$-th roots of unity, that is, the changes in phase are
$$\Bigr(0, 2\pi\frac{1}{N}, 2\pi\frac{1}{N}, \ldots, 2\pi\frac{N-1}{N}\Bigr)$$
all of which are different numbers. Thus, each entry in the phase spectrum changes (except for the entry for $n=0$). For the case $M=2$ and $N=12$, the multipliers are $\exp(-j2\pi 2n/12) = \exp(-j2\pi n/6)$ and so the changes in phase are
$$\Bigr(0, 2\pi\frac{1}{6}, 2\pi\frac{2}{6}, \ldots, 2\pi\frac{5}{6}, 2\pi\frac{6}{6}, 2\pi\frac{7}{6}, \ldots 2\pi\frac{11}{6}\Bigr)$$
But a phase change of $2\pi\frac{6}{6} = 2\pi$ is the same as no phase change, a phase change of $2\pi\frac{7}{6}$ is the same as a phase change of $2\pi\frac{1}{6}$, and so on. So the displayed vector above of phase changes is actually
$$\Bigr(0, 2\pi\frac{1}{6}, 2\pi\frac{2}{6}, \ldots, 2\pi\frac{5}{6}, 0, 2\pi\frac{1}{6}, \ldots 2\pi\frac{5}{6}\Bigr)$$
which is two periods of a sequence of period $6$. Note each entry in the phase spectrum changes except for $n=0$ and $n=6$ which do not change.
Now suppose that each $X[n]$ is a real number and thus the phase spectrum is $0$ everywhere. Then, if you circularly shift by one place ($M=1$ in the time domain, all entries in the phase spectrum will be nonzero except for the $n=0$ entry. If instead you circularly shift by two places ($M=2$) in the time domain, the phase spectrum will have zeroes in both the $n=0$ and $n=6$ entries. Note also that $\exp(-j\pi = -1$, and so for $M=1$, $X[0]$ and $X[6]$ will still be real-valued (though there will have been a phase change of $\pi$ in $X[6]$) and similarly, for $M=2$, $X[0], X[3], X[6]$, and $X[9]$ will still remain real-valued though there will have been a phase change of $\pi$ in $X[3]$ and $X[9]$.
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I am really new to this; hence I could not understand what you meant by "For the case M=1, the multipliers are distinct N-th roots of unity and each"--esp, the multipliers being distinct N-roots of unity. By unity, do you mean the unit circle with the multiplier being the $exp(−j2πMn/N)$? If you could kindly break it down for me, it help make sense, especially how you are accounting for the periods of the multiplier sequence. – dsp_ent Nov 4 '12 at 18:56
@dsp_ent See edited answer. – Dilip Sarwate Nov 4 '12 at 22:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9458562135696411, "perplexity": 266.918974345494}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783396945.81/warc/CC-MAIN-20160624154956-00008-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/112285-functions-help.html | 1. ## functions help
Would you explain How to determine the equation of the function and what does it mean to determine the sign of the function in each interval "created by zeros" of the function?
With these given points:
(-12,-30888), (-10,-12375), (-8,-4032), (-6,-945), (-4,-120), (-2,-3), (0,0), (2,3), (4,120), (6,945), (8,4032), (10,12375), (12,30888)
Thanks
2. hi im new but i''ll try to help, please correct me if i am wrong.
Q1.Would you explain How to determine the equation of the function?
A. you were given a set points and expected to deduce an equation that would out put those values, it looks like there is a pattern so i think this is a geometric sequence problem.
Q2.what does it mean to determine the sign of the function in each interval "created by zeros" of the function?
A. the Zeros of a function are the x values where y=0, that is where the curve of the function intersects the x axis. anything above the x axis is + and below is -. The curve looks like an odd function which means it is flipped across x axis for negative x values.
With these given points:
(-12,-30888), (-10,-12375), (-8,-4032), (-6,-945), (-4,-120), (-2,-3), (0,0), (2,3), (4,120), (6,945), (8,4032), (10,12375), (12,30888)
i cant remember how to do geometric sequences but i'll help on Q2.
with the points you were given, it again looks like f(x) = - f(-x) but im not sure until you figure the equation. y= 0 is a zero, that occurs at x=0 so x= 0 is a 'root' or a 'zero' this splits the graph into two intervals, where if x>= 0 then the sign is positive, and if x< 0 the sign is negative.
a hint on Q1. it looks like its in the form of ax^n where a,n are + integers | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9024330973625183, "perplexity": 805.543634090601}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320023.23/warc/CC-MAIN-20170623063716-20170623083716-00210.warc.gz"} |
https://kr.mathworks.com/help/phased/ref/effearthradius.html | Documentation
## Syntax
```Re = effearthradius Re = effearthradius(RGradient) ```
## Description
`Re = effearthradius` returns the effective radius of spherical earth in meters. The calculation uses a refractivity gradient of `-39e-9`. As a result, `Re` is approximately 4/3 of the actual earth radius.
`Re = effearthradius(RGradient)` specifies the refractivity gradient.
## Input Arguments
`RGradient` Refractivity gradient in units of 1/meter. This value must be a nonpositive scalar. Default: `-39e-9`
## Output Arguments
`Re` Effective earth radius in meters.
collapse all
The effective earth radius is a scaling of the actual earth radius. The scale factor is:
`$\frac{1}{1+r\cdot \text{RGradient}}$`
where r is the actual earth radius in meters and `RGradient` is the refractivity gradient. The refractivity gradient, which depends on the altitude, is the rate of change of refraction index with altitude. The refraction index for a given altitude is the ratio between the free-space propagation speed and the propagation speed in the air band at that altitude.
The most commonly used scale factor is 4/3. This value corresponds to a refractivity gradient of .
## References
[1] Skolnik, M. Introduction to Radar Systems, 3rd Ed. New York: McGraw-Hill, 2001. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9680261611938477, "perplexity": 1574.745980162452}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315681.63/warc/CC-MAIN-20190820221802-20190821003802-00133.warc.gz"} |
http://math.stackexchange.com/questions/146973/expected-value-of-the-maximum-of-two-exponentially-distributed-random-variables | # Expected Value of the maximum of two exponentially distributed random variables
I want to find the expected value of $\text{max}\{X,Y\}$ where $X$ ist $\text{exp}(\lambda)$-distributed and $Y$ ist $\text{exp}(\eta)$-distributed. I figured out how to do this for the minimum of $n$ variables, but i struggle with doing it for 2 with the maximum.
