Datasets:

keirp
/

open-web-math-hq-dev

Dataset card Data Studio Files Files and versions Community

Split (1)

train · 1.36M rows

url stringlengths 15 1.13k	text stringlengths 100 1.04M	metadata stringlengths 1.06k 1.1k
https://math.stackexchange.com/questions/795841/fr%C3%A9chet-derivative-and-local-maximum	Fréchet derivative and local maximum I'm pretty confused with the idea of local maximum in function spaces. Normally having a null Fréchet derivative is a necessary but not sufficient condition for being a local maximum. Computing the derivative let $f:\mathbb{R} \mapsto \mathbb{R}$ be a continuous function. And lets denote the space of such functions $C_{\mathbb{R,R}}$. $$F: C_{\mathbb{R,R}} \mapsto C_{\mathbb{R,R}}$$ $$F: f \mapsto \sin(f)$$ let us compute its derivative at point f: $$D_F(f)h = \lim_{t\to 0} {F(f+th) - F(f) \over t}$$ where: 1. $f \in C_{\mathbb{R,R}}$ 2. $g \in C_{\mathbb{R,R}}$ 3. $t \in \mathbb{R}$ then: $$D_F(f)h = \lim_{t\to 0} {\sin(f+th) - \sin(f) \over t}$$ $$D_F(f)h = \lim_{t\to 0} {\sin(f)\cos(th)+\cos(f)\sin(th) - \sin(f) \over t}$$ $$D_F(f)h = \lim_{t\to 0} {-h\sin(f) (1 - \cos(th))+h\cos(f)\sin(th) \over th}$$ So using: $$\lim_{x\to 0} {\sin(x)\over x} = 1$$ $$\lim_{x\to 0} {1-\cos(x)\over x} = 0$$ it reduces to: $$D_F(f)h = h\cos(f)$$ Local maximum we obviously have: $$0 \leq \|\|F(f)\|\|_\infty \leq 1$$ Hence: if $\|\|F(f)\|\|_\infty = 1$, then f is a local maximum. So any function $f$ such that $f(x) \equiv \pi \pmod \pi$ has a solution is a local maximum. Null Fréchet derivative $$D_F(f) = \cos(f) = 0 \Rightarrow \exists k \in \mathbb{N}, \forall x \in \mathbb{R}, f(x) = k\pi$$ such constant functions are indeed local maximums, but not the only ones. So instead of getting a superset containing all my local maximums, I get a strict subset of it from the nullity of my Fréchet derivative. Question As I'm pretty sure the mathematics I'm taught are right and I'm wrong... Where am I wrong ? • What is the definition of the maximum of a mapping from $C(\mathbb R)$ to $C(\mathbb R)$? You seem to maximize $\\|F(f)\\|_\infty$ - the $\infty$-norm is not that differentiable. And $\cos(f)=0$ implies $\forall x\in \mathbb R$ $\exists k\in N$ such that $f(x)=k\pi$. Now find all such continuous functions... – daw May 15 '14 at 13:10 • @daw "the ∞-norm is not that differentiable", you mean there is a condition about the norm of my banach space for the fréchêt derivative to exists ? Or for a maximum do be defined ? – user2346536 May 15 '14 at 13:22 • @daw "Now find all such continuous functions" $\forall x f(x) = kπ$ seems a good shot to me. Constant functions basically. – user2346536 May 15 '14 at 13:32 • You seem to maximize $\phi(f):=\\|F(f)\\|_\infty$, but you only compute the derivative of $F$, but you do not check differentiability of $\phi$. – daw May 15 '14 at 13:38 • @daw You mean that, finding a maximum of a function $F:C_{\mathbb{R,R}} \mapsto C_{\mathbb{R,R}}$ does not really makes sens as $C_{\mathbb{R,R}}$ is not ordered. And so when I choosed my norm $\|\|\|\|_\infty$, I in fact was trying to maximize $\|\|F(f)\|\|_\infty$ ? Which is $\phi:C_{\mathbb{R,R}} \mapsto \mathbb{R}$ where R is order and $\phi(f) \leq \phi(f)$ does make sens. – user2346536 May 15 '14 at 13:44 The problem is to maximize $$\\|F(f) \\|_\infty.$$ However, the maximum is not differentiable. In order to perform the analysis, the function $f\mapsto\\|F(f) \\|_\infty$ needs to be differentiable	{"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": 0.9330170154571533, "perplexity": 351.84352810230456}, "config": {"markdown_headings": false, "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-2019-39/segments/1568514575168.82/warc/CC-MAIN-20190922053242-20190922075242-00515.warc.gz"}
http://www.zazzle.co.uk/corelli+clothing	Showing All Results 148 results Page 1 of 3 Related Searches: stratford upon avon, marie, card Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo Got it! We won't show you this product again! Undo No matches for Showing All Results 148 results Page 1 of 3	{"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": 0.8284785747528076, "perplexity": 4545.590794873969}, "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/1386163045148/warc/CC-MAIN-20131204131725-00010-ip-10-33-133-15.ec2.internal.warc.gz"}
https://brilliant.org/problems/a-calculus-problem-by-samara-simha-reddy/	Exponential and Powers! (10) Calculus Level 3 $\Large \displaystyle \int_0^{\infty} 3^{-4z^2} \, dz = \, ?$ • Use the approximation $$\pi = \dfrac{22}{7}$$ in your computation of the final value.	{"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": 0.9274441003799438, "perplexity": 2119.1076822715013}, "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-2018-05/segments/1516084891706.88/warc/CC-MAIN-20180123032443-20180123052443-00584.warc.gz"}
https://www.lessonplanet.com/teachers/linear-function-12th-higher-ed	# Linear Function In this linear functions worksheet, students solve and complete 4 different sections to a given problem. First, they determine if the velocity of a car increases at a given speed and explain. Then, students determine the time that this happens and justify their reasoning.	{"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": 0.9678256511688232, "perplexity": 889.0520228554124}, "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-2018-09/segments/1518891816462.95/warc/CC-MAIN-20180225130337-20180225150337-00523.warc.gz"}
https://ufapro888.com/lemon-tree-ezbp/viewtopic.php?id=radius-of-convergence-infinity-4b617c	2/12/2020 11:03 PM (3x-6)^n I know the ratio test says to take the limit as n goes to inf of A_n/A_n+1. The radius of convergence is half the length of the interval of convergence. The ratio test is the best test to determine the convergence, that instructs to find the limit. So our radius of convergence is half of that. (Use Inf For Too And -inf For -0. It is found by adding the absolute values of both endpoints together and dividing by two. Thanks. & {\text{The}}\,\,{\text{radius}}\,\,{\text{of}}\,{\text{convergence}}\,\,\,{\text{for}}\,\,{\text{the}}\,\,{\text{power}}\,{\text{series}}\,\,{\text{is}}\,\,{\text{:}} \cr In the positive case, the power series converges absolutely. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end points of the interval), its length is 2, so the radius of convergence equals 1. So we could say that our radius of convergence is equal to 1. As long as x stays within some value of 0, this thing is going to converge. Examples : 1. The radius of convergence can be zero, which will result in an interval of convergence with a single point, a(the interval of convergence is never empty). radius of convergence of summation from 1 to infinity of 1/(k3^k) (x-5)^k Hence the radius of convergence is infinity, and the interval of convergence is - ∞ \infty ∞ < x < ∞ \infty ∞ (because it converges everywhere). It looks purposely contrived to be solved for x (bring 6 over to one side and divide by 3), but is that just irrelevant information? there is a nite radius of convergence R. Note that the interval of convergence can be open or closed or half-open/half-closed depending on the convergence of the series at the endpoints. This leads to a new concept when dealing with power series: the interval of convergence. No idea! Note that r ≥ 0, because for ˜r = 0 the series +∞ ∑ n=0an˜rn = +∞ ∑ n=0an0n = 1 converges (recall that 00 = 1). So if ak over ak+1 absolute value goes infinity as k goes to infinity, then the radius of convergence r of the power series is infinity, in other words it converges for all z in the complex plane. If you take math in your first year of college, they teach you about Or, for power series which is convergent for all x-values, the radius of convergence is +∞. Im confused. Here is a massive hint: Do you remember that Click on the problem to see the answer, or click here to continue. Remember, for a convergent series, the n-th term goes to 0. This function has a branch point at z = 1, which is one of the possibilities described at Mathematical singularity#Complex analysis. R=27/4 Help would be veeeery much appreciated! 0. reply. Find the Radius and Interval of Convergence for Sum n==1 to infinity of (-2)^n (x+1)^n Question: Find The Radius Of Convergence And Interval Of Convergence For The Given Power Series (note You Must Also Check The Endpoints). 1, then -1, then 3, then -5, then 11 ... we flip-flop back and forth And this is how far-- up to what value, but not including this value. What is the radius of convergence? If The Radius Of Convergence Is Infinity Then Do Not Include Either Endpoint ). So we could ask ourselves a question. Find the radius of convergence and the interval of convergence for each of the series listed below: a.) Close. 1. It will be non negative real number or infinity. Unlike geometric series and p-series, a power series often converges or diverges based on its x value. X=-1. © copyright 2003-2020 Study.com. Find the limit of (2ne (( ln(n^2) + ipin )/(( 16(n^2) + 5i ))^0.5))/((4n 2 + 3in) (1/2)) [From n=1 to infinity] 2. \cr infinite series and the radius of convergence. Question: Radius Of Convergence Of Summation From 1 To Infinity Of 1/(k3^k) (x-5)^k This problem has been solved! \cr Given a real power series +∞ ∑ n=0an(x −x0)n, the radius of convergence is the quantity r = sup{˜r ∈ R: +∞ ∑ n=0an˜rn converges}. Now, let’s get the interval of convergence. Here are some examples. & {\text{Given the power series}}\,:\,\,\,\,\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^n}}}{{n + 1}}} . 2. but we see for e^x, i get \|e\| when using ratio test, which implies that it diverge? Given a real power series #sum_{n=0}^{+infty}a_n(x-x_0)^n#, the radius of convergence is the quantity #r = "sup" \{tilde{r} \in \mathbb{R} : sum_{n=0}^{+infty}a_n tilde{r}^n " converges"\}#. Hello all. answer. Although this fact has useful implications, it’s actually pretty much a no-brainer. now available at Answer to: Find the radius and interval of convergence of the series: Summation_{n=0}^{infinity} (-1)^n x^n/n+1. Answer to: Find the radius and interval of convergence of the series: Summation_{n=0}^{infinity} (-1)^n x^n/n+1. Integral, from 0 to 0.1, of xarctan(3x)dx Please try to show every step so that I can learn. Radius of convergence (3x)^2 from 0 to infinity. When the radius of convergence is infinity, then the interval of convergence is {eq}\left( { - \infty ,\infty } \right) {/eq}. The convergence of the infinite series at X=-1 is spoiled because of a problem far away at X=1, which happens to be at the same distance from zero! [sum z^n/n^2 for n=1 to infinity] defines a function called the dilogarithm. which clearly becomes infinite. Solution for Find the radius of convergence and interval of convergence for the power series from n=0 to infinity of 5^nX^n/n The radius of convergence for this function & \Rightarrow \Im = \mathop {\lim }\limits_{n \to \infty } \left\| {\frac{{ - \left( {1 + \frac{2}{n}} \right)}}{{\left( {1 + \frac{1}{n}} \right)}}} \right\| \cr I think I am supposed to find the convergent point and work some magic, but every attempt has me going way off course, so I need a step by step to see where I am going wrong. Radius of convergence (3x)^2 from 0 to infinity. It is customary to call half the length of the interval of convergence the radius of convergence of the power series. Radius of Convergence of a power series is the radius of the largest disk in which the series converges. Determine the radius of convergence of the power series? What is the radius of convergence of the series #sum_(n=0)^oo(n(x+2)^n)/3^(n+1)#? thanks. This seems very simple but you need to be careful of the notation and wording your textbooks. & \left\| x \right\| < \Im \Rightarrow - \Im < x < \Im . For X smaller than one and bigger than minus one, the Sciences, Culinary Arts and Personal {/eq}, {eq}\displaystyle \eqalign{ My answers: 1. If we differentiate this series term by term we get the new series and compute its radius of convergence with the ratio test: The result looks very similar. Compute the radius of convergence of the power series: (sum from n=1 to infinity) of a n z n, where a n = (2n + 1)! n = 0 to infinity ((x-3)^(2n)) / ((n+2)^)(8n)) If a power series converges on some interval centered at the center of convergence, then the distance from the center of convergence to either endpoint of that interval is known as the radius of convergence which we more precisely define below. What is the radius of convergence of the series: sum over n from 1 to infinity of (n^-1)(z^n), and how do you get it? The function f(x) = \frac{6}{5+x} may be... Find the radius of convergence of the power... PLACE Mathematics: Practice & Study Guide, Praxis Social Studies - Content Knowledge (5081): Study Guide & Practice, TExES Mathematics 7-12 (235): Practice & Study Guide, FTCE Marketing 6-12 (057): Test Practice & Study Guide, FTCE Biology Grades 6-12 (002): Practice & Study Guide, Praxis School Psychologist (5402): Practice & Study Guide, GACE Mathematics (522): Practice & Study Guide, Ohio Assessments for Educators - Mathematics (027): Practice & Study Guide, GACE Marketing Education (546): Practice & Study Guide, WEST Business & Marketing Education (038): Practice & Study Guide, CSET Science Subtest II Physics (220): Test Prep & Study Guide, MTTC English (002): Practice & Study Guide, NMTA Reading (013): Practice & Study Guide, MTTC Speech & Language Impairment (057): Practice & Study Guide, Praxis ParaPro Assessment: Practice & Study Guide, GACE Special Education Adapted Curriculum Test I (083): Practice & Study Guide, Biological and Biomedical Last edited: Mar 22, 2010. Suppose all the Ki are one, but 5 5. So, the radius of convergence is 3. Solution for Find the radius of convergence and interval of convergence for the power series from n=0 to infinity of 5^nX^n/n Compute the radius of convergence of the power series: (sum from n=1 to infinity) of a n z n, where a n = (2n + 1)! All other trademarks and copyrights are the property of their respective owners. The radius of convergence for the power series {eq}\displaystyle f(x) = \sum\limits_{n = 0}^\infty {{C_n}{x^n} = {C_0} + {C_1}x} + {C_2}{x^2} + ... + {C_n}{x^n} + ..., Answer and Explanation: 1 Given: The interval of convergence is never empty. The ratio test is the best test to determine the convergence, that instructs to find the limit. Here is a video clip that explains how to show that a series converges for all x. \cr n = 0 to infinity ((x-3)^(2n)) / ((n+2)^)(8n)) 2. So this is the series z … . The limit does not exist. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end points of the interval), its length is 2, so the radius of convergence equals 1. So, let's look at some examples. The radius of convergence is actually infinity so the series will always converge for any value of x. Question: Radius Of Convergence Of Summation From 1 To Infinity Of 1/(k3^k) (x-5)^k This problem has been solved! [email protected] The interval of convergence for a power series is the set of x values for which that series converges. (n + 2i) n /(3n)! {/eq} is defined the formula: {eq}\displaystyle \Im = \mathop {\lim }\limits_{n \to \infty } \left\| {\frac{{{C_n}}}{{\,{C_{n + 1}}}}} \right\|,\,\,\,\,{\text{where}}\,\,\,\Im \geqslant 0. So our radius of the series listed below: a. including this.! Infinity of xn, which is one of the interval of convergence of power series: ( USA Europe. Will converge non negative real number or infinity set of x values which! 8, 2020. available on hyper typer the radii of convergence using root... Example # 3 at radius of convergence of a power series, example # 3 at radius of convergence the... ( Sometimes we say it diverges ) so the series converges the limit ) dx 3 to 0 Sometimes say! All the always give a sensible answer Get access to this video and our entire Q & a library ). N+1 ): 2 of this question is customary to call half the length of radius of convergence infinity power.... * ( x^k ) b. ) ^n I know the ratio test 1. Take math in your first year of college, they teach you about infinite and... Equivalent to 5 * x actually pretty much a no-brainer.... the! Notice that we now have the radius of convergence of the interval convergence! For this function has a very similar example, example # 3 at radius of convergence radius... Massive hint: do you remember that Click on the 'circle of using! An if Sn gets weird in this case, the n-th term goes to of! 'Circle of convergence # convergence # convergence on the problem to see the,... That we now have the radius of … we ’ ll deal with \... Is 1 ( that is, the series converges for all x-values, the sum can done... Function called the dilogarithm very similar example, let ’ s Get the interval of convergence 3x! 2020. available on hyper typer hyper typer converges because Sn = log ( n+1 ): 2 confuses me that! Because Sn = 1¡ 1 n+1 remember that Click on the \ ( R = 3. Be determined by the ratio test, which implies that it diverge now available Oxford... = 4\ ) find the radius of convergence is usually the distance between the endpoints the. P1 n=1 1 n ( n+1 ) converges because Sn = 1¡ 1 n+1 length the. Video clip that explains how to show that a series converges for \|x\| < 1 ) how show. S say you had the interval of convergence stays the same when we integrate or differentiate power... ( k = 0 ) ^oo ( 3x ) ^2 from 0 to 0.2, of the series.... This test predicts the convergence, that instructs to find the radii of convergence and the interval radius! Interval notation ): 2 entire Q & a library so as as! So this is how far -- up to What value, radius of convergence infinity this thing going... Going to converge minus one, the n-th term goes to inf of A_n/A_n+1 respective owners our. 5X is 1/3 -- up to What value, but not including this value for value... The function blows up or gets weird converges because Sn = 1¡ 1 n+1 skip the multiplication sign so! Keyboard shortcuts the result is zero infinite number of numbers does n't always give a sensible answer say our... Of numbers does n't always give a sensible answer the radius of convergence # powerseries # radius #.. ( n+1 ): radius of convergence ( 3x ) ^k is 1/3 e^x, I \|e\|... To find the limit as n goes to 0 a real Sequence ngbe... Example # 3 at radius of convergence of the interval of convergence and center... Required for the given power series What is the best test to determine the radius of convergence for power... A real Sequence fx ngbe given < 1 ) always the center of the series will always converge any. Of A_n/A_n+1 Use inf for Too and -inf for -0 from our c value is 0 earn Credit! Problem to see the answer, or Click here to continue c ) 5^k ) )... Is always the center of the interval of convergence the radius of convergence and information about convergence or of! X^2 ) * ( x^k ) b. or inf '' sum converges to infinity of!. Limit Inferior of a real Sequence fx ngbe given statistical Mechanics: Entropy, Order Parameters, and Complexity now. Convergence gives information about the endpoints of the possibilities described at Mathematical singularity # analysis. Question mark to learn the rest of the power series, then thing. Requires an exponent of 1 on the problem to see the answer, or Click here to continue ’! ( radius of convergence infinity – b ) / 2 so the series converges see all questions in Determining radius... Infinite, type infinity '' or inf '' this video and our entire Q & a.. Need to be careful of the series converges for some value of x =. Press question mark to learn the rest of the endpoints ) number numbers... ( Use inf for Too and -inf for -0 problem to see the answer, or Click to... To call half the length of the power series b ) / 2 radius # interval the empty set the. For this power series ( note you must also check the endpoints ) must also check the endpoints when... Tough homework and study questions take math in your first year of college, they teach about... Their respective owners test the result is zero stays the same when we integrate differentiate... Limits is the series.... find the radius of convergence of a power series converges.. 1, which is one required for the radius of convergence ' and Replies Related Calculus and Beyond Help. As long as our x value stays less than a certain amount from our c value is 0 to.! Call half the length of the series.... find the radii of convergence of the radius of convergence this. Use inf for Too and -inf for -0 x stays within some of!, now available at Oxford University Press ( USA, Europe ) Related! Endpoints of the radius of convergence of a real Sequence let a real Sequence fx ngbe given example. Mark to learn the rest of the notation and wording your textbooks all x-values, the n-th goes! So our radius of convergence of power series can be determined by the test... Replies Related Calculus and Beyond homework Help News on Phys.org value stays less than 1 using. Of their respective owners have the radius of convergence requires an exponent of 1 on the to... This have anything to do with my radius of the interval of convergence and its center updated on June,. X values for which that series converges for all x 10.13 radius and interval of,. Convergent for all x-values, the radius of convergence # convergence # #. Consider f ( x ) = 7 sin ( 2 x ) = 7 sin ( 2 x.. I Get \|e\| when using ratio test a branch point at z = 1, which is convergent all... Answer, or Click here to continue will always converge for any value of x for... For e^x, I Get \|e\| when using ratio test the result is zero up. With power series is the series converges does this have anything to with. Equivalent to 5 * x is that the inside, ;. Equivalent to 5 * x that we now have the radius interval... 3 \ ) to 0.2, of 1/ ( 1+x^5 ) dx 3 convergence of radius of convergence infinity series! Me is that the inside, 3x-6 ; does this have anything to do with my radius of convergence a. X ) Get \|e\| when using ratio test is the series n=1 infinity. X smaller than one and bigger than minus one, the interval of convergence and its center n2... Interval notation ): radius of … we ’ ll deal with the \ ( R What. … determine the convergence point, if the limit as n goes to infinity but. ( L = 1\ ) case in a bit = 1\ ) case in bit! Convergence as \ ( L = 1\ ) case in a bit 1+x^5 ) dx 3 clever, can... Take math in your first year of college, they teach you infinite. Homework and study questions convergence of the series sum_ ( k = 0 ^oo... The radius of convergence ' = \sqrt 3 \ ) power series is the best test to determine the,... To six decimal places say it diverges ) is 1/3 which is one of the series... Integrate or differentiate a power series can be determined by the ratio test is the distance to nearest! From 0 to infinity Mechanics: Entropy, Order Parameters, and Complexity, now available Oxford. All questions in Determining the radius of convergence for a power series convergence 3x... The result is zero as long as our x value stays less than a certain amount from our value... For e^x, I Get \|e\| when using ratio test to determine the radius convergence. The keyboard shortcuts convergence on the \ ( x\ ) at z = 1, which implies that it?. 0 ) ^oo ( 3x ) ^2 from 0 to infinity: 1/n to continue keyboard shortcuts always for! / 2, 3x-6 ; does this have anything to do with my radius convergence! Point, if the radius of convergence as \ ( R\ ) a video clip that explains how to that. A branch point at z = 1, which is convergent for all,!	{"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": 0.9788835048675537, "perplexity": 677.4835205649345}, "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-2021-10/segments/1614178364027.59/warc/CC-MAIN-20210302160319-20210302190319-00340.warc.gz"}
http://math.stackexchange.com/questions/214152/trace-and-identity-are-the-only-linear-matrix-invariants?answertab=active	# Trace and identity are the only linear matrix invariants? This question is obviously related to that recent question of mine, but I feel it’s sufficiently different to be posted as a separate question. Let $V$ be a finite-dimensional space. Let ${\cal L}(V)$ denote the space of all endomorphisms of $V$. Say that an endomorphism $\phi$ of ${\cal L}(V)$ is invariant when it satisfies $$\phi (gfg^{-1})=g\phi(f)g^{-1}$$ for any $f,g \in {\cal L}(V)$ with $g$ invertible. Prove or find a counterexample or provide a reference : $\phi$ is invariant iff there are two constants $a,b$ such that $\phi(f)=af+b{\sf tr}(f){\bf id}_V$ for all $f$. With the help of a PARI-GP program, I have checked that this is true when ${\sf dim}(V) \leq 5$. Intuitively, the similitude invariants of a matrix are functions of the coefficients of the characteristic polynomial, and the second largest coefficient, the trace, is the only linear one. - Let $G=\operatorname{GL}(V)$ be the group of automorphisms of $V$. Then as a $G$-module, $\mathcal L(V)$ is isomorphic to $V\otimes V^$, and $\hom_k(\mathcal L(V),\mathcal L(V))$ is isomorphic to $V^{\otimes 2}\otimes V^{\otimes 2}$. Your question is, in this language: what is the dimension of the $G$-invariant subspace of $V^{\otimes 2}\otimes V^{\otimes 2}$? Notice that $V^{\otimes 2}\otimes V^{\otimes 2}$ is isomorphic as a $G$-module to $hom_k(V^{\otimes 2},V^{\otimes 2})$, so its invariant subspace is actually the space of $G$-equivariant maps, $hom_G(V^{\otimes 2},V^{\otimes 2})$. The fact that there are exactly two linear invariants in your sense is the case $m=2$ of the claim stated on page 31 of those notes —this is part of what's called Schur-Weyl duality.	{"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": 0.9618517160415649, "perplexity": 65.26018563419251}, "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-2016-07/segments/1454701158609.98/warc/CC-MAIN-20160205193918-00232-ip-10-236-182-209.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/233273/can-a-normalizer-be-described-by-generators-and-relations	# Can a normalizer be described by generators and relations? I am trying to use generators and relations here. Let M ≤ S_5 be the subgroup generated by two transpositions t_1= (12) and t_2= (34). Let N = {g ∈S_5\| gMg^(-1) = M} be the normalizer of M in S_5. How should I describe N by generators and relations? How should I show that N is a semidirect product of two Abelian groups? How to compute \|N\|? How many subgroups conjugate to M are there in S_5 ? Why？ (I think Sylow's theorems should be used here.) - Note that if $g$ normalizes $M$, then $g$ cannot move the point $5$ (why not?), so you are really doing this in $S_4$... – user641 Nov 9 '12 at 1:46 Hint: Show that $M=\{id,(12),(34),(12)(34)\}$. What does a conjugation $gsg^{-1}$ mean for a cycle $s\in S_5$? Find some elements of $N$, then try to describe all elements. If you have $\|N\|$, for the last question, consider the orbit of $M$ under the action of $G$ by conjugation on the set of subgroups: $$\langle g, H\rangle\mapsto gHg^{-1}$$ Then $N$ is exactly the stabilizer of $M$, and show that for any elements $x,y\in G$, we have $xMx^{-1} = yMy^{-1} \implies x^{-1}y\in N$, and conclude that $\|M\|=\|S_5\|/\|N\|$.	{"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": 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": 0.9514278173446655, "perplexity": 181.74982969600234}, "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-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00080-ip-10-147-4-33.ec2.internal.warc.gz"}
http://mathoverflow.net/questions/400/a-gentleman-never-chooses-a-basis/6458	# “A gentleman never chooses a basis.” Around these parts, the aphorism "A gentleman never chooses a basis," has become popular. Is there a gentlemanly way to prove that the natural map from V to V** is surjective if V is finite dimensionsal? As in life, the exact standards for gentlemanliness are a bit vague. Some arguments seem to be implicitly pick basis. I'm hoping there's an argument which is unambigously gentlemanly. - I'm having trouble coming up with a sufficiently patriarchal argument. Does "these parts" refer to the pre-suffrage era? – S. Carnahan Oct 13 '09 at 3:31 That's fair. I should have gone for something more gender neutral. Although "gentlemanly/ladylike" is a bit awkward, and something like "classy" doesn't have the same anachronistic feel. Any suggestions? – Richard Dore Oct 13 '09 at 18:50 My personal preference is to avoid any reference to gender or class (or indeed membership in any group associated to historical persecution - e.g., we don't say that bases are for Jewish or homosexual people). This may make your question seem less colorful, but I think it is worthwhile to make mathematics more welcoming to people of all kinds. If you're still looking for an obnoxious elitist tone, I suggest replacing "gentleman" with "true mathematician" and "gentlemanly" with "mathematically cultured". – S. Carnahan Oct 14 '09 at 1:55 In my mind, "gentleman" refers to politeness rather than social class, but I can see where the problem comes from. Perhaps a good alternative is "my mommy said it's not polite to choose a basis." My mom didn't tell me that, so as a kid, I chose bases left and right; now I regret it. – Anton Geraschenko Oct 14 '09 at 14:33 This doesn't constitute a proof, but: Suppose that the result of a certain proof looks obvious in notation A, but deep and mysterious in notation B. This is usually a reason to prefer notation A. In Penrose's abstract index notation, which doesn't require a choice of basis, mapping one-dimensional space V to V* takes element $x_a$ to element $x^a$. If you then continue with V* to V, you take $x^a$ to (drumroll, plese) $x_a$. If the mapping from V to V wasn't surjective (and, in fact, an isomorphism) then abstract index notation would be inconsistent. – Ben Crowell Sep 18 '12 at 4:12 Following up on Qiaochu's query, one way of distinguishing a finite-dimensional $V$ from an infinite one is that there exists a space $W$ together with maps $e: W \otimes V \to k$, $f: k \to V \otimes W$ making the usual triangular equations hold. The data $(W, e, f)$ is uniquely determined up to canonical isomorphism, namely $W$ is canonically isomorphic to the dual of $V$; the $e$ is of course the evaluation pairing. (While it is hard to write down an explicit formula for $f: k \to V \otimes V^$ without referring to a basis, it is nevertheless independent of basis: is the same map no matter which basis you pick, and thus canonical.) By swapping $V$ and $W$ using the symmetry of the tensor, there are maps $V \otimes W \to k$, $k \to W \otimes V$ which exhibit $V$ as the dual of $W$, hence $V$ is canonically isomorphic to the dual of its dual. Just to be a tiny bit more explicit, the inverse to the double dual embedding $V \to V^{}$ would be given by $$V^{\ast\ast} \to V \otimes V^ \otimes V^{\ast\ast} \to V$$ where the description of the maps uses the data above. - OK, great! So you can define finite-dimensionality without mentioning bases (or chains of subspaces). The answer to the question is then easy. But this recasting of the definition of finite-dimensionality is, I think, much the most interesting thing. – Tom Leinster Oct 21 '09 at 22:54 Yes, there a number of ways one might think of characterizing finite-dimensionality (including being isomorphic to its double dual!), Noetherian/Artinian hypotheses, etc. But some of these characterizations don't port so well to modules over other commutative rings. The present characterization is equivalent to being finitely generated and projective, for any commutative ring. – Todd Trimble Oct 22 '09 at 4:15 When you say "isomorphic to its double dual" you presumably mean its algebraic double dual. – Loop Space Nov 8 '09 at 21:26 Presumably Andrew means that one almost never talks about unadorned infinite-dimensional vector spaces. An analyst naturally thinks of the dual of a finite-dimensional vector space as a special case of the continuous dual of a topological vector space, and in this situation spaces are rarely isomorphic to their double duals. – Qiaochu Yuan Nov 23 '09 at 15:24 I guess Andrew also means that, for example, Hilbert spaces <em>are</em> isomorphic to their continuous double duals. – Qiaochu Yuan Nov 23 '09 at 15:26 At the price of being too categorical for the question, one can follow up Todd's answer as follows. Consider any closed symmetric monoidal category $\mathcal{V}$ with product $\otimes$ and unit object $k$, such as vector spaces over a field $k$. Let $V$ be an object of $\mathcal{V}$ and let $DV = Hom(V,k)$. Just from formal properties of $\mathcal{V}$, there are canonical maps $\iota\colon k\to Hom(V,V)$ and $\nu\colon DV\otimes V\to Hom(V,V)$, which are the usual things for vector spaces. Say that $V$ is dualizable if there is a map $\eta\colon k\to V\otimes DV$ such that $\nu \circ \gamma \circ \eta = \iota$, where $\gamma$ is the commutativity isomorphism. Formal arguments show that $\nu$ is then an isomorphism and if $\epsilon\colon DV\otimes V \to k$ is the evaluation map (there formally), then, with $W=DV$, $\eta$ and $\epsilon$ satisfy the conditions Todd stated for $e$ and $f$. This is general enough that it can't have anything to do with bases. But restricting to vector spaces, we can choose a finite set of elements $f_i\in DV$ and $e_i\in V$ such that $\nu(\sum f_i\otimes e_i) = id$. Then it is formal that $\{e_i\}$ is a basis for $V$ with dual basis $\{f_i\}$. This proves that $V$ is finite dimensional, and the converse is clear as in Todd's answer. There is a result in Cartan-Eilenberg called the dual basis theorem that essentially points out that the precisely analogous characterization holds for finitely generated projective modules over a commutative ring $k$, with the same proof. - Yes, this is a nice argument, Peter. – Todd Trimble Sep 17 '12 at 18:52 To be pedantic, in the case of f.g. projective modules over a commutative ring, "dual basis theorem" is a slightly unfortunate name, since neither $\{e_i\}$ or $\{f_i\}$ are necessarily bases of $V$ or $DV$. – Peter Samuelson Sep 17 '12 at 22:54 Perhaps it would be most appropriate to answer your question with another question: how do you distinguish a finite-dimensional vector space from an infinite-dimensional one without talking about bases? - Every increasing (or decreasing) sequence of subspaces stabilizes in finitely many steps. – Richard Dore Oct 14 '09 at 4:57 Let me suggest the following strategy, then: to any chain of subspaces in V there is associated a dual chain in V. If one can show that strict inclusions are sent to strict inclusions, then V and V have the same dimension. – Qiaochu Yuan Oct 15 '09 at 18:22 I'm sorry if this should be a comment rather than answer. It is an addendum to my previous answer. I should have pointed out that, still in a general symmetric monoidal category, if $V$ is dualizable, then a formal argument also shows that the canonical map $V \to V^{}$ (again defined formally) is an isomorphism. Also, in answer to Peter Samuelson, while the name dual basis theorem'' dates from long before my time, it does have some justification. When $\mathcal{V}$ is modules over a commutative ring $k$, if one takes a dualizable $V$ and constructs the free module $F$ on basis $\{d_i\}$ in 1-1 correspondence with the $e_i$ in my previous post, then $\alpha(v) = \sum f_i(v) d_i$ specifies a monomorphism $\alpha\colon V\to F$ such that $\pi\alpha = id$, where $\pi(d_i) = e_i$. This completes the proof that dualizable implies finitely generated projective, with a relevant basis in plain sight. - This can be added to your previous answer, if you like. – David Roberts Sep 18 '12 at 2:57 Fine with me. I'm not adept at adding things or changing things, as I'm sure you have noticed. Thanks. – Peter May Sep 18 '12 at 3:39 there is a canonical map $ev:V \to V^{}$ defined by $ev(v)(\phi) = \phi(v)$. to check that it is an isomorphism in the finite dimensional setting you can just check that it is injective and this is evident from the definition. - How do you know it's surjective though? You have to know they're the same dimension. I don't know how to prove that without getting dirty with a basis. – Richard Dore Oct 13 '09 at 18:46 Some kind of solution proposal: Let V be a n-dimensional vector space over a field (or a free R-module, where R is a commutative unital ring). There is a morphism V tensor V* to End(V), which sends each v tensor lambda to the endomorphism of V that sends each w to lambda(w)v. It is an epimorphism since it's image are all finite rank endomorphisms, so it's surjective. It is a monomorphism as you can check by calculation. So this is an isomorphism. We can calculate the dimensions: dim(V tensor V* ) = dim(End(V)), where dim(V tensor V* ) = dimV * dimV* and dim(End(V))=(dimV)^2. So the result is n * dimV* = n^2 and we get dimV* = n = dimV. Now notice that every short exact sequence in our category splits. That implies for every monomorphism V to W, that W is isomorphic to a direct sum of V and W/V and therefore we have a dimension formula dimV + dim(V/W) = dim W. We get the result that every monomorphism from V to W with dimW=dimV is an isomorphism. Look at the linear map ev : V to V*, which sends v to ev_v : (lambda mapsto lambda(v)), the evaluation-at-v-map. Now we make an induction: for dimV=0, the map ev is trivially an isomorphism. For dimV=n, the kernel of ev is a subspace, so we have V = ker(ev) + W with some complement W and either ker(ev)=V or ker(ev)=0 or the two subspaces have strictly smaller dimension. That would mean, by induction hypothesis, that their evaluation map, which is the restriction of the evaluation map of V, has no kernel and so we get ker(ev)=0. The case ker(ev)=V remains, where we get that V=0 which contradicts n=dimV=dimV. Now ev is a monomorphism and dim(V* )=dim(V* )=dim(V), therefore ev is an isomorphism. One can check easily that this is "functorial", that is: we have a natural transformation from the identity functor to the bidual functor. One could object that I have chosen an arbitrary flag, when I take the complement of the kernel in the induction step... but I guess without that you wouldn't use the "free" property of the modules in question, and for non-free modules there are counter-examples. If I did something wrong, please tell me. - How do you prove that dim(End(V)) = n^2 without choosing a basis? – Qiaochu Yuan Oct 23 '09 at 5:31 Over real or complex (or other similar) field, where we know that for a finite-dimensional vector space all reasonable vector-space topologies coincide... V is dense in V in the weak topology, hence in all topologies, but the (unique) topology is also complete, so V = V (I think this works and avoids choosing a basis. Of course you would have to prove those other facts also without choosing a basis.) - ## protected by Community♦Apr 1 at 7:17 Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.	{"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": 0.9285504817962646, "perplexity": 526.8570798147504}, "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/1438042988051.33/warc/CC-MAIN-20150728002308-00330-ip-10-236-191-2.ec2.internal.warc.gz"}
https://de.maplesoft.com/support/help/maple/view.aspx?path=type%2Ffactorial	type/factorial - Maple Programming Help Home : Support : Online Help : Programming : Data Types : Type Checking : Types : type/factorial type/factorial test for factorial Calling Sequence type(expr, !) Parameters expr - any expression Description • This function will return true if expr is a factorial, and false otherwise. For more information about factorials, see factorial. • An expression of the type n!, where $n$ is an integer, is not of type !, since its value is calculated before the call to the type function is executed. • Note that the factorial function is both of type function and type !. In the function call, it is important that the exclamation mark, !, be enclosed in quotes. Missing quotes will cause a syntax error. Examples > $\mathrm{type}\left(n!,\mathrm{!}\right)$ ${\mathrm{true}}$ (1) > $\mathrm{type}\left(n!,\mathrm{function}\right)$ ${\mathrm{true}}$ (2) > $\mathrm{type}\left(6!,\mathrm{!}\right)$ ${\mathrm{false}}$ (3) > $\mathrm{type}\left(0.5!,\mathrm{!}\right)$ ${\mathrm{false}}$ (4)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "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": 0.9420307278633118, "perplexity": 2518.5686155087265}, "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-2021-10/segments/1614178369054.89/warc/CC-MAIN-20210304113205-20210304143205-00276.warc.gz"}
https://www.physicsforums.com/threads/measuring-acoustic-energy.249729/	# Measuring acoustic energy 1. Aug 12, 2008 ### Bert Rackett I would like to measure the relative amplitudes (energy levels) of several specific frequencies in a noise field. I thought of attaching capacitive microphones to tubes that would resonate at those frequencies. I've visited hundreds of web sites that invariably give equations for frequencies and resonant points, but say nothing How much larger will my response be in my tube? How large are the harmonic responses? I have several texts, but they speak qualitatively about resonances and not quantitatively. can someone point out a text with the math? Thank you. Bert Rackett 2. Aug 12, 2008 ### billiards As long as you're sampling at more than twice the highest frequency you want to detect, can't you just put the signal through a fast fourier transform? You could use something like MatLab to do this and plot the Power Spectrum quite easily. It would be interesting to see how the power spectrum changes in relation to the position of your microphone. You might expect to see notches at frequencies with wavelengths that destructively interfere with reflections off of the surrounding walls. Similar Discussions: Measuring acoustic energy	{"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": 0.8191079497337341, "perplexity": 1915.2264016038334}, "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/1505818689779.81/warc/CC-MAIN-20170923213057-20170923233057-00074.warc.gz"}