(The context in which this was given is waiting for the later of two trains, with their arrival times being exp-distributed).
Thanks!
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Are these two variable independent? – Davide Giraudo May 19 '12 at 8:36
Yes they are. Forgot to add that, sorry. – ifubic May 19 '12 at 8:41
You can find the cumulative distribution function, and for non-negative random variables there is a formula which links the CDF with the expectation. – Davide Giraudo May 19 '12 at 8:43
Let $V=\max\{X,Y\}$, then $$P(V\leq t)=P(X\leq t,Y\leq t)==P(X\leq t)P(Y\leq t)$$ now find $f_V(t)$ and then $\int_{-\infty}^{+\infty}tf_V(t)dt$, which should be $\frac{1}{\lambda}+\frac{1}{\eta}-\frac{1}{\lambda+\eta}$
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Thanks! That worked out nicely! – ifubic May 19 '12 at 9:59
That would be appropriate to accept the answer then. – Julius May 19 '12 at 10:21
The sample $(X,Y)$ have a density given by $f_X(x)f_Y(y)$ since $X$ and $Y$ are independent. You have to compute $$\iint_{\Bbb R^2}\max\{x,y\}f_X(x)f_Y(y)dxdy.$$ Cut this integral in two parts.
The minimum of two independent exponential random variables with parameters $\lambda$ and $\eta$ is also exponential with parameter $\lambda+\eta$.
Also $\mathbb E\big[\min(X_1+X_2)+\max(X_1,X_2)\big]=\mathbb E\big[X_1+X_2\big]=\frac{1}{\lambda}+\frac{1}{\eta}$. Rearranging and using $\mathbb E\big[\min(X_1,X_2)\big]=\frac{1}{\lambda+\eta}$, we get $\mathbb E\big[\max(X_1,X_2)\big]=\frac{1}{\lambda}+\frac{1}{\eta}-\frac{1}{\lambda+\eta}.$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9053059816360474, "perplexity": 347.87549123087524}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999638008/warc/CC-MAIN-20140305060718-00086-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-5-linear-functions-5-8-graphing-absolute-value-functions-practice-and-problem-solving-exercises-page-345/29 | ## Algebra 1
$y=|x-0.5|$
The graph of $y=|x-h|$, where $h$ is a positive number, translates the graph $y=|x|$ to the right $h$ units. By substituting $0.5$ in for $h$, we can translate the graph $0.5$ units to the right. This gives us the equation $y=|x-0.5|$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.963127613067627, "perplexity": 83.61447881024303}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00217.warc.gz"} |
https://percentages.io/ | What is % of ?
is what % of ?
What is the % change from to ?
Example Problems
# What is Percentage?
### An introduction
Percentage is a mathematical number expressed as a fraction with a denominator of 100. It is normally denoted with the percentage (%) symbol. For example, the fraction $$\frac{50}{100}$$ could also be written as $$50\%$$. That example is simple, because the denominator of our fraction is already 100. Another example is that the ratio $$\frac{20}{40}$$ could also be written as $$50\%$$. Note that both are equivalent fractions, but are written differently. Percentage allows us to normalize numbers and ratios that otherwise appear quite different. The concept of percentage is used by many people in day to day life. It can be used for many things such as calculating your score on a test or calculating how much tip to leave your waiter at a restaurant. It can help keep track of performance over time and it can help make sense of complex stock market prices. The concept of normalizing numbers of any size to a simple ratio with a base of 100 is one that has an unlimite number of uses.
The purpose of this website is to provide a quick and simple tool for users to calculate percentages of all types. It can also be used to help with math homework or as a learning aid. This website also aims to provide users with a friendly and simple user interface that makes interaction with the site painless and fast. I hope you enjoy using this tool as much as I enjoyed making it! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8362298607826233, "perplexity": 321.21944159757}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347419639.53/warc/CC-MAIN-20200601211310-20200602001310-00412.warc.gz"} |
https://stats.stackexchange.com/questions/41551/hypothesis-testing-with-the-geometric-distribution | # Hypothesis testing with the geometric distribution
A single observation of a random variable having a geometric distribution is used to test the null hypothesis $\theta=\theta_0$ against the alternative hypothesis $\theta=\theta_1 > \theta_0$.
If the null hypothesis is rejected if and only if the observed value of the random variable is greater than or equal to the positive integer $k$ , find expressions for the probabilities of type I and type II errors.
My idea:
Type I error equals $\alpha$, which is the probability that we reject $H_0$ while it is true. Probability of type II error is $\beta$, which is the probability that we accept $H_0$ while $H_0$ is false.
Let's say the Null Hypothetis is rejected, this means that $X\ge k$ ?
I know that that $P(X\ge k ; \theta_0)= 1- \theta_0(1-\theta_0)^{k-1}$. Is this value equal to $\alpha$?
How can I calculate $\beta$? $P(X\ge k ; \theta_1)= \theta_1(1-\theta_1)^{k-1}$.
Who can help me making this clear ? :-)
Thank you for your trouble :-)
You have the definitions of Type I and Type II errors correct. However, the probability statement you use to represent the Type I error is incorrect. You want $P(X\geq k;\theta_0)=1-F_X(k-1;\theta_0)$, where $F_X(x;\theta_0)$ is the cumulative distribution function of a geometric random variable with parameter $\theta_0$. For the Type II error, the probability statement you've given is also incorrect. You want the probability of failing to reject the null hypothesis given that the alternative is true. This can be written as $P(X<k;\theta_1)=F_X(k-1;\theta_1)$.