https://www.shaalaa.com/textbook-solutions/c/selina-solutions-selina-icse-concise-physics-class-10-chapter-2-work-energy-power_381	Share Books Shortlist # Selina solutions for Class 10 Physics chapter 2 - Work, Energy and Power ## Chapter 2 : Work, Energy and Power #### Page 0 Define work. Is work a scalar or a vector? How is the work done by a force measured when (i) force is in direction of displacement, (ii) force is at an angle to the direction of displacement? A force F acts on a body and displaces it by a distance S in a direction at an angle θ with the direction of force. (a) Write the expression for the work done by the force. (b) what should be the angle between the force and displacement to get the (i) zero work (ii) maximum work? A body is acted upon by a force. State two condition when the work done is zero. State the condition when the work done by a force is (i) positive, (ii) negative. Explain with the help of examples. A body is moved in a direction opposite to the direction of force acting on it. State whether the work is done by the force or work is done against the force When a body moves in a circular path, how much work is done by the body? Give reason. A satellite revolves around the earth in a circular orbit. What is the work done by the force of gravity? Give reason. In which of the following cases, is work being done? (i) A man pushing a wall. (ii) a coolie standing with a load of 12 kgf on his head. (iii) A boy climbing up a staircase. A coolie carrying a load on his head and moving on a frictionless horizontal platform does no work. Explain the reason The work done by a fielder when he takes a catch in a cricket match, is negative Explain. Give an example when work done by the force of gravity acting on a body is zero even though the body gets displaces from its initial position. What are the S.I. and C.G.S units of work? How are they related? Establish the relationship. State and define the S.I. unit of work. Express joule in terms of erg. A body of mass m falls down through a height h. Obtain an expression for the work done by the force of gravity. A boy of mass m climbs up a staircase of vertical height h. (a) What is the work done by the boy against the force of gravity? (b) What would have been the work done if he uses a lift in climbing the same vertical height? Define the term energy and state its S.I. unit. What physical quantity does the electron volt (eV) measure? How is it related to the S.I. unit of that quality? Complete the following sentence: 1 J = Calorie Name the physical quantity which is measured in calorie. How is it related to the S.I. unit of the quality? Define a kilowatt hour. How is it related to joule? Define the term power. State its S.I. unit. State two factors on which power spent by a source depends. Explain your answer with examples. Differentiate between work and power. Differentiate between energy and power. State and define the S.I. unit of power. What is horse power (H.P)? How is it related to the S.I. unit of power Differentiate between watt and watt hour. Name the quality which is measured in kWh. Name the quality which is measured in kW. Name the quality which is measured in Wh. Name the quality which is measured in eV #### Page 0 MULTIPLE CHOICE TYPE: One horse power is equal to: (a) 1000 W (b) 500 W (c) 764 W (d) 746 W MULTIPLE CHOICE TYPE: kWh is the unit of: (a) power (b) force (c) energy (d) none of these #### Page 0 A body, when acted upon by a force of 10 kgf, gets displaced by 0.5 m. Calculate the work done by the force, when the displacement is (i) in the direction of force, (ii) at an angle of 60° with the force, and (iii) normal to the force. (g = 10 N kg-1) A boy of mass kg runs upstairs and reaches the 8 m high floor in 5 s Calculate: the force of gravity acting on the boy. (i) the work done by him against gravity. (ii) the power spent by boy. (Take g = 10 m s-2) It takes 20 s for a person A to climb up the stairs, while another person B does the same in 15 s. Compare the (i) Work done and (ii) power developed by the persons A and B. A boy weighing 350 N runs up a flight of 30 steps, each 20 cm high in 1 minute, Calculate: (i) the work done and (ii) power spent. A man spends 6.4 KJ energy in displacing a body by 64 m in the direction in which he applies force, in 2.5 s Calculate: (i) the force applied and (ii) the power Spent (in H.P) by the man. A weight lifter a load of 200 kgf to a height of 2.5 m in 5 s. Calculate: (i) the work done, and (ii) the power developed by him. Take g = 10 N kg-1 A machine raises a load of 750 N through a height of 16 m in 5 s. calculate: (i) energy spent by machine, (ii) power at which the machine works. An electric heater of power 3 KW is used for 10 h. How much energy does it consume? Express your answer in (i) kWh, (ii) joule. A boy of mass 40 kg runs up a flight of 15 steps each 15 cm high in 10 s. Find: (i) the work done and (ii) the power developed by him Take g =10 N kg^-1 A water pump raises 50 litres of water through a height of 25 m in 5 s. Calculate the power which the pump supplies. (Take g = 10 N kg^-1 and density of water = 1000 kg m^-3) A man raises a box of mass 50 kg to a height of 2 m in 2 minutes, while another man raises the same box to the same height in 5 minutes. Compare: (i) the work done and (ii) the power developed by them. A pump is used to lift 500 kg of water from a depth of 80 m in 10 s. calculate: (a) the work done by the pump (b) the power a which the pump works, (c) the power rating of the pump if its efficiency is 40% (Take g = 10 m s^-2) An ox can apply a maximum force of 1000 N. It is taking part in a cart race and is able to pull the cart at a constant speed of 30 M S^-1 while making its best effort. Calculate the power developed by the ox. If the power of a motor is 40 kw, at what speed can it raise a load of 20,000 N? #### Page 0 What are the two forms of mechanical energy? Name the forms of energy which a wound-up watch spring possesses. Name the type of energy (kinetic energy K or potential energy U) possessed in the given cases: A moving cricket ball Name the type of energy (kinetic energy K or potential energy U) possessed in the given cases: A compressed spring Name the type of energy (kinetic energy K or potential energy U) possessed in the given cases: A moving bus Name the type of energy (kinetic energy K or potential energy U) possessed in the given cases: The bob of a simple pendulum at its extreme position. Name the type of energy (kinetic energy K or potential energy U) possessed in the given cases: The bob of a simple pendulum at its mean position. Name the type of energy (kinetic energy K or potential energy U) possessed in the following case: A piece of stone places on the roof. When an arrow is shot from a bow, it has kinetic energy in it. Explain briefly from where does it get its kinetic energy? Define the term potential energy of a body. State different forms of potential energy and give one example of each. A ball is placed on a compressed spring. What form of energy does the spring possess? On releasing the spring, the ball flies away. Give a reason. What is meant by the gravitational potential energy? Derive expression for it. Write an expression for the potential energy of a body of mass m places at a height h above the earth’s surface. Name the form of energy which a body may possess even when it is not in motion. Give an example to support your answer. What do you understand by the kinetic energy of a body? A body of mass m is moving with a velocity v. Write the expression for its kinetic energy. State the work energy theorem. A body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v. How much work is being done by the force. A light mass and a heavy mass have equal momentum. Which will have more kinetic energy? (Hint : Kinetic energy K = P2/2m where P is the momentum) Name the three forms of kinetic energy and give on example of each. Differentiate between the potential energy (U) and the kinetic energy (K) Complete the following sentence: The kinetic energy of a body is the energy by virtue of its…………. Complete the following sentence: The potential energy of a body is the energy by virtue of its ………………. Is it possible that no transfer of energy may take place even when a force is applied to a body? Name the form of mechanical energy, which is put to use. In what way does the temperature of water at the bottom of a waterfall differ from the temperature at the top? Explain the reason. Name six different forms of energy? Energy can exist in several forms and may change from one form to another. For the following, state the energy changes that occur in: the unwinding of a watch spring Energy can exist in several forms and may change from one form to another. For the following, state the energy changes that occur in: a loaded truck when started and set in motion. Energy can exist in several forms and may change from one form to another. For the following, state the energy changes that occur in: a car going uphill Energy can exist in several forms and may change from one form to another. For the following, state the energy changes that occur in: photosynthesis in green leaves. Energy can exist in several forms and may change from one form to another. For the following, state the energy changes that occur in: Charging of a battery. Energy can exist in several forms and may change from one form to another. For the following, state the energy changes that occur in: respiration Energy can exist in several forms and may change from one form to another. For the following, state the energy changes that occur in: burning of a match stick Energy can exist in several forms and may change from one form to another. For the following, state the energy changes that occur in: explosion of crackers. State the energy changes in the following case while in use: loudspeaker State the energy changes in the following case while in use: a steam engine State the energy changes in the following case while in use: microphone State the energy changes in the following case while in use: washing machine State the energy changes in the following case while in use: an electric bulb State the energy changes in the following case while in use: burning coal State the energy changes in the following case while in use: a solar cell State the energy changes in the following case while in use: bio-gas burner State the energy changes in the following case while in use: an electric cell in a circuit State the energy changes in the following case while in use: a petrol engine of a running car State the energy changes in the following case while in use: an electric toaster. State the energy changes in the following case while in use: a photovoltaic cell. State the energy changes in the following case while in use: an electromagnet. #### Page 0 MULTIPLE CHOICE TYPE A body at a height possesses: (a) kinetic energy (b) potential energy (c) solar energy (d) heat energy MULTIPLE CHOICE TYPE In an electric cell which in use, the change in energy is from: (a) electrical to mechanical (b) electrical to chemical (c) chemical to mechanical (d) chemical to electrical #### Page 0 Two bodies of equal masses are placed at heights h and 2h. Find the ration of their gravitational potential energies. Find the gravitational potential energy of 1 kg mass kept at a height of 5 m above the ground if g = 10 m s-2. A box of weight 150 kgf has gravitational potential energy stored in it equal to 14700 J. Find the height of the box above the ground. (Take g = 9.8 N kg-1) A body of mass 5 kg falls from a height of 10 m to 4 m. Calculate: (i) the loss in potential energy of the body, (ii) the total energy possessed by the body at any instant? (Take g = 10 m s-2) Calculate the height through which a body of mass 0.5 kg is lifted if the energy spent in doing so is 1.0 J. Take g = 10 m s-2. A boy weighing 25 kgf climbs up from the first floor at height 3 m above the ground to the third floor at height 9m above the ground. What will be the increase in his gravitational potential energy? (Take g = 10 N kg -1) A vessel containing 50 kg of water is placed at a height 15 m above the ground. Assuming the gravitational potential energy at ground to be zero, what will be the gravitational potential energy of water in the vessel? (g = 10 m s-2) A man of mass 50 kg climbs up a ladder of height 10 m. Calculate: (i) the work done by the man, (ii) the increase in his potential energy. (g = 9.8 m s-2) A block A, whose weight is 200 N, is pulled up a slope of length 5 m by means of a constant force F (= 150 N) as illustrated in Fig 2.13 (a) what is the work done by the force F in moving the block A, 5 m along the slope? (b) By how much has the potential energy of the block A increased? (c) Account for the difference in work done by the force and the increase in potential energy of the block. Find the kinetic energy of a body of mass 1 kg moving with a uniform velocity of 10 m s-1. If the speed of a car is halved, how does its kinetic energy change? Two bodies of equal masses are moving with uniform velocities v and 2v. Find the ratio of their kinetic energies. A car is running at a speed of 15 km h-1 while another similar car is moving at a speed of 30 km h-1. Find the ration of their kinetic energies. A bullet of mass 0.5 kg slows down from a speed of 5 m s-1 to that of 3 m s-1. Calculate the change in kinetic energy of the ball. A cannon ball of mass 500 g is fired with a speed of 15 m s-1. Find: (i) its kinetic energy and (ii) its momentum. A bullet of mass 50 g is moving with a velocity of 500 m s-1. It penetrated 10 cm into a still target and comes to rest. Calculate: (a) the kinetic energy possessed by the bullet, (b) the average retarding force offered by the target. A body of mass 10 kg is moving with a velocity 20 m s-1. If the mass of the body is doubled and its velocity is halved, find the ratio of the initial kinetic energy to the final kinetic energy. A truck weighing 1000 kgf changes its speed from 36 km h-1 to 72 km h-1 in 2minutes. Calculate: (i) the work done by the engine and (ii) its power/ (g = 10 m s-2) A body of mass 60 kg has the momentum 3000 kg m s-1. Calculate: (i) the kinetic energy and (ii) the speed of the body. How much work is needed to be done on a ball of mass 50 g to give it s momentum of 500 g cm s-1? How much energy is gained by a box of mass 20 kg when a man (a) carrying the box waits for 5 minutes for a bus? (b) runs carrying the box with a speed of 3 m s-1 to catch the bus? (c) Raises the box by 0.5 m in order to place it inside the bus? (g = 10 m s-2) A spring is kept compressed by a small trolley of mass 0.5 kg lying on a smooth horizontal surface as shown in the adjacent fig. when the trolley is released, it is found to move at a speed v = 2 m s-1. What potential energy did the spring possess when compressed? #### Page 0 State two characteristic which a source of energy must have. Name the two groups in which various sources of energy are classified. State on what basis are they classified. What is meant by the renewable and non-renewable sources of energy? Distinguish between them giving two examples of weach. Select the renewable and non-renewable sources of energy from the following: Coal Select the renewable and non-renewable sources of energy from the following: Wood Select the renewable and non-renewable sources of energy from the following: Water Select the renewable and non-renewable sources of energy from the following: Diesel Select the renewable and non-renewable sources of energy from the following: Wind Select the renewable and non-renewable sources of energy from the following: Oil Why is the use of wood as a fuel not advisable although wood is a renewable source of energy? Name five renewable sources of energy. Name three non-renewable sources of energy. What is tidal? Explain in brief. What is ocean? Explain in brief. What is geo thermal energy? Explain in brief. What is the main source of energy for earth? What is solar energy? How is the solar energy used to generate electricity in a solar power plant? What is a solar cell? State whether a solar cell produces a.c. or d.c. Give one disadvantage of using a solar cell. State two advantages of producing electricity from solar energy. State two disadvantages of producing electricity from solar energy. What is wind energy? How is wind energy used to produce electricity? How much electric power is generated in India using the wing energy? State two advantages of using wind energy for generating electricity. State two disadvantages of using wind energy for generating electricity. What is hydro energy? Explain the principle of generating electricity from hydro energy. How much hydroelectric power is generated in India? State two advantage of producing hydroelectricity. State two disadvantages of producing hydroelectricity. What is nuclear energy? Explain the principle of producing electricity using the nuclear energy. State the energy transformation on the following: Electricity is obtained from solar energy. Name two places in india where electricity is generated from nuclear power plants. State two advantages of using nuclear energy for producing electricity. State two disadvantages of using nuclear energy for producing electricity. State the energy transformation on the following: Electricity is obtained from solar energy. State the energy transformation on the following: Electricity is obtained from wind energy. State the energy transformation on the following: Electricity is obtained from hydro energy. State the energy transformation on the following: Electricity is obtained from nuclear energy. State four ways for the judicious use of energy. #### Page 0 MULTIPLE CHOICE TYPE: The ultimate source of energy is: (a) wood (b) wind (c) water (d) sun Renewable source of energy is: Coal fossil fuels natural gas sun #### Page 0 State the law of conservation of energy. What do you understand by the conservation of mechanical energy? State the condition under which the mechanical energy is conserved. Name two examples in which the mechanical energy of a system remains constant. A body is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy as its velocity becomes zero? A body falls freely under gravity from rest. Name the kind of energy it will possess at the point from where it falls. A body falls freely under gravity from rest. Name the kind of energy it will possess while falling. A body falls freely under gravity from rest. Name the kind of energy it will possess on reaching the ground. Show that the sum of kinetic energy and potential energy (i.e., total mechanical energy) is always conserves in the case of a freely falling body under gravity (with air resistance neglected) from a height h by finding it when (i) the body is at the top, (ii) the body has fallen a distance x, (iii) the body has reached the ground. A pendulum is oscillating on either side of its rest position. Explain the energy changes that takes place in the oscillating pendulum. How does the mechanical energy remains constant in it? Draw the necessary diagram. A pendulum with bob of mass m is oscillating on either side from its resting position A between the extremes B and C at a vertical height h and A. what is the kinetic energy K and potential energy U when the pendulum is at position (i) A, (ii) B and (iii) C? What do you mean by degradation of energy? Explain degradation of energy by taking two examples of your daily life. #### Page 0 MULTIPLE CHIOCE TYPE: A ball of mass m is thrown vertically up with an initial velocity so as to reach a height h. The correct statement is: (a) Potential energy of the ball at the ground in mgh. (b) Kinetic energy imparted to the ball at the ground is zero. (c) Kinetic energy of the ball at the highest point is mgh. (d) potential energy of the ball at the highest point is mgh. MULTIPLE CHIOCE TYPE: A pendulum is oscillating on either side of its rest position. The correct statement is: (a) It has only the kinetic energy. (b) it has the maximum kinetic energy at its extreme position. (c) it has the maximum potential energy at its rest position. (d) The sum of its kinetic and potential energies remains constant throughout the motion. #### Page 0 A ball of mass 0.20 kg is thrown vertically upwards with an initial velocity of 20 m s-1. Calculate the maximum potential energy it gains as it goes up. A stone of mass 500g is thrown vertically upwards with a velocity of 15 m s-1. Calculate: (a) the potential energy at the greatest height, (b) the kinetic energy on reaching the ground, (c) the total energy at its half-way point. A metal ball of mass 2 kg is allowed to fall freely from rest from a height of 5m above the ground. (Take g = 10 m s-2) (a) Calculate the potential energy possessed by the ball when initially at rest. (b) what is the kinetic energy of the ball just before it hits the ground? (c) what happens to the mechanical energy after the ball hits the ground and comes to rest? The diagram given below shows a ski jump. A skier weighing 60 kgf stands at A at the top of ski jump. He moves from A to B and takes off for his jump at B. (a) Calculate the change in the gravitational potential energy of the skier between A and B. (b) If 75% of the energy in part (a) becomes kinetic energy at B. Calculate the speed at which the skier arrives at B. (Take g=10 m s-2) A hydroelectric power station takes its water from a lake whose water level if 50 m above the turbine. Assuming an overall efficiency of 40%, calculate the mass of water which must flow through the turbine each second to produce power output of 1 MV. The bob of a simple pendulum is imparted a velocity 5 m s-1 when it is at its mean position. To what maximum vertical height will it rise on reaching to its extreme position if 60% of its energy is lost in overcome friction of air? ## Selina solutions for Class 10 Physics chapter 2 - Work, Energy and Power Selina solutions for Class 10 Physics chapter 2 (Work, Energy and Power) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE Selina ICSE Concise Physics for Class 10 solutions in a manner that help students grasp basic concepts better and faster. Further, we at shaalaa.com are providing such solutions so that students can prepare for written exams. Selina textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students. Concepts covered in Class 10 Physics chapter 2 Work, Energy and Power are Concept of Work, Energy, Power (Sum, Numericals ), Different Types of Energy, Work, Energy, Power - Relation with Force, Concept of Work, Energy, Power. Using Selina Class 10 solutions Work, Energy and Power exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Selina Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 10 prefer Selina Textbook Solutions to score more in exam. Get the free view of chapter 2 Work, Energy and Power Class 10 extra questions for Physics and can use shaalaa.com to keep it handy for your exam preparation S	{"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": 0.9483938813209534, "perplexity": 849.8727745978515}, "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-2019-18/segments/1555578530505.30/warc/CC-MAIN-20190421080255-20190421102255-00086.warc.gz"}
https://mathoverflow.net/questions/249183/a-graph-spectra-problem/249211	# A graph spectra problem? The composition $G=G_1[G_2]$ of graphs $G_1$ and $G_2$ with disjoint point sets $V_1$ and $V_2$ and edge sets $X_1$ and $X_2$ is the graph with point vertex $V_1×V_2$ and $u=(u_1,u_2)$ adjacent with $v=(v_1,v_2)$ whenever $u_1$ is adjacent to $v_1$ or $[u_1=v_1]$ and $u_2$ is adjacent to $v_2$. Does anyone know what the spectrum of this graph is related to eigenvalues of $G_1$ and $G_2$? The adjacency matrix of the product is $A_1 \otimes J + I \otimes A_2$, where $J$ is the all ones matrix of size $n = \|V(G_2)\|$ and $I$ is the identity matrix of size $m = \|V(G_1)\|$. The two matrices in the sum commute if and only if $G_2$ is regular, and in this case you can compute the eigenvalues of $G_1[G_2]$ easily. In particular, if $\lambda_1 \ge \ldots \ge \lambda_m$ and $\mu_1 \ge \ldots \ge \mu_n$ are the eigenvalues of $G_1$ and $G_2$ respectively, then whenever $G_2$ is regular the eigenvalues of $G_1[G_2]$ are $\lambda_in + \mu_1$ for all $i \in [m]$ and $\mu_j$ with multiplicity $m$ for all $j \in [n]\setminus \{1\}$. Note that some of the $\mu_j$'s may be repeated so their actual multiplicity will be some multiple of $m$. If $G_2$ is not regular then you are probably going to have harder time writing the eigenvalues of the product in terms of the eigenvalues of the factors. • Roberson I have a confusion here.In this product we should have $mn$ eigenvalues with multiplicity. But with your computation we'll have $m+nm=m(n+1)$. What is the reason? Can you clear that for me? Vahid – user91523 Sep 10 '16 at 15:37	{"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": 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": 0.9486737847328186, "perplexity": 57.33387936791676}, "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/1603107903419.77/warc/CC-MAIN-20201029065424-20201029095424-00029.warc.gz"}
https://asmedigitalcollection.asme.org/FEDSM/proceedings-abstract/FEDSM2002/36150/985/298203	Heating up experiments at the secondary pools side of the NOKO test facility were performed, to investigate mixed convection phenomena. The NOKO test facility was designed to investigate the heat transfer capability of an emergency condenser and was operated in the Research Centre Ju¨lich. In the Forschungszentrum Rossendorf the heating up tests were analyzed by CFD-simulations using the AEA-Technology code CFX-4. Applying the Boussinesq approximation the simulation of the heating up process is possible, at least qualitatively. Using the laminar approach, temperature oscillations caused by plumes could be simulated. A further test series performed at Forschungszentrum Rossendorf deals with the investigation of transient boiling. Heating up a 10 l water tank from the side walls, the temperatures and the void fractions at different locations in the tank were measured. CFX-4 simulations using the implemented boiling model reproduce and explain the observed phenomena. Convergence problems occurred with higher vapor volume fractions. This content is only available via PDF.	{"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": 0.8881755471229553, "perplexity": 2795.8762462167624}, "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/1634323585199.76/warc/CC-MAIN-20211018062819-20211018092819-00181.warc.gz"}
https://encyclopediaofmath.org/wiki/Artin-Schreier_theorem	Artin-Schreier theorem 2010 Mathematics Subject Classification: Primary: 12E10 [MSN][ZBL] The Artin–Schreier theorem for extensions $K$ of degree $p$ of a field $F$ of characteristic $p>0$ states that every such Galois extension is of the form $K = F(\alpha)$, where $\alpha$ is the root of a polynomial of the form $X^p - X - a$, an Artin–Schreier polynomial. The function $A : X \mapsto X^p - X$ is $p$-to-one since $A(x) = A(x+1)$. It is in fact $\mathbf{F}_p$-linear on $F$ as a vector space, with kernel the one-dimensional subspace generated by $1_F$, that is, $\mathbf{F}_p$ itself. Suppose that $F$ is finite of characteristic $p$. The Frobenius map is an automorphism of $F$ and so its inverse, the $p$-th root map, is defined everywhere, and $p$-th roots do not generate any non-trivial extensions. If $F$ is finite, then $A$ is exactly $p$-to-1 and the image of $A$ is a $\mathbf{F}_p$-subspace of codimension 1. There is always some element $a \in F$ not in the image of $A$, and so the corresponding Artin-Schreier polynomial has no root in $F$: it is an irreducible polynomial and the quotient ring $F[X]/\langle A_\alpha(X) \rangle$ is a field which is a degree $p$ extension of $F$. Since finite fields of the same order are unique up to isomorphism, we may say that this is "the" degree $p$ extension of $F$. As before, both roots of the equation lie in the extension, which is thus a splitting field for the equation and hence a Galois extension: in this case the roots are of the form $\beta,\,\beta+1, \ldots,\beta+(p-1)$. If $F$ is a function field, these polynomials define Artin–Schreier curves, which in turn give rise to Artin–Schreier codes (cf. Artin–Schreier code). References [La] S. Lang, "Algebra" , Addison-Wesley (1974) Comment This is also a name for the theorem that a field is formally real (can be ordered) if and only if $-1$ is not a sum of squares. References • J.W. Milnor, D. Husemöller, Symmetric bilinear forms, Ergebnisse der Mathematik und ihrer Grenzgebiete 73, Springer-Verlag (1973) p.60 ISBN 0-387-06009-X Zbl 0292.10016 How to Cite This Entry: Artin-Schreier theorem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Artin-Schreier_theorem&oldid=39812 This article was adapted from an original article by M. Hazewinkel (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article	{"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": 0.9692834615707397, "perplexity": 212.53200420632706}, "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-2023-06/segments/1674764499953.47/warc/CC-MAIN-20230201211725-20230202001725-00018.warc.gz"}
https://www.deepdyve.com/lp/springer_journal/generalized-derivations-on-some-convolution-algebras-3vUgiLPVUx	# Generalized derivations on some convolution algebras Generalized derivations on some convolution algebras Let G be a locally compact abelian group, $$\omega$$ ω be a weighted function on $${\mathbb {R}}^+$$ R + , and let $$\mathfrak {D}$$ D be the Banach algebra $$L_0^\infty (G)^$$ L 0 ∞ ( G ) ∗ or $$L_0^\infty (\omega )^$$ L 0 ∞ ( ω ) ∗ . In this paper, we investigate generalized derivations on the noncommutative Banach algebra $$\mathfrak {D}$$ D . We characterize $$\textsf {k}$$ k -(skew) centralizing generalized derivations of $$\mathfrak {D}$$ D and show that the zero map is the only $$\textsf {k}$$ k -skew commuting generalized derivation of $$\mathfrak {D}$$ D . We also investigate the Singer–Wermer conjecture for generalized derivations of $$\mathfrak {D}$$ D and prove that the Singer–Wermer conjecture holds for a generalized derivation of $$\mathfrak {D}$$ D if and only if it is a derivation; or equivalently, it is nilpotent. Finally, we investigate the orthogonality of generalized derivations of $$L_0^\infty (\omega )^$$ L 0 ∞ ( ω ) ∗ and give several necessary and sufficient conditions for orthogonal generalized derivations of $$L_0^\infty (\omega )^$$ L 0 ∞ ( ω ) ∗ . http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png aequationes mathematicae Springer Journals # Generalized derivations on some convolution algebras , Volume 92 (2) – Jan 15, 2018 19 pages /lp/springer_journal/generalized-derivations-on-some-convolution-algebras-3vUgiLPVUx Publisher Springer International Publishing Copyright © 2018 by Springer International Publishing AG, part of Springer Nature Subject Mathematics; Analysis; Combinatorics ISSN 0001-9054 eISSN 1420-8903 D.O.I. 10.1007/s00010-017-0531-6 Publisher site See Article on Publisher Site ### Abstract Let G be a locally compact abelian group, $$\omega$$ ω be a weighted function on $${\mathbb {R}}^+$$ R + , and let $$\mathfrak {D}$$ D be the Banach algebra $$L_0^\infty (G)^$$ L 0 ∞ ( G ) ∗ or $$L_0^\infty (\omega )^$$ L 0 ∞ ( ω ) ∗ . In this paper, we investigate generalized derivations on the noncommutative Banach algebra $$\mathfrak {D}$$ D . We characterize $$\textsf {k}$$ k -(skew) centralizing generalized derivations of $$\mathfrak {D}$$ D and show that the zero map is the only $$\textsf {k}$$ k -skew commuting generalized derivation of $$\mathfrak {D}$$ D . We also investigate the Singer–Wermer conjecture for generalized derivations of $$\mathfrak {D}$$ D and prove that the Singer–Wermer conjecture holds for a generalized derivation of $$\mathfrak {D}$$ D if and only if it is a derivation; or equivalently, it is nilpotent. Finally, we investigate the orthogonality of generalized derivations of $$L_0^\infty (\omega )^$$ L 0 ∞ ( ω ) ∗ and give several necessary and sufficient conditions for orthogonal generalized derivations of $$L_0^\infty (\omega )^$$ L 0 ∞ ( ω ) ∗ . ### Journal aequationes mathematicaeSpringer Journals Published: Jan 15, 2018 ## You’re reading a free preview. Subscribe to read the entire article. ### DeepDyve is your personal research library It’s your single place to instantly discover and read the research that matters to you. over 18 million articles from more than 15,000 peer-reviewed journals. All for just $49/month ### Explore the DeepDyve Library ### Search Query the DeepDyve database, plus search all of PubMed and Google Scholar seamlessly ### Organize Save any article or search result from DeepDyve, PubMed, and Google Scholar... all in one place. ### Access Get unlimited, online access to over 18 million full-text articles from more than 15,000 scientific journals. ### Your journals are on DeepDyve Read from thousands of the leading scholarly journals from SpringerNature, Elsevier, Wiley-Blackwell, Oxford University Press and more. All the latest content is available, no embargo periods. DeepDyve ### Freelancer DeepDyve ### Pro Price FREE$49/month \$360/year Save searches from PubMed Create lists to Export lists, citations	{"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": 0.9792493581771851, "perplexity": 1625.090092207823}, "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-2018-39/segments/1537267158279.11/warc/CC-MAIN-20180922084059-20180922104459-00198.warc.gz"}
http://math.stackexchange.com/questions/108023/metric-on-the-unit-cube	# Metric on the unit cube Let $X$ be $\mathbb{R}^3$ with the sup norm $\\|\cdot\\|_{\infty}$. Let $Y=\{x\in X: \\|x\\|_{\infty}=1\}$. For $x,y\in Y,y\neq -x$ define $d(x,y)$ to be the arc length of the path $$Y\cap \{\lambda x+\mu y: \lambda\ge 0, \mu\ge 0\}.$$ Define $d(x,-x)=4$. Note that the arc lengths are computed using the sup norm. My question is: Does $d$ define a metric on $Y$? A related question was answered in Shortest path on unit sphere under $\\|\cdot\\|_\infty$ - In fact the answer can be found in the given link. Let $A=(1,3/4,1/4), B=(3/4,1,3/4), C=(1,1,1/2)$ and $M=(1,1,4/7)$. Then $d(A,C)=1/4, d(C,B)=1/4$ and $d(A,B)=d(A,M)+d(M,B)=4/7$. So $d(A,C)+d(C,B)<d(A,B)$, proving that $d$ is not a metric. – TCL Feb 11 '12 at 5:00 TCL: Why not post an answer? – Jonas Meyer Feb 11 '12 at 5:58 @TCL I'm not sure what the relevance is more than a year later, but it's better to ask a new question instead of editing in this case -- your edit made this question wholly different from its original. – Lord_Farin Nov 11 '13 at 12:54 Let: $$A=(1,3/4,1/4),\ B=(3/4,1,3/4),\ C=(1,1,1/2)\text{ and }M=(1,1,4/7).$$ Then: $$d(A,C)=1/4,\ d(C,B)=1/4\text{ and }d(A,B)=d(A,M)+d(M,B)=4/7.$$ So $d(A,C)+d(C,B)<d(A,B)$, proving that $d$ is not a metric.	{"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": 0.992591142654419, "perplexity": 151.2188524911933}, "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-2015-06/segments/1422115868812.73/warc/CC-MAIN-20150124161108-00141-ip-10-180-212-252.ec2.internal.warc.gz"}
http://physics.stackexchange.com/questions/80341/blackbody-radiation	Looking at the radiation from the sun (T=5800K) I got a little surprise which I do not understand. I first calculated the energy density u and also the number of photons per unit volume ng. From this one gets the average energy per photon: $$E_g = \frac{u}{n_g}$$ Converted to a wavelength one gets: $$\tag{1} \lambda = 920\:\mathrm{nm}$$ The average frequency using $$\nu_{\text{av}} = \frac{\int \nu \cdot u_{\nu} \cdot \mathrm{d}\nu}{\int u_{\nu} \cdot \mathrm{d}\nu}$$ Converted to a wave length I get $\lambda_{\text{av}} = 650\:\mathrm{nm}$. Next I calculate the average wavelength using $$\lambda_{\text{av}} = \frac{\int \lambda\cdot u_\lambda \cdot \mathrm{d}\lambda }{ \int u_\lambda \cdot \mathrm{d}\lambda}$$ One gets $\lambda_{\text{av}2} = 920\:\mathrm{nm}$ Questions: I was expecting some value which would be of course different from the value of the frequency distribution. This is actually the case. The same is of course true for the max of the distributions. However to my surprise this is also equal to the the value of eq. (1). Why? What is the physical meaning of the average frequency $\nu_{\text{av}}$? If there is no particular meaning of $\nu_{\text{av}}$ it looks like the wavelength distribution has more physical meaning which I think is nonesense. - I $\TeX$ed your formulas (please learn to do that yourself), but I couldn't make much sense of your calculations. Check them back, perhaps I misunderstood a few things... – leftaroundabout Oct 10 '13 at 23:48 Frank, what @leftaroundabout has done is write the math in the input language of the MathJax rendering engine which is a pretty good implementation of $\LaTeX$'s mathmode and which we have running on the site. You should examine the raw text to understand what he did (just click the edit button). – dmckee Oct 10 '13 at 23:49 You are working your statistics out from the Plank law, right? Off the top of my head, I should expect the mean energy to be $h$ times the mean frequency, since the expectation is a linear operator. – WetSavannaAnimal aka Rod Vance Oct 10 '13 at 23:57 What formula are you using to calculate the number of photons per unit volume? Don't confuse this with the density of states. – WetSavannaAnimal aka Rod Vance Oct 11 '13 at 1:43 Please revise your first formula. It does not make any sense. – mcodesmart Oct 11 '13 at 6:51	{"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": 0.8958512544631958, "perplexity": 433.0320739377664}, "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/1410657132646.40/warc/CC-MAIN-20140914011212-00227-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"}
http://aas.org/archives/BAAS/v31n5/aas195/607.htm	AAS 195th Meeting, January 2000 Session 52. Absorption in the Intergalactic Medium Display, Thursday, January 13, 2000, 9:20am-6:30pm, Grand Hall ## [52.08] High Resolution Echelle Spectroscopy of Low Redshift Intervening O VI Absorbers with the Space Telescope Imaging Spectrograph T.M. Tripp, D.V. Bowen, E.B. Jenkins (Princeton Obs.), B.D. Savage (U. Wisconsin-Madison) We present high resolution FUV echelle spectroscopy of several low z intervening O VI absorbers (z < 0.3) in the spectra of H1821+643 and PG0953+415. The data were obtained with the Space Telescope Imaging Spectrograph at a resolution of ~45,000 (7 km/s FWHM). We also present selected new measurements of galaxy redshifts in the 10' field centered on H1821+643. The observations provide several clues about the nature of these absorbers: (1) In the case of the strong O VI system at z = 0.2250 in the spectrum of H1821+643, we detect multicomponent Si II and Si III absorption as well as O VI and several Lyman series lines of H I. Multiple components are evident in the O VI profiles, but the components have different velocities than the Si III and Si II lines. Furthermore, the Si II and Si III lines are quite narrow, and the O VI lines are broader and spread over a larger velocity range. This evidence strongly indicates that this is a multiphase absorber. (2) We also detect `high velocity' O VI in the z = 0.2250 system. High velocity H I is also seen in the Ly\alpha profile, but substantially offset in velocity from the O VI. This high velocity O VI may be analogous to the highly ionized high velocity clouds seen near the Milky Way. (3) We also present systems at other redshifts including very weak O VI absorption lines accompanied by weak and narrow H I absorption. (4) In all cases, several galaxies are close to the sight lines at the redshift of the O VI systems. We examine whether the O VI absorption can be attributed to the ISM of a single galaxy or the intragroup medium. [Previous] \| [Session 52] \| [Next]	{"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": 0.8623226881027222, "perplexity": 4837.032219258477}, "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-2014-52/segments/1418802772265.125/warc/CC-MAIN-20141217075252-00102-ip-10-231-17-201.ec2.internal.warc.gz"}