When you write that $P(X\geq k;\theta_0)=1-\theta_0(1-\theta_0)^{k-1}$, you're actually saying that the Type I error is the probability of $X\neq k$ rather than $X\geq k$. Similarly, when you write that $P(X\geq k;\theta_1)=\theta_1(1-\theta_1)^{k-1}$, the inequality is incorrect (it should be $<$) and you've written that it is the probability of $X=k$ rather than $X<k$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9721764922142029, "perplexity": 72.67408169674303}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00274.warc.gz"} |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-a-molecular-approach-3rd-edition/chapter-4-sections-4-1-4-9-exercises-problems-by-topic-page-187/35b | ## Chemistry: A Molecular Approach (3rd Edition)
$8.25g\ HNO_{3}$
Divide the mass of the base by its molar mass to find the moles of base to react. Multiply this by the mole ratio from the balanced chemical equation to get the moles of acid. Finally, multiply the moles of acid by the molar mass of the acid to get the mass of acid. $4.85g\ Ca(OH)_{2}\times\frac{1mol\ Ca(OH)_{2}}{74.0927g\ Ca(OH)_{2}}\times\frac{2mol\ HNO_{3}}{1mol\ Ca(OH)_{2}}\times\frac{63.0128g\ HNO_{3}}{1mol\ HNO_{3}}=8.25g\ HNO_{3}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9767606258392334, "perplexity": 1121.2284882578178}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160641.81/warc/CC-MAIN-20180924185233-20180924205633-00491.warc.gz"} |
https://sbseminar.wordpress.com/2009/07/28/topology-that-algebra-cant-see/ | # Topology that Algebra can’t see
Let $X$ be an algebraic variety over $\mathbb{C}$; that is to say, the zero locus of a bunch of polynomials with complex coefficients. We will consider this zero locus as a topological space using the ordinary topology on $\mathbb{C}$. One of the main themes of algebraic geometry in the last century has been learning how to study the topology of $X$ in terms of the algebraic properties of the defining equations.
In this post, I will explain that there are intrinsic limits to this approach; things that cannot be computed algebraically. In particular, I want to explain how from a categorical point of view, we can’t even compute the homology $H_1( \ , \mathbb{Z})$. And, even if you don’t believe in categories, you’ll still have to concede that we can’t compute $\pi_1( \ )$. This is a very pretty example and it should be more widely known.
Absolutely none of the ideas in this post are original; I think most of them are due to Serre. (Thanks to Attila Smith in comments for the reference.)
What does it mean that something cannot be computed algebraically? I could give a general definition, but let me just explain the kind of examples we will be looking at. Let $K$ be a finite Galois extension of $\mathbb{Q}$. Suppose that all of the equations defining $X$ have coefficients in $K$. Let $X^{\sigma}$ be the new variety obtained by replacing all the coefficients of $X$ by their Galois conjugates, for some $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$. Since the Galois action preserves the truth of every algebraic statement; clearly we will not be able to use algebraic methods to distinguish $X$ and $X^{\sigma}$. Thus, if we can come up with cases where $X$ and $X^{\sigma}$ have different topology, then the manner in which they differ will be something that cannot be seen by algebraic methods.
If we were looking at topology of real points, it would be easy to build examples. For example, let $K = \mathbb{Q}[\sqrt{10}]$ and let $X$ be the elliptic curve $y^2 = x^3 - \sqrt{10} x +1$, so $X^{\sigma}$ is the elliptic curve $y^2 = x^3 + \sqrt{10} x +1$. Looking at real solutions, $X$ has two connected components and $X^{\sigma}$ has only one. In the figure below, $X$ is the solid black curve and $X^{\sigma}$ is the dashed blue curve.
Looking at complex solutions, however, both $X$ and $X^{\sigma}$ are two dimensional tori. (More precisely, tori with a single point removed; the “point at infinity”.) So one might think that determining the real geometry of a variety requires some subtle ideas, but that complex geometry is purely algebraic. In fact, this is not so.
Consider two lattices in the complex plane. The first, $M$, will be spanned by $1$ and $\sqrt{-5}$. The second, $N$, Will be spanned by $2$ and $1+\sqrt{-5}$.
Let $E$ be the elliptic curve $\mathbb{C}/M$ and let $F = \mathbb{C}/N$. Notice that both $M$ and $N$ are taken into themselves under multiplication by $\sqrt{-5}$. This multiplication descends to holomorphic endomorphisms of $E$ and $F$, which we will denote $s$ and $t$.
Since $E$ is an elliptic curve, it can be written as $y^2 = x^3 + ax + b$ for some $a$ and $b$. (Plus a point at infinity.) It turns out that the endomorphism $s$ imposes enough rigidity on the situation to see that $a$ and $b$ can be taken to be algebraic numbers; I will sketch an explanation for this below. In fact, we can take $a$ and $b$ to be in $\mathbb{Q}(\sqrt{5})$. Moreover, the map $s$ can be written as $(x,y ) \mapsto (f(x,y), g(x,y))$, where $f$ and $g$ are rational functions whose coefficients are algebraic — explicitly, in $\mathbb{Q}(\sqrt{5}, \sqrt{-5})$.
I wanted to compute $a$, $b$, $f$ and $g$ for you*, but I couldn’t find them in any books or websites and it seemed a bit difficult. I was able to work out that the $j$ invariant of $E$ is
$\displaystyle{ -632000 + 282880 \sqrt{5} =2^8 * \sqrt{5}^3 * (4+\sqrt{5})^3 * (2-\sqrt{5})^4}.$
There is a theory of elliptic curves with unusual endomorphisms — the technical term is curves with complex multiplication. Here is one consequence of this theory: Let $K = \mathbb{Q}(\sqrt{5}, \sqrt{-5})$ and let $\sigma$ be the Galois automorphism where $\sqrt{5} \mapsto - \sqrt{5}$ and $\sqrt{-5} \mapsto \sqrt{-5}$. Then the curve $F$ is given by $y^2 = x^3 + a^{\sigma} x + b^{\sigma}$. Moreover, the automorphism $t$ is given by applying $\sigma$ to the coefficients of $s$.
So, why do I say that “from a categorical point of view, we can’t even compute the homology $H_1*( \ , \mathbb{Z})$.” Of course, $H_1(E, \mathbb{Z})$ is naturally isomorphic to $M$, and unnaturally isomorphic to $\mathbb{Z}^2$; similarly, $H_1(F, \mathbb{Z})$ is also unnaturally isomorphic to $\mathbb{Z}^2$. So you could say that $H_1(E, \mathbb{Z})$ and $H_1(F, \mathbb{Z})$ were the same, if you are willing to engage in unnatural acts.
But homology is a functor! Computing it should means computing what it does to morphisms, not just what it does to objects! Looking at how multiplication by $\sqrt{-5}$ acts on $M$ and $N$. we see that the $\mathbb{Z}[\sqrt{-5}]$ module $M$ is generated by a single element, where as the $\mathbb{Z}[\sqrt{-5}]$ module $N$ is not. So there is an element $\gamma$ of $H_1(E, \mathbb{Z})$ so that $\gamma$ and $s_* \gamma$ span $H_1(E, \mathbb{Z})$, but there is no $\delta$ so that $\delta$ and $t_* \delta$ span $H_1(F, \mathbb{Z})$. In short, the action of an endomorphism on $H_1$ cannot be computed algebraically.
A clever trick, for those who mistrust categories: If you believe in the categorical philosophy, you are done now. But, if you are a classical sort, you might want to see two varieties $X$ and $X^{\sigma}$ which are not homeomorphic.
If we were living in the topological category, we could take the mapping cylinder of $s: E \to E$ and get a map whose topology reflects the map $s$. But in the algebraic category, mapping cylinders don’t exist. In the remainder of this post, I’ll sketch a trick to get around that, due to again to Serre.
Just like abelian curves come from lattices in $\mathbb{C}$, abelian varieties come from lattices in higher dimensional complex vector spaces. Just as $\mathbb{Z}[\sqrt{-5}]$ has nonisomorphic modules, so does $\mathbb{Z}[e^{2 \pi i/n}]$ for certain values of $n$. A more complicated version of the above argument constructs an abelian variety $A$, with endomorphism $z$, such that $z^n = \mathrm{Id}$, and such that there is a Galois automorphism $\sigma$ for which $z^{\sigma} : A^{\sigma} \to A^{\sigma}$ is topologically different from $z: A \to A$. I’ll write $g$ for the dimension of $A$; for the curious, $g = \phi(n)/2$.
Now, let $B$ be simply connected, with a free action of $\mathbb{Z}/n$. (And let this variety and action be given by rational coefficients, so it is unaffected by $\sigma$.) Consider $X := A \times_{\mathbb{Z}/n} B$. This is $(A \times B)/(\mathbb{Z}/n)$, where $(\mathbb{Z}/n)$ acts diagonally. It is extremely plausible, and not hard to prove, that $X^{\sigma} = A^{\sigma} \times_{\mathbb{Z}/n} B$. Now, some basic topology tells us that $\pi_1(X)$ and $\pi_1(X^{\sigma})$ are both semidirect products of $\mathbb{Z}/n$ acting on $\mathbb{Z}^{2g}$. However, in the case of $A$, this action is by the action of $z_*$ on $H_1(A, \mathbb{Z})$, and in the case of $A^{\sigma}$ it is by the other action. These are nonisomorphic groups, so we see that even those who mistrust categories must admit that $\pi_1$ cannot be computed algebraically.
Sketch of why these numbers are algebraic. We can write down the condition that $f$ and $g$ give an automorphism of $y^2 = x^3 + ax +b$, whose square is $[-5]$. This is a whole lot of equations in $a$, $b$ and the coefficients of $f$ and $g$; all the coefficients of these equations are rational. Now, an elliptic curve $\mathbb{C}/\Lambda$ will only have such an automorphism if $\Lambda$ is closed under multiplication by $\sqrt{-5}$. It is a good exercise to show that any such lattice is, up to dilation and rotation, either $M$ or $N$. So these equations will only have two solutions. (Actually, they will have two solutions up to automorphism, see the footnote. But this is a sketch!) Any set of equations with finitely many solutions, and coefficients in a field $K$, has all of its solutions in $K^{\mathrm{alg}}$; this is a corollary of the Nullstellensatz.
*There is a technical point I am glossing over: $a$ and $b$ are not quite well defined. Specifically, the elliptic curve $y^2 = x^3 + ax +b$ is, as a complex variety, isomorphic to $y^2 = x^3 + D^2 ax + D^3 b$. Moreover, even if $a$, $b$ and $D$ are all in $\mathbb{Q}$, the isomorphism will involve coefficients in $\mathbb{Q}[\sqrt{D}]$. This is called twisting, and is important if you want to get all the details right, but I will not deal with it.
## 15 thoughts on “Topology that Algebra can’t see”
1. Attila Smith says:
Dear David,
the relevant article by Serre might be:
Exemple de variétés projectives conjuguées non homéomorphes.
Your explanations are friendly and clear: bravo!
2. Thanks for the reference and the complement; I’ll update the article.
3. Jesse Kass says:
Dear David,
Nice post!
You may have already seen the paper, but James Milne and Junecue Suh have some recent work on this topic too.
http://arxiv.org/abs/0804.1953
4. Wesley Calvert says:
Russell Miller and I treated a rather different formalization of the same intuitive problem (can one compute topological invariants from elementary data) in a paper soon to appear in the proceedings of Unconventional Computation 2009.
We showed that if a manifold is “effectively given,” in the sense that we have algebraically computable functions (in the sense of Blum-Shub-Smale) to describe the transition functions, then
a) We cannot, in most cases, decide if a loop is nullhomotopic, but
b) There is a canonical system of loops which suffices to compute a presentation of pi_1.
I don’t have the preprint available online (yet), but can provide it upon request.
Thanks for the interesting post!
5. If you view your curve as a real algebraic set in $\mathbb{R}^4$ then you certainly can compute its Betti numbers algebraically (well, semi-algebraically). See e.g. http://arxiv.org/abs/0806.3911 for abundant references on this.
6. I’m not sure whether you are disagreeing or just discussing a different, but related, idea. So I’ll explain the difference in perspective between Serre’s result and the one you link to.
To my mind, pure algebra does not include testing inequalities, and Galois theory is designed to study things that can be computed algebraically. So semialgebraic methods are outside this paradigm. For example, in Galois theory we consider the field automorphism which takes $+ \sqrt{2}$ to $- \sqrt{2}$. This automorphism preserves field structure but not inequalities: it takes a positive number to a negative one. Even worse, we are allowed to take a real number like $2^{1/4}$ to an imaginary one like $i 2^{1/4}$.
The amazing thing, then is that there is so much that can be computed purely algebraically — including betti numbers! (You only need to read the first section of the linked reference to get my point, although the whole thing is excellent.) It is data like the action of endomorphisms on cohomology, or like the fundamental group, which requires nonalgebraic techniques.
Finally, although this is not my field, I’ll point out that there are lots of subtle issues in semi-algebraic computation. For example, it is not known whether there is an efficient way to test whether an algebraic number is positive!
7. Danny Calegari says:
Dear Wesley – from the sound of it, you might be interested in the papers of Nabutovsky and Weinberger, which have a similar flavor (if you are not already aware of them . . .) For example, see Weinberger’s nice book:
http://www.ams.org/mathscinet-getitem?mr=2109177
Best,
Danny
8. David,
to me it seems that computations involving real algebraic numbers can still be counted as algebraic, although it’s a matter of taste.