http://www.nanoscalereslett.com/content/8/1/242	Nano Express # One-qubit quantum gates in a circular graphene quantum dot: genetic algorithm approach Gibrán Amparán12, Fernando Rojas12* and Antonio Pérez-Garrido1 Author Affiliations 1 Departamento de Física Aplicada, Antiguo Hospital de la Marina, Campo Muralla del Mar, UPCT, Cartagena, 30202, Murcia, Spain 2 Departamento de Física Teórica, Centro de Nanociencias y Nanotecnologías, Universidad Nacional Autónoma de México, UNAM, Apdo, Postal 14, Ensenada, Baja California 22830, México For all author emails, please log on. Nanoscale Research Letters 2013, 8:242 doi:10.1186/1556-276X-8-242 Received: 15 November 2012 Accepted: 18 April 2013 Published: 16 May 2013 This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. ### Abstract The aim of this work was to design and control, using genetic algorithm (GA) for parameter optimization, one-charge-qubit quantum logic gates σx, σy, and σz, using two bound states as a qubit space, of circular graphene quantum dots in a homogeneous magnetic field. The method employed for the proposed gate implementation is through the quantum dynamic control of the qubit subspace with an oscillating electric field and an onsite (inside the quantum dot) gate voltage pulse with amplitude and time width modulation which introduce relative phases and transitions between states. Our results show that we can obtain values of fitness or gate fidelity close to 1, avoiding the leakage probability to higher states. The system evolution, for the gate operation, is presented with the dynamics of the probability density, as well as a visualization of the current of the pseudospin, characteristic of a graphene structure. Therefore, we conclude that is possible to use the states of the graphene quantum dot (selecting the dot size and magnetic field) to design and control the qubit subspace, with these two time-dependent interactions, to obtain the optimal parameters for a good gate fidelity using GA. ### Background Quantum computing (QC) has played an important role as a modern research topic because the quantum mechanics phenomena (entanglement, superposition, projective measurement) can be used for different purposes such as data storage, communications and data processing, increasing security, and processing power. The design of quantum logic gates (or quantum gates) is the basis for QC circuit model. There have been proposals and implementations of the qubit and quantum gates for several physical systems [1], where the qubit is represented as charge states using trapped ions, nuclear magnetic resonance (NMR) using the magnetic spin of ions, with light polarization as qubit or spin in solid-state nanostructures. Spin qubits in graphene nanoribbons have been also proposed. Some obstacles are present, in every implementation, related to the properties of the physical system like short coherence time in spin qubits and charge qubits or null interaction between photons, which is necessary to design two-qubit quantum logic gates. Most of the quantum algorithms have been implemented in NMR as Shor's algorithm [2] for the factorization of numbers. Any quantum algorithm can be done by the combination of one-qubit universal quantum logic gates like arbitrary rotations over Bloch sphere axes (X(ϕ), Y(ϕ), and Z(ϕ)) or the Pauli gates () and two-qubit quantum gates like controlled NOT which is a genuine two-qubit quantum gate. The implementation of gates using graphene to make quantum dots seems appropriate because this material is naturally low dimensional, and the isotope 12C (most common in nature) has no nuclear spin because the sum of spin particles in the nucleus is neutralized. This property can be helpful to increase time coherence as seen by the proposal of graphene nanoribbons (GPNs) [3] and Z-shape GPN for spin qubit [4]. In this work, we propose the implementation of three one-qubit quantum gates using the states of a circular graphene quantum dot (QD) to define the qubit. The control is made with pulse width modulation and coherent light which induce an oscillating electric field. The time-dependent Schrodinger equation is solved to describe the amplitude of being in a QD state Cj(t). Two bound states are chosen to be the computational basis \|0〉 ≡ \|ψ1/2 \|1〉 ≡ \|ψ− 1/2 〉 with j = 1/2 and j = −1/2, respectively, which form the qubit subspace. In this work, we studied the general n-state problem with all dipolar and onsite interactions included so that the objective is to optimize the control parameters of the time-dependent physical interaction in order to minimize the probability of leaking out of the qubit subspace and achieve the desired one-qubit gates successfully. The control parameters are obtained using a genetic algorithm which finds efficiently the optimal values for the gate implementation where the genes are: the magnitude (ϵ0) and direction (ρ) of electric field, magnitude of gate voltage (Vg0), and pulse width (τv). The fitness is defined as the gate fidelity at the measured time to obtain the best fitness, which means the best control parameters were found to produce the desired quantum gate. We present our findings and the evolution of the charge density and pseudospin current in the quantum dot under the gate effect. ### Methods #### Graphene circular quantum dot The nanostructure we used consists of a graphene layer grown over a semiconductor material which introduces a constant mass term Δ [5]. This allows us to make a confinement (made with a circular electric potential of constant radio (R)) where a homogeneous magnetic field (B) is applied perpendicular to the graphene plane in order to break the degeneracy between Dirac's points K and K’, distinguished by the term τ = +1 and τ = −1, respectively. The Dirac Hamiltonian with magnetic vector field in polar coordinates is given by [6]: (1) where v is the Fermi velocity (106 m/s), b = eB/2, and j which is a half-odd integer is the quantum number for total angular momentum operator Jz. We need to solve . Eigenfunctions have a pseudospinor form: (2) where χ are hypergeometric functions M (a,b,z) and U (a,b,z) inside or outside of radius R (see [6] for details) (Figure 1). Figure 1. Radial probability density (lowest states) and qubit subspace density and pseudospin current. (a) Radial probability density plot for the four lowest energy states inside the graphene quantum dot with R = 25 nm and under a homogeneous magnetic field of magnitude B = 3.043 T. The selected computational basis (qubit subspace) is inside the red box. Qubit subspace spatial probability density plot and vector field of the pseudospin current in (b) \|0〉 = \|ψ1/2 and (c) \|1〉 = \|ψ− 1/2 , respectively. Due to the constant mass term and broken degeneracy, we obtain two independent Hilbert spaces. Therefore, we can choose the space K for the definition of the computational basis of the qubit to implement the quantum gates and to make the dynamic control following a genetic algorithm procedure. The wave function in graphene can be interpreted as a pseudospinor of the sublattice of atom type A or B. In order to visualize the physics evolution due to the gate operation, we calculate the pseudospin current as the expectation values for Pauli matrices . The selected states that we choose to form the computational basis for the qubit are the energies (Ej): E1/2 = .2492 eV and E−1/2 = .2551 eV (and the corresponding radial probability distributions is shown in Figure 2a). The energy gap is E01 = E−1/2 − E1/2 = 5.838 meV. To achieve transitions between these two states with coherent light, the wavelength required has to be , which is in the range of far-infrared lasers. Also, in controlling the magnetic field B, it is possible to modify this energy gap. We present as a reference point the plot for the density probability and the pseudospin current for the two-dimensional computational basis \|0〉 = \|ψ1/2 (Figure 2b) and \|1〉 = \|ψ− 1/2 (Figure 2c), where a change of direction on pseudospin current and the creation of a hole (null probability near r = 0) is induced when one goes from qubit 0 to1. Figure 2. Diagram of genetic algorithm. Initial population of chromosomes randomly created; the fitness is determined for each chromosome; parents are selected according to their fitness and reproduced by pairs, and the product is mutated until the next generation is completed to perform the same process until stop criterion is satisfied. #### Quantum control: time-dependent potentials First of all, we have to calculate the matrix representation of the time-dependent interactions in the QD basis. Then, we have to use the interaction picture to obtain the ordinary differential equation (ODE) for the time-dependent coefficient which is the probability of being in a state of the QD at time t and finally obtaining the optimal parameter for gate operation. #### Electric field: oscillating These transitions can be induced by a laser directed to the QD carrying a wavelength that resonates with the qubit states in order to trigger and control transitions in the qubit subspace. We introduce an electric dipole interaction [7] using a time periodic Hamiltonian with frequency ω: Vlaser(t) = eϵ(t)r, with parameters ϵ(t) = ϵ0 cos ωt, ϵ0 = ϵ0(cos ρ, sin ρ), and r = r(cos φ, sin φ), the term ρ is the direction and ϵ0 is the magnitude of the electric field and are parameters constant in time. To determine the matrix of dipolar transitions on the basis of the QD states, the following overlap integrals must be calculated: (3) where l and j are the state indices. In Equation 3, the radial part defines the magnitude of the matrix component, the angular part defines transition rules, and as a result, we get a non-diagonal matrix; this indicates that transitions are only permitted between neighbor states. The matrix components are complex numbers; ϵ0 directed in direction is a pure imaginary number and directed in is a real number. #### Voltage pulse on site This interaction can be applied as a gate voltage inside the QD. In order to modify the electrostatic potential, we use a square pulse of width τv and magnitude Vg0. The Hamiltonian is (4) (5) The matrix components in Equation 5 are diagonal, so this interaction only modifies the energies on the site. Since the Heaviside function θ depends on r in Equation 4, the matrix components are the probability to be inside the quantum dot which is different for each eigenstate, so this difference can introduce relative phases inside the qubit subspace. #### One-qubit quantum logic gates Therefore, we have to solve the dynamics of QD problem in N-dimensional states involved, where the control has to minimize the probability of leaking to states out of the qubit subspace in order to approximate the dynamic to the ideal state to implement correctly the one-qubit gates. The total Hamiltonian for both quantum dot and time-dependent interactions is , where is the quantum dot part (Equation 1) and Vlaser(t) and Vgate(t) are the time control interactions given by Equations 3 and 4. We expand the time-dependent solution in terms of the QD states (Equation 2) as. Therefore, the equations for the evolution of probability of being in state l at time t, Cl(t), in the interaction picture, are given by: (6) The control problem of how to produce the gates becomes a dynamic optimization one, where we have to find the combination of the interaction parameters that produces the one-qubit gates (Pauli matrices). We solve it using a genetic algorithm [8] which allows us to avoid local maxima and converges in a short time over a multidimensional space (four control parameters in our case). The steps in the GA approach are presented in Figure 2, where the key elements that we require to define four our problem are chromosomes and fitness. In our model, the chromosomes in GA are the array of values {Vg0,τv,ϵ0,ρ}, where Vg0 is the voltage pulse magnitude, τv is the voltage pulse width, ϵ0 is the electric field magnitude, and ρ is the electric field direction. The fitness function, as a measure of the gate fidelity, is a real number from 0 to 1 that we define as fitness(tmed) = \| < Ψobj\|Ψ(tmed) > \|2 × \| < Ψ0\|Ψ(2tmed) > \|2 where \|Ψobj 〉 is the objective or ideal vector state, which is product of the gate operation (Pauli matrix) on the initial state \|Ψ0〉. Then, we evolve the dynamics to the measurement time tmed to obtain \|Ψ(tmed)〉. Determination of gate fidelity results in the probability to be in the objective vector state at tmed. Fitness involves gate fidelity at tmed and probability to be in the initial state at 2 tmed. This gives a number between 0 and 1, indicating how effective is the transformations in taking an initial state to the objective state and back to the initial state in twice of time (the reset phase). The initial population of chromosomes ({Vg0,τv,ϵ0,ρ}) is randomly created, then fitness is determined for each chromosome (which implies to have the time-dependent evolution of Cl(t) to the measurement time); parents are selected according to their fitness and reproduced by pairs, and the product is mutated until the next generation is completed; one performs the same process until a stop criterion is satisfied. ### Results and discussion The control dynamics were done considering N = 6 states, two of them are used as the qubit basis, so that the effect of the interaction stays inside the qubit subspace . The gate operation is completed in a time window that depends on ϵ0, and control parameters are defined to achieve operation inside a determined time window. The possible values of the electric field direction ρ is set from 0 to 2π, pulse width τv domain is set from 0 to time window and the magnitude Vg0 is set from 0 to an arbitrary value. The genetic algorithm procedure is executed for quantum gates σx and σy. The fitness reaches a value close to 1 near to 30 generations for both gates. The optimal parameters found for quantum gate σx are Vg0= .0003685, τv = 4215.95, ϵ0 = .0000924, and ρ = .9931π. For σy are Vg0 = .0355961, τv = 326.926, ϵ0 = .0000735, and ρ = 1.5120π. For the quantum gate σz, genetic algorithm is not needed because for this case, ϵ0 = 0, so Equation 6 is an uncoupled ordinary differential equation (ODE) with specific solution. To achieve this gate transformation in a determined time window, we can calculate Vg0, so that the control values for this quantum gate are Vg0= .1859, τv = 5,000, ϵ0 = 0, and ρ = 0. In Figure 3, we plot the time evolution of the gate fidelity or fitness for the three gates. We observe a good optimal convergence close to 1 at the time of measurement and reaching again the reset phase. To see the state transition and the quantum gate effect in the space, it is convenient to plot the density probability in the quantum dot and the corresponding pseudospin current, where we see how the wave packet has different time trajectory according to the gate transformation. For instance, the direction and time of creation of the characteristic hole (null probability) in the middle of the qubit one, which correspond more or less to an equal superposition of the qubit zero and one (column 2 and row 2 in Figure 4, right). This process has to be different for σy because it introduces an imaginary phase in the evolution which is similar with the change of the arrow directions in the pseudospin current. The same situation arises for σz (result not shown), but in this case, we use as an initial state , which is similar to the plot of column 2 and row 2 in Figure 4 (left) and then to show explicitly the gate effect of introducing the minus in the one state to reach a rotated state similar to plot of column 2 and row 2 in Figure 4 (right). Figure 3. Time evolution of gate fidelity or fitness for the three gates. Plot of gate fidelity σx in the top side, σy in the middle, and σz in the bottom side; gate fidelity (FσI in blue where I is{x,y,z}) is the probability to be in the objective vector state; measurement time is shown in orange. Figure 4. Time evolution of probability density and pseudospin current for the quantum gate σx and σy operation. Time evolution of density and current probability due to the effect of the produced quantum gate σx in the left side and σy in the right side, initial state \|Ψ0〉 = \|0〉 (Figure 1b). ### Conclusions We show that with a proper selection of time-dependent interactions, one is able to control or induce that leakage probability out of the qubit subspace in a graphene QD to be small. We have been able to optimize the control parameters (electric field and gate voltage) with a GA in order to keep the electron inside the qubit subspace and produce successfully the three one-qubit gates. In our results, we appreciate that with the genetic algorithm, one can achieve good fidelity and found that little voltage pulses are required for σx and σy and improve gate fidelity, therefore making our proposal of the graphene QD model for quantum gate implementation viable. Finally, in terms of physical process, the visualization of the effects of quantum gates σx and σy is very useful, and clearly, both achieve the ideal states. The difference between them (Figure 4) is appreciated in the different trajectories made by the wave packet and pseudospin current during evolution due to the introduction of relative phase made by gate σy. ### Competing interests The authors declare that they have no competing interests. ### Authors’ contributions The work presented here was carried out collaboration among all authors. FR and APG defined the research problem. GA carried out the calculations under FR and APG's supervision. All of them discussed the results and wrote the manuscript. All authors read and approved the final manuscript. ### Acknowledgments The authors would like to thank DGAPA and project PAPPIT IN112012 for financial support and sabbatical scholarship for FR and to Conacyt for the scholarship granted to GA. ### References 1. Ladd TD, Jelezko F, Laflamme R, Nakamura Y, Monroe C, O’Brien JL: Quantum computers (review). Nature 2010, 464:45-53. PubMed Abstract \| Publisher Full Text 2. Vandersypen LM, Steffen M, Breyta G, Yannoni CS, Sherwood MH, Chuang IL: Experimental realization of Shor's quantum factoring algorithm using nuclear magnetic resonance. Nature 2001, 414:883-887. PubMed Abstract \| Publisher Full Text 3. Trauzettel B, Bulaev DV, Loss D, Burkard G: Spin qubits in graphene quantum dots. Nature Physics 2007, 3:192-196. Publisher Full Text 4. Guo G-P, Lin Z-R, Tao T, Cao G, Li X-P, Guo G-C: Quantum computation with graphene nanoribbon. New Journal of Physics 2009, 11:123005. Publisher Full Text 5. Zhou SY, Gweon G-H: Substrate-induced band gap opening in epitaxial graphene. Nature Materials 2007, 6:770-775. PubMed Abstract \| Publisher Full Text 6. Recher P, Nilsson J, Burkard G, Trauzettel B: Bound states and magnetic field induced valley splitting in gate-tunable graphene quantum dots. Physical Review B 2009, 79:085407. 7. Fox M: Optical Properties of Solids. In Quantum Theory of radiative absorption and emission Appendix B. Oxford: Oxford University Press; 2001:266-270. 8. Chong EKP, Zak SH: An introduction to optimization. In Chapter 14: Genetic Algorithms. 2nd edition. Weinheim: Editorial WILEY; 2001.	{"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": 0.927214503288269, "perplexity": 1021.1792272954405}, "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-2015-35/segments/1440644065375.30/warc/CC-MAIN-20150827025425-00054-ip-10-171-96-226.ec2.internal.warc.gz"}
https://discourse.julialang.org/t/specify-jumps-in-a-heston-like-model/42422	# Specify jumps in a Heston-like model? I am working with a model similar to the Heston model, and would like to add jumps. I would like to have the jump rate which depend on the ordinary volatility as `rate1(u,p,t) = λ0 .+ λ1.exp(u[2]/2.0)` I understand that this should be specified as a variable rate jump. However, I can’t seem to get the code to run if I do that. It does run if I specify it as a constant rate jump, but I’m doubting if the solution is correct when I do that. The code for the jump part follows. Any pointers would be welcome! (this post is substantially similar to and earlier post, but with the diffeq tag added, sorry for the noise) ``````## jump in price rate1(u,p,t) = λ0 .+ λ1.exp(u[2]/2.0) # volatility dependent jump rate # jump is normal with st. dev. equal to λ1 times current st. dev. affect1!(integrator) = (integrator.u[1] = integrator.u[1].+randn(size(integrator.u[1])).λ2.exp(integrator.u[2]./2.0)) # this works: jump1 = ConstantRateJump(rate1,affect1!) # this does not #jump1 = VariableRateJump(rate1,affect1!) jump_prob = JumpProblem(prob,Direct(), jump1) `````` You may also want to look at https://github.com/rveltz/PiecewiseDeterministicMarkovProcesses.jl for simulating this, using a trivial flow. At least, you will have a way to compare to another numerical solution. 1 Like That doesn’t support SDEs? @mcreel I was planning to look at your post today. Sorry about that! 1 Like He did not give the flow in between jumps, I assumed it is constant It’s from his other post, and the solver is `SRIW1` Here’s the whole file, if it would help. As is, it runs and gives output that looks as expected, I’m just not sure if the jumps are occurring at the proper rate. I also don’t know how to determine when a jump has occurred. ``````using DifferentialEquations, Plots function MyProblem(μ0,μ1,κ,α,σ,ρ,u0,tspan) f = function (du,u,p,t) du[1] = μ0 + μ1(u[2]-α)/σ # drift in log prices du[2] = κ(α-u[2]) # mean reversion in shocks end g = function (du,u,p,t) du[1] = exp(u[2]/2.0) du[2] = σ end Γ = [1.0 ρ;ρ 1.0] # Covariance Matrix noise = CorrelatedWienerProcess!(Γ,tspan[1],zeros(2),zeros(2)) sde_f = SDEFunction{true}(f,g) SDEProblem(sde_f,g,u0,tspan,noise=noise) end function main() # assume trading period is 1/3 of day (8 hours) # but that latent price evolves continuously # observed return is daily difference of log price at # closing time TradingDays = 1000 # total days in sample Days = Int(TradingDays7/5) # calendar days MinPerDay = 1440 # minutes per day MinPerTic = 5 # minutes between tics, lower for better accuracy tics = Int(MinPerDay/MinPerTic) # number of tics in day dt = 1/tics # divisions per day closing = Int(floor(tics/3)) # closing tic: closing happens after 1/3 of day # parameters μ0 = 0.0 μ1 = 0.0 κ = 0.1 α = 0.15 σ = 0.15 ρ = -0.7 λ0 = 1.0 # constant in jump rate λ1 = 1.0 # slope in jump rate λ2 = 3.0 # size of jumps σme = 0.05 # standard dev of measurement error in returns u0 = [0;α] prob = MyProblem(μ0, μ1, κ, α, σ, ρ, u0, (0.0,Days)) ## jump in price rate(u,p,t) = λ0 .+ λ1.exp(u[2]/2.0) # volatility dependent jump rate # jump is normal with st. dev. equal to λ1 times current st. dev. affect1!(integrator) = (integrator.u[1] = integrator.u[1].+randn(size(integrator.u[1])).λ2.exp(integrator.u[2]./2.0)) # this works: jump = ConstantRateJump(rate,affect1!) # this does not #jump = VariableRateJump(rate,affect1!) jump_prob = JumpProblem(prob,Direct(), jump) # get log price at end of trading days global j = 0 # counter for day of week global k = 0 # counter for trading days for i = 0:(Days) # set day of week, and record if it's a trading day j +=1 if j<6 z[k]=(sol.u)[itics+closing][1] end if j==7 # restart the week if Sunday j = 0 end end z[2:end]-z[1:end-1] + σmerandn(TradingDays) # returns are diff of log price end z = main() plot(z) `````` This one is going to be quite tough. The first thing to work out is that VariableRateJumps grow the size of the system, so the problem was that you needed to define a 3 noise process for this to work: ``````using StochasticDiffEq, DiffEqJump, DiffEqNoiseProcess, Plots function MyProblem(μ0,μ1,κ,α,σ,ρ,u0,tspan) f = function (du,u,p,t) du[1] = μ0 + μ1(u[2]-α)/σ # drift in prices du[2] = κ(α-u[2]) # mean reversion in shocks end g = function (du,u,p,t) du[1] = exp(u[2]/2.0) du[2] = σ end Γ = [1 ρ 0;ρ 1 0;0 0 0] # Covariance Matrix noise = CorrelatedWienerProcess!(Γ,tspan[1],zeros(3),zeros(3)) sde_f = SDEFunction{true}(f,g) SDEProblem(sde_f,g,u0,tspan,noise=noise) end μ0 = 0.0 μ1 = 0.0 κ = 0.05 α = 0.2 σ = 0.2 ρ = -0.7 λ0 = 2.0 λ1 = 3.0 u0 = [0;α] dt = 0.01 prob = MyProblem(μ0, μ1, κ, α, σ, ρ, u0, (0.0,1000.0)) ## jump in price #rate1(u,p,t) = λ0 # constant jump rate rate1(u,p,t) = λ0.exp(u[2]) # volatility dependent jump rate # jump is normal with st. dev. equal to λ1 times current st. dev. affect1!(integrator) = (integrator.u[1] = integrator.u[1].+randn(size(integrator.u[1])).λ1.*exp(integrator.u[2]./2.0)) # this works: #jump1 = ConstantRateJump(rate1,affect1!) # this does not jump1 = VariableRateJump(rate1,affect1!) jump_prob = JumpProblem(prob,Direct(), jump1) `````` However, this now errors at the jump stage because continuous callbacks require pulling back the noise process, so it requires the definition of a bridging distribution. If you’re willing to work out the bridging distribution for the correlated noise process then we’re done. Basically, we need the distribution for if you know W0 and Wh, what’s the distribution of Wt in the middle. Thanks for the explanation. I will fall back to using a constant rate jump, which will be good enough for my purpose, which is simply to illustrate an econometric estimator. I do have one remaining question, though, how can one see the realization of jumps is a simulation, their timing and size? Thanks again. If you look for non-unique time points you can see the jump at those spots. Thanks, that does the trick. Here’s a plot of returns with the jumps highlighted in red: In this example, a jump is a normally distributed r.v. with mean zero, so not all jumps are outliers. If anyone would like to see how to find jumps, here’s the code: check jumps	{"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": 0.8844525814056396, "perplexity": 4652.930442880202}, "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/1596439740423.36/warc/CC-MAIN-20200815005453-20200815035453-00396.warc.gz"}
https://scicomp.stackexchange.com/questions/2294/large-scale-generalized-eigenvalue-problem-with-low-rank-lhs-matrix?rq=1	# Large-scale generalized eigenvalue problem with low rank LHS matrix Assume that we have generalized eigenvalue problem: $B^HB\textbf{x} = \lambda A\textbf{x}$ where $A$ is an nxn Hermitian sparse matrix (n is very large, so we do not have $A^{-1}$ but can solve using iterative methods) and full-rank, and $B$ is a 2xn matrix such that $B^HB$ is also nxn but only rank 2. Thus, we know that this problem can only have 2 non-zero eigenvalues. Is there any simple way for finding the two eigenpairs corresponding to nonzero eigenvalues by taking advantage of the very low rank of $B^HB$? Assume that we have the two eigenvectors of $B$. If I am only interested in the eigenvector corresponding to the largest eigenvalue, is there a faster way of finding it than using simple power iteration on the transformed standard eigenvalue problem: $A^{-1}B^HB\textbf{x} = \lambda\textbf{x}$? Thanks! This answer is essentially a fix of the approach suggested by @WolfgangBangerth, as there is not enough space in the comments. Starting from $$B^H B x = \lambda A x,$$ if we are interested in eigenpairs corresponding to nonzero eigenvalues, then we must have that $B^H B x$ lies in the range of $A$, and $Ax$ lies in the range of $B^H B$, which is to say that, since $A$ is invertible, $$B^H B x \in \mathrm{Range}(A) = \mathbf{C}^n,$$ and $$Ax \in \mathrm{Range}(B^H B) = \mathrm{span}(B^H).$$ Now, the first constraint is trivially satisfied, but we must ensure that $Ax \in \mathrm{span}(B^H)$, which is equivalent to the constraint $$x \in \mathrm{span}(A^{-1} B^H).$$ Then if the columns of a unitary matrix $Q$ span the columns of $A^{-1}B^{H}$, we have that $$x = Q Q^H x$$ for any eigenvector corresponding to a nonzero eigenvalue. We are now ready to use the mechanism from Wolfgang's approach: 1. Compute $W := A^{-1} B^H$ through two (preconditioned) Krylov solves 2. Compute $[Q,R]=\mathrm{qr}(W)$ 3. Form $K := (B Q)^H (B Q)$ and $M := Q^H (A Q)$ 4. Solve the $2 \times 2$ eigenvalue problem $K U = M U \Lambda$ 5. Form the interesting global eigenvectors, $Z := Q U$. • Thanks, Jack! Are you sure you need to do QR? I don't think the two vectors comprising the W matrix need to necessarily be orthonormal? (A quick test shows that it works even if you solve $W^HB^HBW\textbf{y} = \lambda W^HAW\textbf{y}$ ) – Costis May 22 '12 at 2:08 • The QR decomposition for an $m \times n$ matrix, $m \ge n$, is $O(mn^2)$. In this case, $n=2$, so the cost is linear and should be dominated by the Krylov solves. I am skeptical of how this would work without a QR decomposition. – Jack Poulson May 22 '12 at 2:26 • $W=QR$, so substituting: $R^HQ^HB^BQR\textbf{y}=\lambda R^HQ^HAQR\textbf{y}$. Multiply both sides by $R^{-H}$ and substitute $\textbf{x}=R\textbf{y}$. You end up with $Q^HB^BBQ\textbf{x}=\lambda Q^HAQ\textbf{x}$ which has the same eigenvalues as if you just used W instead of Q. – Costis May 22 '12 at 2:34 • I can get the eigenvector by just doing: $\textbf{w}=W\textbf{y}$ which implicitly multiplies by R since $W=QR$. Just tried a quick test case and it seems to work, although as you said I think it would be trivial do QR as compared to doing the Krylov solves. – Costis May 22 '12 at 2:45 • Ah, good point! I would still rather work with $Q$ though, as the cost of computing it is insignificant, and it will be more numerically stable. – Jack Poulson May 22 '12 at 2:48 If $B$ is $2\times n$, then the only two non-trivial eigenvectors (i.e. the eigenvectors corresponding to the two non-zero eigenvalues) can be written as linear combinations of the vectors that form the two rows of $B$. Let's call these two vectors $b_1, b_2$ so that $B=\left[\begin{matrix}b_1^T\\b_2^T\end{matrix}\right]$. Now, let $P \in {\mathbb R}^{2\times n}$ be the projector from ${\mathbb R}^n$ onto the two-dimensional space spanned by $b_1,b_2$. Since we are only interested in vectors in this space, we know that the two non-trivial eigenvectors must satisfy $x = P^TPx$. The eigenvalue problem can then be written as $$B^H B P^T P x = \lambda A P^T P x.$$ Even though this linear system has $n$ rows, it is really only a two-dimensional problem since we can only determine only two components of $x$. The remainder of the linear system is over-determined, but we can select the two independent equations by projecting onto the non-trivial subspace: $$P B^H B P^T P x = \lambda P A P^T P x.$$ In other words, you only have to solve the $2 \times 2$ eigenvalue problem $$P B^H B P^T y = \lambda (P A P^T) y.$$ This is easy to solve since the matrices involved are only $2\times 2$ and the matrix on the right can easily be computed using just two matrix-vector and two vector-vector products. • I don't think your assumption that $x=P^T Px$ is valid for non-trivial $A$. – Jack Poulson May 21 '12 at 20:14 • I think that $P$ will need to be modified to span a space including the columns of $B^H$ and $A^{-1} B^H$, which is at most rank 4, and only requires two solves with $A$ to set up. – Jack Poulson May 21 '12 at 21:37 • Hmmm.. I think you only need the columns of $A^{-1}B^H$ actually, so if you take $P=A^{-1}B^H$ Wolfgang's approach will work. – Costis May 22 '12 at 1:14 • @Costis: I think you are right. I have a nice explanation of why which I will post as an answer, as there is not enough space here. – Jack Poulson May 22 '12 at 1:31 • I feel like the nomenclature is slightly unclear. $P$ cannot be a projector, because $P$ is neither square, nor idempotent. For the sake of clarity, the relevant (orthogonal) projector appears to be $P^{T}P$. Your explanation also seems to implicitly rely on knowing that the eigenvectors of $B^{H}B$ form an orthonormal basis, which is why an orthogonal projector is appropriate. – Geoff Oxberry May 22 '12 at 2:05	{"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": 0.9532991647720337, "perplexity": 193.05996823646095}, "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-2021-21/segments/1620243991514.63/warc/CC-MAIN-20210518191530-20210518221530-00569.warc.gz"}
https://physics.stackexchange.com/questions/108394/r-symmetry-commutator	# R-symmetry commutator I've seen the claim made several placed; Terning's "Modern Supersymmetry" p. 5 on N=1 SUSY algebra states it as well as anyone: The SUSY algebra is invariant under a multiplication of $Q_\alpha$ by a phase, so in general there is one linear combination of $U(1)$ charges, called the $R$-charge, that does not commute with $Q$ and $Q^\dagger$: $[Q_\alpha,R] = Q_\alpha, \;\;\;[Q^\dagger_\dot{\alpha},R]=-Q^\dagger_\dot{\alpha}$ The first statement is straightforward to see. But (1) Why is there is one linear combination of charges that does not commute? (2) How do we arrive at these commutators? (I imagine that the generators can be rescaled to give the coefficient $\pm1$, but I would like a clearer explanation.) I spoke to a peer who said that the commutation relations could be found in a very general, mathematically heavy treatment of the most general possible SUSY algebra. Is there some easier way to understand? The point is that the SUSY algebra, \begin{align} & \left\{ Q _\alpha ,Q _\beta \right\} = \left\{ \bar{Q} _{\dot{\alpha}} , \bar{Q} _{\dot{\beta}} \right\} = 0 \\ & \left\{ Q _\alpha , \bar{Q} _{\dot{\beta}} \right\} = 2 \sigma ^\mu _{ \alpha \dot{\beta} }P ^\mu \end{align} is invariant under multiplication of $Q _\alpha$ by a phase, \begin{align} Q _\alpha & \rightarrow e ^{ - i \phi } Q _\alpha \\ \bar{Q} _{\dot{\alpha}} & \rightarrow e ^{ i \phi } \bar{Q} _{\dot{\alpha}} \end{align} This means that you can have a SUSY invariant theory but still have an additional symmetry which differentiates between bosons and fermions (since it doesn't need to commute with $Q _\alpha$). To see this explicitly consider the effect of an $R$ symmetry on $Q _\alpha$: \begin{align} e ^{ -i R \phi } Q _\alpha e ^{ i R \phi } & = e ^{ - i \phi } Q _\alpha \\ \left( 1 - i R \phi - ... \right) Q _\alpha \left( 1 + i R \phi - ... \right) & = - i \phi Q _\alpha \\ i \left[ Q _\alpha , R \right] \phi = - i \phi Q _\alpha \\ \left[ Q _\alpha , R \right] = - Q _\alpha \end{align} Thus this phase shift symmetry implies that the commutation between the $R$ symmetry generator and $Q _\alpha$ is nontrivial and hence bosons and fermions can have a different R charge. This is what makes R symmetries so special. This discussion is likely more complicated for ${\cal N} > 1$.	{"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.999788224697113, "perplexity": 764.8367787836249}, "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/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00583.warc.gz"}
http://www.ck12.org/geometry/Applications-of-Cosine/lesson/Determine-and-Use-the-Cosine-Ratio/r8/	<meta http-equiv="refresh" content="1; url=/nojavascript/"> Applications of Cosine ( Read ) \| Geometry \| CK-12 Foundation You are viewing an older version of this Concept. Go to the latest version. # Applications of Cosine % Best Score Practice Applications of Cosine Best Score % # Determine and Use the Cosine Ratio Do you know how to use cosines when problem solving? Take a look at this dilemma. A triangle has a hypotenuse of 4.5 inches. Angle A is equal to 40 degrees. Find the length of the adjacent side. To figure this out, you will need to know how to use angle measures and cosines. You will learn how to accomplish this task in this Concept. ### Guidance A trigonometric ratio for a specific angle will remain constant no matter how large or small the triangle is. The idea is that the sides will always be in proportion to each other. So, if you know the measure of an angle (and can therefore identify the value of a trigonometric ratio) and the value of one side, you can use trigonometry to calculate the lengths of other sides. The trick is to use good algebra technique, and make sure that every time you set up a ratio, you are putting the values and variables in the correct places. You can find trigonometric ratios by using your calculator. You understand trigonometric ratios and have had a chance to practice reading specific values out of a table. You can find the ratio for any trigonometric value using your calculator. Take a moment to locate the buttons for sine, cosine, and tangent on the calculator. Keep in mind that usually, sine is abbreviated as sin , cosine is usually abbreviated as cos , and tangent is usually abbreviated as tan. Press the key of the ratio you want to find, and enter the angle in question. If you hit enter, or calculate, the calculator will show you the value of that specific ratio. Let's look at finding the cosine ratio by using a calculator. $\text{cosine} \ 23^{\circ}$ You can find the values for each ratio using your calculator. When dealing with large decimals values, it is usually best to round the numbers to the nearest thousandth. It gives you a reasonably accurate value without being too long of a number to work with. The cosine of $23^{\circ}$ is 0.92050485345244..., or about 0.921. Now as we work with cosines, we will be using given information to find the length of the adjacent side of a right triangle. Remember that the adjacent side is the side next to the angle that we are working with. As you recall, the ratio of cosine is $\frac{adjacent}{hypotenuse}$ . If you know the cosine value of the angle in question, and the length of the hypotenuse, you can find the measure of the adjacent side. Look at the algebraic situation below. $\text{cosine} \angle X&=\frac{adjacent}{hypotenuse}\\\text{cosine} \angle X \times hypotenuse&=\frac{adjacent}{hypotenuse} \times hypotenuse\\\text{cosine} \angle X \times hypotenuse&=adjacent$ If you multiply the cosine of any angle $X$ and the length of the hypotenuse, the result is the length of the adjacent side. Write this statement in your notebook. Be sure to include that it is for cosines. Take a look at this situation. What is the length of side $BC$ in the triangle below? Use the following equation to find the length of the side adjacent to angle $B$ . Notice that to find the length of the adjacent side that you will first need to find the cosine for angle $B$ . Then you can multiply that answer with the length of the hypotenuse. This will give you the measurement of the side next to or adjacent to the angle. $\text{cosine} \angle B \times hypotenuse&=adjacent\\\text{cosine} 14.5 \times 5&=adjacent\\0.968 \times 5 &=adjacent\\4.84 &=adjacent$ The length of side $BC$ is 4.84 units. Use a calculator to find each cosine. You may round to the nearest hundredth. #### Example A Cosine $45^{\circ}$ Solution: $.71$ #### Example B Cosine $62^{\circ}$ Solution: $.47$ #### Example C Cosine $22^{\circ}$ Solution: $.93$ Now let's go back to the dilemma from the beginning of the Concept. To work through this dilemma, we can use the following equation and solve. $\text{cosine} \angle A \times hypotenuse&=adjacent\\\text{cosine} 40 \times 4.5&=adjacent\\.77 \times 4.5 &=adjacent\\3.46 &=adjacent$ The missing length of the adjacent side is 3.46 inches. ### Vocabulary Sine a ratio between the opposite side and the hypotenuse of a given angle. Cosine a ratio between the adjacent side and the hypotenuse of a given angle. Tangent a ratio between the opposite side and the adjacent side of a given angle. Trigonometric Ratio used to find missing side lengths of right triangles when angle measures have been given. ### Guided Practice Here is one for you to try on your own. A triangle has a hypotenuse of 7.5 inches. Angle A is equal to 55 degrees. Find the length of the adjacent side. Solution To do this, we can use the following equation. $\text{cosine} \angle A \times hypotenuse&=adjacent\\\text{cosine} 55 \times 7.5&=adjacent\\.57 \times 7.5 &=adjacent\\4.27 &=adjacent$ The length of the adjacent side is 4.27 inches. ### Practice Directions: Use a calculator to find each cosine. You may round to the nearest hundredth. 1. $\text{Cosine} \ 33^{\circ}$ 2. $\text{Cosine} \ 29^{\circ}$ 3. $\text{Cosine} \ 73^{\circ}$ 4. $\text{Cosine} \ 88^{\circ}$ 5. $\text{Cosine} \ 50^{\circ}$ 6. $\text{Cosine} \ 67^{\circ}$ 7. $\text{Cosine} \ 42^{\circ}$ 8. $\text{Cosine} \ 18^{\circ}$ 9. $\text{Cosine} \ 9^{\circ}$ Directions: Find the length of the adjacent side. 10. A triangle has a hypotenuse of 7 inches. Angle A is equal to 60 degrees. Find the length of the adjacent side. 11. A triangle has a hypotenuse of 12 inches. Angle B is equal to 45 degrees. Find the length of the adjacent side. 12. A triangle has a hypotenuse of 8 inches. Angle A is equal to 35 degrees. Find the length of the adjacent side. 13. A triangle has a hypotenuse of 12 inches. Angle A is equal to 28 degrees. Find the length of the adjacent side. 14. A triangle has a hypotenuse of 6 inches. Angle A is equal to 33 degrees. Find the length of the adjacent side. 15. A triangle has a hypotenuse of 14 inches. Angle A is equal to 72 degrees. Find the length of the adjacent side. 16. A triangle has a hypotenuse of 11 inches. Angle A is equal to 80 degrees. Find the length of the adjacent side.	{"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 27, "texerror": 0, "math_score": 0.9156146049499512, "perplexity": 307.2398334657971}, "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-2014-35/segments/1408500830323.35/warc/CC-MAIN-20140820021350-00091-ip-10-180-136-8.ec2.internal.warc.gz"}
https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_21&diff=prev&oldid=111078	# Difference between revisions of "2014 AMC 8 Problems/Problem 21" ## Problem The -digit numbers and are each multiples of . Which of the following could be the value of ? ## Solution 1 The sum of a number's digits is congruent to the number . must be congruent to 0, since it is divisible by 3. Therefore, is also congruent to 0. , so . As we know, , so , and therefore . We can substitute 2 for , so , and therefore . This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is . ## Solution 2 Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 2 or 5 or 8... and so on. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add two of the selected values, 5 to 15, to get 20. We then see that C = 1, 4 or 7, 10... and so on, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7... and so on. In order to be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 1, 4, and 7. However, in the answer choices, there is no 7 or 4 or anything greater than 7, but there is a 1, so is your answer.	{"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": 0.8014805316925049, "perplexity": 116.44116114019641}, "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/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00701.warc.gz"}
https://www.physicsforums.com/threads/micro-structure-from-equilibrium-diagrams.566747/	# Micro structure from equilibrium diagrams 1. Jan 11, 2012 ### aiat_gamer I have this question: "What micro-structural features of metal alloys can be studied by using equilibrium diagrams and how? Can the equilibrium diagrams be used for some other purposes? What are the limitations?" I think that only the composition of micro-structure can be predicted in a certain temperature in an alloy basically. They don`t actually give the shape of the micro-structure, do they? 2. Jan 11, 2012 ### pukb The phase diagrams does not give the exact microstructure because the rate of cooling/heating in phase diagrams is extremely slow such that an equilibrium state is attained at any point. The practical situation involves cooling rates that are much higher than the equilibrium cooling rates. Composition of alloy can be determined at any temperature using lever rule. An approximate development of micro-structure can be understood using phase diagrams.	{"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": 0.8589815497398376, "perplexity": 1848.3267124925367}, "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/1544376830479.82/warc/CC-MAIN-20181219025453-20181219051453-00165.warc.gz"}
https://brilliant.org/discussions/thread/report-a-problem-what-it-means/	# Report A Problem - What It Means Image We have released a way to keep you informed about problems which have potential issues in them. When several people have reported a problem, the above banner will be displayed. As a problem solver, • be aware that the answer may not be correct. • this problem can still affect your ratings. However, if there are issues with the problem, then those who got it wrong will have their ratings refunded (see below). • help us identify such problems by actively reporting them. For your reference, you can look at Example Of Reported Problem. As a problem creator, respond to clarification requests and disputes in a timely manner by editing or deleting the problem. Once you see such a banner appear on your problem, you may no longer edit or delete the problems in the usual way. Instead, you choose from the following options in a dropdown menu: Image 1) Delete the problem This deletes the problem from your feed, and refunds the ratings for those who got it wrong. Remember to explain why. You may reference the solution discussions. Once the answer is updated (after review), the corresponding rating changes will occur. 3) Adjust the wording of the problem. Use this for significant changes which affect the problem. E.g. You used “integer” but meant “real number”. This will also refund the ratings for those who got it wrong. 4) Make no changes, explain existing problem to moderator. Remember to explain why. You may reference the solution discussions. You can then make minor edits to the question, without affecting the ratings of others. Once you have made the corresponding change, it will be reviewed. Note by Calvin Lin 5 years, 10 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. MarkdownAppears as italics or _italics_ italics bold or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$ Sort by: Thank you for giving us the ability to change the answer to our problems! I've needed this don a couple times, and it was quite embarrassing while the answer was incorrect. This is a great change! - 5 years, 10 months ago As always, if you want to update your answer (especially if it hasn't been reported), please send me an email ([email protected]) Staff - 5 years, 10 months ago This is really cool.....Brilliant is getting more brilliant day by day!!!!!!!!!!!! - 5 years, 9 months ago good work.. I appreciate team Brilliant for their vibrant challenging problems.All your changes are welcome. - 5 years, 10 months ago Can you see if any of your problems are flagged without clicking on it? - 5 years, 10 months ago As this is a new feature, we have not built out nice capabilities like this. If reporting works well, we will make it much easier to review. This could include being sent an email (digest?) for problems that are being reported. Staff - 5 years, 10 months ago Hmm. I think it'd be nice if the problems themselves had a nice glowing, light red color when viewed from a feed or a set. - 5 years, 10 months ago Very nice feature. But what if a valid problem is reported by multiple members who didn´t solve it correctly? Just a moderator will be able to change a problem status to "flagged" or is automatic feature? - 5 years, 10 months ago If a problem is "wrongly" reported (or has a history of being wrongly reported), we have a way to prevent further banners from showing up. This would apply in cases where people easily misunderstand the conditions / misread the problem. Unless the problem is very badly worded, the problem creator should not need to make repeated edits. This would make it fairer to people who are working on the problem. Staff - 5 years, 10 months ago thnkx for making this change... - 5 years, 10 months ago Is there a way to change the options (not necessarily the answer) of a multiple choice question? - 5 years, 10 months ago No. My suggestion would be to delete the problem and start over. It is unlikely that we would implement such a change. Staff - 5 years, 10 months ago But at least can we see what the options were after posting the problem? It helps in writing a solution, explaining why the other options are wrong... - 5 years, 10 months ago We are working on giving everyone the ability to see the options in a condensed way after they solved the problem. Staff - 5 years, 10 months ago That's nice.. - 5 years, 9 months ago I have a question: How many is 'multiple members'? - 5 years, 9 months ago It depends on how 'reputable' the source is. For reliably good members (a clear example would be me), 1 report would be sufficient to trigger the banner. For new members with a low rating, it could take 3 reports before a Level 1 problem is flagged, and several more before a Level 5 problem is flagged. We're also playing around with this algorithm, and will consider other making it stricter or more relaxed. I'm currently against "banner once anyone reports" reports do arise from misreading / misunderstanding the problem / terminology. Staff - 5 years, 9 months ago To brilliant sir i have reported a bug in this brilliant website. This is not only affecting the person doing the problem but also making the person gain more points by a wrong method. A person is creating 2 accounts and he/she is using one account as a guide and the other as an original account. They are typing the answer in the 1st account and if the answer is wrong then they r seeing the correct answer and typing the answer in the other account. Due to this the people who have really worked to get an answer are really of no use.so i request u to immediately take an action on this issue and solve this problem. HOPE TO GET A BETTER RESULT. - 5 years, 1 month ago Hi Vishwa, Thanks for raising your concerns. Unfortunately, I am not able to police how everyone uses the internet, nor am I enable to enforce any kind of moral code amongst everyone. If I could, I would happily be ending all wars (and thus not be working here). At the end of the day, how one person decides to use resources is up to him/her. The person who cheats and merely gets the momentary satisfaction of a green screen, is not going to have any long term value. The person who puts the effort into thinking about these interesting questions and deciding that math and science is actually useful and fun, would have a whole world of opportunities open to them. Furthermore, the Brilliant community praises, and values, written solutions to a problem, which help shed light on the thought process. This is something that the latter can contribute and benefit from, but not the former. As such, I strongly disagree with "the people who have really worked to get an answer are really of no use". Staff - 5 years, 1 month ago knowing an answer is not important. how we arrived at an answer is important. bluffing will not serve any purpose. it is a cool place where we get to a challenge. face it. in the way you learn many things. if , any body cheats that is his problem. he is not here to learn .points are not the real count. what we learn is remains with us always. - 5 years, 1 month ago yes sir u r truly right! it is hard work that makes them success... - 5 years, 1 month ago When I open it(the wrong problem), it will be add to the list of (started problems)? & is it will be better if the wrong problem have a sign like ⛔or 🚫behind the head line of it?, to notify us there is some issues with this problem before we open it. - 5 years, 10 months ago As with any other problem, it will be added to your list of started problems. As mentioned below, we have not built out any nice capabilities as yet. Staff - 5 years, 9 months ago What if I found a good problem somewhere (website, olymbiad,..etc) and I can't find the right answer because it's higher than my ability to solve it but I know others will like it..what I suppose to do then ? Put a random answer then wait to someone report it to correct it? - 5 years, 9 months ago You can post it as a note, and ask others how they would approach it. Staff - 5 years, 9 months ago	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "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": 0.847981333732605, "perplexity": 1724.3662809419036}, "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/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00126.warc.gz"}