By the way, computing the fundamental group becomes doable, as one would be able, starting from a variety, to construct a finite simplical complex with the same fundamental group. Then you can run something along the lines of http://www.maths.qmul.ac.uk/~leonard/fundamental/
To comment on your claim that testing positivity is not known to be efficient, it seems to me that you misread the part of the Lipton’s post you are referring to. Indeed, he talks about SSP, i.e. testing inequalities of the form $\sum_{m=1}^n \sqrt{a_m}\leq k.$ This way you’re effectively leaving the polynomial “universe”. An algebraic way to encode this would need $n$ variables $x_m,$ and a system involving nonlinear polynomial equations and inequalities in all of them. Needless to say, already the task of testing solvability of a system of $n$-variate polynomial equations is also NP-hard, even if you allow complex solutions.
On the other hand, if you work with the encoding of a real algebaric number by a defining univariate polynomial $f\in\mathbb{Z}[t]$ and a rational interval to locate the particular root (or what is called Thom encoding, i.e. the signs of the derivatives of $f$ at the root) then everything becomes efficient, and these procedures are dating back from 19th century.
(The complexity of SSP is related to the complexity semidefinite programming, and as such is indeed quite important. But this is another story.)
9. David,
I’ve been thinking about a problem, only tangentially related to this post, but maybe you’ll enjoy it.
Suppose I hand you an algorithm. It can be run on your household computer. [Formalize this as you will.] This algorithm spit out, one at a time, digits for a positive real number (say less than 1).
Is there another algorithm (call it the Decider) which will decide whether or not the first algorithm is describing 0?
I believe the answer is no. However, I’m a little weak on how one defines an algorithm, in ZFC set theory say.
Application: Given an explicit function (say an L-function attached to an explicit elliptic curve with integer coefficients), is there an algorithm to determine if there is more than a double pole at the origin?
Hi David,
Is it true that when you are looking at “real” or “complex” points in a scheme defined over rationals this incompatibility creeps in.
Consider the following reformulation :
If K1 and K2 are conjugate schemes defined over some algebraic extension of rationals, and L is some Galois extension containing K1 and K2, then for any locally constant sheaf F the cohomologies over K1 and K2 are isomorphic.
But this looks like a obvious base change…
11. David Speyer says:
For a locally constant sheaf with torsion coefficients, sure, that’s base change. If you are going to consider locally constant sheaves like $\mathbb{Z}$, then you have to worry about the fact that the etale and analytic topologies don’t give the same results for such sheaves.
Indeed, the singular cohomology of $X(\mathbb{C})$ with coefficients in $\mathbb{Z}$ is naturally isomorphic to the cohomology of the locally constant sheaf $\mathbb{Z}$ in the analytic topology. The examples above show that $H_{an}(K_1, \mathbb{Z})$ and $H_{an}(K_2, \mathbb{Z})$ need not be naturally isomorphic, but of course you only said isomorphic. I’d need to think a bit to see if I could find some $F$ for which these groups were not isomorphic at all.
13. If my understanding of your question is correct, take the family $y^2z=x(x-z)(x-\lambda z)$ on $\mathbb{P}^2\times \mathbb{A}^1$. For $\lambda\neq 0,1$, you have a smooth elliptic curve, which is topologically $S^1\times S^1$. For $\lambda=0,1$, you have a nodal cubic, which is a sphere with the north and south poles identified, which is not the same topology. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 180, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9423325657844543, "perplexity": 170.11211818404914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549425193.20/warc/CC-MAIN-20170725122451-20170725142451-00046.warc.gz"} |
http://www.aimath.org/WWN/qptsurface2/articles/html/7a/ | # Colliot-Thelene 2: Rational points on surfaces with a pencil of curves of genus one
We had , , and
separable, all irreducible, even degree, , , up to . , , .
and is the composition with the map
where .
Fact. .
Theorem. [Theorem B] Assume Schinzel's hypothesis, and the finiteness of . Let , . Let . Assume: , and .
Then is infinite, and is Zariski dense.
Here,
Schinzel's hypothesis: Let for be distinct irreducible polynomials with leading coefficient positive (plus technical condition, to exclude polynomials like ); then there exist infinitely many values such that each is a prime.
Remark. The assumption that is satisfied for general . For the assumption that , in general we have so this reduces to .
The proof of this theorem will take up the rest of these notes.
We shall define a finite set of bad places'. For each , we have some , and we look for , with very close to for , and find one such that:
• ;
• , spanned by and .
We have , containing -adic places and those over (note we are being sloppy for real places), and places of bad reduction of , of , and such that . If , then is separable (finite étale cover).
We look at and look at its prime decomposition; it will have some part in and another part of primes of multiplicity , and one prime , the Schinzel prime'. We realize for . Then has bad reduction in .
First we find such that . For , we introduce the algebra , where is the quadratic extension connected to . Then . A priori, . Since is even, .
Let , with projection . We know that . For almost all places of , is trivial.
Fix as before plus places where some is not trivial on . Now . Fix for with this property. Note . Suppose is very close to for and the decomposition of has all primes in split in .
Claim. For such , .
Proof. The only places where it will fail to have a point are those of bad reduction. For , , as is close to . For , since the two rational curves are defined. For , write
the second because splits in ; therefore the prime that is left over forces the prime to split in , we again have points locally.
For the second part, we now need to control the Selmer groups uniformly in the family , for satisfying , namely, is very close to for , decomposes into primes in , primes splitting in , and the Schinzel prime .
Let , and . We have
Here
and the Cartesian square from . Then is the kernel of a symmetric pairing on .
We find two `constant' subgroups . First, we find , fixed because very close to for . For the second pair:
Proposition. For each , there exists a unique such that for any with , , and for each , its component in belongs to .
We define the subgroup
Note .
Proposition.
Proposition. On , the restriction of the pairing is independent of .
To prove this, use various reciprocity laws.
To conclude, write , where , , and is the supplement. Use the assumption that to get rid of ...
Now use finiteness of and Cassels-Tate pairing, implies , so the rank is .
Back to the main index for Rational and integral points on higher dimensional varieties. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9908494353294373, "perplexity": 875.3284596219305}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738661296.34/warc/CC-MAIN-20160924173741-00272-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://web.sas.upenn.edu/maam2/seminar-abstracts/ | # Seminar Abstracts
• October 23rd, 3pm EDT
Speaker: Theresa Anderson, Purdue University
Title: Dyadic analysis meets number theory
Abstract: We explore one of many ways that analysis and number theory interact by showing what goes on in the background to construct a measure that is $p$-adic doubling for any finite set of primes $p$ yet not doubling. This is recent work joint with Bingyang Hu. I will follow this by a short talk on the adjacent topic of “Complete classification of adjacent dyadic systems”. There will be ample question time and questions pertaining to professional development and careers in the mathematical sciences are especially welcome.
• October 30th, 3pm EDT
Speaker: Kornélia Héra, University of Chicago
Title: Hausdorff dimension of Furstenberg-type sets
Abstract: We say that a planar set F is a (t,s)-Furstenberg set, if there exists an s-dimensional family of lines in the plane such that each line of this family intersects F in an at least t-dimensional set. We present Hausdorff dimension estimates for (t,s)-Furstenberg sets and for more general Furstenberg type sets in higher dimensions.
The talk is based on joint work with Tamás Keleti and András Máthé, and with Pablo Shmerkin and Alexia Yavicoli.
• November 6th, 3pm EST (Note: Daylight Savings Time ends November 1st)
Speaker: Kyle Hambrook, San Jose State University
Title: Explicit Salem Sets of Arbitrary Dimension in Euclidean Space
Abstract: A set in $R^n$ is called Salem if it supports a probability measure whose Fourier transform decays as fast as the Hausdorff dimension of the set will allow. We construct the first explicit (i.e., non-random) examples of Salem sets in $R^n$ of arbitrary prescribed Hausdorff dimension. This completely resolves a problem proposed by Kahane more than 60 years ago. The construction is based on a form of Diophantine approximation in number fields. This is joint work with Robert Fraser.
• November 13th, 3pm EST
Speaker: Polona Durcik, Chapman University
Title: A triangular Hilbert transform with curvature
Abstract: The triangular Hilbert transform is a two-dimensional bilinear singular integral originating in time-frequency analysis. No Lebesgue space bounds are currently known for this operator. In this talk, we discuss recent joint work with Michael Christ and Joris Roos on a variant of the triangular Hilbert transform involving curvature. As an application, we also discuss a quantitative nonlinear Roth type theorem on patterns in the Euclidean plane.
• November 20th, 3pm EST
Speaker: Alex Barron, University of Illinois Urbana-Champaign
Title: A sharp global Strichartz estimate for the Schrodinger equation on the cylinder
Abstract: The classical Strichartz estimates show that a solution to the linear Schrodinger equation on Euclidean space is in certain Lebesgue spaces globally in time provided the initial data is in L^2. On compact manifolds one can no longer have global control, and some loss of derivatives is necessary in interesting cases (meaning the initial data needs to be in a Sobolev space rather than L^2). On non-compact manifolds it is a challenging problem to understand when one can have good space-time estimates with no loss of derivatives.
In this talk we discuss a global-in-time Strichartz-type estimate for the linear Schrodinger equation on the infinite cylinder. Our estimate is sharp, scale-invariant, and requires only L^2 data. Joint work with M. Christ and B. Pausader.
• December 4th, 3:30pm EST
Speaker: Bruno Poggi, University of Minnesota
Title: Theory of $A_{\infty}$ weights for elliptic measures and generalized Carleson perturbations for elliptic operators
Abstract: We present Carleson perturbations for elliptic operators on domains for which there exists a robust elliptic PDE theory. Such domains include, in particular, (a) 1-sided NTA domains satisfying the capacity density condition (thus we extend some recent results of Akman-Hofmann-Martell-Toro), (b) domains with low-dimensional Ahlfors-David regular boundaries, and (c) certain domains with boundaries with pieces of distinct dimensions. Our Carleson perturbations are generalized in the sense that, in addition to the classical additive perturbations, we allow for scalar-multiplicative perturbations, which admit non-trivial differences on the boundary between the perturbed matrix and the original matrix. Finally, we investigate corollaries of our techniques, with implications to free boundary problems an a characterization of $A_{\infty}$ among elliptic measures. This is joint work with Joseph Feneuil.
• December 4th, 4pm EST
Speaker: Kan Jiang, Ningbo University
Title: Some thickness theorems and their applications
Abstract: In this talk, I will introduce some old and new thickness theorems. In terms of these results, we are able to give many interesting applications.
• December 11th, 3pm EST
Speaker: Tyler Bongers, Harvard University
Title: Energy techniques for nonlinear projections and Favard curve lengths
Abstract: There are many classical results relating the structure, dimension, and measure of a set to the structure of its orthogonal projections, including theorems of Marstrand and Besicovitch. It turns out that many nonlinear projection-type operators also have special geometry that allow us to build similar relationships between a set and its “projections,” just as in the linear setting. In this work, we will show how energy techniques of Mattila can be strengthened and generalized to projection-type operators satisfying a transversality condition, and apply these results to study visibility and Favard curve lengths of sets. This work is joint with Krystal Taylor. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 4, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8008618354797363, "perplexity": 1288.949450313815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585450.39/warc/CC-MAIN-20211022021705-20211022051705-00295.warc.gz"} |
https://amsi.org.au/ESA_Senior_Years/SeniorTopic4/4f/4f_3answers.html | ##### Exercise 1
Let $$T$$ be the waiting time (in days) until the first call. Then $$T \stackrel{\mathrm{d}}{=} \exp(1.8)$$, and therefore $$F_T(t) = 1 - \exp(-1.8t)$$, for $$t > 0$$. We need to be careful about the units of time used here.
1. $$15$$ minutes equals $$\dfrac{15}{24 \times 60}$$ days, or 0.01042 days. Hence, the chance of a call in the first 15 minutes equals $$F_T(0.01042) = 1 - \exp(-1.8 \times 0.01042) = 0.0186$$.
2. Due to the lack of memory property, the probability is the same as that in part a, namely $$0.0186$$.
3. $$10$$ hours equals $$\dfrac{10}{24}$$ days, or 0.41667 days. So the probability of no calls during a shift is $$\Pr(T > 0.41667) = \exp(-1.8 \times 0.41667) = 0.4724$$.
4. Assuming independence between days, the probability of no calls in four successive days equals $$0.4724^4 = 0.0498$$.
5. Solving for the time $$y$$ in days: \begin{alignat*}{2} && \Pr(T \leq y) &= 0.1 \\\\ &\implies\quad& F_T(y) &= 0.1 \\\\ &\implies& 1 - \exp(-1.8y) &= 0.1 \\\\ &\implies& -1.8y &= \ln(0.9) \\\\ &\implies& y &= 0.0585 \text{ days}. \end{alignat*} Hence, $$x$$ is 1.4 hours, which is 1 hour 24 minutes.