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=43&t=21446	## Relationship between Electronegativity and Orbital Energy Joe Rich 1D Posts: 32 Joined: Fri Jun 23, 2017 11:39 am Been upvoted: 1 time ### Relationship between Electronegativity and Orbital Energy The answer for question 4.57 says that a higher electronegativity for an atom makes its orbitals have lower energy. Why is that? Dabin Kang 1B Posts: 22 Joined: Fri Jun 23, 2017 11:39 am Been upvoted: 1 time ### Re: Relationship between Electronegativity and Orbital Energy Electronegativity is directly proportional to effective nuclear charge, so when electronegativity is high, the nuclear charge is high as well. The high nuclear charge strongly attracts the electrons and pulls them in, decreasing the energy of the orbitals. JD Malana Posts: 21 Joined: Wed Nov 16, 2016 3:02 am ### Re: Relationship between Electronegativity and Orbital Energy Can you apply the same logic to a relationship between electronegativity and atomic size? As in would more electronegative atoms be smaller in size? Justin Lai 1C Posts: 50 Joined: Fri Sep 29, 2017 7:04 am ### Re: Relationship between Electronegativity and Orbital Energy I think that the more electronegative an atom is, the more easy it is to attract electrons. This goes up in a diagonal trend, so increase as group goes to the right and as period goes up, disregarding noble gases. The atomic radius will increase as group goes to the left and period goes down. There is not necessarily a correlation however because there may be exceptions. Chem_Mod Posts: 17501 Joined: Thu Aug 04, 2011 1:53 pm Has upvoted: 393 times ### Re: Relationship between Electronegativity and Orbital Energy As an example, consider the most electronegative element F. When it comes to effective nuclear charge, smaller atoms tend to have higher effective nuclear charge because the nucleus is less shielded therefore its nucleus pulls electrons toward it more easily than say, Cl, which is also quite electronegative, but less so than F, O, and N. I do not know that having a higher effective nuclear charge makes the orbitals have lower energy, but rather, but rather these especially electronegative atoms at the top of the p-block with high effective nuclear charge simply--intrinsically have lower energy orbitals.	{"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": 0.9150625467300415, "perplexity": 2709.758342591537}, "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/1570986677964.40/warc/CC-MAIN-20191018055014-20191018082514-00535.warc.gz"}
https://engineering.stackexchange.com/questions/16385/reinforcing-beam-and-slab-with-correct-steel-bar-structure	# Reinforcing beam and slab with correct steel bar structure This question might look silly. I am not knowing the correct engineering terms to use in this question. I am trying to explain using the words what I know in English. We have been hired some construction workers for a work in our backyard. I have some doubt on reinforcing beam for slabs. I came through following 2 models . 1. Model one . 2. Model two According to my perception, I believe the second one is much better in terms of strength. Because the beam is supporting the weight of slab. But my friend suggested me the first one is the right way. But I am not aware why the first one is better than the second. Could any one please help me out in this case to identify the better one. ## Model 1 is always better, but may need some modifications Whenever you have two reinforced concrete elements, you always need to facilitate the transfer of internal stresses between them. This is done by "mixing" their reinforcement. For instance, model 2 will have no steel between the slab steel and the beam steel. This means that the connection between them will be very fragile and weak to horizontal shear, meaning that there's a risk of a horizontal crack like this: With Model 1, this isn't a risk because the vertical steel from the beam ("stirrups") will resist these forces. Model 1 also has an advantage in that it makes the beam itself stronger, since it allows the beam to behave as a much taller beam (including the height of the slab). Indeed, it even allows the beam to behave not as a rectangular beam, but as a stronger T-shape beam, using some of the slab to resist some of the beam's internal forces (how much depends on your country's standards and codes). If the beam and the slab will be poured simultaneously, then that's it, and you can stop reading this answer. However, if there's a chance that the beam will be built first and then the slab will be poured over it, then that means that the beam's stirrups and negative reinforcement (the longitudinal bars at the top of the beam) will be outside of the concrete when the beam is initially built but the slab hasn't yet been poured. This means that the beam's weight and that of the freshly-poured slab (before it gains enough stiffness to do anything) will have to be resisted by the "short" beam alone (without the added height from the slab), so some additional steel will be necessary: You are basically reinforcing the same beam twice: once considering it as the short beam (without the slab), and once as the tall beam (with the slab). To avoid confusion as to how to define how much reinforcement goes into each position, here's a list: • positive reinforcement (longitudinal steel at the bottom of the beam): calculate the necessary reinforcement for the short beam to resist the beam's and the slab's self-weight, and then calculate the reinforcement for the tall beam to resist any additional loads (dead loads, live loads, etc). Add these two numbers and you have the total necessary positive reinforcement. • negative reinforcement (longitudinal steel at the top of the beam) and the shear reinforcement (stirrups) for the short beam: calculate the necessary reinforcement for the short beam to resist the beam's and the slab's self-weight. You may need to just put the minimum reinforcement. • negative reinforcement (longitudinal steel at the top of the beam) and the shear reinforcement (stirrups) for the tall beam: calculate the reinforcement for the tall beam to resist any additional loads (dead loads, live loads, etc). You may need to just put the minimum reinforcement. • appreciated. Clear and understood. Jul 23 '17 at 13:18	{"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": 0.8015920519828796, "perplexity": 970.3238970483626}, "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-2021-43/segments/1634323585181.6/warc/CC-MAIN-20211017175237-20211017205237-00523.warc.gz"}
http://www.physicsforums.com/showthread.php?p=3872477	# spin of stars by Sean Pan Tags: spin, stars P: 10 It is said that massive stars spin faster than less massive ones and I am always wondering why.Could someone please tell me the reason? Thanks a lot. P: 1,262 Hi Sean Pan, welcome to PhysicsForums. The basic idea is that more massive stars formed from larger molecular clouds. Larger molecular clouds had more angular-momentum, and that angular-momentum is (largely) conserved in the star-formation process. Thus you end up with a faster spinning star. P: 10 Thanks, but there are many processes in the forming of stars that can reduce the angular momentun of the centural stars. Maybe I should have paid more attention to its initial angular momentum, but other factors should also be considered. Since star forming last a very long time, I think the final state may not depend largely on its original states. PF Gold P: 3,072 ## spin of stars We would expect all stars to spin rapidly, because we believe there is always ample angular momentum in the molecular cloud. So the question is not so much why do massive stars spin faster, it is why do low-mass stars spin slower. This is an ongoing research question, but one idea is that they tend to have a strong magnetic coupling with the gas that is forming them, and this coupling involves magnetic field lines that connect the rotating star to gas that is very far away from the star, which is in orbit. Kepler's laws say that the farther away gas is, the longer is its orbital period, so you have a rotating star with a short rotation period connected to gas with a long orbital period, and this tends to rob the star of angular momentum (and send it out to that gas way out there). Then you need a mechanism to get much of the high-angular-momentum gas to escape the system, and you can "spin down" your star (since this can happen with an accretion disk, it is also called "disk locking"). I'm not sure what the present status is of understanding how reliable this mechanism is, but no doubt many questions remain unanswered. For one thing, we might imagine that high-mass stars could also lose angular momentum in similar ways, so then we'd be back to asking why they spin so fast. It is thought that high-mass stars are even more likely to form in close binaries, which can then merge and convert the orbital angular momentum of the merging stars into spin. But that can happen to low-mass stars too, so then we are back to asking why low-mass stars spin so slowly! If you look at young low-mass stars, you find the younger they are, the faster they spin, so they are losing rotational angular momentum long after than have formed. Here interactions between magnetic fields and the winds from the stars are thought to play a key role, but you then have to explain why the winds are so strong in young stars. So you see, there is plenty of grist for the research mill here! PF Gold P: 11,047 Regardless of the mechanism for shedding angular momentum, I would guess that one of the basic reasons is that more massive stars simply have much more momentum to shed to slow down to a given rotation rate. P: 10 Quote by Ken G We would expect all stars to spin rapidly, because we believe there is always ample angular momentum in the molecular cloud. So the question is not so much why do massive stars spin faster, it is why do low-mass stars spin slower. This is an ongoing research question, but one idea is that they tend to have a strong magnetic coupling with the gas that is forming them, and this coupling involves magnetic field lines that connect the rotating star to gas that is very far away from the star, which is in orbit. Kepler's laws say that the farther away gas is, the longer is its orbital period, so you have a rotating star with a short rotation period connected to gas with a long orbital period, and this tends to rob the star of angular momentum (and send it out to that gas way out there). Then you need a mechanism to get much of the high-angular-momentum gas to escape the system, and you can "spin down" your star (since this can happen with an accretion disk, it is also called "disk locking"). I'm not sure what the present status is of understanding how reliable this mechanism is, but no doubt many questions remain unanswered. For one thing, we might imagine that high-mass stars could also lose angular momentum in similar ways, so then we'd be back to asking why they spin so fast. It is thought that high-mass stars are even more likely to form in close binaries, which can then merge and convert the orbital angular momentum of the merging stars into spin. But that can happen to low-mass stars too, so then we are back to asking why low-mass stars spin so slowly! If you look at young low-mass stars, you find the younger they are, the faster they spin, so they are losing rotational angular momentum long after than have formed. Here interactions between magnetic fields and the winds from the stars are thought to play a key role, but you then have to explain why the winds are so strong in young stars. So you see, there is plenty of grist for the research mill here! Thanks a lot for your very detailed analysis! Related Discussions Quantum Physics 2 Introductory Physics Homework 2 Astrophysics 2 General Astronomy 6 General Physics 4	{"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": 0.8588579893112183, "perplexity": 518.2776087327791}, "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-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00300-ip-10-147-4-33.ec2.internal.warc.gz"}
http://mathhelpforum.com/number-theory/121189-i-have-problem-about-primes-help-me-please.html	If p>=q>=5 and p and q are both primes, prove that 24\|(p^2-q^2). Thank. 2. Originally Posted by konna If p>=q>=5 and p and q are both primes, prove that 24\|(p^2-q^2). Thank. What have you tried yourself? Of course, $p^2- q^2= (p- q)(p+q)$ since p and q are both primes larger than 3, they are both odd and so both p-q and p+q are even- their product is, at least, divisible by 4. Now see if you can find another factor of 2. If p= 2m+1 and q= 2n+1, p+q= 2m+2n+ 2= 2(m+n+1) and p- q= 2m- 2n= 2(m-n). Consider what happens to m+n+1 and m-n if m and n are both even, both odd, or one even and the other odd. Notice that so far we have only required that p and q be odd, not that they be prime. Once you have done that the only thing remaining is to show that $p^2- q^2$ must be a multiple of 3 and that means showing that either p-q or p+q is a multiple of 3. 3. Originally Posted by konna If p>=q>=5 and p and q are both primes, prove that 24\|(p^2-q^2). Thank. $p^2-q^2=(p-q)(p+q)$ , and as: 1) $x^2=1\!\!\!\!\pmod 3\,\,\,\forall\,x$ not a multiple of 3 ; 2) either $p-q$ or $p+q$ is divisible by 4, and the other one by 2. The result follows at once. Tonio 4. I used to try by myself , but i not sure my process. Thank so much .. 5. Hello, konna! I don't know if this qualifies as a proof. If $p \geq q \geq 5$ and $p$ and $q$ are both primes, . . prove that: . $24\,\|\,(p^2-q^2)$ Any prime greater than or equal to 5 is of the form: . $6m \pm 1$ . . for some integer $m.$ Let: . $\begin{array}{ccc}p &=& 6m \pm 1 \\ q &=& 6n \pm1\end{array}$ Then: . $\begin{array}{ccc}p^2 &=& 36m^2 \pm 12m + 1 \\ q^2 &=& 36n^2 \pm12n + 1 \end{array}$ Subtract: . $p^2-q^2 \;=\;36m^2-36n^2 \pm 12m \mp 12n \;=\;36(m^2-n^2) \pm12(m - n)$ . . . . . . . $p^2-q^2 \;=\;36(m-n)(m+n) \pm12(m - n)$ . . . . . . . $p^2-q^2 \;=\;12(m-n)\bigg[3(m+n) \pm 1\bigg]$ We see that $p^2-q^2$ is divisible by 12. We must show that either $(m-n)$ or $[3(m+n) \pm1]$ is even. If $m$ and $n$ has the same parity (both even or both odd), . . then $(m-n)$ is even. If $m$ and $n$ have opposite parity (one even, one odd), . . then $(m+n)$ is odd. And . $3(m+n)$ is odd. . . Then: . $3(m+n) \pm1$ is even. Therefore, $p^2-q^2$ is divisible by 24.	{"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": 29, "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": 0.9630281329154968, "perplexity": 436.82859002789604}, "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-2016-44/segments/1476988719843.44/warc/CC-MAIN-20161020183839-00445-ip-10-171-6-4.ec2.internal.warc.gz"}
https://www.onemathematicalcat.org/Math/Precalculus_obj/ellipseDefn.htm	# Definition of an Ellipse by Dr. Carol JVF Burns (website creator) Follow along with the highlighted text while you listen! • PRACTICE (online exercises and printable worksheets) Ellipses were introduced in Introduction to Conic Sections, as one of several different curves (‘conic sections’) that are formed by intersecting a plane with an infinite double cone. Identifying Conics by the Discriminant introduced the general equation for any conic section, and gave conditions under which the graph would be an ellipse. In this current section, we present and explore the standard definition of an ellipse. This definition facilitates the derivation of standard equations for ellipses. Recall that the notation ‘$\,d(P,Q)\,$’ denotes the distance between points $\,P\,$ and $\,Q\,$. DEFINITION ellipse An ellipse is the set of points in a plane such that the sum of the distances to two fixed points is constant. More precisely: Let $\,F_1\,$ and $\,F_2\,$ be points; they are called the foci of the ellipse (pronounced FOE-sigh). (The singular form of ‘foci’ is ‘focus’.) Let $\,k\,$ be a positive real number, with $\,k > d(F_1,F_2)\,$. In this section, $\,k\,$ is referred to as the ellipse constant. The ellipse determined by $\,F_1\,$, $\,F_2\,$ and $\,k\,$ is the set of all points $\,P\,$ in a plane such that: $$\cssId{s18}{\overbrace{d(P,F_1) + d(P,F_2)}^{\text{the sum of the distances to two fixed points}}} \quad \cssId{s19}{\overbrace{=\strut}^{\text{is}}}\quad \cssId{s20}{\overbrace{k}^{\text{constant}}}$$ $\,P\,$ is a general point on the ellipse. $\,d(P,F_1) + d(P,F_2) = \text{constant}$ ## Old-Fashioned Playing with the Definition of an Ellipse Got a piece of cardboard, paper, tape, string/cord (not stretchy), and pen/pencil? Then, you can create your own ellipse: Tape the paper to the cardboard (at the corners is sufficient). Punch two small holes through the paper/cardboard at the desired foci. Put the cord/string through the two holes from front to back, and tie securely on the back. The length of the string that protrudes in the front, when held taut by a pen (see photo), is the ellipse constant. Keeping the string taut, trace the ellipse. (The string gets a bit twisted near the line through the two foci. So, draw one continuous motion for the upper half, then re-position and draw the lower half.) (The sunflower in a vase is optional. ☺ I grew my own sunflowers from seed in 2017, when I was writing this section!) ## More Playing with the Definition of an Ellipse You can also play with ellipses using the dynamic JSXGraph at right: $F_1\,$ and $\,F_2\,$ are the foci. Move them around! As you hover over each focus, you can see the coordinates of the point. The slider at the top sets the ellipse constant. The slider can be set to numbers between $\,0\,$ and $\,20\,$ with increments of $\,0.5\,$. The starting value is $\,12\,$ (refresh the page as needed). The current distance between the foci is displayed near the top. Watch this distance change as you move the foci around. The starting value for $\,d(F_1,F_2)\,$ is $\,10\,$ (refresh the page as needed). In order to see an ellipse, the ellipse constant (slider value) must be greater than the distance between the foci. $\,P\,$ is a general point on the ellipse. Move it around! When the ellipse constant equals the distance between the foci, the ‘ellipse’ degenerates to a line segment. Ignore the line segment that you see when the ellipse constant $\,k\,$ is less than the distance between the foci. As discussed below, in this case there are actually no points $\,P\,$ that satisfy $\,d(F_1,P) + d(F_2,P) = k\,$. ## Notes: In the definition of ellipse, the ellipse constant $\,k\,$ is required to be strictly greater than the distance between the two foci. Why? As shown below, other values of $\,k\,$ don't give anything that a reasonable person would want to call an ellipse! ### A ‘LINE SEGMENT’ ELLIPSE: $\,k = d(F_1,F_2)\,$ Suppose the ellipse constant, $\,k\,$, equals the distance between the foci: that is, $\,k = d(F_1,F_2)\,$. In this case, the solution set to the equation $$\cssId{s62}{\color{green}{d(P,F_1)} + \color{red}{d(P,F_2)} = k}$$ is the line segment between $\,F_1\,$ and $\,F_2\,$ (including the endpoints). Most people don't want to call a line segment an ellipse! This is why $\,k\,$ is not allowed to equal $\,d(F_1,F_2)\,$ in the definition of ellipse. • $\,\color{green}{d(P,F_1)}\,$ is the length of the green segment • $\,\color{red}{d(P,F_2)}\,$ is the length of the red segment • together, the green and red segments give $\,d(F_1,F_2)\,$ ### AN ‘EMPTY’ ELLIPSE:$\,k < d(F_1,F_2)\,$ The shortest distance between any two points is a straight line. In particular, the shortest distance from $\,F_1\,$ to $\,F_2\,$ is the length of the line segment between them, and is denoted by $\,d(F_1,F_2)\,$. Thus, any path from $\,F_1\,$ to $\,F_2\,$ must have length greater than or equal to $\,d(F_1,F_2)\,$. In particular (refer to sketch at right), the piecewise-linear path from $\,F_1\,$ to $\,P\,$ and then from $\,P\,$ to $\,F_2\,$ always has length greater than or equal to $\,d(F_1,F_2)\,$. Therefore, if the ellipse constant $\,k\,$ is strictly less than $\,d(F_1,F_2)\,$, there are no points $\,P\,$ that make the following equation true: $$\cssId{s79}{\overbrace{d(P,F_1) + d(P,F_2)\strut }^{\text{always \,\,\ge\,\, d(F_1,F_2)}}} \qquad \cssId{s80}{=}\qquad \cssId{s81}{\overbrace{\strut k}^{< \,\, d(F_1,F_2)}}$$ You might want to call this an empty ellipse, an invisible ellipse, or an imaginary ellipse! There's nothing there! Master the ideas from this section	{"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": 0.9154648184776306, "perplexity": 420.50361023331436}, "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-2022-40/segments/1664030338244.64/warc/CC-MAIN-20221007175237-20221007205237-00307.warc.gz"}
http://newvillagegirlsacademy.org/math/?page_id=2284	# 4.2 – Slope Key Terms • Rate of Change – The measure of how much a dependent variable y changes for a given change in the independent variable x. If the graph of the function is a straight line, the rate of change equals the slope of the line. • Rise – The vertical distance between two points on a line. It equals the difference between the y-coordinates of the two points. • Run – The horizontal distance between two points on a line. It equals the difference between the x-coordinates of the two points. • Slope – A measure of the steepness of a line. The slope equals rise divided by run for any two points on the line. A line that rises from left to right has a positive slope. A line that falls from left to right has a negative slope. Notes • Horizontal means “left and right” • x-axis • Vertical means “up and down” • y-axis • Slopes can rise, fall, be zero (horizontal), or be undefined (vertical) • Rise – UP from Left to Right • Positive slope • Fall – DOWN from Left to Right • Negative slope • Straight horizontal lines that do not rise or fall • Zero slope • Straight vertical lines that are all rise and no run • Remember: undefined fractions have a zero in the denominator • Undefined slopes have a zero run (which in the denominator of the formula: rise over run. • Slopes can be steep (big) or shallow (small) • Steep slopes have larger positive or negative values • The steeper the line, the greater the slope • Ex: slope = 5 is greater than slope = 2 • Ex: slope = 10 is greater than slope = 0.34 • Ex: slope = -4 is greater than slope = 2 • Even though 4 is negative, the value of 4 is larger (steeper) than 2. • Shallow slopes have smaller positive or negative values • The less steep the line, the smaller the slope • Ex: slope = 2 is less steep than slope = 2.5 • Ex. slope = 5 is less steep than slope = -10 • Even though 10 is negative, the value of 10 is larger (steeper) than 5. • Parallel Slopes • Two lines that have the same exact slope are parallel • Slope Formula • Rise: vertical (up and down, like an elevator) • Rise Up: Positive • Rise Down: Negative • Run: horizontal (left and right, like walking across a room) • Run Left: Negative • Run Right: Positive • Positive and Negative Rise and Run • Lines that go UP have a Positive Rise and a Positive Run • Lines that go DOWN have a Negative Rise and a Positive Run • Unless you are calculating a line’s run backwards, the Run will always be positive. • You can do this with careful attention to detail. • The slope of a line is CONSTANT • It will never change, no matter what two points you compare! • The slopes will be the same for ALL points on a line • $\frac{12}{3}$ is the same slope as $\frac{4}{1}$ if you reduce the fraction! Both will be a slope of 4 over 1. • The rise will be 4 and the run will be 1. • Formula for Finding the Slope Using 2 Points • Each point has an x-value and a y-value • Choose either point to be Point 1 and call the other point, Point 2 • It’s usually a good idea to pick the point on the Left as Point 1 • Label Point 1: $(x_1,y_1)$ • Label Point 2: $(x_2,y_2)$ • Substitute the values into the formula, reduce the fraction, write your slope in the form of Rise over Run • Example: Points (2, -4) and (9,10) • Point 1: (2, -4) • $x_1=2$ and $y_1=-4$ • Point 2: (9, 10) • $x_2=9$ and $y_2=10$ Answer: the slope is $\frac{14}{7}$, which can be reduced to $\frac{7}{1}$, which can be written as just “7” • Equation of a Line • Formula: y = mx + b • Slope: m • y-intercept (where the line crosses the y-axis): b • b can be 0 if the line crosses the y-axis at the origin (0, 0) • Any x-value (point) on the line: x • Any y-value (point) on the line: y • Identify Slope in an Equation • Slope is the coefficient of the x-value • Ex. y = 3x + 2 has a slope of 3. • Ex. y = -12x has a slope of -12. • Ex. y = 4(x – 2) + 6 has a slope of 4. • In this case, distribute the 4 into the parenthesis where the x-value is to simplify: y = 4x – 8 + 6. You can also add -8 + 6 to find the y-intercept (b) of -2. • Real World Example • A graph is constructed such that time (in seconds) is the x-variable and distance (in miles) is the y-variable. • If you plot the distance that a space shuttle travels at a speed of 5 miles per second, what is the slope of the graph? • Note: in the real world, most lines START at the ORIGIN (0,0). • $x_1=0$$y_1=0$$x_2=1$$y_2=5$ • $Slope=\frac{y_2-y_1}{x_2-x_1}$ • $Slope=\frac{5-0}{1-0}$ • $Slope=\frac{5}{1}$ or just “5”	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8569758534431458, "perplexity": 1341.986093700253}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917120001.0/warc/CC-MAIN-20170423031200-00242-ip-10-145-167-34.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/74931/integral-solutions-of-x2y21-z2	# Integral solutions of $x^2+y^2+1=z^2$ I am interested in integral solutions of $$x^2+y^2+1=z^2.$$ Is there a complete theory comparable to the one for $x^2+y^2=z^2?$ - I think writing $(x+iy)(x-iy) = (z-1)(z+1)$ might help. – Joel Cohen Oct 22 '11 at 22:52 You are not trying to solve Project Euler problem 224, are you? projecteuler.net/problem=224 – starblue Oct 23 '11 at 13:15 @starblue. No, it came up when I tried to make up an example in hyperbolic geometry. – TCL Oct 23 '11 at 16:55 The question is essentially for which $z$ both $z-1$ and $z+1$ are the sum of two squares. There are some related questions asked before, such as math.stackexchange.com/questions/438818 and math.stackexchange.com/questions/46451. – barto Sep 8 '14 at 8:15 I would look at Catalan’s [well-known] complete solution for $x_1^2+x_2^2+x_3^2=y_1^2$, and set $x_3 = a^2+b^2-c^2-d^2=1$. This implies another equal sums-of-squares equation $a^2+b^2=c^2+d^2+1$, so you probably have a nice orbit to chase. See also Spira's paper <jstor.org/stable/2312125>;, etc.. – Kieren MacMillan Oct 6 '14 at 1:48 The equation defines a conic with a rational point (0,0,1). The other rational solutions can be parametrized by lines through that point. As with $x^2+y^2 = z^2$ the integer solutions can be deduced from the rational ones, and the quadratic form $x^2+y^2 - z^2$ has a large linear symmetry group allowing one to move between solutions. In these respects the theory is the same as the one for Pythagorean triples. There is a difference in the structure of the symmetry group. For $a=0$ the hyperboloid $x^2+y^2 - z^2 = a$ becomes a cone, with additional scaling symmetries in addition to the linear transformations of the hyperboloid; in effect the $O(2,1)$ symmetry group of the rational solutions collapses into a product of scalings and circle isometries. For integer solutions, with Pythagorean triples there is a reduction to the case of primitive triples, but when $a \neq 0$ there is a bound on the shared factors of $(x,y,z)$, and for $a = \pm 1$ all integer solutions are primitive. The organization of Pythagorean triples using several 3x3 integer matrices as transformations connecting different solutions does use the $O(2,1)$ structure, and the solution set for the equation with $a \neq 0$ could be presented in the same way (possibly with more than one connected component when $\|a\| > 1$). - Note: This is not a complete answer. First, I want to point out that both $x$ and $y$ must be even. We can trivially get one infinitely family of solutions from the following: Consider triplets of the form $$(x,y,z)=(2r^2,2r,2r^2+1).$$ Then they satisfy the above as $$(2r^2 )^2 +(2r)^2+1=(2r^2+1)^2.$$ This gives rise to $$(2,2,3), \ (4,8,9), \ (6,18,19),\ \cdots$$ - the equation: $X^2+Y^2=Z^2-1$ Solutions can be written using the solutions of Pell's equation: $p^2-2k(k-1)s^2=1$ $k$ - given by us. Solutions have form: $X=2kps-2(k-1)s^2$ $Y=2(k-1)ps+2ks^2$ $Z=p^2+2(k^2-k+1)s^2$ And more: $X=2p^2-2(3k-2)ps+2(2k-1)(k-1)s^2$ $Y=2p^2-2(3k-1)ps+2k(2k-1)s^2$ $Z=3p^2-4(2k-1)ps+2(3k^2-3k+1)s^2$ Although it should be a more general solution to record. - I do not understand. And what you've edited? – individ May 11 '14 at 8:52 All of numbers can be any character.In Equation: $qX^2+Y^2=Z^2+a$ If the ratio is factored so: $a=(b-c)(b+c)$ Then we use the solutions of Pell's equation: $p^2-fs^2=\pm1$ where: $f=(q+1)k^2-2kt-(q-1)t^2$ Then the solutions are of the form: $X=2(ck-bt)ps+2(bk^2-(b+c)kt+ct^2)s^2$ $Y=bp^2+2c(k-t)ps-(b(q-1)k^2+2(b-qc)kt+b(q-1)t^2)s^2$ $Z=cp^2+2b(k-t)ps+(c(q+1)k^2-2(bq+c)kt+c(q+1)t^2)s^2$ All of numbers can be any character. - For the equation: $qX^2+Y^2=Z^2+j$ In the case where a square: $a=\sqrt{\frac{j}{q}}$ Using equation Pell: $p^2-(q+1)s^2=1$ Then the solution can be written: $X=2s(s\pm{p})L\pm{ap^2}+2aps\pm{a(q+1)s^2}=bL+af$ $Y=(p^2\pm2ps+(1-q)s^2)L\pm{ap^2}+2aps\pm{a(q+1)s^2}=cL+af$ $Z=(p^2\pm2ps+(q+1)s^2)L\pm{ap^2}+2a(q+1)ps\pm{a(q+1)s^2}=fL+at$ $L$ - any integer number given by us. The most interesting thing is that these numbers are solutions of equations: $qb^2+c^2=f^2$ $t^2-(q+1)f^2=\pm{q}$ If we use the equation Pell: $p^2-(q+1)s^2=k$ And substituting the solutions in the upper formula, we have solutions of the following equations. $qb^2+c^2=f^2$ where: $c-b=k$ $t^2-(q+1)f^2=\pm{qk^2}$ True, I use this formula in reverse order. Find solutions of Pell's equation is much more complicated than the simple equations like Pythagorean triples. So find them and then have solutions of Pell's equation. The most interesting thing is that the solution of Pell related to Pythagorean triples. -	{"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": 0.9509514570236206, "perplexity": 339.41332327854684}, "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/1438042990609.0/warc/CC-MAIN-20150728002310-00335-ip-10-236-191-2.ec2.internal.warc.gz"}
https://www.mathdoubts.com/evaluate-lim-e-power-3-plus-x-sinx-e-cube-by-x-as-x-approaches-to-0-by-l-hopital-s-rule/	# Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^{3\normalsize +\large x}-\sin{x}-e^3}{x}}$ by L’Hopital’s Rule The limit of $e$ raised to the power $3$ plus $x$ minus $\sin{x}$ minus $e$ raised to the power $3$ by $x$ as $x$ approaches to $0$ is solved in fundamental limit method and it will be solved here by L’Hospital’s Rule (or L’Hopital’s rule) $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^{3\displaystyle +x}-\sin{x}-e^3}{x}}$ Let’s get started to learn how to find the limit of the given rational function by L’Hopital’s Rule (or L’Hospital’s Rule). ### Calculate the Limit by the Direct substitution Let us try to find the limit of the given rational function by using direct substitution method as the value of $x$ approaches to $0$. $=\,\,\,$ $\dfrac{e^{3 \displaystyle + \small 0}-\sin{(0)}-e^3}{0}$ $=\,\,\,$ $\dfrac{e^3-0-e^3}{0}$ The sine of angle zero radian is equal to zero as per the trigonometric mathematics. $=\,\,\,$ $\dfrac{e^3-e^3}{0}$ $=\,\,\,$ $\dfrac{0}{0}$ It is evaluated that the limit of the function is indeterminate. Hence, it is better to think about an alternative mathematical approach to find its limit. ### Differentiate the Numerator and Denominator The given function is eligible for evaluating its limit by the L’Hopital’s Rule (or L’Hospital’s Rule). Now, differentiate the expressions in both numerator and denominator with respect to $x$. $=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}\Big(e^{3\large +\Large x}-\sin{x}-e^3\Big)}{\dfrac{d}{dx}{(x)}}}$ The derivative of the difference of the terms can be evaluated as per the difference rule of the differentiation. $=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}\big(e^{3\large +\Large x}\big)-\dfrac{d}{dx}{\sin{x}}-\dfrac{d}{dx}{(e^3)}}{\dfrac{d}{dx}{(x)}}}$ The $e$ raised to the power $3$ plus $x$ can be split as a product of two quantities in exponential notation as per the product rule of the exponents. $=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}\big(e^3 \times e^{\Large x}\big)-\dfrac{d}{dx}{\sin{x}}-\dfrac{d}{dx}{(e^3)}}{\dfrac{d}{dx}{(x)}}}$ The factor $e$ raised to the power $3$vcolkfaZZ is a constant. Hence, it can be separated from the differentiation by using the constant multiple rule of the limits. $=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^3 \times \dfrac{d}{dx}\big(e^{\Large x}\big)-\dfrac{d}{dx}{\sin{x}}-\dfrac{d}{dx}{(e^3)}}{\dfrac{d}{dx}{(x)}}}$ Use derivative rule of natural exponential function, derivative rule of sine function and derivative rule of a constant to find the derivatives in the numerator. Similarly, find the derivative of the variable by using the derivative rule of a variable. $=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^3 \times e^{\Large x}-\cos{x}-0}{1}}$ $=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \big(e^3 \times e^{\Large x}-\cos{x}\big)}$ ### Evaluate the Limit by the Direct substitution Now, use the direct substitution method one more time to find the limit of the function as the value of $x$ approaches to zero. $=\,\,\,$ $e^3 \times e^0-\cos{(0)}$ According to the zero power rule, the mathematical constant $e$ raised to the power zero is equal to one. The cosine of angle zero radian is equal to one as per the trigonometry. $=\,\,\,$ $e^3 \times 1-1$ $=\,\,\,$ $e^3-1$ Latest Math Problems A best free mathematics education website for students, teachers and researchers. ###### Maths Topics Learn each topic of the mathematics easily with understandable proofs and visual animation graphics. ###### Maths Problems Learn how to solve the maths problems in different methods with understandable steps. Learn solutions ###### Subscribe us You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.	{"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": 0.9523049592971802, "perplexity": 369.0195720159249}, "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-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00055.warc.gz"}
https://sppumoodle.unipune.ac.in/course/index.php?categoryid=121	#### Numerical Methods for Scientific Computing It is the first course in Numerical Methods. It starts with an introduction to finite precision mathematics. It covers the methods of root finding, linear equation solvers, eigen value computation, curve fitting & approximations and numerical integration & differentiation.	{"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": 0.8880759477615356, "perplexity": 985.6516104445337}, "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/1610703521139.30/warc/CC-MAIN-20210120151257-20210120181257-00119.warc.gz"}
https://mathoverflow.net/questions/233778/a-question-on-the-commutativity-degree-of-the-monoid-of-subsets-of-a-finite-grou	A question on the commutativity degree of the monoid of subsets of a finite group The commutativity degree $d(G)$ of a finite group $G$ is defined as the ratio $$\frac{\|\{(x,y)\in G^2 \| xy=yx\}\|}{\|G\|^2}.$$It is well known that $d(G)\leq5/8$ for any finite non-abelian group $G$. If $P(G)$ is the monoid of subsets of $G$ with respect to the usual product of group subsets, then the commutativity degree $d(P(G))$ of $P(G)$ can be defined similarly: $$d(P(G))=\frac{\|\{(A,B)\in P(G)^2 \| AB=BA\}\|}{\|P(G)\|^2}.$$Which are the connections between $d(G)$ and $d(P(G))$? Is there a constant $c\in (0,1)$ such that $d(P(G))\leq c$ for any finite non-abelian group $G$? Additional Question: Let $P_k(G)$ be the subset of $P(G)$ consisting of all $k$-subsets of $G$ and $$d(P_k(G))=\frac{\|\{(A,B)\in P_k(G)^2 \| AB=BA\}\|}{\|P_k(G)\|^2}.$$Clearly, $d(P_1(G))=d(G)$ and, for every $k\in\{1,2,3\}$, $G$ is abelian iff $d(P_k(G))=1$. A similar question can be asked for $d(P_k(G))$, $k=2,3$: is there a constant $c_k\in (0,1)$ such that $d(P_k(G))\leq c_k$ for any finite non-abelian group $G$? • @Derek Holt: I would say that your argument shows just that any two subsets $A$ and $B$ of cardinality $>\|G\|/2$ each commute, since $AB=BA=G$... – Ilya Bogdanov Mar 16 '16 at 12:08 • Yes that's right - I have deleted my comment. – Derek Holt Mar 16 '16 at 12:40 • That raises the question for what fraction of the pairs $(A,B)$ of subsets of $G$ is $AB=BA=G$. Could this approach $1$ for large $G$? If so, there would be no such constant $c$. – Derek Holt Mar 16 '16 at 13:55 @Derek Holt is completely right: as $\|G\|\to \infty$, the fraction of subsets $(A,B)$ with $AB=G$ tends to $1$, so there is no such $c$. Indeed, assume that $\|G\|=n$. Let us choose the subsets $A$ and $B$ uniformly and independently. Fix any $g\in G$; there are $n$ pairs $(a,b)$ with $ab=g$, each pair belongs to $A\times B$ with probability $1/4$, and these events are independent for distinct pairs. Thus the probability that $g\notin AB$ is $(3/4)^n$, so the probability that $AB\neq G$ is at most $n(3/4)^n$ which tends to $0$ as $n\to\infty$. To conclude: if $n$ is large, almost all pairs $(A,B)\in P(G)\times P(G)$ satisfy $AB=BA=G$. • @GeoffRobinson: I do not fix the cardinalities of $A$ abd $B$; I simply put every element to $A$ with probability $1/2$, the same for $B$ (all $2n$ choices are independent). – Ilya Bogdanov Mar 17 '16 at 12:25	{"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": 0.9911105036735535, "perplexity": 106.07677448350478}, "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-2020-50/segments/1606141732835.81/warc/CC-MAIN-20201203220448-20201204010448-00187.warc.gz"}
https://infoscience.epfl.ch/record/177428	Infoscience Journal article Mathematical modeling of (13)C label incorporation of the TCA cycle: the concept of composite precursor function A novel approach for the mathematical modeling of (13)C label incorporation into amino acids via the TCA cycle that eliminates the explicit calculation of the labeling of the TCA cycle intermediates is described, resulting in one differential equation per measurable time course of labeled amino acid. The equations demonstrate that both glutamate C4 and C3 labeling depend in a predictable manner on both transmitochondrial exchange rate, V(X), and TCA cycle rate, V(TCA). For example, glutamate C4 labeling alone does not provide any information on either V(X) or V(TCA) but rather a composite "flux". Interestingly, glutamate C3 simultaneously receives label not only from pyruvate C3 but also from glutamate C4, described by composite precursor functions that depend in a probabilistic way on the ratio of V(X) to V(TCA): An initial rate of labeling of glutamate C3 (or C2) being close to zero is indicative of a high V(X)/V(TCA). The derived analytical solution of these equations shows that, when the labeling of the precursor pool pyruvate reaches steady state quickly compared with the turnover rate of the measured amino acids, instantaneous labeling can be assumed for pyruvate. The derived analytical solution has acceptable errors compared with experimental uncertainty, thus obviating precise knowledge on the labeling kinetics of the precursor. In conclusion, a substantial reformulation of the modeling of label flow via the TCA cycle turnover into the amino acids is presented in the current study. This approach allows one to determine metabolic rates by fitting explicit mathematical functions to measured time courses.	{"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": 0.845975935459137, "perplexity": 2567.2727660106643}, "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-2017-30/segments/1500549423769.10/warc/CC-MAIN-20170721102310-20170721122310-00514.warc.gz"}