##### Exercise 2
1. $$f_X(\mu) = \dfrac{1}{\sigma \sqrt{2 \pi}} \approx \dfrac{0.40}{\sigma}$$.
2. We have $$f_X(x) = f_X(\mu)\,\exp(k(x))$$, where $$k(x) \leq 0$$ for all values of $$x$$. So $$\exp(k(x)) \leq 1$$, and the result follows.
3. To find points of inflexion, we need the second derivative of $$f_X(x)$$. Using the chain rule, we have \begin{align*} f_X'(x) &= \dfrac{d}{dx}\Bigl[\dfrac{1}{\sigma \sqrt{2\pi}}\,\exp\bigl(\dfrac{-(x-\mu)^2}{2\sigma^2}\Bigr)\Bigr] \\\\ &= \dfrac{1}{\sigma \sqrt{2\pi}}\Bigl(\dfrac{-(x-\mu)}{\sigma^2}\Bigr)\;\exp\Bigl(\dfrac{-(x-\mu)^2}{2\sigma^2}\Bigr). \end{align*} Now, using the product rule, we have $f_X''(x) = \dfrac{1}{\sigma \sqrt{2\pi}}\Bigl[\dfrac{(x - \mu)^2}{\sigma^4}-\dfrac{1}{\sigma^2}\Bigr]\;\exp\Bigl(\dfrac{-(x-\mu)^2}{2\sigma^2}\Bigr).$ At a point of inflexion, $$f_X''(x) = 0$$. This gives \begin{alignat*}{2} && f_X''(x) &= 0 \\\\ &\implies\quad& \dfrac{(x-\mu)^2}{\sigma^4} - \dfrac{1}{\sigma^2} &= 0 \\\\ &\implies& (x - \mu)^2 &= \sigma^2 \\\\ &\implies& x &= \mu \pm \sigma. \end{alignat*} Hence, the points of inflexion are at $$x = \mu - \sigma$$ and $$x = \mu + \sigma$$; these are the points either side of $$\mu$$ at which the curve changes from convex to concave.
4. $$f_X(\mu + k\sigma) = \dfrac{1}{\sigma \sqrt{2\pi}} \exp(-\dfrac{1}{2}k^2)$$.
$$k$$ $$\exp(-\dfrac{1}{2}k^2)$$ 1 1 2 3 4 5 1 0.607 0.135 0.011 0.0003 4e-06
Note that $$f_X(\mu - k\sigma) = f_X(\mu + k\sigma)$$, by symmetry.
There are a couple of useful interpretations:
• Firstly, when sketching a Normal pdf, the height of the curve at $$\mu \pm \sigma$$ is 61% of the height of the central peak, and so on.
• Second, recall that the height of a pdf reflects relative probabilities, so that if $$f_X(b) = 2 f_X(a)$$, then the chance of an observation near $$b$$ is approximately twice as likely as an observation near $$a$$. This means, for example, that observations near $$\mu$$ are approximately $$250\ 000$$ times more likely that observations near $$\mu + 5 \sigma$$, since $$\dfrac{1}{0.000004} = 250\ 0000$$.
##### Exercise 3
Let $$D$$ be the difference between the forecast maximum temperature and the actual maximum temperature (in degrees Celsius). Then $$D \stackrel{\mathrm{d}}{=} \mathrm{N}(0,1.2^2)$$.
1. $$\Pr(-1.0 < D < 1.0) = 0.595$$.
2. $$\Pr(D < -0.5) = 0.338$$ and $$\Pr(-0.5 < D < 0.5) = 0.323$$, so it is very slightly more probable that there is an underestimate of 0.5 degrees or more.
3. We want to find the 0.01 quantile of the distribution; that is, we want $$c_{0.01}$$ satisfying $$F_D(c_{0.01}) = 0.01$$. We find that $$c_{0.01} = -2.79$$ degrees. So 1% of forecast maximums are 2.79 degrees or more lower than the actual maximum. By symmetry, 1% of forecast maximums are 2.79 degrees or more higher than the actual maximum.
##### Exercise 4
1. The pdfs of the random variables $$A$$ and $$L$$ are shown on the same axes in figure 6. The green distribution, on the left, is for the dose of anaesthetic required to render the animal unconscious. The average dose is 120 mg, and most values are in the range from about 60 mg to 180 mg. The red distribution, on the right, is for the lethal dose. The mean is 400 mg — much higher than the mean of the green distribution. There is little overlap of the two distributions. (Which is how we want things to be!) In fact, you might think that they do not overlap at all, based on a visual assessment of figure 6.
Figure 6: The pdfs of anaesthetic and lethal doses.
2. This question is about the anaesthetic dose administered, so we need to consider the distribution that renders animals suitably unconscious (the green distribution in figure 6). We need to find the value $$d^*$$ that corresponds to 99.9% of the animals being rendered suitably unconscious; this means a cumulative probability of 0.999. This is shown in figure 7.
Figure 7: The pdf of anaesthetic dose, showing $$d^*$$.
We want to find $$d^*$$ such that $$\Pr(A \leq d^*) = 0.999$$; this is the 0.999 quantile of the distribution. This can be achieved in Excel using the function $$\sf \text{NORM.INV}$$. If you enter $$\sf \text{=NORM.INV(0.999, 120, 20)}$$ in a cell, you should find that the dose required to render 99.9% of animals unconscious is $$d^* = 181.80$$ mg.
3. We now consider the distribution of the lethal dose, and what happens if a dose of 181.80 mg is administered. If a dose of 181.80 mg is used, there will be a small proportion of animals for whom this dose is lethal: those for whom the lethal dose is less than or equal to 181.80 mg. We are considering the pdf of $$L$$ (the red distribution in figure 6), and need to find the area under the curve corresponding to a dose of 181.80 or less. This is shown on the left in the following figure.
Figure 8: The pdf of lethal dose, showing $$d^*$$.
The tail area is extremely small and hard to see, so the graph on the right is zoomed in to show the detail of the pdf of $$L$$ near $$d^*$$.