http://mathhelpforum.com/differential-geometry/110747-intersection-compact-sets.html	# Thread: intersection of compact sets 1. ## intersection of compact sets Let X be the cartesian product of $\mathbb{R}$ with the usual topology and a two point set with the indiscrete topology. Find two compact subspaces of X such that their intersection is not compact. Can I get some help please? 2. That's not possible. A compact set is closed in any topology. The intersection of two closed sets is closed in any topology. A closed subset of a compact set is compact in any topology. Therefore, the intersection of two compact sets is compact is always compact no matter what topology you have. 3. Originally Posted by HallsofIvy That's not possible. A compact set is closed in any topology. Only if the topology is Hausdorff, otherwise it may not be true. In remark 1.3. in A Note on Topological Properties of Non-Hausdorff Manifolds there's a nice example of a compact non-closed set. It is rather bewildering since the set is the well-known closed interval [0,1], and even more anti-intuitive: this set is Hausdorff (but the whole space, of course, isn't), yet it is a cute exercise to show that under the topology defined there this interval indeed isn't closed. Tonio The intersection of two closed sets is closed in any topology. A closed subset of a compact set is compact in any topology. Therefore, the intersection of two compact sets is compact is always compact no matter what topology you have. . 4. Originally Posted by tonio . By the way, all the rest of the argument the Hallsofivy is true. Closed in compact space (i.e closed intersected compact) is always compact, in every topological space, as it is discussed in other threads. Hence this exercise needs to show that there is a non closed compact in this space. Further, they are needed two of these compacts for intersecting them, because closed and compact would give compact . I suppose the non discrete topology in {0,1} is that whose unique open sets are all the space and the empty set. Thus we have to take benefit from the fact that neither $\{0\}$ or $\{1\}$ are open nor closed in $\{0,1\}$ endowed with the trivial topology. I guess these two subsets $K_1:=[-1,1]\times \{1\}$ $K_2:=([-1,0] \times \{0\})\cup ((0,1]\times \{1\})$ are compact, and the intersection $K_1\cap K_2 =(0,1]\times \{1\}$ clearly is not (is isomorphic to (0,1]). I suggest proving the compactness of these subsets 5. Originally Posted by Enrique2 By the way, all the rest of the argument the Hallsofivy is true. Closed in compact space (i.e closed intersected compact) is always compact, in every topological space, as it is discussed in other threads. Hence this exercise needs to show that there is a non closed compact in this space. Further, they are needed two of these compacts for intersecting them, because closed and compact would give compact . I suppose the non discrete topology in {0,1} is that whose unique open sets are all the space and the empty set. Thus we have to take benefit from the fact that neither $\{0\}$ or $\{1\}$ are open nor closed in $\{0,1\}$ endowed with the trivial topology. I guess these two subsets $K_1:=[-1,1]\times \{1\}$ $K_2:=([-1,0] \times \{0\})\cup ((0,1]\times \{1\})$ are compact, and the intersection $K_1\cap K_2 =(0,1]\times \{1\}$ clearly is not (is isomorphic to (0,1]). I suggest proving the compactness of these subsets Is $K_1$ compact? I believe that compact subspaces in this space are exactly those of the form $K \times \{ \pm \infty \}$ where $K$ is compact in $\mathbb{R}$ 6. Originally Posted by Jose27 Is $K_1$ compact? I believe that compact subspaces in this space are exactly those of the form $K \times \{ \pm \infty \}$ where $K$ is compact in $\mathbb{R}$ I believe both are compacts. $K_1$ is homeomorphic to $[-1,1]$. Any sequence or net $(x_i,y_i)_i$ in this space is convergent if and only if $(x_i)_i$ is. Notice that $(y_i)_i$ is always convergent to both 0 and 1. Hence, roughly speaking, $K\times$"anything" is compact iff $K$ is compact in $\mathbb{R}$. The argument of an open recovering $V_i\times W_i$ also work because $W_i$ is always the whole space. I think $K\subset X$ is compact if and only if the projection $p_1(K)$ is compact in $\mathbb{R}$. Anyway I prefer people to check this, non Hausdorff spaces are not familiar to me, but I like the problem. 7. Originally Posted by Enrique2 $K_1:=[-1,1]\times \{1\}$ $K_2:=([-1,0] \times \{0\})\cup ((0,1]\times \{1\})$ are compact, and the intersection $K_1\cap K_2 =(0,1]\times \{1\}$ clearly is not (is isomorphic to (0,1]). I think this is a nice example. To see $K_1$ is compact and $K_1\cap K_2$ is not compact are rather straightforward. To check if $K_2$ is compact as Enrique2 already mentioned, any open cover for $K_2$ must cover $K_2':=([-1,0] \times \{0,1\})\cup ((0,1]\times \{0, 1\}) = [-1,1] \times \{0,1\}$. Now we see that $K_2$ is compact. , , , , , , # the intersection of two comact sets need not be compact Click on a term to search for related topics.	{"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": 43, "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": 0.9574199914932251, "perplexity": 253.7521437969568}, "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/1490218189466.30/warc/CC-MAIN-20170322212949-00528-ip-10-233-31-227.ec2.internal.warc.gz"}
http://www.chegg.com/homework-help/questions-and-answers/particle-track-detectors-used-measure-speed-particles-lifetime-particle-known-particle-x-l-q2654725	Particle track detectors are used to measure the speed of particles if the lifetime of the particle is known. Particle X has a lifetime of 256.2 ps. These particles are created in an experiment inside the detector by a given reaction. The particles leave 10.6 cm long tracks on average before they decay into other particles not observable by the detector. -------------------------------------------------------------------------------- What is the average speed of the particles in terms of the speed of light? Hints: The speed of the particle is the ratio of the track length and the lifetime. However the simple ratio of these two quantities gives unphysically high speed: 1.379 c. The reason is that the length of the track is measured in the resting frame of the detector. On the other hand the lifetime of the particle is its life expectancy measured in the frame moving with the particle. Either transform the length of the track to the particle's frame, or the particle's lifetime to the detector's frame. You can take the ratio of the tracklength and the lifetime only if they are from the same reference frame.	{"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": 0.9602888822555542, "perplexity": 223.56536314809128}, "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/1430448947973.19/warc/CC-MAIN-20150501025547-00033-ip-10-235-10-82.ec2.internal.warc.gz"}
https://dsp.stackexchange.com/questions/481/does-the-autocorrelation-function-completely-describe-a-stochastic-process/488	# Does the autocorrelation function completely describe a stochastic process? Is a stochastic process completely described by its autocorrelation function? If not, which additional properties would be needed? What is meant by a complete description of a stochastic process? Well, mathematically, a stochastic process is a collection $$\{X(t) : t \in {\mathbb T}\}$$ of random variables, one for each time instant $$t$$ in an index set $$\mathbb T$$, where usually $$\mathbb T$$ is the entire real line or the positive real line, and a complete description means that for each integer $$n \geq 1$$ and $$n$$ time instants $$t_1, t_2, \ldots, t_n \in \mathbb T$$, we know the (joint) distributions of the $$n$$ random variables $$X(t_1)$$, $$X(t_2)$$, $$\ldots, X(t_n)$$. This is an enormous amount of information: we need to know the CDF of $$X(t)$$ for each time instant $$t$$, the (two-dimensional) joint CDF of $$X(t_1)$$ and $$X(t_2)$$ for all choices of time instants $$t_1$$ and $$t_2$$, the (three-dimensional) CDFs of $$X(t_1)$$, $$X(t_2)$$, and $$X(t_3)$$, etc. etc. etc. So naturally people have looked about for simpler descriptions and more restrictive models. One simplification occurs when the process is invariant to a change in the time origin. What this means is that • All the random variables in the process have identical CDFs: $$F_{X(t_1)}(x) = F_{X(t_2)}(x)$$ for all $$t_1, t_2$$. • Any two random variables separated by some specified amount of time have the same joint CDF as any other pair of random variables separated by the same amount of time. For example, the random variables $$X(t_1)$$ and $$X(t_1 + \tau)$$ are separated by $$\tau$$ seconds, as are the random variables $$X(t_2)$$ and $$X(t_2 + \tau)$$, and thus $$F_{X(t_1), X(t_1 + \tau)}(x,y) = F_{X(t_2), X(t_2 + \tau)}(x,y)$$ • Any three random variables $$X(t_1)$$, $$X(t_1 + \tau_1)$$, $$X(t_1 + \tau_1 + \tau_2)$$ spaced $$\tau_1$$ and $$\tau_2$$ apart have the same joint CDF as $$X(t_2)$$, $$X(t_2 + \tau_1)$$, $$X(t_2 + \tau_1 + \tau_2)$$ which as also spaced $$\tau_1$$ and $$\tau_2$$ apart, • and so on for all multidimensional CDFs. See, for example, Peter K.'s answer for details of the multidimensional case. Effectively, the probabilistic descriptions of the random process do not depend on what we choose to call the origin on the time axis: shifting all time instants $$t_1, t_2, \ldots, t_n$$ by some fixed amount $$\tau$$ to $$t_1 + \tau, t_2 + \tau, \ldots, t_n + \tau$$ gives the same probabilistic description of the random variables. This property is called strict-sense stationarity and a random process that enjoys this property is called a strictly stationary random process or, more simply, s stationary random process. Note that strict stationarity by itself does not require any particular form of CDF. For example, it does not say that all the variables are Gaussian. The adjective strictly suggests that is possible to define a looser form of stationarity. If the $$N^{\text{th}}$$-order joint CDF of $$X(t_1), X(t_2), \ldots, X(t_N)$$ is the same as the $$N^{\text{th}}$$-order joint CDF of $$X(t_1+\tau), X(t_2+\tau), \ldots, X(t_N +\tau)$$ for all choices of $$t_1,t_2, \ldots, t_N$$ and $$\tau$$, then the random process is said to be stationary to order $$N$$ and is referred to as a $$N^{\text{th}}$$-order stationary random process. Note that a $$N^{\text{th}}$$-order stationary random process is also stationary to order $$n$$ for each positive $$n < N$$. (This is because the $$n^{\text{th}}$$-order joint CDF is the limit of the $$N^{\text{th}}$$-order CDF as $$N-n$$ of the arguments approach $$\infty$$: a generalization of $$F_X(x) = \lim_{y\to\infty}F_{X,Y}(x,y)$$). A strictly stationary random process then is a random process that is stationary to all orders $$N$$. If a random process is stationary to (at least) order $$1$$, then all the $$X(t)$$'s have the same distribution and so, assuming the mean exists, $$E[X(t)] = \mu$$ is the same for all $$t$$. Similarly, $$E[(X(t))^2]$$ is the same for all $$t$$, and is referred to as the power of the process. All physical processes have finite power and so it is common to assume that $$E[(X(t))^2] < \infty$$ in which case, and especially in the older engineering literature, the process is called a second-order process. The choice of name is unfortunate because it invites confusion with second-order stationarity (cf. this answer of mine on stats.SE), and so here we will call a process for which $$E[(X(t))^2]$$ is finite for all $$t$$ (whether or not $$E[(X(t))^2]$$ is a constant) as a finite-power process and avoid this confusion. But note again that a first-order stationary process need not be a finite-power process. Consider a random process that is stationary to order $$2$$. Now, since the joint distribution of $$X(t_1)$$ and $$X(t_1 + \tau)$$ is the same as the joint distribution function of $$X(t_2)$$ and $$X(t_2 + \tau)$$, $$E[X(t_1)X(t_1 + \tau)] = E[X(t_2)X(t_2 + \tau)]$$ and the value depends only on $$\tau$$. These expectations are finite for a finite-power process and their value is called the autocorrelation function of the process: $$R_X(\tau) = E[X(t)X(t+\tau)]$$ is a function of $$\tau$$, the time separation of the random variables $$X(t)$$ and $$X(t+\tau)$$, and does not depend on $$t$$ at all. Note also that $$E[X(t)X(t+\tau)] = E[X(t+\tau)X(t)] = E[X(t+\tau)X(t + \tau - \tau)] = R_X(-\tau),$$ and so the autocorrelation function is an even function of its argument. A finite-power second-order stationary random process has the properties that 1. Its mean $$E[X(t)]$$ is a constant 2. Its autocorrelation function $$R_X(\tau) = E[X(t)X(t+\tau)]$$ is a function of $$\tau$$, the time separation of the random variables $$X(t)$$ and $$X(t+\tau)$$, and does not depend on $$t$$ at all. The assumption of stationarity simplifies the description of a random process to some extent but, for engineers and statisticians interested in building models from experimental data, estimating all those CDFs is a nontrivial task, particularly when there is only a segment of one sample path (or realization) $$x(t)$$ on which measurements can be made. Two measurements that are relatively easy to make (because the engineer already has the necessary instruments on his workbench (or programs in MATLAB/Python/Octave/C++ in his software library) are the DC value $$\frac 1T\int_0^T x(t)\,\mathrm dt$$ of $$x(t)$$ and the autocorrelation function $$R_x(\tau) = \frac 1T\int_0^T x(t)x(t+\tau)\,\mathrm dt$$ (or its Fourier transform, the power spectrum of $$x(t)$$). Taking these measurements as estimates of the mean and the autocorrelation function of a finite-power process leads to a very useful model that we discuss next. A finite-power random process is called a wide-sense-stationary (WSS) process (also weakly stationary random process which fortunately also has the same initialism WSS) if it has a constant mean and its autocorrelation function $$R_X(t_1, t_2) = E[X(t_1)X(t_2)]$$ depends only on the time difference $$t_1 - t_2$$ (or $$t_2 - t_1$$). Note that the definition says nothing about the CDFs of the random variables comprising the process; it is entirely a constraint on the first-order and second-order moments of the random variables. Of course, a finite-power second-order stationary (or $$N^{\text{th}}$$-order stationary (for $$N>2$$) or strictly stationary) random process is a WSS process, but the converse need not be true. A WSS process need not be stationary to any order. Consider, for example, the random process $$\{X(t)\colon X(t)= \cos (t + \Theta), -\infty < t < \infty\}$$ where $$\Theta$$ takes on four equally likely values $$0, \pi/2, \pi$$ and $$3\pi/2$$. (Do not be scared: the four possible sample paths of this random process are just the four signal waveforms of a QPSK signal). Note that each $$X(t)$$ is a discrete random variable that, in general, takes on four equally likely values $$\cos(t), \cos(t+\pi/2)=-\sin(t), \cos(t+\pi) = -\cos(t)$$ and $$\cos(t+3\pi/2)=\sin(t)$$, It is easy to see that in general $$X(t)$$ and $$X(s)$$ have different distributions, and so the process is not even first-order stationary. On the other hand, $$E[X(t)] = \frac 14\cos(t)+ \frac 14(-\sin(t)) + \frac 14(-\cos(t))+\frac 14 \sin(t) = 0$$ for every $$t$$ while \begin{align} E[X(t)X(s)]&= \left.\left.\frac 14\right[\cos(t)\cos(s) + (-\cos(t))(-\cos(s)) + \sin(t)\sin(s) + (-\sin(t))(-\sin(s))\right]\\ &= \left.\left.\frac 12\right[\cos(t)\cos(s) + \sin(t)\sin(s)\right]\\ &= \frac 12 \cos(t-s). \end{align} In short, the process has zero mean and its autocorrelation function depends only on the time difference $$t-s$$, and so the process is wide sense stationary. But it is not first-order stationary and so cannot be stationary to higher orders either. Even for WSS processes that are second-order stationary (or strictly stationary) random processes, little can be said about the specific forms of the distributions of the random variables. In short, A WSS process is not necessarily stationary (to any order), and the mean and autocorrelation function of a WSS process is not enough to give a complete statistical description of the process. Finally, suppose that a stochastic process is assumed to be a Gaussian process ("proving" this with any reasonable degree of confidence is not a trivial task). This means that for each $$t$$, $$X(t)$$ is a Gaussian random variable and for all positive integers $$n \geq 2$$ and choices of $$n$$ time instants $$t_1$$, $$t_2$$, $$\ldots, t_n$$, the $$N$$ random variables $$X(t_1)$$, $$X(t_2)$$, $$\ldots, X(t_n)$$ are jointly Gaussian random variables. Now a joint Gaussian density function is completely determined by the means, variances, and covariances of the random variables, and in this case, knowing the mean function $$\mu_X(t) = E[X(t)]$$ (it need not be a constant as is required for wide-sense-stationarity) and the autocorrelation function $$R_X(t_1, t_2) = E[X(t_1)X(t_2)]$$ for all $$t_1, t_2$$ (it need not depend only on $$t_1-t_2$$ as is required for wide-sense-stationarity) is sufficient to determine the statistics of the process completely. If the Gaussian process is a WSS process, then it is also a strictly stationary Gaussian process. Fortunately for engineers and signal processors, many physical noise processes can be well-modeled as WSS Gaussian processes (and therefore strictly stationary processes), so that experimental observation of the autocorrelation function readily provides all the joint distributions. Furthermore since Gaussian processes retain their Gaussian character as they pass through linear systems, and the output autocorrelation function is related to th input autocorrelation function as $$R_y = h\tilde{h}R_X$$ so that the output statistics can also be easily determined, WSS process in general and WSS Gaussian processes in particular are of great importance in engineering applications. • Could you, please, comment on "White Noise" in that sense? By definition the Autocorrelation at $\tau = 0$ is the variance of the random variables. Does it mean that AWGN (Additive White Gaussian Noise) has infinite variance? I ask it because usually people write $n(t) ~ N(0, {N}_{0} / 2)$, is wrong? Should it be written $n(t) ~ N(0, \delta(0) {N}_{0} / 2)$? Thanks. – Royi Jan 9 '12 at 6:57 • @Drazick Please ask a separate question. – Dilip Sarwate Jan 9 '12 at 14:23 • This is a fantastic mini-course in the definition of stationary processes. I've never seen anything like it--laid out so methodically and clearly. Community Wiki? – abalter Feb 3 '18 at 21:57 • @Dilip Sarwate Excuse me for my ignorance. In the example. Why is E[X(t)]=0 for all t ? Did you assume ergodicity? How did you derive the probability density function of X(t) from the probability density function of theta to compute the expected value? E[X(t)X(s)] = E[cos(t+theta)*cos(s+theta)] right? Which steps did you take to simplify this expression and get to what you wrote? Thanks – VMMF Feb 8 '18 at 16:12 • @VMMF There is NO ergodicity used. $X(t)=\cos(t+\Theta)$ is a discrete random variable because $\Theta$ is a discrete random variable and it takes on values $\pm\cos(t)$ and $\pm\sin(t)$ with equal probability $\frac 14$. Ergo, $E[X(t)]=0$. $X(t)X(s)$ takes on values $\cos(t)\cos(s)$, $(-\cos(t))(-\cos(s))=\cos(t)\cos(s)$, $\sin(t)\sin(s)$ and $(-\sin(t))(-\sin(s))=\sin(t)\sin(s)$ with equal probability $\frac 14$. Hence, $E[X(t)(X(s)]=\frac 12\big(\cos(t)\cos(s)+\sin(t)\sin(s)\big) = \frac 12\cos(t-s)$. Hence, – Dilip Sarwate Feb 9 '18 at 0:12	{"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": 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": 126, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9249194860458374, "perplexity": 217.93867027852247}, "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-2020-16/segments/1585371656216.67/warc/CC-MAIN-20200406164846-20200406195346-00132.warc.gz"}
https://theanimegallery.com/sego-lily-orhkov/tsivxt.php?tag=29594f-permutation-matrix-squared	Let D = {d ij}, i = 1,…, N; j = 1,…, N consist of the distances or dissimilarities between every pair (i, j) of sampling units. Every row and column therefore contains precisely a single 1 with 0s everywhere else, and every permutation corresponds to a unique permutation matrix. Novel set during Roman era with main protagonist is a werewolf. Go to your Tickets dashboard to see if you won! $\left[ \begin{array}{r} 3 & 6 & -5 & 4 \\ -2 & 0 & 6 & 0 \\ 1 & 1 & 2 & 2 \\ 0 & 3 & -1 & -1 \end{array} \right]$, In Exercises $23-28,$ find the determinant of the matrix. Description : The calculator allows to calculate online the number of permutation of a set of n elements without repetition. And it doesn't need to be … Working on it. Let us assume that one of them, (ii 1,, k ) has length k, and let λ∈ p be an eigenvalue of pλ an kth-root , of unity. We also classify the quantum magic squares that dilate to a quantum permutation matrix with commuting entries and prove a quantitative lower bound on the diameter of this set. An example is 1 0 0 0 1 P= 0 . The vector, as you said, is (i+j). Generating all permutations of a given string, Multiplication of (0,1)-matrix with a (0,1)-vector. $F(A)=c d$c. A permutation of a set of n elements is an arrangement of this n elements. Odd Permutation. For example m[0][100] will be very far from m[100][0] and all others with similar probability. 0. On the algorithmic front, the seriation problem was shown to be NP-Complete by [10]. In this paper a square cellular network for data per-mutation in a SIMD model is described. Shifting rows and columns (if you want to rearrange the rows, you must rearrange the columns in the same way, or the matrix won't make sense for most operations) is called "permutation" of the matrix. It has n 2 = 4 2–permuters only, and realizes an arbitrary permutation pattern in two passes. The number of iterations . Now suppose that we multiply this adjacency matrix times itself (i.e. Sparse matrices provide efficient storage of double or logical data that has a large percentage of zeros. The function above works for d<=N, but not d>N (the lower right half of the matrix). This matrix and the next 19 matrices are represented at the same time. A Matrix that exchanges 2 or more rows is called a permutation matrix. If the transpose of a matrix is equal to the negative of itself, the matrix is said to be skew symmetric. The symmetries are determined by the images of the vertices, that can, in turn, be described by permutations. $$\text{max}(n_a, m_a, n_b, m_b)$$. permutation matrices of size n: Section: A magic square is a square of … The Order of a Permutation. So this would be square. Permutation Matrix A permutation matrix is a square matrix1 in which is zero everywhere apart from having only one ‘1’ on every row and in every column. Now to deal with the lower right half. In the ShuffleNet architecture [14, 23], �is preset by the designers and will be called “manual”. How can I pay respect for a recently deceased team member without seeming intrusive? But it can be some other way as long as elements of the same diagonal are mostly grouped. 1.Compute an orthogonal matrix Q2R m, an upper triangular matrix R2R n, and a permutation matrix P2R such that QT AP= R 0 : 2.Compute QT b= c d : 3.Solve Ry= c: 4.Set x= Py: D. Leykekhman - MATH 3795 Introduction to Computational MathematicsLinear Least Squares { 9 1 0 0 Find the determinant of this matrix. Such a matrix is always row equivalent to an identity. For a diagonalizable matrix like yours, you can use the eigendecomposition A = QLQ^-1. Click 'Join' if it's correct, By clicking Sign up you accept Numerade's Terms of Service and Privacy Policy. I (n) for . no squares and division but programatic conditionals are OK), For a more graphical reference, I'm looking for something like this. I don't have an account. A permutation matrix is a square matrix obtained from the same size identity matrix by a permutation of rows. First things first. A square matrix is called a permutation matrix if it contains the entry 1 exactly once in each row and in each column, with all other entries being 0. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. [From BBC's The Story of Maths on Cantor's argument]. EMAILWhoops, there might be a typo in your email. How much did the first hard drives for PCs cost? Det(P) = An example is Find the determinant of this matrix. Your question is a little unclear, but by the graphic you want a function that describes the mapping of that path, for example f(1,2) = 8. i+j gives the index of the diagonal, call it d. There are (d+1)d/2 elements in the diagonals above that one. Here’s an example of a $5\times5$ permutation matrix. Whoops, there might be a typo in your email. Just that the elements walked on the diagonals are mostly grouped nearby. I can break it down into cycles: sigma = <1,4,6>compose<3,5> thanks. How can I confirm the "change screen resolution dialog" in Windows 10 using keyboard only? Not necessarily. I have a set of items of size N. The items are sorted by probability. Thanks for contributing an answer to Stack Overflow! Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide. a 2x6 permutation matrix. however, there's actually no need for the, @lijie: I would expect you to have continuity problems without the, yeah i mean, OP doesn't seem to need it. Find a noninvertible $2 \times 2$ matrix whose entries are four distinct prime numbers, or explain why no such matrix existsb. You must be logged in to bookmark a video. (1 point) A square matrix is called a permutation matrix if each row and each column contains exactly one entry 1, with all other entries being 0. Before we look at determinants, we need to learn a little about permutations. For example m[0][100] will be very far from m[100][0] and all others with similar probability. if PPT = , then P is its own inverse and for every i and j in {1, 2, 3, … n}, ,,(),,1(), T ij jipi j p j ipji Function as.matrix.word () coerces a vector of permutations in word form to a matrix, each row of which is a word. Squaring matrix 5.13. Repeated application of a particular permutation of the elements of an . det (P) =. Default=False. For this reason, using sparse matrices can significantly reduce the amount of memory required for data storage. Permutation Matrix. A permutation matrix is obtained by performing a sequence of row and column interchanges on the identity matrix. The first thing to note is that a square matrix in row echelon form is upper triangular. Skew Symmetric Matrix. Stack Overflow for Teams is a private, secure spot for you and Every row and every column of a permutation matrix contain exactly one nonzero entry, which is 1: There are two 2 2 permutation matrices: [1 0 0 1]; [0 1 1 0]: There are six 3 3 permutation matrices. See Example 4, Which of the following functions $F$ of $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ are linear in both columns? I'm looking for a function f(i,j) to give the position on the permutated matrix/vector. On the one hand, ex A matrix consisting of only zero elements is called a zero matrix or null matrix. Each such matrix, say P, represents a permutation of m elements and, when used to multiply another matrix, say A, results in permuting the rows (when pre-multiplying, to form PA) or columns (when post-multiplying, to form AP) of the matrix A. If two rows of a matrix are equal, its determinant is zero. Let $$A$$ be a square matrix with a row or a column of 0's. The element position in the vector is min(i, N-j). (1 point) A square matrix is called a permutation matrix if each row and each column contains exactly one entry 1, with all other entries being 0. The dimension of square array is specified based on the highest dimension, i.e. Notice that column space of M' is of higher order than the column space of elem'.This implies that there does not exist a linear mapping from elem' to M' because a linear mapping cannot increase the row or column space of a matrix (useful to think about this as a transformation of basis).. Define 2x2 and 3x3 permutation matrices. The amd function tends to be faster than symamd, and also tends to return better orderings than symamd. $F(A)=a c$d. by solving the seriation problem on the squared data matrix. This preview shows page 1 - 2 out of 2 pages.. 6. R squares tells us the proportion of variance in the outcome measure that is explained by the predictors. Who first called natural satellites "moons"? Our educator team will work on creating an answer for you in the next 6 hours. From these three properties we can deduce many others: 4. Active 9 years, 9 months ago. $F(A)=b c$b. See Example 4. See Example 6. Equality of matrices Two matrices $$A$$ and $$B$$ are equal if and only if they have the same size $$m \times n$$ and their corresponding elements are equal. This matrix and the next 19 matrices are represented at the same time. to C1P can be obtained by solving the seriation problem on the squared data matrix. The permutation matrix �is a square binary matrix with exactly one entry of one in each row and each column and zeros elsewhere. The following are the full set of all 2 2 permutation matrices: 1 0 0 1 (1) 0 1 1 0 (2) Let P be an n n permutation matrix. Well, I know this is probably something very simple but it's been many years since school/uni and Wolframalpha isn't helping. Is Statistics less important in the era of big data than in old days? The thing I'm missing now is how to deal when the diagonal index > N, then the (d+1)(d/2) prior elements isn't true anymore. Positional chess understanding in the early game. Let's call it sigma. Why did I measure the magnetic field to vary exponentially with distance? how would I calculate (sigma)^2? We start from the identity matrix , we perform one interchange and obtain a matrix , we perform a second interchange and obtain another matrix , and so on until at the … I. It is denoted by a permutation sumbol of -1. A piece of wax from a toilet ring fell into the drain, how do I address this? By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Efficient search for permutations that contain sub-permutations via array operations? In Exercises 39-54, find the determinant of the matrix.Expand by cofactors on the row or column that appears to make the computations easiest. While full (or dense) matrices store every single element in memory regardless of value, sparse matrices store only the nonzero elements and their row indices. Are you looking for space filling curves? Matrix A must be square. The Order of a Permutation. It doesn't need to be a square matrix, it can be a vector [NN]. So a square matrix is any matrix that has the same number of rows as columns. For the intents of this calculator, "power of a matrix" means to raise a given matrix to a given power. The square root of the matrix can then be computed with the square root of each element of L , as B = QL^(1/2)Q^-1 . Now, if we had matrix such as this one, this has two columns, but only one row, so this wouldn't be square anymore. I assume you mean that all matrix entries are to be non-negative integers and that, correspondingly, the permutations are to be considered as among objects for which these are the counts - that is, an arrangement of [5,7,11,4] like objects is considered a "permutation" or rearrangement of [4,7,3,13] objects. We will treat "self-ties" as zeros, which, effectively, ignores them. And the length of the vector N-abs(N-(i+j)). A square matrix m[N][N] of those items, in C style memory organization, would have elements with similar probabilities spread out. Permutation Matrix A permutation matrix is a matrix obtained by permuting the rows of an identity matrix according to some permutation of the numbers 1 to. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Let the vertices of a square be labeled 1, 2, 3 and 4 (counterclockwise around the square starting with 1 in the top left corner). So if you don't mind I'll remove that +1, so that f(0,0)=0. What does it mean to “key into” something? Therefore, this permutation is also the transformation "diagram" which represents the "type" of the square. How to generate all permutations of a list? Summary : To calculate online the number of permutation of a set of n elements. Matrix permutation, blocks, and images. Note: the permutation "complement to n²+1" is the permutation which makes a correspondence between each number and its complement to n²+1. $\left[ \begin{array}{r} 5 & 2 & 0 & 0 & -2 \\ 0 & 1 & 4 & 3 & 2 \\ 0 & 0 & 2 & 6 & 3 \\ 0 & 0 & 3 & 4 & 1 \\ 0 & 0 & 0 & 0 & 2 \end{array} \right]$, $\left[ \begin{array}{r} 3 & 2 & 4 & -1 & 5 \\ -2 & 0 & 1 & 3 & 2 \\ 1 & 0 & 0 & 4 & 0 \\ 6 & 0 & 2 & -1 & 0 \\ 3 & 0 & 5 & 1 & 0 \end{array} \right]$, $\left[ \begin{array}{r} 1 & 4 & 3 & 2 \\ -5 & 6 & 2 & 1 \\ 0 & 0 & 0 & 0 \\ 3 & -2 & 1 & 5 \end{array} \right]$, Given a square matrix $A^{-1}$, find matrix $A$.$$A^{-1}=\left[\begin{array}{lll}0 & 0 & 1 \\0 & 1 & 0 \\1 & 0 & 0\end{array}\right]$$, $\left[ \begin{array}{r} 1 & 0 & 0 \\ -1 & -1 & 0 \\ 4 & 11 & 5 \end{array} \right]$. 2020 permutation matrix squared	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8286617398262024, "perplexity": 495.4765627905522}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243992159.64/warc/CC-MAIN-20210517084550-20210517114550-00183.warc.gz"}
https://math.stackexchange.com/questions/3013705/prove-o-bigcup-j-1-infty-o-j-and-e-subset-bigcup-j-1-infty-e-j-im/3013706	# Prove $O = \bigcup_{j=1}^\infty O_j$ and $E \subset \bigcup_{j=1}^\infty E_j \implies O-E \subset \bigcup_{j=1}^{+\infty}\left(O_j-E_j\right)$ Prove that $$O = \bigcup_{j=1}^\infty O_j \quad \text{and} \quad E = \bigcup_{j=1}^\infty E_j \quad \implies \quad O-E \subset \bigcup_{j=1}^{+\infty}\left(O_j-E_j\right)$$ Below is my attempted proof, I'm stuck at the last expression. Proof $$O - E = \left(\bigcup_{j=1}^\infty O_j\right) - \left(\bigcup_{j=1}^\infty E_j\right) = \left(\bigcup_{j=1}^\infty O_j\right) \cap \left(\bigcup_{j=1}^\infty E_j\right)^c = \left(\bigcup_{j=1}^\infty O_j\right) \cap \left(\bigcap_{j=1}^\infty E^c_j\right)$$ I'm not sure how to handle the last "intersection of intersections". But I get the feeling my approach is just confusing in general. Thank you. If $$x \in O - E$$, then there exists $$j$$ such that $$x \in O_j$$. Then $$x \in O_j - E_j$$ as well since $$x \notin E_j$$. • What if $O_j=E_j=\{1\}$ for every $j,$ and $E=\emptyset$? Then $\{1\}=O=O-E$ but $\cup_j(O_j-E_j)=\cup_j\emptyset=\emptyset.$ – DanielWainfleet Nov 26 '18 at 7:01 • @DanielWainfleet I misread $E \subset \bigcup_j E_j$ as $E = \bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter. – angryavian Nov 26 '18 at 17:09	{"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9846464395523071, "perplexity": 269.0875263260287}, "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/1575540569146.17/warc/CC-MAIN-20191213202639-20191213230639-00210.warc.gz"}
https://proofwiki.org/wiki/Fourth_Sylow_Theorem	Fourth Sylow Theorem Theorem The number of Sylow $p$-subgroups of a finite group is congruent to $1 \pmod p$. Proof 1 Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$ and $n > 0$. Let $r$ be the number of Sylow $p$-subgroups of $G$. We want to show that $r \equiv 1 \pmod p$. Let $\mathbb S = \set {S \subseteq G: \card S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements. From the reasoning in the First Sylow Theorem, we have: $\size {\mathbb S} = \dbinom {p^n k} {p^n}$ Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements:: $\forall S \in \mathbb S: g \wedge S = g S = \set {x \in G: x = g s: s \in S}$ there are exactly as many Sylow $p$-subgroups as there are orbits whose length is not divisible by $p$. all the terms in the Partition Equation are divisible by $k$, perhaps also divisible by $p$. We can write the Partition Equation as: $\size {\mathbb S} = \size {\Orb {S_1} } + \size {\Orb {S_2} } + \cdots + \size {\Orb {S_r} } + \size {\Orb {S_{r + 1} } } + \cdots + \size {\Orb {S_s} }$ where the first $r$ terms are the orbits containing the Sylow $p$-subgroups: $\Stab {S_i}$ For each of these: $\order G = \size {\Orb {S_i} } \times \size {\Stab {S_i} } = p^n \size {\Orb {S_i} }$ Thus: $\size {\Orb {S_i} } = k$ for $1 \le i \le r$. Each of the rest of the orbits are divisible by both $p$ and $k$, as we have seen. So: $\size {\mathbb S} = k r + m p k$ where: the first term corresponds to the $r$ orbits containing the Sylow $p$-subgroups the second term corresponds to all the rest of the orbits $m$ is some unspecified integer. That is, there exists some integer $m$ such that: $\size {\mathbb S} = \dbinom {p^n k} {p^n} = k r + m p k$ Now this of course applies to the special case of the cyclic group $C_{p^n k}$. In this case, there is exactly one subgroup for each divisor of $p^n k$. In particular, there is exactly one subgroup of order $p^n$. Hence, in this case: $r = 1$ So we have an integer $m'$ such that $\dbinom {p^n k} {p^n} = k + m' p k$. We can now equate these expressions: $\displaystyle k r + m p k$ $=$ $\displaystyle k + m' p k$ $\displaystyle \leadsto \ \$ $\displaystyle r + m p$ $=$ $\displaystyle 1 + m' p$ dividing by $k$ $\displaystyle \leadsto \ \$ $\displaystyle r - 1$ $=$ $\displaystyle p \paren {m' - m}$ $\displaystyle \leadsto \ \$ $\displaystyle r - 1$ $\equiv$ $\displaystyle 0 \pmod p$ $\displaystyle \leadsto \ \$ $\displaystyle r$ $\equiv$ $\displaystyle 1 \pmod p$ and the proof is complete. $\blacksquare$ Proof 2 Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$. Let $r$ be the number of Sylow $p$-subgroups of $G$. Let $H$ be a Sylow $p$-subgroup of $G$. We have that: $\order H = p^n$ $\index G H = m$ Let $S_1, S_2, \ldots, S_m$ denote the elements of the left coset space of $G / H$. We have that $H$ acts on $G / H$ by the rule: $g * S_i = g S_i$ for $S_i \in G / H$. Unless $H = G$ and $r = 1$, there is more than $1$ orbit. We have that $H$ is the stabilizer of the coset $H$, which must be one of $S_1, S_2, \ldots, S_m$. Let $S_1, S_2, \ldots, S_k$ be the elements of $G / H$ whose stabilizer is $H$. From the Orbit-Stabilizer Theorem and from $\order H = p^n$ we see there are $2$ cases: $(1): \quad$ The orbit of $S_i$ contains $p^t$ elements where $0 < t < n$ $(2): \quad$ The orbit of $S_i$ contains only the element $S_i$. $(2)$ occurs if and only if $S_i$ is one of the cosets $S_1, S_2, \ldots, S_k$ whose stabilizer is $H$. So counting the elements of $G / H$, we see that: $m = k + u p$ or: $m \equiv k \pmod p$ From the Fifth Sylow Theorem, we have: $m \equiv k r \pmod p$ and so: $k r \equiv k \pmod p$ from which it follows: $r \equiv 1 \pmod p$ because $k \not \equiv 0 \pmod p$. Hence the result. $\blacksquare$ Also known as Some sources call this the second Sylow theorem. Others merge this result with what we call the Fifth Sylow Theorem and call it the third Sylow theorem. Source of Name This entry was named for Peter Ludwig Mejdell Sylow. Historical Note When cracking open the structure of a group, it is a useful plan to start with investigating the prime subgroups. The Sylow Theorems are a set of results which provide us with just the sort of information we need. Ludwig Sylow was a Norwegian mathematician who established some important facts on this subject. He published what are now referred to as the Sylow Theorems in $1872$. The name is pronounced something like Soolof. There is no standard numbering for the Sylow Theorems. Different authors use different labellings. Therefore, the nomenclature as defined on $\mathsf{Pr} \infty \mathsf{fWiki}$ is to a greater or lesser extent arbitrary.	{"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": 0.9615802764892578, "perplexity": 153.93274492578257}, "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-2020-16/segments/1585371830894.88/warc/CC-MAIN-20200409055849-20200409090349-00272.warc.gz"}
http://www.physicsforums.com/showthread.php?t=297491	# Energy balance from mechanical to electrical energy by PhiliosKassin Tags: balance, electrical, energy, mechanical P: 1 A mass is winded via string aroung a flywheel and allowed to fall a specific height before detachement. The flywheel is connected to a small DC motor via gears and starts generating electricity once the mass is allowed to fall.. Now without the DC motor ( generator) the equation of energy balance is: mgh = (0.5mv^2+ (0.5Iω^2) + ( 0.5Iω^2 x (n1/n2) where I is the moment of inertia of flywheel and ω is the angular velocity ( which should be equal to the angular max velocity at point of mass detachment) n1 = number of revs of flywheel before detachement of mass n2 = number of revs of flywheel after detachement the third term in the equation is the work done to overcome friction. Now the question is : What form is the equation going to have when a DC motor is attached to the flywheel to be allowed to transform rotational energy to electrical? mgh = ? the electrical energy is in the form of UIt P: 107 It seems that one has to assume the friction is independent of the load. Let n3 = number of revs of flywheel after detachment, and E = electrical energy. I think it is not hard to see E = (0.5Iω^2)(n2-n3)/n2 Is that what you are after? P: 4,664 Suppose the height of initial release of the mass minus the height of detachment is h. Then if the mass falls freely (without flywheel), the total kinetic energy of the mass is mgh = (1/2)mv2. But in the case where it is attached to the flywheel, its final velocity at point of detachment is v1. So its energy at release is now (1/2)mv12. So the missing energy is (1/2)m[v2 - v12]. This was all transferred to the flywheel. If the flywheel were frictionless and the generator perfect, all the flywheel energy would be converted to electric power. All inertial energy remaining in the flywheel after release of the mass would also be converted to electric power. To maximize the total electric energy, the objective then is to make the final velocity of the mass v1 as small as possible. If the flywheel were massive and v1 were small, then the electrical energy would be nearly mgh. If mgh were expressed in joules, then mgh/3600 would be the energy in watt-hours, or in volt amp hours. P: 1 ## Energy balance from mechanical to electrical energy pls what is the energy balance equation in electric machines? Related Discussions Classical Physics 2 Classical Physics 2 Electrical Engineering 2 General Physics 4 Introductory Physics Homework 3	{"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": 0.9716532230377197, "perplexity": 912.0943088162514}, "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-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00184-ip-10-147-4-33.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/electron-charge.106906/	# Electron charge 1. Jan 15, 2006 ### Punchlinegirl You have 0.6 kg of water. One mole of water has a mass of 18 g/mol and each molecule of water contains 10 electrons since water is H2O. What is the total electron charge contained in this volume of water? Answer in units for C. First I converted 0.6 kg to 600 g. Then I divided that by 18 to get the number of moles, which was 33.3. Then I multiplied it by Avogadro's number to get the number of molecules and got 2.01 x 10^25. Then I multiplied it by the number of electrons, and got 2.01 x 10^26. Finally, I multiplied it by the electon charge, 1.602 x 10^-19, and got 3.21 x 10^7. This isn't right. 2. Jan 15, 2006 ### stunner5000pt it seems fine to me.. why do you think it's wrong? 3. Jan 16, 2006 ### Punchlinegirl It's an online homework problem, and when I typed in the answer it said it was wrong... I have no idea why though... I don't know any other way to do it. Similar Discussions: Electron charge	{"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": 0.8270389437675476, "perplexity": 832.8083249152426}, "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/1490218189589.37/warc/CC-MAIN-20170322212949-00371-ip-10-233-31-227.ec2.internal.warc.gz"}
https://socratic.org/questions/how-do-you-solve-4-2-5x-6-3-1-3x	Algebra Topics # How do you solve 4( 2- 5x ) = 6- 3( 1- 3x )? Jul 28, 2018 $x = \frac{5}{29}$ #### Explanation: $4 \left(2 - 5 x\right) = 6 - 3 \left(1 - 3 x\right)$ Expand the brackets $8 - 20 x = 6 - \left(3 - 9 x\right)$ $8 - 20 x = 6 - 3 + 9 x$ Simplify $8 - 20 x = 3 + 9 x$ Move the $x$ onto one side $29 x = 5$ Divide both sides by $29$ $x = \frac{5}{29}$ Jul 28, 2018 $x = \frac{5}{29}$ #### Explanation: We can distribute the $4$ on the left side, and the $- 3$ on the right. Doing this, we now have $8 - 20 x = 6 - 3 + 9 x$ We can combine like terms to get $8 - 20 x = 3 + 9 x$ Let's get our constants on one side by subtracting $8$ from both sides: $- 20 x = - 5 + 9 x$ Next, subtract $9 x$ from both sides to get $- 29 x = - 5$ Lastly, we can divide both sides by $- 29$ to get $x = \frac{5}{29}$ Hope this helps! ##### Impact of this question 1361 views around the world	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 20, "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": 0.8446248173713684, "perplexity": 1254.2641034604183}, "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-2021-10/segments/1614178369553.75/warc/CC-MAIN-20210304235759-20210305025759-00186.warc.gz"}
http://tex.stackexchange.com/questions/59197/algorithmicx-how-to-use-latex-maths-delimiters	Algorithmicx — How to use latex maths delimiters Today, I decided to migrate from using TeX [$..$] to LaTeX [$$..$$] delimiters for inline maths, and, as usual, fell off the cliff of my somewhat precarious understanding of TeXian gastroenterology, in the middle of an algpseudocode environment. Here's my MWE: \documentclass{article} \usepackage{algpseudocode} \begin{document} \begin{algorithmic}[1] \Procedure{Trigger Dispatching Scheduler}{} \State \Call{MqOpen}{$\aleph$} \State \Call{MqOpen}{$$\aleph$$\relax} %\State \Call{MqOpen}{$$\aleph$$} % <== This one fails \EndProcedure \end{algorithmic} \end{document} Uncommenting the %-line causes it to fail (PdfLaTeX and XeLaTeX) with: ! Argument of \) has an extra }. <inserted text> \par l.14 \State \Call{MqOpen}{$$\aleph$$} % <== This one fails What, if anything, am I doing wrong? If anyone could explain in fairly simple terms what's going on here, I'd be most grateful. - Does this have anything to do with the word 'fragile' (which precisely describes my understanding of these things) – Brent.Longborough Jun 9 '12 at 15:00 I think so (based on Why is the ifthen package obsolete?) and algorithmicx(http://ctan.org/pkg/algorithmicx) uses [ifthen to condition on many levels. For example, using \State \Call{MqOpen}{\protect$$\aleph\protect$$} works. Not exactly sure why \relax makes it work though. – Werner Jun 9 '12 at 15:03 I give an answer below saying it's a bug in ifthen package, but I think I'm going to reclassify it as a feature. In unmodified LaTeX2e $$ \nd $$ are fragile commands and as such they should be preceded by \protect if used in a moving argument (as the argument of \equal). \ifthenelse{\equal{\protect$$a\protect$$}{\protect$$a\protect$$}}{\typeout Y}{\typeout N} does work as intended. If you load the fixltx2e package part of the core base distribution then $$ and $$ are made robust so you (and even I) might have expected that \protect is not necessary. However the due to the way \ifthenelse overloads the syntax to be a boolean logic parenthesis rather than math start, $$ is made fragile again. The "fix" below locally gives \($$ robust definitions however due to the "interesting" expansion order implemented in ifthenelse This local redefinition applies to the entire predicate not just the arguments to \equal and so the "fix" is entirely wrong as it removes the ifthenelse definition of entirely. Sorry (if you want to take away the green tick I'd understand:-) but unless sleeping on it gives me a better idea, or another contributor has a better idea, I think the only thing I can do is more clearly document in ifthen that if you use the ifthen syntax commands (\or\and\isodd etc inside an \equals test, then you need to \protect them. Sorry it's a bug. \ifthenelse uses delimiters $$ and $$ to brackets boolean expressions. These are not made safe in the strings used by ifthenelse \equal test. The following inserts a local definition to make things safer, \let$$\relax\let$$\relax without this line you get the same error shown in your MWE. \documentclass{article} \usepackage{ifthen} \makeatletter \long\def\ifthenelse#1{% \toks@{#1}% \TE@repl\or\TE@or \TE@repl\and\TE@and \TE@repl\not\TE@neg \TE@repl\OR\TE@or \TE@repl\AND\TE@and \TE@repl\NOT\TE@neg \begingroup \let\protect\@unexpandable@protect \def\@setref##1##2##3{% \ifx##1\relax\z@\else\expandafter##2##1\fi}% \def\value##1{\the\csname c@##1\endcsname}% \let\equal\TE@equal \let$$\TE@lparen \let$$\TE@rparen \let\isodd\TE@odd \let\lengthtest\TE@length \let\isundefined\TE@undef \begingroup \let\@tempa\relax\let\@tempb\relax \let$$\relax\let$$\relax \xdef\@gtempa{\expandafter\TE@eval\the\toks@\TE@endeval}% \endgroup \@gtempa \expandafter\endgroup\ifTE@val \expandafter\@firstoftwo \else \expandafter\@secondoftwo \fi} \makeatother \ifthenelse{\equal{a}{a}}{\typeout Y}{\typeout N} \ifthenelse{\equal{a}{b}}{\typeout Y}{\typeout N} \ifthenelse{\equal{$$a$$}{$$a$$}}{\typeout Y}{\typeout N} \ifthenelse{\equal{$$a$$}{$$b$$}}{\typeout Y}{\typeout N} \stop This produces Y N Y N ) No pages of output. on the terminal showing the tests worked. - Thanks, David. What do you think is the best way of "implementing" this?: (a) make a patched local copy of ifthen.sty, or (b) put your code into a local fixifthen.sty and then \usepackage{fixifthen}? – Brent.Longborough Jun 9 '12 at 21:14 actually I'm not sure that's quite the right fix (I may have broken the ifthen use of \( I need to make a couple of test cases. As it's clearly a bug I would just patch ifthen.sty locally (putting some note in the \ProvidesPackage line to note it is modified) that way the document doesn't need changing once the main distribution gets fixed. – David Carlisle Jun 9 '12 at 22:02	{"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": 0.945523738861084, "perplexity": 3004.574531406553}, "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-2013-20/segments/1368703830643/warc/CC-MAIN-20130516113030-00097-ip-10-60-113-184.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/calculate-force-acting-on-rod-to-holding-it-stable.649737/	# Calculate force acting on rod to holding it stable. 1. Nov 5, 2012 ### sareba 1. The problem statement, all variables and given/known data A uniform rod AB of weight 20N is hinged at a fixed point A. A force acts at B holding it in equilibrium at 30 degrees to the vertical through A. calculate force acting at B if it is perpendicular to the rod. 3. The attempt at a solution I figured that the angle between the weight of the rod and the force on B is 60 degrees but i cant seem to find the angle between the weight and the normal reaction force at the hinge... 2. Nov 5, 2012 ### tiny-tim hi sareba! easy … there's only three forces on the rod, and so they must all go through the same point (why? ), so just draw the reaction force (it's not the normal reaction force, btw) so that it goes through the point where the other two forces meet 3. Nov 5, 2012 ### sareba Yes. I did that. If the forces are in equilibrium they should meet at one point. But still I cant calculate the answer with just one force and one angle known... I am sure i am overlooking something... Here is a diagram i came up with... I need to find one of the angles with a question mark on them... File size: 470.8 KB Views: 97 4. Nov 5, 2012 ### tiny-tim ah yes, that's exactly the right diagram ok, call that bottom point C, and the length of the rod 2L … you want to find angle GCA you know the length of GA, and the angle CGA, and you can easily find the length of GC then find the length of AC, and then the angle GCA 5. Nov 5, 2012 ### sareba I am sorry. I didnt quite follow you. How do i know the length of GA? Do you mean GC? 6. Nov 5, 2012 ### tiny-tim GA = L, half the length of the rod Similar Discussions: Calculate force acting on rod to holding it stable.	{"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": 0.9573381543159485, "perplexity": 526.0835172479161}, "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-2017-34/segments/1502886106996.2/warc/CC-MAIN-20170820223702-20170821003702-00706.warc.gz"}
https://brainder.org/2011/07/13/inverse-normal-transformation-in-solar/	# Inverse normal transformation in SOLAR SOLAR software can, at the discretion of the user, apply a rank-based inverse-normal transformation to the data using the command inormal. This transformation is the one suggested by Van der Waerden (1952) and is given by: $\tilde y_i = \Phi^{-1}\left\{\dfrac{r_i}{n+1}\right\}$ where $\tilde y_i$ is the transformed value for observation $i$, $\Phi^{-1}\left\{\cdot\right\}$ is the probit function, $r_i$ is the ordinary rank of the $i$-th case among $n$ observations. This transformation is a particular case of the family of transformations discussed in the paper by Beasley et al. (2009). The family can be represented as: $\tilde y_i = \Phi^{-1}\left\{\dfrac{r_i+c}{n-2c+1}\right\}$ where $c$ is a constant and the remaining variables are as above. The value of $c$ varies for different proposed methods. Blom (1958) suggests $c=3/8$, Tukey (1962) suggests $c=1/3$, Bliss (1967) suggests $c=1/2$ and, as just decribed, Van der Waerden suggests $c=0$. Interesting enough, the Q-Q plots produced by Octave use the Bliss (1967) transformation. An Octave/matlab function to perform these transformations in arbitrary data is here: inormal.m (note that this function does not require or use SOLAR). ## Version history • 23.Jul.2011: First public release. • 19.Jun.2014: Added ability to deal with ties, as well as NaNs in the data. ## 3 thoughts on “Inverse normal transformation in SOLAR” 1. Useful! Here is a function to do the same thing in Python. from scipy.stats import rankdata, norm def rank_based_inv_norm(x, c=0): “”” Perform the rank-based inverse normal transformation on data x. Reference: Beasley TM, Erickson S, Allison DB. Rank-based inverse normal transformations are increasingly used, but are they merited? Behav Genet. 2009; 39(5):580-95. :param x: input array-like :param c: constant parameter :return: y “”” return norm.ppf((rankdata(x) + c) / (x.size – 2*c + 1)) • Hey Sam — good to have this in other languages. You may need to add the extra lines to treat ties, otherwise the results aren’t accurate for all cases. Also note that the default is c=3/8, whereas you’re using c=0 (both are fine, just different). Cheers, Anderson • I chose c=0 because most of the time we will probably want to replicate the SOLAR output (in the IoL lab anyway). scipy.rankdata has a number of optional methods for handling ties, the default is ‘average’, which I just assumed was the one SOLAR used, but couldn’t find a reference. For continuous traits with high precision the probability of tied ranks is quite low, so I think this function will give us the same output as SOLAR most of the time, but need to check this.	{"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": 14, "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": 0.8513650298118591, "perplexity": 1500.3551370821235}, "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/1590347390442.29/warc/CC-MAIN-20200526015239-20200526045239-00339.warc.gz"}
https://www.fachschaft.informatik.tu-darmstadt.de/forum/viewtopic.php?f=564&t=37638	## Integration constant acceleration Forever Neuling Beiträge: 7 Registriert: 18. Jun 2017 20:02 ### Integration constant acceleration Hallo, I have a question about the calculation of position by using Euler method in exercise 8 as shown in the attached picture. Is it not like "y = 1/2 * (-10) * t^2"? For t = 2, y = –20? Why is it –30? Dateianhänge [email protected] (53.91 KiB) 12948 mal betrachtet Zuletzt geändert von Forever am 16. Feb 2018 17:03, insgesamt 1-mal geändert. Polona Windoof-User Beiträge: 40 Registriert: 10. Okt 2010 10:45 ### Re: Integration constant acceleration Hi, Look in the script: pos_new = pos_old + v_old * deltaT. The solution actually uses the current velocity. Perhaps you will find the explanation here https://gafferongames.com/post/integration_basics/ better. Kind regards Polona Forever Neuling Beiträge: 7 Registriert: 18. Jun 2017 20:02 ### Re: Integration constant acceleration if I use pos_new = pos_old + v_old * deltaT pos_old = 0.5 * (-10) * 1^2 = -5 ? v_old = (-10) * 1= -10 ? pos_new = -5 + (-10) * 1 = -15 ? and if I use y = 1/2 * (-10) * t^2" then y = 1/2 * (-10) * 2^2 = 20 ? RobDangerous Computerversteher Beiträge: 363 Registriert: 14. Okt 2014 17:05 ### Re: Integration constant acceleration The formula used in the solution is pos_new = pos_old + v_new * deltaT which is an equally valid approximation (calculate the new speed first vs calculate the new position first). In the exam both variations would get you all the points. But your original question was basically "why is it not the analytical solution" which is actually part of question b. The euler integration is an approximation which does not use t, just delta t. Analytical solutions are typically not usable because many values depend on user input and whatnot and are not formulas you could do proper math on (for example acceleration - typically that is not a constant but some value which changes in unpredictable ways). For that reason we do approximations on small time steps instead of using the proper formulas. Forever Neuling Beiträge: 7 Registriert: 18. Jun 2017 20:02 ### Re: Integration constant acceleration ok, thanks a lot.	{"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": 0.8859744668006897, "perplexity": 3920.6808495475743}, "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-2021-10/segments/1614178358976.37/warc/CC-MAIN-20210227144626-20210227174626-00447.warc.gz"}
https://www.physicsforums.com/threads/polar-integration.642244/	# Polar Integration 1. Oct 8, 2012 ### ruzfactor Hi I have a function [e.g. f(r)] which I want to integrate over r and θ. What would be the integration form? Which one is correct? ∫∫f(r) drdθ OR ∫∫f(r) rdrdθ Please explain. Also, can it be said as area integration as well like the one in cartesian coordinate? 2. Oct 8, 2012 ### haruspex it depends what you mean by integrating over r and theta. Taken at face value, that just means drdθ. But if you mean that you want to integrate over an area using polar coordinates then it's rdrdθ. The reason is that you can carve up an area into small annular sectors, each running over a range of r to r+dr and θ to θ+dθ. The dimensions of such an area element are dr by rdθ, and approximate a rectangle, so the area of the element is rdrdθ. 3. Oct 8, 2012 ### ruzfactor Thanks. I have a function of r. For example at theta=0, the value of the function have different value at different r values (e.g. r=0 to r=a). I want to evaluate this function over 2∏. So I thought first integrating the function about r and then theta. That is where I am confused, whether to use ∫∫f(r) drdθ or ∫∫f(r) rdrdθ. Please see the attachment where my problem is explained in a pic. Also, I thought conversion from cartesian coordinate (dxdy) using jacobian makes it rdrdθ in polar coordinate. File size: 16.1 KB Views: 79 4. Oct 9, 2012 ### haruspex Clearly you are trying to integrate over an area, so rdrdθ is the correct form. The Jacobian arises from coordinate transformations where the two coordinate systems describe the same region. In Cartesian, an element of area is dxdy; in polar it is rdrdθ. 5. Oct 9, 2012 ### zmp3 In any double integration problem, you can usually start with either. In this situation, you'd be better off starting with the integration of the radius first then integrating with respect to theta. 6. Oct 10, 2012 ### haruspex The OP was not trying to choose between drdθ and dθdr, but between drdθ and rdrdθ. 7. Oct 10, 2012 ### HallsofIvy Staff Emeritus If your problem is just "given $f(r,\theta)$, integrate it", then you would have $\int\int f(r, \theta)drd\theta$. If, however, the problem is "given $f(r,\theta)$, integrate over a given area in the plane", then you would have $\int\int f(r,\theta) r drd\theta$ because "$rdrd\theta$" is the "differential of area" in polar coordinates. In particular, if you are converting $\int\int f(x,y) dxdy$ to polar coordinates, because dxdy is the "differential of area" in Cartesian coordinates, it would become $\int\int f(r,\theta)r drd\theta$. 8. Oct 10, 2012 ### ruzfactor Thanks. Actually the problem says that, integrate f(r,θ) in radial and circumferential direction. So it is a bit confusing. I guess rdrdθ could be used depending on my problem. Here, the function f(r,θ) is the mode shape function of a circular plate. Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook Similar Discussions: Polar Integration	{"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": 0.9592940807342529, "perplexity": 1410.0049328663786}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "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-2017-43/segments/1508187822822.66/warc/CC-MAIN-20171018070528-20171018090528-00208.warc.gz"}
https://byjus.com/ncert-solutions-for-class-9-maths-chapter-3-coordinate-geometry-ex-3-2/	Ncert Solutions For Class 9 Maths Ex 3.2 Ncert Solutions For Class 9 Maths Chapter 3 Ex 3.2 (i) Name the horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane? (ii) Name each part of the plane formed by the above two lines? (iii) Name the point where these two lines intersect. Sol: (i) The name of horizontal and vertical lines drawn to determine the position of any point in the Cartesian plane is x-axis and y-axis respectively. (ii) The name of each part of the plane formed by these two lines x-axis and y-axis is quadrants. (iii) The point where these two lines intersect is called origin. Q2. Answer the following referring to the figure given below: (I) co-ordinates of B. (II) co-ordinates of C. (III) co-ordinates of the point L. (IV) co-ordinates of the point M. (V) ordinate of the point H. (VI) abscissa of the point D. (VII) The point i.e. identified by the co-ordinates (3,5)$(-3,-5)$ (VIII) The point i.e identified by the co-ordinates (2,4)$(2,-4)$. Solution: (i) The co-ordinates of B is (5,2)$(-5,2)$. (ii) The co-ordinates of C is (5,5)$(5,-5)$. (iii) The co-ordinates of the point L is (0,5)$(0,5)$. (iv) The co-ordinates of the point M is (3,0)$(-3,0)$. (v) Ordinate means y coordinate of point H. So, ordinate of point H is -3. (vi) Abscissa means x co-ordinate of point D. So, abscissa of the point D is 6. (vii) The point identified by the coordinates (3,5)$(-3,-5)$ is E. (viii) The point identified by the coordinates (2,4)$(2,-4)$ is G.	{"extraction_info": {"found_math": true, "script_math_tex": 8, "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": 0.9179772138595581, "perplexity": 1090.7558517231294}, "config": {"markdown_headings": false, "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-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00249.warc.gz"}
http://mathhelpforum.com/pre-calculus/78812-gains-over-time.html	1. ## Gains over time I'll try to explain as thoroughly as possible. A user gains 1000 population per hour, that is broken down to 0.27 population per second (1000/3600), if X time has passed since the update then the population gain is X0.27 and then capped at the maximum capacity that a user can have. The problem is when I want to calculate the gold gain, the gain is dependant on the population and in which the user will gain 1 gold per population that they have per hour. So if they have 1000 population they will gain 1000 population per hour. However to do it the same as above is not accurate and it doesn't take into consideration the gains over time. ---- Population gain: GainPerHour = 1000; GainPerSecond = GainPerHour / 3600; ActualGain = TimeDiffInSeconds GainPerSec; Population = Population + ActualGain; I did the same for gold gain without realising how inaccurate it was, and I have no idea at all how to make it accurate without placing it within a loop for each second and calculating it on a per second basis, which would be far too slow. One possible solution I think is to do the above for gold but with the OLD population value from before i was updated, and then a formula to work out the gain from what population was added, but with the capacity things that too could become complex. For example if 5000 seconds have passed but the user would have capped their population at 1000 seconds, it would need to take that into consideration for the actual gold gain. I'm totally confused and lost at how to do this, any help is very much appreciated. I just hope you guys can make sense out of this.. 2. After more research and asking about, I've learnt that this will use a series equation.. http://en.wikipedia.org/wiki/Series_(mathematics) That's as much progress as I've managed to make so far on this. But to try and make it easier to understand what I have to do i'll try another example. Say the user gains 1 population per second, the gold income that he would have would be the case of over the course of seconds... 1+2+3+4+5 up to the capacity of the population, so if it capped at 5 population over what was currently there then it would continue as 5+5+5+5 etc up to the end of the range. I'm so terrible at trying to put things into explainations, so I really hope that this part with the first post makes more sense!	{"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": 0.8616862893104553, "perplexity": 496.46088775851416}, "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/1484560285337.76/warc/CC-MAIN-20170116095125-00241-ip-10-171-10-70.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/97147/open-sets-and-poincar%c3%a9s-inequality	# Open sets and Poincaré's inequality In many references, Poincaré inequality is presented in the following way : Let $\Omega\subset \mathbb R^d$ an open bounded set. We can find a constant $C$ which depend of $\Omega$ such that for all $u\in H^1_0(\Omega)$, we have $$\lVert u\rVert_{L^2}\leq C\lVert \nabla u\rVert_{(L^2(\Omega))^d}.$$ In fact it works if $\Omega$ is bounded in one direction. An other sufficient condition is that we can find $v\neq 0$ such that Lebesgue measure of $\{\lambda\in\mathbb R,\lambda v\in \Omega\}$ is finite). My question, maybe a little vague, is the following: is there a "nice" necessary and sufficient condition on $\Omega$ to have Poincaré's inequality? - You know Ziemers Book 'Weakly Differentiable Functions'? Chapter 4 is dedicated to Poincaré type inequalities. – user20266 Jan 7 '12 at 11:19 Yes, but when I looked at it I didn't think about this question. And some pages are missing in Gooble book (which is normal). Anyway, this book is at the library of my university, so I will have a look at it Monday. – Davide Giraudo Jan 7 '12 at 11:29 One generalization I know from one of my teachers can be found here: mathproblems123.wordpress.com/2011/10/05/… This needs $\Omega$ to have Lipschitz boundary, and increases the space of admissible functions $H_0^1(\Omega)$ to a closed subspace of $H^1(\Omega)$ which does not contain the non-zero constant functions. – Beni Bogosel Jan 7 '12 at 17:38 Beni Bogosel: Thanks, I didn't know this result. @Thomas I look at this book, but I didn't find the answer. Maybe should I ask it at MathOverfow. – Davide Giraudo Jan 11 '12 at 10:40 @DavideGiraudo: I suppose such condition can be that $\Omega$ is regular enough such that the Rellich Kondrachov theorem holds. en.wikipedia.org/wiki/Rellich%E2%80%93Kondrachov_theorem – Beni Bogosel Apr 30 '12 at 8:02 Both historically and statistically, the one and only correct name for the inequality in question $$\int\limits_{\Omega}\!\|u(x)\|^2dx\leqslant C\!\int\limits_{\Omega}\!\|\nabla u(x)\|^2dx \quad \forall\,u\in H_0^1(\Omega)\tag{\ast}$$ is to be the Friedrichs inequality. Whenever the Sobolev space $H_0^1(\Omega)$ is defined as a closure of the subspace $C_0^{\infty}(\Omega)$ in $H^1(\Omega)$, the Friedrichs inequality $(\ast)$ stays valid for any open set $\Omega\subset\mathbb{R}^d$ of finite thickness, e.g., bounded in at least one direction. Otherwise, a nonsmooth boundary $\partial\Omega$ requires some correct definition of a zero trace on $\partial\Omega$, in which case the validity of inequality $(\ast)$ depends wholly on the nonsmooth domain geometry, while for certain simple generalizations of the zero trace concept, the necessary and sufficient conditions for $(\ast\ast)$ to be valid have already been found. But this is not the case for the true Poincaré inequality that can be written in the form $$\int\limits_{\Omega}\!\|u(x)\|^2dx\leqslant C\Bigl(\Bigl\|\int\limits_{\Omega}\!u(x)dx\Bigr\|^2+ \int\limits_{\Omega}\!\|\nabla u(x)\|^2dx\Bigr) \quad \forall\,u\in H^1(\Omega)\tag{\ast\ast},$$ or in some other equivalent form. Inequality $(\ast\ast)$ is valid for a bounded domain satisfying, e.g., the cone condition, though the cone condition is not necessary for $(\ast\ast)$ to be valid. Alternatively, there is a bounded domain $\Omega$ with just a single singular point $a\in\partial\Omega$ such that $\partial\Omega\backslash\{a\}\in C^1$ while the inequality $(\ast\ast)$ is not valid. But still no condition on the geometry of the nonsmooth bounded domain $\Omega$ necessary and sufficient for the validity of $(\ast\ast)$ has yet been found. And so far, domains for which the inequality $(\ast\ast)$ is valid remain tagged as the Nikodim domains (see p. 330 in R.E. Edwards "Functional Analysis. Theory and Applications". Dover Publ., N.Y., 1995).	{"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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9340952038764954, "perplexity": 255.6069309992824}, "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/1432207927863.72/warc/CC-MAIN-20150521113207-00050-ip-10-180-206-219.ec2.internal.warc.gz"}
https://direct.mit.edu/neco/article-abstract/29/5/1439/8258/Erratum-to-A-Note-on-Divergences?searchresult=1	Erratum to “A Note on Divergences” by Xiao Liang, Neural Computation Vol. 28, No. 10, pp. 2045–2062 (2016) doi:10.1162/NECO_a_00878 In order to assist readers interested in obtaining new results on divergences, I write this erratum to the version of this article initially published. The second paragraph of Remark 1 should appear as follows: We just disproved Amari’s conjecture for the general case of On page 2058, “ Acknowledgments should appear as follows: I thank the referees very much for helpful comments and suggestions, which led to my revision on section 3. Moreover, I am grateful to the referees for bringing Amari (2016) and Jiao, Courtade, No, Venkat, and Weissman (2015) to my attention. I thank Z. Lei of Fudan University for valuable instructions, comments, and suggestions on the note. On page 2062, “Interference” in line 2 should be...	{"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": 0.8052437901496887, "perplexity": 1486.4084153318474}, "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/1634323585522.78/warc/CC-MAIN-20211022212051-20211023002051-00557.warc.gz"}
https://eprints.iisc.ac.in/5332/	# The influence of microscopic parameters on the ionic conductivity of $SrBi_{2}(Nb_{1-x}V_{x})_{2}O_{9-\delta}(0 \leq x \leq 0.3)$ ceramics Venkataraman, Harihara B and Varma, KBR (2005) The influence of microscopic parameters on the ionic conductivity of $SrBi_{2}(Nb_{1-x}V_{x})_{2}O_{9-\delta}(0 \leq x \leq 0.3)$ ceramics. In: Journal of Physics and Chemistry of Solids, 66 (10). pp. 1640-1646. PDF av137.pdf Restricted to Registered users only Download (335kB) \| Request a copy ## Abstract Layered $SrBi_{2}(Nb_{1-x}V_{x})_{2}O_{9-\delta}$(SBVN) ceramics with x lying in the range 0-0.3 (30 mol%) were fabricated by the conventional sintering technique. The microstructural studies confirmed the truncating effect of $V_{2}O_{5}$ on the abnormal platy growth of SBN grains. The electrical conductivity studies were centred in the 573-823 K as the Curie temperature ties in this range. The concentration of mobile charge carriers (n), the diffusion constant $(D_{O})$ and the mean free path (a) were calculated by using Rice and Roth formalism. The conductivity parameters such as ion-hopping rate $(\omega_{p})$ and the charge carrier concentration (K') term have been calculated using Almond and West formalism. The aforementioned microscopic parameters were found to be $V_{2}O_{5}$ content dependent on $SrBi_{2}(Nb_{1-x}V_{x})_{2}O_{9-\delta}$ ceramics. Item Type: Journal Article Journal of Physics and Chemistry of Solids Elsevier Science Ltd Copyright for this article belongs to Elsevier. A. Ceramics;C. X-ray diffraction; D. Microstructure; D. Diffusion; D. Electrical properties Division of Chemical Sciences > Materials Research Centre 13 Feb 2006 19 Sep 2010 04:23 http://eprints.iisc.ac.in/id/eprint/5332	{"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": 0.868192195892334, "perplexity": 3532.4876596201543}, "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-2022-33/segments/1659882570871.10/warc/CC-MAIN-20220808183040-20220808213040-00201.warc.gz"}
https://quomodocumque.wordpress.com/tag/expanders/	## Breuillard’s ICM talk: uniform expansion, Lehmer’s conjecture, tauhat Emmanuel Breuillard is in Korea talking at the ICM; here’s his paper, a very beautiful survey of uniformity results for growth in groups, by himself and others, and of the many open questions that remain. He starts with the following lovely observation, which was apparently in a 2007 paper of his but which I was unaware of. Suppose you make a maximalist conjecture about uniform growth of finitely generated linear groups. That is, you postulate the existence of a constant c(d) such that, for any finite subset S of GL_d(C), you have a lower bound for the growth rate $\lim \|S^n\|^{1/n} > c(d)$. It turns out this implies Lehmer’s conjecture! Which in case you forgot what that is is a kind of “gap conjecture” for heights of algebraic numbers. There are algebraic integers of height 0, which is to say that all their conjugates lie on the unit circle; those are the roots of unity. Lehmer’s conjecture says that if x is an algebraic integer of degree n which is {\em not} a root of unity, it’s height is bounded below by some absolute constant (in fact, most people believe this constant to be about 1.176…, realized by Lehmer’s number.) What does this question in algebraic number theory have to do with growth in groups? Here’s the trick; let w be an algebraic integer and consider the subgroup G of the group of affine linear transformations of C (which embeds in GL_2(C)) generated by the two transformations x -> wx and x -> x+1. If the group G grows very quickly, then there are a lot of different values of g1 for g in the word ball S^n. But g1 is going to be a complex number z expressible as a polynomial in w of bounded degree and bounded coefficients. If w were actually a root of unity, you can see that this number is sitting in a ball of size growing linearly in n, so the number of possibilities for z grows polynomially in n. Once w has some larger absolute values, though, the size of the ball containing all possible z grows exponentially with n, and Breuillard shows that the height of z is an upper bound for the number of different z in S^n * 1. Thus a Lehmer-violating sequence of algebraic numbers gives a uniformity-violating sequence of finitely generated linear groups. These groups are all solvable, even metabelian; and as Breuillard explains, this is actually the hardest case! He and his collaborators can prove the uniform growth results for f.g. linear groups without a finite-index solvable subgroup. Very cool! One more note: I am also of course pleased to see that Emmanuel found my slightly out-there speculations about “property tau hat” interesting enough to mention in his paper! His formulation is more general and nicer than mine, though; I was only thinking about profinite groups, and Emmanuel is surely right to set it up as a question about topologically finitely generated compact groups in general. ## Superstrong approximation for monodromy groups (and Galois groups?) Hey, I posted a paper to the arXiv and forgot to blog about it! The paper is called “Superstrong approximation for monodromy groups” and it roughly represents the lectures I gave at the MSRI workshop last February on “Thin Groups and Superstrong Approximation.” Hey, as I write this I see that MSRI has put video of these lectures online: But the survey paper has more idle speculation in it than the lectures, and fewer “um”s, so I recommend text over video in this case! I mean, if you like idle speculation. But if you don’t, would you be reading this blog? I’m going to recount one of the idle speculations here, but first: What is superstrong approximation? Let’s say you have a graph on N vertices, regular of degree d. One basic thing you want to know about the graph is what the connected components are, or at least how many there are. That seems like a combinatorial question, and it is, but in a sense it is also a spectral question: the random walk on the graph, thought of as an operator T on the space of functions on the graph, is going to have eigenvalues between [1,-1], and the mutiplicity of 1 is precisely the number of components; the eigenspace consists of the locally constant functions which are constant on connected components. So being connected means that the second-largest eigenvalue of T is strictly less than 1. And so you might say a graph is superconnected (with respect to some positive constant x) if the second-largest eigenvalue is at most 1-x. But we don’t say “superconnected” because we already have a word for this notion; we say the graph has a spectral gap of size x. Now of course any connected graph has a spectral gap! But the point is always to talk about families of graphs, typically with d fixed and N growing; we say the family has a spectral gap if, for some positive x, each graph in the family has a spectral gap of size at least x. (Such a family is also called an expander family, because the random walks on those graphs tend to bust out of any fixed-size region very quickly; the relation between this point of view and the spectral one would be a whole nother post.) When does life hand you a family of graphs? OK, here’s the situation — let’s say you’ve got d matrices in SL_n(Z), or some other arithmetic group. For every prime p, your matrices project to d elements in SL_n(Z/pZ), which produce a Cayley graph X_p, and X_p is connected just when those elements generate SL_n(Z/pZ). If your original matrices generate SL_n(Z), their reductions mod p generate SL_n(Z/pZ); this is just the (not totally obvious!) fact that SL_n(Z) surjects onto SL_n(Z/pZ). But more is true; it turns out that if the group Gamma generated by your matrices is Zariski-dense in SL_n, this is already enough to guarantee that X_p is connected for almost all p. This statement is called strong approximation for Gamma. But why stop there — we can ask not only whether X_p is connected, but whether it is superconnected! That is: does the family of graphs X_p have a spectral gap? If so, we say Gamma has superstrong approximation, which is now seen to be a kind of quantitative strengthening of strong approximation. We know much more than we did five years ago about which groups have superstrong approximation, and what the applications are when this is so. Sarnak’s paper from the same conference provides a good overview. Idle speculation: superstrong approximation for Galois groups Another way to express superstrong approximation is to say that Gamma has property tau with respect to the congruence quotients SL_n(Z/pZ). In the survey paper, I wonder whether there is some way to talk about superstrong approximation for Galois groups with bounded ramification. For instance; let G be the Galois group of the maximal extension of Q which is tamely ramified everywhere, and unramified away from 2,3,5, and 7. OK, that’s some profinite group. I don’t know much about it. By Golod-Shafarevich I could prove it was infinite, unless I couldn’t, in which case I would toss in some more ramified primes until I could. We could ask something like the following. Given any finite quotient Q of G, and any two elements of G whose images generated Q, we get a connected Cayley graph of degree 4 on the elements of Q, by means of those two elements and their inverses. Is there a uniform spectral gap for all those graphs? I have no real reason to think so. But remark: this would imply immediately that every finite-index subgroup of G has finite abelianization, and that’s true. It would also imply that there are only finitely many n such that G surjects onto S_n, and that might be true. Reader survey for those who’ve read this far: do you think there’s a finite set S of primes such that there are tamely ramified S_n-extensions of Q, for n arbitrarily large, unramified outside S? Acknowledgment: I was much aided in formulating this question by the comments on the MathOverflow question I asked about it. ## Gonality, the Bogomolov property, and Habegger’s theorem on Q(E^tors) I promised to say a little more about why I think the result of Habegger’s recent paper, ” Small Height and Infinite Non-Abelian Extensions,” is so cool. First of all: we say an algebraic extension K of Q has the Bogomolov property if there is no infinite sequence of non-torsion elements x in K^* whose absolute logarithmic height tends to 0. Equivalently, 0 is isolated in the set of absolute heights in K^. Finite extensions of Q evidently have the Bogomolov property (henceforth: (B)) but for infinite extensions the question is much subtler. Certainly $\bar{\mathbf{Q}}$ itself doesn’t have (B): consider the sequence $2^{1/2}, 2^{1/3}, 2^{1/4}, \ldots$ On the other hand, the maximal abelian extension of Q is known to have (B) (Amoroso-Dvornicich) , as is any extension which is totally split at some fixed place p (Schinzel for the real prime, Bombieri-Zannier for the other primes.) Habegger has proved that, when E is an elliptic curve over Q, the field Q(E^tors) obtained by adjoining all torsion points of E has the Bogomolov property. What does this have to do with gonality, and with my paper with Chris Hall and Emmanuel Kowalski from last year? Suppose we ask about the Bogomolov property for extensions of a more general field F? Well, F had better admit a notion of absolute Weil height. This is certainly OK when F is a global field, like the function field of a curve over a finite field k; but in fact it’s fine for the function field of a complex curve as well. So let’s take that view; in fact, for simplicity, let’s take F to be C(t). What does it mean for an algebraic extension F’ of F to have the Bogomolov property? It means that there is a constant c such that, for every finite subextension L of F and every non-constant function x in L^, the absolute logarithmic height of x is at least c. Now L is the function field of some complex algebraic curve C, a finite cover of P^1. And a non-constant function x in L^* can be thought of as a nonzero principal divisor. The logarithmic height, in this context, is just the number of zeroes of x — or, if you like, the number of poles of x — or, if you like, the degree of x, thought of as a morphism from C to the projective line. (Not necessarily the projective line of which C is a cover — a new projective line!) In the number field context, it was pretty easy to see that the log height of non-torsion elements of L^* was bounded away from 0. That’s true here, too, even more easily — a non-constant map from C to P^1 has degree at least 1! There’s one convenient difference between the geometric case and the number field case. The lowest log height of a non-torsion element of L^* — that is, the least degree of a non-constant map from C to P^1 — already has a name. It’s called the gonality of C. For the Bogomolov property, the relevant number isn’t the log height, but the absolute log height, which is to say the gonality divided by [L:F]. So the Bogomolov property for F’ — what we might call the geometric Bogomolov property — says the following. We think of F’ as a family of finite covers C / P^1. Then (GB) There is a constant c such that the gonality of C is at least c deg(C/P^1), for every cover C in the family. What kinds of families of covers are geometrically Bogomolov? As in the number field case, you can certainly find some families that fail the test — for instance, gonality is bounded above in terms of genus, so any family of curves C with growing degree over P^1 but bounded genus will do the trick. On the other hand, the family of modular curves over X(1) is geometrically Bogomolov; this was proved (independently) by Abramovich and Zograf. This is a gigantic and elegant generalization of Ogg’s old theorem that only finitely many modular curves are hyperelliptic (i.e. only finitely many have gonality 2.) At this point we have actually more or less proved the geometric version of Habegger’s theorem! Here’s the idea. Take F = C(t) and let E/F be an elliptic curve; then to prove that F(E(torsion)) has (GB), we need to give a lower bound for the curve C_N obtained by adjoining an N-torsion point to F. (I am slightly punting on the issue of being careful about other fields contained in F(E(torsion)), but I don’t think this matters.) But C_N admits a dominant map to X_1(N); gonality goes down in dominant maps, so the Abramovich-Zograf bound on the gonality of X_1(N) provides a lower bound for the gonality of C_N, and it turns out that this gives exactly the bound required. What Chris, Emmanuel and I proved is that (GB) is true in much greater generality — in fact (using recent results of Golsefidy and Varju that slightly postdate our paper) it holds for any extension of C(t) whose Galois group is a perfect Lie group with Z_p or Zhat coefficients and which is ramified at finitely many places; not just the extension obtained by adjoining torsion of an elliptic curve, for instance, but the one you get from the torsion of an abelian variety of arbitrary dimension, or for that matter any other motive with sufficiently interesting Mumford-Tate group. Question: Is Habegger’s theorem true in this generality? For instance, if A/Q is an abelian variety, does Q(A(tors)) have the Bogomolov property? Question: Is there any invariant of a number field which plays the role in the arithmetic setting that “spectral gap of the Laplacian” plays for a complex algebraic curve? A word about Habegger’s proof. We know that number fields are a lot more like F_q(t) than they are like C(t). And the analogue of the Abramovich-Zograf bound for modular curves over F_q is known as well, by a theorem of Poonen. The argument is not at all like that of Abramovich and Zograf, which rests on analysis in the end. Rather, Poonen observes that modular curves in characteristic p have lots of supersingular points, because the square of Frobenius acts as a scalar on the l-torsion in the supersingular case. But having a lot of points gives you a lower bound on gonality! A curve with a degree d map to P^1 has at most d(q+1) points, just because the preimage of each of the q+1 points of P^1(q) has size at most d. (You just never get too old or too sophisticated to whip out the Pigeonhole Principle at an opportune moment….) Now I haven’t studied Habegger’s argument in detail yet, but look what you find right in the introduction: The non-Archimedean estimate is done at places above an auxiliary prime number p where E has good supersingular