To find the left-tail area in this distribution, we can use the cdf function in Excel; we enter $$\sf \text{=NORM.DIST(181.80, 400, 50, 1)}$$ in a cell. The value returned, and hence the proportion dying, is 0.0000064, or 0.00064%. This corresponds to 6 in a million, which is very small, as we suspect from the diagrams: a good result. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9947602152824402, "perplexity": 454.899292889905}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335124.77/warc/CC-MAIN-20220928051515-20220928081515-00116.warc.gz"} |
https://era.library.ualberta.ca/items/e90b84ab-ed33-40d8-92b3-7b0ed0f94fcb | Usage
• 2 views
Intrinsic Disorder Effects and Persistent Current Studies of YBCO Thin Films and Superconducting Tunnel Junctions
• Author / Creator
• This thesis studies the intrinsic disorder effects and the transport and magnetic properties of ring-shaped epitaxial thin films and superconducting tunnel junctions (STJs) of the high temperature superconductor YBa$_2$Cu$_3$O$_{7-\delta}$. We used an unconventional contactless technique that allows us to directly measure the persistent current of superconducting rings. In order to study the disorder effects on the persistent current, we slowly increased oxygen vacancies in YBa$_2$Cu$_3$O$_{7-\delta}$ by changing $\delta$ from 0.03 to 0.55 in steps of $\sim$0.021. Monitoring the corresponding changes in the temperature dependence of the persistent current revealed an anomaly in its flow within a certain range of disorder. We found that this anomaly is directly related to the occurrence of a spinodal decomposition of oxygen vacancies in YBCO, which we explain as a competition between two coexisting phases, oxygen rich and oxygen deficient. The analysis of the time dependence of the persistent current revealed that increasing oxygen vacancies transforms the vortex structure from quasi-lattice into a glass and subsequently into a pinned liquid phase. Our results also exhibited the first evidence of self-organization of the vortex structure with increasing disorder. We also performed the first direct measurement of the temperature dependence of the $c$-axis persistent current ($J_c$) that is purely due to tunnelling Cooper-pairs through intrinsic Josephson junctions (IJJs) of YBCO. This is made possible by incorporating IJJs of YBCO into ring-shaped films. Then, we studied the temperature dependence of the persistent current of YBCO nanowires embedded in SrTiO$_3$-barrier integrated between two semi-ring-shaped YBCO thin films and systematically varied the nanowires length. Our observations revealed that $J_c$ has two different temperature dependences: a GL-dependence ($J_c \propto (T_c - T)^{3/2}$) at low temperatures which we found the same in all studied samples, and another power law dependence ($J_c \propto (T_c - T)^{\alpha > 3/2}$) at high temperatures which turned out to depend on the length of the nanowires. We attribute the cross-over between these two temperature dependences to the depinning and the dissipative motion of vortices. These experimental approaches and findings not only provide new information, but more importantly open new avenues of investigating the transport and magnetic properties of superconducting films, junctions, and nanowires.
• Subjects / Keywords
2009-11
• Type of Item
Thesis
• Degree
Doctor of Philosophy
• DOI
https://doi.org/10.7939/R3N01048S
This thesis is made available by the University of Alberta Libraries with permission of the copyright owner solely for non-commercial purposes. This thesis, or any portion thereof, may not otherwise be copied or reproduced without the written consent of the copyright owner, except to the extent permitted by Canadian copyright law.
• Language
English
• Institution
University of Alberta
• Degree level
Doctoral
• Department
• Department of Physics
• Supervisor / co-supervisor and their department(s)
• Chow, Kim H. (Physics)
• Examining committee members and their departments
• Etsell, Thomas H. (Chemical and Materials Engineering)
• Heimpel, Moritz (Physics)
• Fenrich, Frances (Physics)
• Jung, Jan A. (Physics)
• Christen, David K. (Superconductive and Energy Efficient Materials, Materials Science and Technology Division,Oak Ridge National Laboratory, Oak Ridge, TN 37831 USA) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8045891523361206, "perplexity": 2965.1277356346204}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946011.29/warc/CC-MAIN-20180423125457-20180423145457-00229.warc.gz"} |
http://interglobalmedianetwork.com/tags/spectacle4/ | ## Spectacle V4
This past Saturday I presented at React Camp, which was a part of Open Camps NYC. I used a presentation deck called Spectacle, created by Formidable Labs. I chose to use the Spectacle boilerplate because of its flexibility. What was so cool about Spectacle is that it is written with React. The presentation went really well. The slide deck looked great. I (and others) had been concerned that my font would not be large enough, but there was totally no problem there. [Read More] | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8953669667243958, "perplexity": 4408.008238908609}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376828501.85/warc/CC-MAIN-20181217091227-20181217113227-00637.warc.gz"} |
https://dsp.stackexchange.com/questions/17734/estimate-the-discrete-fourier-transform-series-of-a-signal-with-missing-sample | # Estimate the Discrete Fourier Transform / Series of a Signal with Missing Samples
Assuming we have a discrete signal $${ \left\{ x \left[ n \right] \right\}}_{n = 1}^{N}$$.
Which has a Discrete Fourier Transform / Series.
Now, assume I'd like to estimate its Discrete Fourier Series coefficient yet some samples of $$x \left[ n \right]$$ are missing (The indices are known).
How could that be done efficiently without computing the Pseudo Inverse of the adapting Fourier Series matrix?
• How many samples are missing? Aug 19 '14 at 19:06
• Let's say $K < N$. Something like $N = 3500$ and $K = 500$.
– Royi
Aug 19 '14 at 19:28
• – Royi
Jan 16 at 12:20
Given $$\left\{ x \left[ n \right] \right\}_{n \in M}$$ where $$M$$ is the set of indices given for the samples of $$x \left[ n \right]$$.
The trivial solution (Which it would be great to have a faster more efficient solution is what I'm looking for) would be:
$$\arg \min_{y} \frac{1}{2} \left\| \hat{F}^{T} y - x \right\|_{2}^{2}$$
Where $$\hat{F}$$ is formed by subset of columns of the DFT Matrix $$F$$ matching the given indices of the samples, $$x$$ is the vector of the given samples and $$y$$ is the vector of the estimated DFT of the full data of $$x \left[ n \right]$$.
The solution is then given by the Pseudo Inverse (Least Squares Solution):
$$y = { ( \hat{F} \hat{F}^{T} ) }^{-1} \hat{F} x$$
In practice, the matrix will be very poorly conditioned hence solution must be generated using the LS Solution using the SVD.
A sample code is shared on GitHub Repository.
Result of the code: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9599288702011108, "perplexity": 381.4557360895398}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00349.warc.gz"} |