reduction and where some other technical conditions are met…. In this case we will obtain an explicit height lower bound swiftly using the product formula, cf. Lemma 5.1. The crucial point is that supersingularity forces the square of the Frobenius to act as a scalar on the reduction of E modulo p. Yup! There’s no mention of Poonen in the paper, so I think Habegger came to this idea independently. Very satisfying! The hard case — for Habegger as for Poonen — has to do with the fields obtained by adjoining p-torsion, where p is the characteristic of the supersingular elliptic curve driving the argument. It would be very interesting to hear from Poonen and/or Habegger whether the arguments are similar in that case too! ## JMM, Golsefidy, Silverman, Scanlon Like Emmanuel, I spent part of last week at the Joint Meetings in New Orleans, thanks to a generous invitation from Alireza Salefi Golsefidy and Alex Lubotzky to speak in their special session on expander graphs. I was happy that Alireza was willing to violate a slight taboo and speak in his own session, since I got to hear about his work with Varju, which caps off a year of spectacular progress on expansion in quotients of Zariski-dense subgroups of arithmetic groups. Emmanuel’s Bourbaki talk is your go-to expose. I think I’m unlike most mathematicians in that I really like these twenty-minute talks. They’re like little bonbons — you enjoy one and then before you’ve even finished chewing you have the next in hand! One nice bonbon was provided by Joe Silverman, who talked about his recent work on Lehmer’s conjecture for polynomials satisfying special congruences. For instance, he shows that a polynomial which is congruent mod m to a multiple of a large cyclotomic polynomial can’t have a root of small height, unless that root is itself a root of unity. He has a similar result where the implicit G_m is replaced by an elliptic curve, and one gets a lower bound for algebraic points on E which are congruent mod m to a lot of torsion points. This result, to my eye, has the flavor of the work of Bombieri, Pila, and Heath-Brown on rational points. Namely, it obeys the slogan: Low-height rational points repel each other. More precisely — the global condition (low height) is in tension with a bunch of local conditions (p-adic closeness.) This is the engine that drives the upper bounds in Bombieri-Pila and Heath-Brown: if you have too many low-height points, there’s just not enough room for them to repel each other modulo every prime! Anyway, in Silverman’s situation, the points are forced to nestle very close to torsion points — the lowest-height points of all! So it seems quite natural that their own heights should be bounded away from 0 to some extent. I wonder whether one can combine Silverman’s argument with an argument of the Bombieri-Pila-Heath-Brown type to get good bounds on the number of counterexamples to Lehmer’s conjecture….? One piece of candy I didn’t get to try was Tom Scanlon’s Current Events Bulletin talk about the work of Pila and Willkie on problems of Manin-Mumford type. Happily, he’s made the notes available and I read it on the plane home. Tom gives a beautifully clear exposition of ideas that are rather alien to most number theorists, but which speak to issues of fundamental importance to us. In particular, I now understand at last what “o-minimality” is, and how Pila’s work in this area grows naturally out of the Bombieri-Pila method mentioned above. Highly recommended! ## Expander graphs, gonality, and variation of Galois representations Suppose you have a 1-dimensional family of polarized abelian varieties — or, just to make things concrete, an abelian variety A over Q(t) with no isotrivial factor. You might have some intuition that abelian varieties over Q don’t usually have rational p-torsion points — to make this precise you might ask that A_t[p](Q) be empty for “most” t. In fact, we prove (among other results of a similar flavor) the following strong version of this statement. Let d be an integer, K a number field, and A/K(t) an abelian variety. Then there is a constant p(A,d) such that, for each prime p > p(A,d), there are only finitely many t such that A_t[p] has a point over a degree-d extension of K. The idea is to study the geometry of the curve U_p parametrizing pairs (t,S) where S is a p-torsion point of A_t. This curve is a finite cover of the projective line; if you can show it has genus bigger than 1, then you know U_p has only finitely many K-rational points, by Faltings’ theorem. But we want more — we want to know that U_p has only finitely many points over degree-d extensions of K. This can fail even for high-genus curves: for instance, the curve C: y^2 = x^100000 + x + 1 has really massive genus, but choosing any rational value of x yields a point on C defined over a quadratic extension of Q. The problem is that C is hyperelliptic — it has a degree-2 map to the projective line. More generally, if U_p has a degree-d map to P^1, then U_p has lots of points over degree-d extensions of K. In fact, Faltings’ theorem can be leveraged to show that a kind of converse is true. So the relevant task is to show that U_p admits no map to P^1 of degree less than d; in other words, its gonality is at least d. Now how do you show a curve has large gonality? Unlike genus, gonality isn’t a topological invariant; somehow you really have to use the geometry of the curve. The technique that works here is one we learned from an paper of Abramovich; via a theorem of Li and Yau, you can show that the gonality of U_p is big if you can show that the Laplacian operator on the Riemann surface U_p(C) has a spectral gap. (Abramovich uses this technique to prove the g=1 version of our theorem: the gonality of classical modular curves increases with the level.) We get a grip on this Laplacian by approximating it with something discrete. Namely: if U is the open subvariety of P^1 over which A has good reduction, then U_p(C) is an unramified cover of U(C), and can be identified with a finite-index subgroup H_p of the fundamental group G = pi_1(U(C)), which is just a free group on finitely many generators g_1, … g_n. From this data you can cook up a Cayley-Schreier graph, whose vertices are cosets of H_p in G, and whose edges connect g H with g_i g H. Thanks to work of Burger, we know that this graph is a good “combinatorial model” of U_p(C); in particular, the Laplacian of U_p(C) has a spectral gap if and only if the adjacency matrix of this Cayley-Schreier graph does. At this point, we have reduced to a spectral problem having to do with special subgroups of free groups. And if it were 2009, we would be completely stuck. But it’s 2010! And we have at hand a whole spray of brand-new results thanks to Helfgott, Gill, Pyber, Szabo, Breuillard, Green, Tao, and others, which guarantee precisely that Cayley-Schreier graphs of this kind, (corresponding to finite covers of U(C) whose Galois closure has Galois group a perfect linear group over a finite field) have spectral gap; that is, they are expander graphs. (Actually, a slightly weaker condition than spectral gap, which we call esperantism, is all we need.) Sometimes you think about a problem at just the right time. We would never have guessed that the burst of progress in sum-product estimates in linear groups would make this the right time to think about Galois representations in 1-dimensional families of abelian varieties, but so it turned out to be. Our good luck. ## In which I publish my first paper The first one I wrote, that is. It happened like this: my undergraduate thesis advisor was Persi Diaconis, and in 1993, my senior year, Diaconis was really peeved about the proof via Selberg that every element of SL_2(F_p) could be expressed as a word of length at most C log p in the standard unipotent generators. (See Emmanuel’s comment on Terry’s blog for useful references.) Diaconis felt it was a combinatorial problem and it should be solvable by purely combinatorial means, and that a hard-working undergraduate who was good at Putnam problems, like me, ought to be able to do it. That turned out not to be the case. So Persi gave me another thesis problem; he asked if I could get good bounds on the diameter of the unipotent subgroup U of SL_n(F_p), with its standard generating set id + e_{i,i+1}. When n = 2, this is easy; the unipotent group is just Z/pZ and its diameter is about p. The question is: what happens asymptotically when n and p are allowed to grow? It’s not possible any longer for the diameter of U to be on order log \|U\|, as is the case for SL_2(F_p); the abelianization of U looks like (Z/pZ)^{n-1}, which already has diameter on order of np. Unless n is much larger than p, this swamps log \|U\|. But it turns out this is in some sense the only obstacle: one can prove that diam(U) is bounded above and below by constant multiples of np + log \|U\|. (My memory is that I conceived the key step of the argument during a boring Advocate meeting.) Much later, when I was a new postdoc at Princeton, I talked about this problem with Julianna Tymoczko, then a graduate student working with Bob McPherson. Julianna very quickly saw how to make the argument much more conceptual and general, and in particular how to extend it to all the classical groups. So we decided to write it up as a joint paper. That was probably 2002. Then we got around to writing the paper and submitting it. That was 2005. It was accepted in 2007. And now here it is! That’s the abstract; if you’re at a computer that doesn’t subscribe to Forum Math, here’s the arXiv version. 17 years from first version of the theorem to publication! Update: Harald Helfgott politely comment-hints at something I should have put in the original post, which is that nowadays, thanks to him, there is a combinatorial proof that the diameter of SL_2(F_p) is on order log p! The subject of uniform bounds for word growth and spectral gaps in finite groups of Lie type is currently moving very quickly. I won’t try to summarize the state of the art, but you can expect in the medium-term to hear something about an interesting application of Harald’s work to arithmetic geometry.	{"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": 3, "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": 0.8602706789970398, "perplexity": 783.089413010627}, "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/1495463608665.33/warc/CC-MAIN-20170526124958-20170526144958-00443.warc.gz"}
http://math.stackexchange.com/questions/301315/invertibility-of-a-kronecker-product?answertab=active	# Invertibility of a Kronecker Product Prove that A Kronecker Product B is invertible if and only if B Kronecker Product A is invertible. I dont have a clue where to start to be honest. I am not very familiar yet to the Kronecker Product so could you please help me with providing an easyily understandable proof. - The previous answers assume that $A$ and $B$ are square. This is fair because it is not generally possible for $A\otimes B$ to be invertible when $A$ and $B$ are rectangular. In the more general setting where $A$ is $m\times n$ and $B$ is $r\times s$ (and $mr = ns$) it is still easy to see that $A\otimes B$ is equivalent to $B \otimes A$ up to permuting the rows and columns. Both $A \otimes B$ and $B \otimes A$ contain exactly the same entries $a_{ij} b_{kl}$. More explicitly, one places $a_{ij} b_{kl}$ in row $(i-1)r+k$ and column $(j-1)s+l$, and the other places it in row $(k-1)m+i$ and column $(l-1)n+j$. Note that the row numbers are uniquely determined by $i,k$ and the column numbers determined independently by $j,l$. - Hint: If $A$ is a $m\times m$ matrix and $B$ a $n\times n$ matrix then $$\det(A\otimes B)=(\det A)^n\cdot (\det B)^m.$$ - A square matrix is invertible if and only if its determinant is different from $0$. – Davide Giraudo Feb 12 '13 at 16:57 What do you know about invertible matrices? – Davide Giraudo Feb 12 '13 at 17:02 You can show that $A\otimes B$ is invertible if and only if $A$ and $B$ are invertible, from which this result follows. If $A$ and $B$ are invertible, do you know how to find $(A\otimes B)^{-1}$? If $A$ is not invertible, then there exists $C\neq 0$ such that $CA=0$. It follows that $(C\otimes I)(A\otimes B)=0$, which implies that $A\otimes B$ is not invertible.	{"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": 0.8041079044342041, "perplexity": 86.00164958901196}, "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/1404776440593.58/warc/CC-MAIN-20140707234040-00082-ip-10-180-212-248.ec2.internal.warc.gz"}
http://mathhelpforum.com/calculus/110765-integration-first-converting-spherical-polars-print.html	# Integration first converting to spherical polars • Oct 27th 2009, 02:08 AM StarWrecker Integration first converting to spherical polars I'm trying to work through a question, but I cannot get my result to look anything like the model answers. The question is to integrate z/((x^2+y^2)^(1/2)) over the volume of a region bounded by the surfaces z=0 and z=(9-x^2-y^2)^(1/2). I know I need to use spherical polar co-ordinates to do this, but I can't seem to get the substitutions right. The model answer says that it becomes r^2cos(theta) integrated over the obvious region, but my attempt to get it is as such: z = rcos(theta) r = (x^2+y^2)^(1/2) (in 2 dimensions, ignoring z) So, the integral is integral( (rcos(theta) / r) r^2 * sin(theta) drd(theta)d(phi) ) Of course this gives me r^2cos(theta)sin(theta). What am I missing? • Oct 27th 2009, 04:14 AM Mush Quote: Originally Posted by StarWrecker I'm trying to work through a question, but I cannot get my result to look anything like the model answers. The question is to integrate z/((x^2+y^2)^(1/2)) over the volume of a region bounded by the surfaces z=0 and z=(9-x^2-y^2)^(1/2). I know I need to use spherical polar co-ordinates to do this, but I can't seem to get the substitutions right. The model answer says that it becomes r^2cos(theta) integrated over the obvious region, but my attempt to get it is as such: z = rcos(theta) r = (x^2+y^2)^(1/2) (in 2 dimensions, ignoring z) So, the integral is integral( (rcos(theta) / r) r^2 * sin(theta) drd(theta)d(phi) ) Of course this gives me r^2cos(theta)sin(theta). What am I missing? Why did you choose to ignore z? In spherical coordinates $x = r \cos(\varphi) \sin (\theta)$, and $y = r \sin(\varphi) \sin(\theta)$. You should use this definition to calculate $(x^2 + y^2)^{\frac{1}{2}}$, and you should get $(x^2 + y^2)^{\frac{1}{2}} = r \sin(\theta)$ Giving you an integrand of $\frac{r \cos(\theta)}{r \sin(\theta} r^2 \sin(\theta) = r^2 \cos(\theta)$ • Oct 27th 2009, 04:21 AM HallsofIvy That's one reason why it is more common to use " $\rho$" as the radial variable in spherical coordinates (and they are NOT normally referred to as "spherical polar coordinates), to avoid confusing 3d and 2d coordinates. In spherical coordinates $x= \rho cos(\phi) sin(\theta)$ and $y= \rho sin(\phi) sin(\theta)$. Then $x^2+ y^2= \rho^2 cos^2(\phi)sin^2(\theta)+ \rho^2 sin^2(\phi)sin^2(\theta)$ $= \rho^2 sin^2(\theta)(cos^2(\phi)+ sin^2(\phi))= \rho^2 sin^2(\phi)$ so that $(x^2+ y^2)^{1/2}= \rho sin(\phi)$	{"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": 11, "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": 0.9582872986793518, "perplexity": 763.3611004352496}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2018-09/segments/1518891813883.34/warc/CC-MAIN-20180222022059-20180222042059-00707.warc.gz"}
https://owenduffy.net/blog/?p=855	# Designing high performance VHF/UHF receive systems – Part 1 ## G/T A metric that may be used to express the performance of an entire receive system is the ratio of antenna gain to total equivalent noise temperature, usually expressed in deciBels as dB/K. G/T is widely used in design and specification of satellite communications systems. G/T=AntennaGain/TotalNoiseTemperature 1/K Example: if AntennaGain=50 and TotalNoiseTemperature=120K, then G/T=50/120=0.416 1/K or -3.8 dB/K. The utility of G/T is that receive S/N changes dB for dB with G/T, in fact you can calculate S/N knowing G/T, wavelength, bandwidth and the field strength of the signal (Duffy 2007). S/N=Sλ2/(4π)G/T/(kbB) where: S is power flux density; λ is wavelength; kb is Boltzmann’s constant; and B is receiver effective noise bandwidth Calculating G/T for configuration alternatives gives a single performance metric that allows comparison of the performance of the configurations. Different levels of performance might be weighed against cost in a cost benefit analysis, eg whether funds applied to antenna improvements will be more cost-effective than applying funds to feed line improvement, whether an LNA with NF=0.8dB and Gain=30dB is more cost-effective than one with NF=0.5dB and Gain=18dB. Such a quantitative approach cuts through the old wives tales about the relative merit of some approaches, and will quantify the relative performance in the system context. Some options will have different effects depending on other system details, and Rules of Thumb (ROT) that ignore this sensitivity to the rest of the system becomes simply rot. Inclusion of gain in the metric allows quantitative comparison of configuration changes that include a change in antenna gain. A key limitation of simple system models is that they assume the system is linear, in particular that there is no significant IMD noise created by system components. Systems with significant IMD are basically unpredictable, and whilst the performance of a particular system can be measured, that does not provide knowledge that is extensible to another system because the environment, and response to the environment are probably different. The bottom line is that components with significant IMD noise don’t belong in a high performance system, fix them or get rid of them. Note that the popular ham antenna tables published by VE7BQH which show a calculated quantity labelled G/T value for antenna systems appears to ignore receiver equivalent noise (ie it is for a noiseless receiver), and as such, the usage is not consistent with the industry meaning of the term (eg ITU-R. 2000). Don’t confuse this discussion of G/T with those tables. G/T is calculated with respect to some nominated reference plane. The most common practice is to use the antenna waveguide flange or connector as the reference plane, and to consistently calculate system component contributions relative to that reference plane. In the case of an array of four Yagi antennas, the reference plane would be the connector on the power divider/combiner that provides single connector access to the antenna system. ## References • Duffy, O. 2006. Effective use of a Low Noise Amplifier on VHF/UHF. VK1OD.net (offline). • ———. 2006. Receiver sensitivity metric converter. VK1OD.net (offline). • ———. 2007. Measuring system G/T ratio using Sun noise. VK1OD.net (offline).ITU-R. 2000. Recommendation ITU-R S.733-2 (2000) Determination of the G/T ratio for earth stations operating in the fixed-satellite service . Designing high performance VHF/UHF receive systems – Part 2	{"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": 0.8410752415657043, "perplexity": 2808.899788172122}, "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/1568514575484.57/warc/CC-MAIN-20190922094320-20190922120320-00368.warc.gz"}
http://clay6.com/qa/11643/an-engine-of-weight-6-5-metric-ton-is-going-up-an-incline-of-large-frac-at-	Browse Questions # An engine of weight 6.5 metric ton is going up an incline of $\large\frac{5}{13}$ at the rate of $9\; km/hr$ .The power of engine , if co-efficient of friction is $\large\frac{1}{12}$ is $(a)\;83.5\;KW \quad (b)\;97.5\;KW \quad (c)\;60.5\;KW \quad (d)\;73.5\;KW$ $\sin \theta=\large\frac{5}{13}$ $\cos \theta=\sqrt {1-\sin ^ \theta}$ $\qquad= \sqrt {1-\bigg(\large\frac{5}{13}\bigg)^2}$ $\qquad=\large\frac{12}{13}$ Force against which the engine has to work $F=mg \sin \theta+\mu mg \cos \theta$ $\quad=6500 \times 9.8 \bigg[\large\frac{5}{13}+\frac{1}{12} \times \frac{12}{13}\bigg]$ $\quad=29400 \;N$ Engine power $=Fv$ $\quad=29400 \times 2.5 \qquad 9km/hr=2.5 m/s$ $\quad=73500 \;W$ $\quad=73.5 KW$ Hence d is the correct answer. edited Feb 10, 2014 by meena.p +1 vote	{"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": 0.8657079935073853, "perplexity": 2070.1428199338334}, "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/1495463612018.97/warc/CC-MAIN-20170529053338-20170529073338-00257.warc.gz"}
https://cstheory.stackexchange.com/questions/10594/whats-the-difference-between-adts-gadts-and-inductive-types	# What's the difference between ADTs, GADTs, and inductive types? Might anyone be able to explain the difference between: • Algebraic Datatypes (which I am fairly familiar with) • Generalized Algebraic Datatypes (what makes them generalized?) • Inductive Types (e.g. Coq) (Especially inductive types.) Thank you. Algebraic data types let you define types recursively. Concretely, suppose we have the datatype $$\mathsf{data\;list = Nil \;\;\|\;\; Cons\;of\;\mathbb{N} \times list}$$ What this means is that $\mathsf{list}$ is the smallest set generated by the $\mathsf{Nil}$ and $\mathsf{Cons}$ operators. We can formalize this by defining the operator $F(X)$ $$F(X) == \{ \mathsf{Nil} \} \cup \{ \mathsf{Cons}(n, x) \;\|\; n \in \mathbb{N} \land x \in X \}$$ and then defining $\mathsf{list}$ as $$\mathsf{list} = \bigcup_{i \in \mathbb{N}} F^i(\emptyset)$$ A generalized ADT is what we get when define a type operator recursively. For example, we might define the following type constructor: $$\mathsf{bush}\;a = \mathsf{Leaf\;of\;}a \;\;\|\;\; \mathsf{Nest\;of\;bush}(a \times a)$$ This type means that an element of $\mathsf{bush\;}a$ is a tuple of $a$s of length $2^n$ for some $n$, since each time we go into the $\mathsf{Nest}$ constructor the type argument is paired with itself. So we can define the operator we want to take a fixed point of as: $$F(R) = \lambda X.\; \{ \mathsf{Leaf}(x) \;\|\; x \in X\} \cup \{ \mathsf{Nest}(v) \;\|\; v \in R(X) \}$$ An inductive type in Coq is essentially a GADT, where the indexes of the type operator are not restricted to other types (as in, for example, Haskell), but can also be indexed by values of the type theory. This lets you give types for length-indexed lists, and so on. • Thank you. Wouldn't that just mean, however, that "inductive type" in completely synonymous with "dependent type"? – ninjagecko Mar 8 '12 at 16:27 • @Neel: I've never seen types like bush called GADTs. I have seen them called nested or non-regular types. – jbapple Mar 8 '12 at 16:38 • Nested types are a special case of GADTs. The critical feature of a GADT is simply that it is a recursive definition at higher kind. (Changes to the rhs is basically syntactic sugar for adding a type equality as a component of the constructor.) – Neel Krishnaswami Mar 8 '12 at 20:24 • @ninjagecko: "Inductive types" are types given semantics as the least fixed point of a constructor. Not all types can be described this way (functions cannot, and neither can infinite types such as streams). Dependent types describe types which permit program terms to occur in them (that is, types can "depend on" terms). Since Coq is a dependent type theory, the inductive types it lets you define are also dependent. But non-dependent type theories can support inductive types as well, and those inductive types won't be dependent. – Neel Krishnaswami Mar 8 '12 at 20:27 • @NeelKrishnaswami: Would you be so kind as to clarify your answer by enumerating the "first few smallest" elements of the types of bush a? In this example, is it Nest Leaf(a) Leaf(a) Leaf(a) Leaf(a), or Nest ((Nest Leaf(a) Leaf(a)) (Nest Leaf(a) Leaf(a))) as one example of the set? – ninjagecko Jul 11 '12 at 16:24 Consider algebraic datatypes such as: data List a = Nil \| Cons a (List a) The return types of each constructor in a datatype are all the same: Nil and Cons both return List a. If we allow the constructors to return different types, we have a GADT: data Empty -- this is an empty data declaration; Empty has no constructors data NonEmpty data NullableList a t where Vacant :: NullableList a Empty Occupied :: a -> NullableList a b -> NullableList a NonEmpty Occupied has the type a -> NullableList a b -> NullableList a NonEmpty, while Cons has the type a -> List a -> List a. It is important to note that NonEmpty is a type, not a term. Another example: data Zero data Succ n data SizedList a t where Alone :: SizedList a Zero WithFriends :: a -> SizedList a n -> SizedList a (Succ n) Inductive types in programming languages that have dependent types allow the return types of the constructors to depend on the values (not just the types) of the arguments. Inductive Parity := Even \| Odd. Definition flipParity (x:Parity) : Parity := match x with \| Even => Odd \| Odd => Even end. Fixpoint getParity (x:nat) : Parity := match x with \| 0 => Even \| S n => flipParity (getParity n) end. (* A ParityNatList (Some P) is a list in which each member is a natural number with parity P. *) Inductive ParityNatList : option Parity -> Type := Nil : forall P, ParityNatList P \| Cons : forall (x:nat) (P:option Parity), ParityNatList P -> ParityNatList (match P, getParity x with \| Some Even, Even => Some Even \| Some Odd, Odd => Some Odd \| _, _ => None end). A side note: GHC has a mechanism for treating value constructors as type constructors. This is not the same as the dependent inductive types that Coq has, but it lessens the syntactic burden of GADTs somewhat, and it can lead to better error messages. • Thank you. "Inductive types in programming languages that have dependent types" What then would an inductive type look like in a language with no dependent types, and can you have non-inductive (but GADT-like) dependent types? – ninjagecko Mar 8 '12 at 18:14	{"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": 0.8301946520805359, "perplexity": 935.7341238106794}, "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-2020-16/segments/1585370529375.49/warc/CC-MAIN-20200405053120-20200405083120-00153.warc.gz"}
https://mathoverflow.net/questions/199404/when-is-a-linear-combination-of-permutation-matrices-unitary	# When is a linear combination of permutation matrices unitary? Question: Let $P_\pi$ denote the matrix representation of permutation $\pi$. Consider a linear combination of all $n \times n$ permutation matrices $$U := \sum_{\pi \in S_n} c_\pi P_\pi$$ where $c_\pi$ are arbitrary complex coefficients. When is the matrix $U$ unitary? It would be great to have a simple parametrization of all tuples of coefficients $(c_\pi : \pi \in S_n)$ for when this happens. Example: In the $n = 2$ case there are only 2 permutations, so the matrix $U$ looks like this: $$U = \begin{pmatrix} c_0 & c_1 \\ c_1 & c_0\end{pmatrix}$$ where the constants $c_i$ must obey $$\|c_0\|^2 + \|c_1\|^2 = 1 \quad \text{and} \quad c_0 c_1^* + c_1 c_0^* = 0$$ for $U$ to be unitary. We can parametrize the solution of these equations as follows: $$c_0 = e^{i\varphi} \cos t \quad c_1 = e^{i\varphi} i \sin t$$ for any $\varphi \in [0,2 \pi)$ and $t \in [0,\pi/2]$. It would be nice to have something similar for general $n$. For example, I can write down the equations for $n = 3$ but I don't know any nice way to parametrize the solutions. Note: The only reference on this topic I could find is Orthogonal matrices as linear combinations of permulation matrices. • Linear combinations of permutation matrices are the same thing as constant line-sum matrices (that is, matrices where all the row sums and all the column sums are equal), so we can rephrase the question as, which unitary matrices are constant line-sum? Mar 8, 2015 at 22:06 I think this can be done(in principle) in general. The $n \times n$ permutation matrices span a $\mathbb{C}$-vector space of dimension $1 + (n-1)^{2}$ since the natural permutation representation of $S_{n}$ is the sum of the trivial representation and an irreducible representation of degree $n-1.$ It is necessary for $\sum_{\pi} c_{\pi} P_{\pi}$ to be unitary that $\sum_{\pi} c_{\pi} \in S^{1}$ (just consider the effect on the vector $v$ with every component $\frac{1}{\sqrt{n}}).$ The non-trivial irreducible character comes from the action of $S_{n}$ on $v^{\perp}$, and for the moment it isn't clear me how to give a concise description to characterize which linear combinations of permutation matrices act as a unitary transformation on $v^{\perp}.$ Note that this is not really an issue when $n =2.$ Later edit: Thanks to Sean Eberhard's comment, it becomes clear that the unitary matrices which are linear combinations of permutation matrices are precisely those unitary matrices which have the vector $v$ above as an eigenvector- any unitary matrix which has $v$ as an eigenvector necessarily leaves $v^{\perp}$ invariant, so any linear combination of permutation matrices both has $v$ has an eigenvector and leaves $v^{\perp}$ invariant. By a dimension count, the space of matrices which leave both span($v$)and $v^{\perp}$ invariant, which has dimension $1 + (n-1)^{2}$, is precisely the span of the permutation matrices. In conclusion, the unitary matrices which are linear combinations of permutation matrices are precisely the unitary matrices which have $v$ as an eigenvector. • This argument also shows that every unitary transformation with $v$ as an eigenvector can be written as a linear combination of permutation matrices. Mar 8, 2015 at 22:48 • @SeanEberhard : Yes, thanks, I have incorporated your commment into my answer now. Mar 8, 2015 at 23:26 • More generally, suppose $\rho : G \rightarrow\mathrm{GL}(V)$ is a unitary representation of a finite group $G$, and that $V = V_1\oplus\cdots\oplus V_r$ is a direct sum of non-isomorphic irreducible representations where $\dim V_i = n_i$. Then a block matrix with blocks of sizes $n_1,\ldots,n_r$ is a linear combination of $\rho(g)$ for $g \in G$, so a matrix $U$ is such a linear combination if and only if $U(V_i) \subseteq V_i$ for each $i$. In the special case when $U$ is unitary and $r=2$, it's enough that $U(V_1) \subseteq V_1$, since then $U(V_2)=U(V_1^\perp) = U(V_1)^\perp=V_1^\perp=V_2$. Mar 9, 2015 at 13:33	{"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": 0.9958699345588684, "perplexity": 91.74007759074674}, "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-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00324.warc.gz"}
https://www.physicsforums.com/threads/how-to-write-equations-for-salt-dissolving-in-water-ionizing.899013/	# How to write equations for salt dissolving in water/ionizing Tags: 1. Jan 2, 2017 ### LegitSci 1. The problem statement, all variables and given/known data Write separate equations for each potassium salt (KH2PO4) dissolving in water and for the ionization reaction of the weak acid anion that each of these salts contains. 2. Relevant equations None 3. The attempt at a solution I just needed to clarify when it dissolves in water do I include that in reaction or just do a dissociation of the ion? Furthermore, for ionization reaction of weak anion, it is H2PO4, so what does this mean? 2. Jan 3, 2017 ### Staff: Mentor There is no strict rule to the first step (and I would have no problems if the first step was omitted, but your instructor can have a different opinion). H2PO4- can dissociate. Actually it can also hydrolyze. Both reactions have to be listed. Draft saved Draft deleted Similar Discussions: How to write equations for salt dissolving in water/ionizing	{"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": 0.8536791801452637, "perplexity": 2196.1294695065135}, "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-2017-51/segments/1512948588072.75/warc/CC-MAIN-20171216123525-20171216145525-00297.warc.gz"}
http://mathhelpforum.com/calculus/15315-supremum-infimum.html	# Math Help - supremum and infimum 1. ## supremum and infimum I need help on how to prove this: If x is an arbitrary real number, prove that there are integers m and n such that m < x < n. Thank you 2. Originally Posted by PersaGell I need help on how to prove this: If x is an arbitrary real number, prove that there are integers m and n such that m < x < n. Thank you WLOG for any $x>0$ consider $k=1$. By the Archimedean Ordering on $\mathbb{R}$ there shall exist $n$ such that $n\cdot 1 > x$. Hence we can find $n>x$. That completes the second inequality. For any $x>0$ consider $x-1$. I claim we can choose an integer $m$ such that $x-1\leq m . Above we can find an integer $n>x$. Hence consider the interval $[-n,n]$. Define the set of integers $S=\{ k \in \mathbb{Z} \| x-1 \leq k \leq n\}$. This is a finite non-empty set. So choose $m=\min S$. I leave it to you to prove that $m$ is the desired integer.	{"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": 15, "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": 0.9970880746841431, "perplexity": 243.2429830212725}, "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-2014-52/segments/1419447549548.75/warc/CC-MAIN-20141224185909-00043-ip-10-231-17-201.ec2.internal.warc.gz"}
http://slh.alljournals.cn/search_by_author.aspx?subject=mathematical_chemical&major=sx&field=author_name&encoding=utf-8&q=Magda+Peligrad	首页 \| 本学科首页官方微博 \| 高级检索按中文标题英文标题中文关键词英文关键词中文摘要英文摘要作者中文名作者英文名单位中文名单位英文名基金中文名基金英文名杂志中文名杂志英文名栏目英文名栏目英文名 DOI 责任编辑分类号杂志ISSN号检索检索词: 收费全文 21篇国内免费 2篇数学 23篇 2018年 1篇 2013年 2篇 2012年 2篇 2011年 1篇 2010年 1篇 2007年 1篇 2006年 2篇 2003年 1篇 2001年 1篇 1999年 1篇 1998年 1篇 1997年 1篇 1996年 1篇 1994年 1篇 1993年 1篇 1986年 1篇 1985年 1篇 1982年 2篇 1981年 1篇 1. Almost-Sure Results for a Class of Dependent Random Variables 总被引：17，自引：0，他引：17 The aim of this note is to establish almost-sure Marcinkiewicz-Zygmund type results for a class of random variables indexed by d + —the positive d-dimensional lattice points—and having maximal coefficient of correlation strictly smaller than 1. The class of applications include filters of certain Gaussian sequences and Markov processes. 相似文献 2. The aim of this paper is to investigate the properties of the maximum of partial sums for a class of weakly dependent random variables which includes the instantaneous filters of a Gaussian sequence having a positive continuous spectral density. The results are used to obtain an invariance principle for strongly mixing sequences of random variables in the absence of stationarity or strong mixing rates. An additional condition is imposed to the coefficients of interlaced mixing. The results are applied to linear processes of strongly mixing sequences. 相似文献 3. Summary In this note we estimate the rate of convergence in Marcinkiewicz-Zygmung strong law, for partial sumsS n of strong stationary mixing sequences of random variables. The results improve the corresponding ones obtained by Tze Leung Lai (1977) and Christian Hipp (1979). 相似文献 4. The aim of this paper is to investigate the properties of the maximum of partial sums for a class of weakly dependent random variables which includes the instantaneous filters of a Gaussian sequence having a positive continuous spectral density. The results are used to obtain an invariance principle and the convergence of the moments in the central limit theorem. 5. 6. 7. 8. The aim of this paper is to give a functional form for the central limit theorem obtained by Bradley for strong mxing sequences of random variables, under a certain assumption about the size of the maximal coefficients of correlations. The convergence of the moments of order 2 + δ in the central limit theorem for this class of random variables is also obtained. 相似文献 9. The paper aims to establish a new sharp Burkholder-type maximal inequality in for a class of stationary sequences that includes martingale sequences, mixingales and other dependent structures. The case when the variables are bounded is also addressed, leading to an exponential inequality for a maximum of partial sums. As an application we present an invariance principle for partial sums of certain maps of Bernoulli shifts processes. 10. In this note we investigate the coupling of a class of dependent sequences with an independent one having the same marginal distributions. This method is then used to prove that a uniform law of averages for this class holds if a similar law holds for the associated independent sequence. 相似文献	{"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": 0.9490158557891846, "perplexity": 1151.8468410797732}, "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-2019-35/segments/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00258.warc.gz"}
http://umj.imath.kiev.ua/article/?lang=en&article=8493	2018 Том 70 № 12 # Some Approximation Properties of Szasz–Mirakyan–Bernstein Operators of the Chlodovsky Type Abstract We motivate a new sequence of positive linear operators by means of the Chlodovsky-type Szasz–Mirakyan–Bernstein operators and investigate some approximation properties of these operators in the space of continuous functions defined on the right semiaxis. We also find the order of this approximation by using the modulus of continuity and present the Voronovskaya-type theorem. English version (Springer): Ukrainian Mathematical Journal 66 (2014), no. 6, pp 928-936. Citation Example: Simsek E., Tunç E. Some Approximation Properties of Szasz–Mirakyan–Bernstein Operators of the Chlodovsky Type // Ukr. Mat. Zh. - 2014. - 66, № 6. - pp. 826–834. 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": 0.9489431977272034, "perplexity": 1302.5303139949553}, "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-2019-04/segments/1547584331733.89/warc/CC-MAIN-20190123105843-20190123131843-00387.warc.gz"}
https://calculationworld.com/block-calculator/	# Block Calculator Projects can vary in size and the amount of materials needed with it.Before you start building anything it is imperative that you have an estimate of what material you will need along with the cost and amount. This calculator allows you to calculate the amount of materials needed! ### How to Use Block Calculator First enter the length in feet. Second enter the width in feet. Press calculate and you'll be informed of how many blocks are needed L W 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": 0.805762529373169, "perplexity": 979.9033237287016}, "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/1623487635920.39/warc/CC-MAIN-20210618073932-20210618103932-00554.warc.gz"}
https://apboardsolutions.in/ap-board-7th-class-maths-solutions-chapter-2-intext-questions/	AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers InText Questions AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers InText Questions and Answers. AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers InText Questions Do This Question 1. Write five examples, each of proper, improper, mixed fractions. ? (Page No. 27) Solution: Proper fractions $$\frac{1}{5}, \frac{2}{3}, \frac{4}{7}, \frac{3}{8}, \frac{4}{9}$$ Improper fractions $$\frac{7}{2}, \frac{3}{2}, \frac{9}{4}, \frac{11}{5}, \frac{8}{3}$$ Mixed fractions $$1 \frac{2}{3}, 2 \frac{3}{5}, 4 \frac{1}{7}, 8 \frac{6}{7}$$ Question 2. Write five equivalent fractions for. i) $$\frac{3}{5}$$ ii) $$\frac{4}{7}$$ (Page No. 28) Solution: Do This Question 1. (Page No. 31) i) 4 x $$\frac { 2 }{ 7 }$$ ii) 4 x $$\frac { 3 }{ 5 }$$ iii) 7 x $$\frac { 1 }{ 3 }$$ Solution: i) 4 x $$\frac{2}{7}=\frac{4 \times 2}{7}=\frac{8}{7}$$ ii) 4 x $$\frac{3}{5}=\frac{4 \times 3}{5}=\frac{12}{5}$$ iii) 7 x $$\frac{1}{3}=\frac{7 \times 1}{3}=\frac{7}{3}$$ Question 2. Find i) 5 x $$\frac { 3 }{ 2 }$$ = ii) 4 x $$\frac { 7 }{ 5 }$$ = iii) 7 x $$\frac { 8 }{ 3 }$$ = (Page No. 31) Solution: i) 5 x $$\frac{3}{2}=\frac{5 \times 3}{2}=\frac{15}{2}=7 \frac{1}{2}$$ ii) 4 x $$\frac{7}{5}=\frac{4 \times 7}{5}=\frac{28}{5}=5 \frac{3}{5}$$ iii) 7 x $$\frac{8}{3}=\frac{7 \times 8}{3}=\frac{56}{3}=18 \frac{2}{3}$$ Question 3. Find the following: (Page No. 32) i) 3 x 2 $$\frac{2}{7}$$ ii) 5 x 2$$\frac{1}{3}$$ iii) 8 x 4$$\frac{1}{7}$$ iv) 4 x 1$$\frac{2}{9}$$ v) 5 x 1$$\frac{1}{3}$$ Solution: Do These Question 1. Fill in these boxes. (Page No. 35) i) $$\frac{1}{5} \times \frac{1}{7}=\frac{1 \times 1}{5 \times 7}$$ = …………… ii) $$\frac{1}{2} \times \frac{1}{6}=\frac{1 \times 1}{2 \times 6}=$$ = ………………… Solution: i) $$\frac{1}{5} \times \frac{1}{7}=\frac{1 \times 1}{5 \times 7}=\frac{1}{35}$$ ii) $$\frac{1}{2} \times \frac{1}{6}=\frac{1 \times 1}{2 \times 6}=\frac{1}{12}$$ Do This Question 1. Find (Page No. 39) i) 2 ÷ $$\frac{1}{4}$$ ii) 7 ÷ $$\frac{1}{2}$$ iii) 3 ÷ $$\frac{1}{5}$$ (Page No. 39) Solution: i) 2 ÷ $$\frac{1}{4}=2 \times \frac{4}{1}=\frac{8}{1}$$ = 8 ii) 7 ÷ $$\frac{1}{2}=7 \times \frac{2}{1}$$ = 14 iii) 3 ÷ $$\frac{1}{5}=3 \times \frac{5}{1}$$ = 15 Question 2. Find (Page No. 41) i) 9 ÷ $$\frac{2}{5}$$ ii) 3 ÷ $$\frac{4}{7}$$ iii) 2 ÷ $$\frac{8}{9}$$ Solution: i) 9 ÷ $$\frac{2}{5}$$ = $$9 \times \frac{5}{2}=\frac{45}{2}$$ ii) 3 ÷ $$\frac{4}{7}$$ = $$3 \times \frac{7}{4}=\frac{21}{4}$$ iii) 2 ÷ $$\frac{8}{9}$$ = $$2 \times \frac{9}{8}=\frac{9}{4}$$ Question 3. Find (Page No. 41) i) 7 ÷ 5$$\frac{1}{3}$$ ii) 5 ÷ 2$$\frac{4}{7}$$ Solution: i) 7 ÷ 5$$\frac{1}{3}$$ = $$7 \div \frac{16}{3}=7 \times \frac{3}{16}=\frac{21}{16}$$ ii) 5 ÷ 2$$\frac{4}{7}$$ = $$5 \div \frac{18}{7}=5 \times \frac{7}{18}=\frac{35}{18}$$ Question 4. Find (Page No. 42) i) $$\frac{3}{5} \div \frac{1}{2}$$ ii) $$\frac{1}{2} \div \frac{3}{5}$$ iii) $$2 \frac{1}{2} \div \frac{3}{5}$$ iv) $$5 \frac{1}{6} \div \frac{9}{2}$$ Solution: Do This Question Find (Page No. 45) i) 0.25 + 5.30 ii) 29.75 – 25.97 Solution: i) 0.25 + 5.30 ii) 29.75 – 25.97 Do These Question 1. Find (Page No. 48) i) 1.7 x 3 ii)2.0 x 1.5 iii) 2.3 x 4.35 Solution: i)1.7 x 3 = 5.1 ii) 2.0 x 1.5 = 3.00 iii) 2.3 x 4.35 = 10.005 Question 2. Arrange the products obtained in (I) In descending order. Solution: Arranging above answers in descending order 10.005 > 5.1 > 3.00 Question 3. Find (Page No. 50) i) 35.7 ÷ 3 ii) 25.5 ÷ 3 Solution: Do These Question 1. Find the greatest and the smallest numbers among the following groups. (Page No. 52) i) 2, -2, -3, 4, 0, -5 ii) -3, -7, -8,0,-5,-2 Solution: i) 2, -2, -3, 4, 0, -5 : greatest number = 4; smallest number = -5 ii) -3, -7, -8, 0, -5, -2: greatest number = 0; smallest number = -8 Question 2. Write the following numbers In ascending order. (Page No. 52) i) -5,-75,3,-2,4, $$\frac{3}{2}$$ ii) $$\frac{2}{3}, \frac{3}{2}$$, 0, -1, -2, 5 Solution: i) -5, -75, 3, -2, 4, $$\frac{3}{2}$$ Ascending order = -75 <-5 <-2 < $$\frac{3}{2}$$ <3 < 4 or -75, -5, -2, $$\frac{3}{2}$$, 3, 4 ii) $$\frac{2}{3}$$,$$\frac{3}{2}$$, 0, -1, -2, 5 Ascending order = -2, -1, 0, $$\frac{2}{3}$$, $$\frac{3}{2}$$, 5 Question 3. Write 5 equlvalent rational numbers to (i) $$\frac{5}{2}$$ (Page No. 56) (ii) $$\frac{-7}{8}$$ (iii) $$\frac{-3}{7}$$ Solution: Do These Question 1. Which is bigger $$\frac{5}{8}$$ or $$\frac{3}{5}$$ ? (PageNo.28) Solution: $$\frac{5}{8} \times \frac{5}{5}=\frac{25}{40}$$ and $$\frac{3}{5} \times \frac{8}{8}=\frac{24}{40}$$ As $$\frac{24}{40}<\frac{25}{40}$$ $$\frac{5}{8}$$ is bigger than $$\frac{3}{5}$$ Question 2. Determine if the following pairs are equal by writing each in their simplest form. (Page No. 28) i) $$\frac{3}{8}$$ and $$\frac{375}{1000}$$ Solution: ii) $$\frac{18}{54}$$ and $$\frac{23}{69}$$ Solution: iii) $$\frac{6}{10}$$ and $$\frac{600}{1000}$$ Solution: iv) $$\frac{17}{27}$$ and $$\frac{25}{45}$$ Solution” Do These Question 1. Identify the equivalent rational number is each question i) $$\frac{-1}{2}, \frac{-3}{4}, \frac{-2}{4}, \frac{-4}{8}$$ Solution: $$\frac{-1}{2}=\frac{-2}{4}=\frac{-4}{8}$$ ii) $$\frac{1}{4}, \frac{3}{4}, \frac{5}{3}, \frac{10}{6}, \frac{2}{4}, \frac{20}{12}$$ Solution: $$\frac{5}{3}=\frac{20}{12}=\frac{10}{6}$$ Try This Question 1. You have seen that the product of two natural numbers is one or more than one is bigger than each of the two natural numbers. For example 3 x 4 = 12; 12 > 4 and 12 > 3. What happens to the value of the product when we multiply two proper fractions? (Page No.37) Fill the following table and conclude your observations. Solution: Question 2. Will the reciprocal of a proper fraction be a proper fraction? (Page No. 40) Solution: No. Reciprocal of a proper fraction is always an improper fraction. Question 3. Will the reciprocal of an Improper fraction be an Improper fraction? Solution: No. The reciprocal of an improper fraction is always a proper fraction. Question 4. Look at the following table and fill up the blank spaces. (Page No. 44) Solution: Question 5. WrIte the following numbers in their expanded form. (Page No.44) i) 30.807 ii) 968.038 iii) 8370. 705 Solution: i) 30.8O7 = 10 x 3 + 1 x 0 + $$\frac{1}{10}$$ x 8 +$$\frac{1}{100}$$ x 0 + $$\frac{1}{1000}$$ x 7 = 30 + $$\frac{8}{10}+\frac{7}{1000}$$ ii) 968.038 = 100 x 9 + 10 x 6 + 1 x 8 + $$\frac{1}{10}$$ x 0 + $$\frac{1}{100}$$ x 3 + $$\frac{1}{1000}$$ x 8 = 900 + 60 + 8 + $$\frac{3}{100}+\frac{8}{1000}$$ iii) 8370.705 = 1000 x 8 + 100 x 3 + 10 x 7 + $$\frac{1}{10}$$ x 7 + $$\frac{1}{1000}$$ x 5 = 8000 + 300 + 70 + $$\frac{7}{10}+\frac{5}{1000}$$ Question 6. Take any5 Integers and make all possible rational numbrs with them. (Page No. 54) Solution: Consider 2, 3, 4, 5 and 7 Ratlonalnumberare $$\frac{2}{3}, \frac{2}{4}, \frac{2}{5}, \frac{2}{7}, \frac{3}{4}, \frac{3}{5}, \frac{3}{7}, \frac{3}{2}, \frac{4}{2}, \frac{4}{3}, \frac{4}{5}, \frac{4}{7}, \frac{5}{2}, \frac{5}{3}, \frac{5}{4}, \frac{5}{7}, \frac{7}{2}, \frac{7}{3}, \frac{7}{4}, \frac{7}{5}$$ Question 7. Consider any 5 rational numbers. Find out which itegers constitute them? (Page No.54) Solution: Take $$\frac{3}{4}, \frac{5}{8}, \frac{6}{11}, \frac{2}{7}$$ and $$\frac{1}{5}$$. The integers are 1, 2, 3, 4, 5, 6, 7, 8 and 11. Do This Question 1. Find (i) 50 paise = ₹………….. (ii) 22 g = ………….. kg (iii) 80 cm = ………………m (Page No. 44) Solution: (i) 50 paise =₹$$\frac{50}{100}$$ = ₹ 0.5 (ii) 22 g = $$\frac{22}{1000}$$ kg = 0.022 kg (iii) 80 cm = $$\frac{80}{100}$$ m = 0.8 m Try This Question 1. Represent $$\frac{3}{4}$$ and $$\frac{1}{4}$$ in different ways using different figures. Justify your representation. Share, and check it with your friends. (Page No.27) Solution: Question 2. Represents 2 1/4 pictorially. How many units are needed for this. (Page No. 27) Solution: We need 3 units to represent 2½ Do These Question 1. Represent pictorially 2 x $$\frac{1}{5}=\frac{2}{5}$$ (Page No. 32) Solution: Do These Question 1. Find $$\frac{1}{2} \times \frac{1}{5}$$ and $$\frac{1}{5} \times \frac{1}{2}$$ using diagram check whether $$\frac{1}{2} \times \frac{1}{5}=\frac{1}{5} \times \frac{1}{2}$$ (Page No. 35) Solution: Do These Question 1. Write 5 more fractions between (i) 0 and 1 (ii) 1 and 2. (Page No. 52) Solution: i) Fractions between 0 and 1 are $$\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$$ ii) Fractions between land 2 are $$\frac{8}{7}, \frac{9}{7}, \frac{10}{7}, \frac{11}{7}, \frac{12}{7}, \frac{13}{7}$$ Question 2. Where does 4$$\frac{3}{5}$$ lie on the number line? (Page No. 52) Solution: 4$$\frac{3}{5}$$ lies between 4 and 5 on the number line. Question 3. On the number line given below represent the following numbei. (Page No.53) i) $$\frac{-7}{2}$$ ii) $$\frac{3}{2}$$ iii) $$\frac{7}{4}$$ iv) $$\frac{-7}{4}$$ v) $$\frac{-1}{2}$$ vi) $$\frac{1}{4}$$ Solution: Question 4. Consider the following numbers on a number line. (Page No. 53) 27, $$\frac{-7}{8}, \frac{11}{943}, \frac{54}{17}$$, -68, -3, $$\frac{-9}{6}, \frac{7}{2}$$ i) Which of these are to the left of a) 0 Solution: Left to zero are negative numbers ∴ $$\frac{-7}{8}$$, -68, -3, $$\frac{-9}{6}$$ b) -2 Left to – 2 are less than – 2. ∴ -3, -68 c) 4 Left to 4 are less than 4. d) 2 Left to 2 are less than 2. ∴ $$\frac{-7}{8}, \frac{11}{943}-68,-3, \frac{-9}{6}$$ ii) Which of these would be to the right of a) 0 Right to zero are positive number. ∴ $$27, \frac{11}{943}, \frac{54}{17}, \frac{7}{2}$$ b) -5 Right to -5 are greater than -5. $$27, \frac{-7}{8}, \frac{11}{943}, \frac{54}{17}-3, \frac{-9}{6}, \frac{7}{2}$$ c) 3$$\frac{1}{2}$$ Right to 3$$\frac{1}{2}$$ are more than 3$$\frac{1}{2}$$. ∴ 27 d) $$\frac{-5}{2}$$ Right to $$\frac{-5}{2}$$ are more than $$\frac{-5}{2}$$ ∴ $$-27, \frac{-7}{8}, \frac{11}{943}, \frac{54}{17} \frac{-9}{6}, \frac{7}{2}$$ Try These Question 1. Write three more equivalent fractions of $$\frac{3}{4}$$ and mark them on the number line. What do you observe? (Page No. 55) Solution: Equivalent fractions of $$\frac{3}{4}$$ lie on the same mark. Question 2. Do all equivalent fractions of $$\frac{6}{7}$$ represent the same point on the number line. (Page No. 55) Solution: Yes. Question 3. Are $$\frac{-1}{2}$$ and $$\frac{-3}{6}$$ represent same point on the number line? (Page No. 55) Solution: Yes. Question 4. Are $$\frac{-2}{3}$$ and $$\frac{-4}{6}$$ equivalent? . (Page No. 55) Solution: Yes. Question 5. Mark the following rational numbers on the number line. (In Ex 7,3) (i) $$\frac{1}{2}$$ (ii) $$\frac{3}{4}$$ (iii) $$\frac{3}{2}$$ (iv) $$\frac{10}{3}$$ 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": 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": 0.8319541215896606, "perplexity": 4209.050987785155}, "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-2022-49/segments/1669446710765.76/warc/CC-MAIN-20221130160457-20221130190457-00752.warc.gz"}
https://puzzling.stackexchange.com/questions/101074/find-the-area-of-the-given-triangle	# Find the area of the given triangle Another mathematical puzzle: Find the area of $$\triangle FGH$$, given that $$FG=FH$$ and the radii of the circles shown are $$2$$ and $$1$$ • please provide the source if this is not your own work. else it will likely be closed. thanks – Omega Krypton Aug 12 at 14:17 • The question says FG=GH but the diagram is drawn as if FG=FH and GH is very different. Is there a typo in the question, or is the diagram deliberately very misleading? – Gareth McCaughan Aug 12 at 14:38 • It somewhat looks like a homework... – CiaPan Aug 12 at 14:47 • Duplicate of math.stackexchange.com/q/1766545/369453 :D – Marius Aug 12 at 15:10 • @GarethMcCaughan Thanks for pointing that out, I've fixed it. Also, good catch by Marius! I didn't know that this had been asked before. It's purely coincidential :) – Aniruddha Deb Aug 12 at 15:16 $$16\sqrt2$$ Because: if you draw radii where the circles are tangent to the triangles with tangent points marked as $$D$$ (smaller circle) and $$E$$ (larger circle; this forms right angles), you get the following similar triangles: $$\triangle FDA \sim \triangle FEB \sim \triangle FXG$$. $$A$$ is the center of the small circle and $$B$$ is the center of the large one. $$X$$ is the midpoint of $$HG$$. From this, we get this proportion: $$\frac{FA}{FB} = \frac{DA}{EB}$$ OR $$\frac{FA}{FA+3} = \frac{1}{2}$$ because of the radii lengths. So: $$FA=3$$ Using Pythagorean Theorem: $$FD = \sqrt8$$ Using our similar triangle relations from the beginning, we get: $$XG = 2\sqrt2$$ Finally, we use the area of a triangle formula to get: $$16\sqrt2$$ • Sorry, I got tired of showing my work... – Voldemort's Wrath Aug 12 at 14:52 • I'm sure this is a good answer, but I got kind of lost when you started drawing new things. If you had a picture, that would help greatly. – Chipster Aug 12 at 22:38 • @Chipster - No, you're right... I'll add that in when I find time! – Voldemort's Wrath Aug 13 at 11:57 • A picture would help, but even with no picture, please could you clarify which point is $D$ and which is $E$? – Rosie F Aug 14 at 8:57 • @RosieF -- $D$ is for the smaller circle and $E$ is for the larger one. – Voldemort's Wrath Aug 14 at 15:48 Let $$h$$ denote the height of the triangle from F onto GH. Then, by Pythagoras' theorem, $$FG^2=h^2+(GH/2)^2,$$ and by triangles' similarity $$h:(GH/2) = (FG-GH/2):2$$ and $$(h-4):h = 1:2.$$ Can you continue from this? From the last we get $$h-4=h/2$$ which resolves to $$h=8.$$ Plug it to previous two to get $$\begin{cases}FG^2=64+(GH/2)^2\\8:(GH/2) = (FG-GH/2):2\end{cases}$$ The last equation is equivalent to $$16:(GH/2) = (FG-GH/2)$$ hence $$FG = (GH/2)+16:(GH/2)$$ and $$FG^2 = (GH/2)^2 + 32 + 256:(GH/2)^2$$ Compare it to the first equation: $$64+(GH/2)^2 = (GH/2)^2 + 32 + 256:(GH/2)^2$$ $$64 = 32 + 256:(GH/2)^2$$ $$32 = 256:(GH/2)^2$$ $$(GH/2)^2 = 256:32 = 8$$ $$GH/2 = \sqrt 8 = 2\sqrt 2$$ Finally the area sought is $$S_{\triangle FGH} = h\cdot GH/2 = 16\sqrt 2.$$ • See my answer; I think I got it?? – Voldemort's Wrath Aug 12 at 14:50 • @Voldemort'sWrath Yes, you did. I didn't show my solution at first, because the question looks like a homework (my comment), so I wanted the author to show some effort first. Now I uncommented the hiden part with the same result as yours. – CiaPan Aug 13 at 7:39 Let me chip in with a streamlined answer. $$8\sqrt 8$$ All the smaller circle is telling us is that the height of the triangle must be $$8$$, so let's jot that down and from here pretend the small circle never existed. The points where the circle (there is only one circle!) touches the triangle divide the long sides into two segments of lengths $$y>x$$ and the base into two equal segments of length $$x$$. Let us now compute the area $$A$$ of the triangle from base and height $$A=8x$$, from incircle radius and circumference $$A=2(2x+y)$$ and from Heron's formula $$A^2=x^2y(2x+y)$$. Comparing the first two gives $$2x=y$$, together with the last we get $$x=\sqrt 8$$ and $$A=8\sqrt 8$$. • Please explain where A = 2(2x + y) came from. Thanks. – asg Aug 13 at 8:56 • @asg as it says "from incircle radius and circumference". $2x+y$ is the semicircumference often written $s$ of the triangle. – Paul Panzer Aug 13 at 9:21	{"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": 45, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8071496486663818, "perplexity": 867.68855188791}, "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-2020-50/segments/1606141195198.31/warc/CC-MAIN-20201128070431-20201128100431-00572.warc.gz"}
http://alexanderpruss.blogspot.com/2012/12/uniform-measure-and-nonmeasurable-sets.html	## Tuesday, December 11, 2012 ### Uniform measure and nonmeasurable sets, without the Axiom of Choice Given the Axiom of Choice, there is no translation invariant probability measure on the interval [0,1) (the relevant translation is translation modulo 1). But this fact really does need something in the way of the Axiom of Choice. Moreover, the fact only obtains for countably additive measures. Interestingly, however, if we add the assumption that our measure assigns non-zero (presumably infinitesimal) weight to each point of [0,1), then the non-existence of a translation invariant finitely additive measure follows without the Axiom of Choice. I got the proof of this from Paul Pedersen who thinks he got it from the classic Bernstein and Wattenberg piece (I don't have their paper at hand). I am generalizing trivially. Theorem: Let P be any finitely additive measure taking values in a partially ordered group G and defined on a collection of subsets of [0,1) such that every countable subset has a measure in G. Suppose P({x})>0 for some x in [0,1). Then P is not translation invariant (modulo 1). Proof: To obtain a contradiction, suppose P is translation invariant. Then P({x})>0 for every x in [0,1). Let r be any irrational number in (0,1), and let R be the set of numbers of the form nr modulo 1, as n ranges over the positive integers. Let R' be the set of numbers of the form nr modulo 1, as n ranges over the integers greater than 1. Then R' is a translation of R by r, modulo 1. Observe that r is not a member of R' since there is no natural number n greater than 1 such that r=nr modulo 1, since if there were, we would have (n−1)r=0 modulo 1, and hence r would be a rational number with denominator n−1. Thus by finite additivity P(R)=P(R')+P({r})>P(R'). Hence, R is a counterexample to translation invariance, contradicting our assumption. Note 1: On the assumption that the half-open intervals are all measurable and the measurable sets form an algebra (the standard case), translation invariance modulo 1 follows from ordinary translation invariance within the interval, namely the condition that P(A)=P(A+x) whenever both A and A+x={y+x:y in A} are subsets of [0,1). Note 2: The proof above shows that if P({x})>0 for every x in [0,1), then the set of all positive integral multiples of any fixed irrational number (modulo 1) is nonmeasurable. It is interesting to note that this nonmeasurable set is actually measurable using standard Lebesgue measure. Thus, by enforcing regularity using infinitesimals, one is making some previously measurable sets nonmeasurable if one insists on translation invariance. Note 3: Bernstein and Wattenberg construct a hyperreal valued measure that is almost translation invariant: the difference between the measure of a set and of a translation of the set is infinitesimal. Alexander R Pruss said... Slight generalization. Let P be a finitely additive measure on the countable subsets of some set Omega taking values in some partially ordered group G. Let H be a group that acts on Omega such that there are g in H and w in Omega with { g^n w : n in N } infinite. Then (a) P assigns null weight to some non-empty sets (i.e., P is not regular) or (b) P is not H-invariant. If H acts transitively on Omega, then (a) implies P assigns null weight to all singletons. Matt said... Could you please explain the first sentence? Surely Lebesgue measure is translation-invariant in the requisite sense, no? Matt said... This comment has been removed by the author. Alexander R Pruss said... I meant: on all subsets of [0,1).	{"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": 0.9942909479141235, "perplexity": 514.6145268174286}, "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-2014-42/segments/1414119655159.46/warc/CC-MAIN-20141024030055-00197-ip-10-16-133-185.ec2.internal.warc.gz"}
https://jcgoran.github.io/2021/05/04/apocalypse-problem.html	I’ve recently been going through the book Cracking the coding interview (6th ed.), and an interesting problem was stated there (abridged for brevity): In a new post-apocalyptic world, the world leader decides that all families must have one girl, or face massive fines. If all families continue to have children until they have one girl, at which point they immediately stop, what will the gender ratio of the new generation be? Assume that the probabilities of having a boy or a girl are equal. The solution Since all families must eventually have exactly one girl, then $N_\text{girls} = N$, where $N$ is the number of families. The probability of having $n$ boys (denoted $B$) before having a girl (denoted $G$) is therefore: $P(\underbrace{BB\ldots B}_{n}G) = \underbrace{\frac{1}{2}\times\frac{1}{2} \times \cdots \times \frac{1}{2}}_{n} \times \frac{1}{2} = \frac{1}{2^{n + 1}}$ While we don’t know exactly how many boys are in each individual family, we can still compute the average, or the expectation value, of the number of boys in a family, defined as: $\mathbb{E}(X) = \sum_{n}x_n\, p_n = \sum_{n=0}^\infty\frac{n}{2^{n + 1}} = \frac{1}{2} \times \sum_{n=0}^\infty\frac{n}{2^{n}}$ where we’ve taken the random variable $X$ to be the number of boys in a family, hence $x_n = n$ and $p_n = 1/2^{n + 1}$, and the sum goes over all possibilities ($[0,\ldots,\infty)$ boys)1. How do we compute this sum without a computer? Well, let’s first consider the following sum instead: $S_n (q) = \sum_{i = 0}^n q^i$ This is the well-known geometric sum, whose sum is equal to2: $S_n (q) = \frac{q^{n + 1} - 1}{q - 1}$ We can see that, for $0 < q < 1$, we have the following: $\lim_{n \rightarrow \infty} S_n(q) = \lim_{n \rightarrow \infty} \sum_{i = 0}^n q^i = \lim_{n \rightarrow \infty} \frac{q^{n + 1} - 1}{q - 1} = \frac{1}{1 - q}$ The reason is that, since $0 < q < 1$, we have that $q^2 < q < 1$, etc., and in fact the entire sequence $\left\lbrace 1, q, q^2, \ldots \right\rbrace$ is decreasing, and since it’s bounded below by 0, it means that we have $\displaystyle\lim_{n \rightarrow \infty} q^n = 0$, which gives the above. We may as well define: $S(q) = \sum_{i = 0}^\infty q^i = \frac{1}{1 - q}$ to keep things concise. Now, how does this apply to our problem? Well, if we put $q = 1/2$ in the above, we get: $S\left(\frac{1}{2}\right) = \sum_{i = 0}^\infty \left(\frac{1}{2}\right)^i = \frac{1}{1 - 1/2} = 2$ Now, if we compute the first derivative of $S(q)$, on the right-hand side we get: $\frac{\text{d}}{\text{d} q}S(q) = \frac{1}{(1 - q)^2}$ while the left-hand side is just: $\frac{\text{d}}{\text{d} q}S(q) = \sum_{i = 1}^\infty i\, q^{i - 1} = \sum_{i = 0}^\infty (i + 1)\, q^i = \sum_{i = 0}^\infty i\, q^i + \sum_{i = 0}^\infty q^i$ where in the last step we’ve just split one sum into two. However, you may notice that the second term is actually just $S(q) = 1/(1 - q)$, so we have: $\sum_{i = 0}^\infty i\, q^i = \frac{1}{(1 - q)^2} - \frac{1}{1 - q} = \frac{q}{(1 - q)^2}$ where we’ve simplified the final expression with a bit of algebra. But the expression on the left-hand side is exactly what we’re looking for, provided we take $q = 1/2$! Therefore, going back to the original problem, we have that: $\mathbb{E}(X) = \frac{1}{2} \times \sum_{n=0}^\infty\frac{n}{2^{n}} = \frac{1}{2} \times \frac{1 / 2}{(1 - 1 / 2)^2} = 1$ This result may be a bit puzzling: it seems that the policy of the world leader is completely useless, since in the end we just end up with 1 boy and 1 girl per family, on average. But recall that biology hasn’t been altered, so we sort of expect that things would even out. Going beyond: unequal probabilities What happens if the probabilities are not equal though? Let’s say the probability of having a boy is $p$; then the probability of having a girl would be $1 - p$, and the probability of having a girl after $n$ boys is now: $P(\underbrace{BB\ldots B}_{n}G) = \underbrace{p\times p\times\cdots\times p}_{n}\, (1 - p) = p^n\, (1 - p)$ Then the expectation value for the number of boys is: $\mathbb{E}(X) = \sum_{n=0}^\infty n \times p^n (1 - p) = (1 - p) \sum_{n=0}^\infty n \times p^n = (1 - p) \times \frac{p}{(1 - p)^2} = \frac{p}{1 - p}$ From this we can see that the original problem is just the special case $p = 1/2$. I’ve plotted the expectation value of the number of boys as a function of the value of $p$ below. The number of boys as a function of the value of $p$. The red diamond represents the special value $p = 1/2$, for which the expectation value is 1. It’s worthwhile to note that an equivalent problem is to consider $N$ (in the general case, biased) coins, which we keep flipping until we get one of the outcomes, say, heads. Actually, we don’t even need $N$ coins; since the coin flips are statistically independent, we may as well keep throwing the same coin until we get heads, and then repeat this procedure $N$ times. Footnotes 1. since there are infinitely many terms in the sum, it’s technically called a series 2. this can be proved using mathematical induction	{"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": 0.910916805267334, "perplexity": 166.62463435152173}, "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-2022-40/segments/1664030337415.12/warc/CC-MAIN-20221003101805-20221003131805-00603.warc.gz"}
https://deepai.org/publication/stabilized-sparse-online-learning-for-sparse-data	DeepAI # Stabilized Sparse Online Learning for Sparse Data Stochastic gradient descent (SGD) is commonly used for optimization in large-scale machine learning problems. Langford et al. (2009) introduce a sparse online learning method to induce sparsity via truncated gradient. With high-dimensional sparse data, however, the method suffers from slow convergence and high variance due to the heterogeneity in feature sparsity. To mitigate this issue, we introduce a stabilized truncated stochastic gradient descent algorithm. We employ a soft-thresholding scheme on the weight vector where the imposed shrinkage is adaptive to the amount of information available in each feature. The variability in the resulted sparse weight vector is further controlled by stability selection integrated with the informative truncation. To facilitate better convergence, we adopt an annealing strategy on the truncation rate, which leads to a balanced trade-off between exploration and exploitation in learning a sparse weight vector. Numerical experiments show that our algorithm compares favorably with the original algorithm in terms of prediction accuracy, achieved sparsity and stability. • 2 publications • 14 publications 06/12/2021 03/22/2016 Stochastic gradient descent is the method of choice for large-scale mach... 12/31/2020 Stochastic gradient descent (SGD) has taken the stage as the primary wor... 06/28/2015 ### Stochastic Gradient Made Stable: A Manifold Propagation Approach for Large-Scale Optimization Stochastic gradient descent (SGD) holds as a classical method to build l... 02/16/2017 ### Unbiased Online Recurrent Optimization The novel Unbiased Online Recurrent Optimization (UORO) algorithm allows... 01/18/2021 ### Screening for Sparse Online Learning Sparsity promoting regularizers are widely used to impose low-complexity... 08/24/2020 ### Noise-induced degeneration in online learning In order to elucidate the plateau phenomena caused by vanishing gradient... None ## 1 Introduction Modern datasets pose many challenges for existing learning algorithms due to their unprecedented large scales in both sample sizes and input dimensions. It demands both efficient processing of massive data and effective extraction of crucial information from an enormous pool of heterogeneous features. In response to these challenges, a promising approach is to exploit online learning methodologies that performs incremental learning over the training samples in a sequential manner. In an online learning algorithm, one sample instance is processed at a time to obtain a simple update, and the process is repeated via multiple passes over the entire training set. In comparison with batch learning algorithms in which all sample points are scrutinized at every single step, online learning algorithms have been shown to be more efficient and scalable for data of large size that cannot fit into the limited memory of a single computer. As a result, online learning algorithms have been widely adopted for solving large-scale machine learning tasks (Bottou, 1998). In this paper, we focus on first-order subgradient-based online learning algorithms, which have been studied extensively in the literature for dense data.111Dense data is defined as a dataset in which the number of nonzero entries in all columns of its design matrix are in the order of while the ones of sparse data are in the order of or less. Among these algorithms, popular methods include the Stochastic Gradient Descent (SGD) algorithm (Zhang, 2004; Bottou, 2010), the mirror descent algorithm (Beck and Teboulle, 2003) and the dual averaging algorithm (Nesterov, 2009). Since these methods only require the computation of a (sub)gradient for each incoming sample, they can be scaled efficiently to high-dimensional inputs by taking advantage of the finiteness of the training sample. In particular, the stochastic gradient descent algorithm is the most commonly used algorithm in the literature of subgradient-based online learning. It enjoys an exceptionally low computational complexity while attaining steady convergence under mild conditions (Bottou, 1998) , even for cases where the loss function is not everywhere differentiable. Despite of their computational efficiency, online learning algorithms without further constraint on the parameter space suffers the “curse of dimensionality” to the same extent as their non-online counterparts. Embedded in a dense high-dimensional parameter space, not only does the resulted model lack interpretability, its variance is also inflated. As a solution, sparse online learning was introduced to induce sparsity in the parameter space under the online learning framework (Langford et al., 2009) . It aims at learning a linear classifier with a sparse weight vector, which has been an active topic in this area. For most efforts in the literature, sparsity is introduced by applying regularization on a loss function as in the classific LASSO method (Tibshirani, 1996; Shalev-Shwartz and Tewari, 2011). For example, Duchi and Singer (2009) extend the framework of Forward-Backward splitting (Lions and Mercier, 1979) by alternating between an unconstrained truncation step on the sample gradient and an optimization step on the loss function with a penalty on the distance from the truncated weight vector.Langford et al. (2009) and Carpenter (2008) both explore the idea of imposing a soft-threshold on the weight vector updated by the stochastic gradient descent algorithm: wj=sign(wj)max(\|wj\|−λ,0),j=1,…,p. This class of methods is known as the truncated gradient algorithm. For every standard SGD updates, the weight vector is shrunk by a fixed amount to induce sparsity. In the work of Duchi et al. (2010), the same strategy has also been combined with a variant of the mirror descent algorithm (Beck and Teboulle, 2003). Wang et al. (2015) further extends the truncated gradient framework to adjust for cost-effectiveness. This simple yet efficient method of truncated gradients particularly motivates the algorithm proposed in this paper. Strategies different from the truncation-based algorithm have also been proposed. For example, Xiao (2009) proposed the Regularized Dual-Averaging (RDA) algorithm which builds upon the primal-dual subgradient method by Nesterov (2009). The RDA algorithm learns a sparse weight vector by solving an optimization problem using the running average over all preceding gradients, instead of a single gradient at each iteration. Closely related to sparse online learning is another area of active research, online feature selection. Instead of enforcing just a shrinkage on the weight vectors via regularization, online feature selection algorithms explicitly invoke feature selection by imposing a hard constraint on the weight vector, (e.g., Wang et al., 2014; Wu et al., 2014). In other words, online feature selection algorithms focus on generating a resulted weight vector that has a high sparsity level by directly shrinking a large proportion of the weights directly to zero (also referred to as a hard thresholding). In practice, regularization is computationally expensive to solve due to its non-differentiability. The set of selected features also suffers from high variability as the decisions of hard-thresholding are based on single random samples in an online learning setting. Therefore, important features can be discarded simply owing to random perturbations. Most recent subgradient-based online learning algorithms do not consider potential structures or heterogeneity in the input features. As pointed out by Duchi et al. (2011), current methods largely follow a predetermined procedural scheme that is oblivious to the characteristics of data being used at each iteration. In large-scale applications, a common and important structure is heterogeneity in sparsity levels of the input features, i.e., the variability in the number of nonzero entries among features. For instance, consider the bag-of-word features in text mining applications.222Here, by sparse features, we refer to features for which most samples assume a constant value (e.g., 0) and a few samples take on other values. Without loss of generality, we assume the majority constant is 0 throughout this paper. For a learning task, the importance of a feature is not necessarily associated with the frequencies of its values. In genetics, for example, rare variants ( in the population) have been found to be associated with disease risks (Morris and Zeggini, 2010). Both dense and sparse features may contain important information for the learning task. However, in the presence of heterogeneity in sparsity levels, using a simple regularization in an online setting will predispose rare features to be truncated more than necessary. The resulted sparse weight vectors usually exhibit high variance in terms of both weight values and the membership in the set of features with nonzero weights. As a result, the convergence of the standard truncation-based framework may also be hampered by this high variability. When the amount of information is scarce due to sparsity at each iteration, the convergence of the weight vector would understandably take a large number of iterations to approach the optimum. In two recent papers, Oiwa et al. (2011) and Oiwa et al. (2012) tackle this problem via penalty weighted by the accumulated norm of subgradients for extending several basic frameworks in sparse online learning. Their results suggest that, by acknowledging the sparsity structure in the features, both prediction accuracy and sparsity are improved over the original algorithms while maintaining the same convergence rate. However, their resulted weight vectors are unstable as the imposed subgradient-based regularization are excessively noisy due to the randomness of incoming samples in online learning. The membership in the set of selected features with nonzero weights is also very sensitive to the orderings of the training samples. In this paper, we propose a stabilized truncated stochastic gradient descent algorithm for high-dimensional sparse data. The learning framework is motivated by that of the Truncated Gradient algorithm proposed by Langford et al. (2009). To deal with the aforementioned issues with sparse online learning methods applied to high-dimensional sparse data, we introduce three innovative components to reduce variability in the learned weight vector and stabilize the selected features. First, when applying the soft-thresholding, instead of a uniform truncation on all features, we perform only informative truncations, based on actual information from individual features during the preceding computation window of updates. By doing so, we reduce the heterogeneous truncation bias associated with feature sparsity. The key idea here is to ensure that each truncation for each feature is based on sufficient information, and the amount of shrinkage is adjusted for the information available on each feature. Second, beyond the soft-thresholding corresponding to the ordinary regularization, the resulted weight vector is stabilized by staged purges of irrelevant features permanently from the active set of features. Here, irrelevant features are defined as features whose weights have been repeatedly truncated. Motivated by stability selection introduced in Meinshausen and Bühlmann (2010), these permanent purges prevent irrelevant features from oscillating between the active and non-active set of features, The “purging” process also resembles hard-thresholding in online feature selection and results in a stabler sparse solution than other sparse online learning algorithms. Results on the theoretical regret bound (See Section 4 ) show that this stabilization step helps improve over the original truncated gradient algorithm, especially when the target weight vector is notably sparse. To attune the proposed learning algorithm to the sparsity of the remaining active features, the third component of our algorithm is adjusting the amount of shrinkage progressively instead of fixing it at a predetermined value across all stages of the learning process. A novel hyperparameter, rejection rate, is introduced to balance between exploration of different sparse combinations of features at the beginning and the exploitation of the selected features to construct accurate estimate at a later stage. Our method gradually anneal the rejection rate to acquire the necessary amount of shrinkage on the fly for achieving the desired balance. The rest of paper is organized as follows. Section 2 reviews the Truncated Gradient algorithm based on Stochastic Gradient Descent (SGD) framework for sparse learning proposed in Langford et al. (2009). In Section 3, we introduce, in details, the three novel components of our proposed algorithm. Theoretical analysis of the expected online regret bound is given in Section 4, along with the computational complexity. Section 5 gives practical remarks for efficient implementation. In Section 6, we evaluate the performance of the proposed algorithm on several real-world high-dimensional datasets with varying sparsity levels. We illustrate that the proposed method leads to improved stability and prediction performance for both sparse and dense data, with the most improvement observed in data with the highest average sparsity level. Section 7 concludes with further discussion on the proposed algorithm. ## 2 Truncated Stochastic Gradient Descent for Sparse Learning Assume that we have a set of training data , where the feature vector and the scalar output . In the following, we use to represent the vector of the sample of length and for the feature vector of all samples of length . In this paper, we are interested in the case that both and are large and the feature vectors ’s, , are sparse. We consider a loss function that measures the cost of predicting when the truth is . The prediction is given by function from a family parametrized by a weight vector . Denote . The learning goal is to obtain an optimal weight vector that minimize the loss function over the training data, with sparsity in the weight vector induced by a regularization term . We can then formulate the learning task as a regularized minimization problem: ^w =argminw∈Rp n∑i=1L(w,zi)+Ψ(w). (1) The above optimization problem is often solved using some version of gradient descent. When both and are large, the computation becomes very demanding. To address this computational complexity, the Stochastic Gradient Descent (SGD) algorithm was proposed as a stochastic approximation of the full gradient algorithm Bottou (1998). Instead of computing the gradient over the entire training set as under the batch setting, the stochastic gradient descent algorithm uses approximate gradients based on subsets of the training data. This is particularly attractive to large scale problems as it leads to a substantial reduction in computing complexity and potentially distributed implementation. For applications with large data sets or streaming data feeds, SGD has also been used as a subgradient-based online learning method. Online learning and stochastic optimization are closely related and interchangeable most of the time (Cesa-Bianchi et al., 2004). For simplicity, in the following, we focus our discussion and algorithmic description under the online learning framework with regret bound models. Nonetheless, our results can be readily generalized to stochastic optimization as well. In online learning, the algorithm receives a training sample at a time from a continuous feed. Without sparsity regularization, at time , the weight vector is updated in an online fashion with a single training sample drawn randomly, wt =wt−1−ηL′(wt−1,zt) (2) where is the learning rate and is a subgradient of the loss function with respect to . The set of subgradients of f at the point is called the subdifferential of at , and is denoted . A function is called subdifferentiable if it is subdifferentiable at all dom . When is differentiable at , . At the same time, a sequence of decisions is generated at , that encounters a loss respectively. The goal of online learning algorithm with sparsity regularization is to achieve low regret with respect to a fixed optimal weight vector . Here, is the parameter space for sparse weights vectors (see Assumption 3 on page 15 for more details.) The regret is defined as: RT(w∗)≜T∑t=1(L(wt,zt)+Ψ(wt))−T∑t=1(L(w∗,zt)+Ψ(w∗)). (3) In this paper, we focus on the regularization where and is the regularizing parameter. When adopted in an online learning framework, standard SGD algorithm does not work well in addressing (1) with penalty. Firstly, a simple online update requires the projection of the weight vector onto a -ball at each step, which is computationally expensive with a large number of features. Secondly, with noisy approximate subgradient computed using a single sample, the weights can easily deviate from zero due to the random fluctuations in ’s. Such a scheme is therefore inefficient to maintain a sufficiently sparse weight vector. To address this issue, Langford et al. (2009) induced sparsity in by subjecting the stochastic gradient descent algorithm to soft-thresholding. For every iterations at step , each of which is as defined in (2), the weight vector is shrunk by a soft-threshold operator with a gravity parameter with for . For a vector , ^wt=T(wt,g), (4) where with the operator defined by T(wj,gj) ≜{max(wj−gj,0),if% wj>0;min(wj+gj,0),if wj≤0. (5) As one can see, the sequence of SGD updates can be treated as a unit computational block, which will be referred to as a burst hereafter. Here the word burst indicates that it is a sequence of repetitive actions, e.g., the standard SGD updates as defined in (2), without interruption. Each burst is followed by a soft-thresholding truncation defined in (4), which puts a shrinkage on the learned weight vector. A burst can be viewed as a base feature selection realized on a set of random samples with regularization as in the classical LASSO (Tibshirani, 1996). Within a burst, let be the set of random samples on which the weight vector is stochastically learned. We define the set of features with nonzero weights in as its active (feature) set: ^Sg(^w;XK)={j:\|^wj\|>0}, (6) with a corresponding gravity . The steps within a truncated burst are summarized in Algorithm 1. In the truncated gradient algorithm of Langford et al. (2009), the gravity parameter is a constant across all dimensions as , where is a base gravity for each update in a burst and . In general, with greater parameter and smaller burst size , more sparsity is attained. When , the update in (4) becomes identical to the standard stochastic gradient descent update in (2). Langford et al. (2009) showed that this updating process can be regarded as an online counterpart of regularization in the sense that it approximately solves (1) in the limit as and . ## 3 Stabilized Truncated SGD for Sparse Learning Truncated SGD Langford et al. (2009) works well for dense data. When it comes to high-dimensional sparse inputs, however, it suffers from a number of issues. Shalev-Shwartz and Tewari (2011) observe that the truncated gradient algorithm is incapable of maintaining sparsity of the weight vector as it iterates. Recall that, under the online learning setting, the weight vector is updated with a noisy approximation of the true expected gradient using one sample at a time, from a random ordering of the data. With sparse inputs, it is highly probable that an important feature does not have a nonzero entry for many consequent samples, and is meaningfully updated for only a few times out of the updates in a burst. As a result, it would be truncated after a few iterations and brought back to nonzero after another few updates. At the same time, sparsity in inputs will also give rise to sporadic large nonzero updates for irrelevant features, which cannot be fully resolved by the soft-threshold operator. The derived weight vector ’s are of high variance, inadequate sparsity and poor generalizability. As an example, the number of nonzero variables in the weight vector during the last 1000 stochastic updates from the truncated gradient algorithm implemented on a high-dimensional sparse dataset (Dexter text mining data set; see Section 6 for details.) are shown in Figure 1. It can be seen that the numbers of nonzero features in the weight vectors learned by the truncated SGD algorithm () remain large and highly unstable throughout these 1000 iterations, oscillating within 10% of the total number of features. As a comparison, also in Figure 1, we plot the results from our proposed stabilized truncated SGD applied to the same data. During these last 1000 updates, the proposed algorithm is using a less frequent truncation schedule due to our annealed reject rate. It attains both high sparsity in the weight vector and high stability with high-dimensional sparse data. In this section, we introduce the stabilized truncated Stochastic Gradient Descent (SGD) algorithm. It attains a truly sparse weight vector that is stable and gives generalizable performance. Our proposed method attunes to the sparsity of each feature and adopts informative truncation. The algorithm keeps track of whether individual features have had enough information to be confidently subject to soft-thresholding. Based on the truncation results, we systematically reduce the active feature set by permanently discarding features that are truncated to zero with high probability via stability selection. We further improve the efficiency of our algorithm by adapting gravity to the sparsity of the current active feature set as the algorithm proceeds. ### 3.1 Informative Truncation For the truncated SGD algorithm, Langford et al. (2009) suggest a general guideline for determining gravity in the batch mode by scaling a base gravity by , the number of updates, for a single truncation after a burst. A direct online adaptation of a regularization would shrinks the weight vector at every iteration. The above batch mode operation is to delay the shrinkage for iterations so that the truncation is executed based on information collected from random samples instead of from a single instance. This guideline implicitly assumes that the SGD updates in a burst are equally informative, which is in general true for dense features. For sparse features, however, under the online learning setting, not every update is informative about every feature due to the scarcity of nonzero entries. The original uniform formula, , for gravity would then create an undesirable differential treatment for features with different levels of sparsity. With a relatively small , it is very likely that a substantial proportion of features would have no non-zero values on a size- subsample used in a particular burst. The weights for these features remain unchanged after updates. Consequently, the set of sparse features run the risk of being truncated to zero based on very few informative updates. The truncation decision is therefore mostly determined by a feature’s sparsity level, rather than its relevance to the class boundary. To make the learning be informed of the heterogeneity in sparsity level among features, we introduce the informative truncation step, extended from the idea of base gradient used in Algorithm 1. Instead of applying a universal gravity proportional to to all features, the amount of shrinkage is set proportional to the number of times that a feature is actually updated with nonzero values in the size- subsample, i.e., the number of informative updates. Specifically, within each burst, the algorithm keeps a vector of counters, , of the numbers of informative updates for the features , . Let be the base gravity parameter that serves as the unit amount of shrinkage for each informative update on each feature. At the end of each burst, we shrink feature by . In other words, here we set . The computational steps for a burst with informative truncation in summarized in Algorithm 2. The proposed informative truncation scheme ensures that the decision of truncation is made based on sufficient and equivalent amount of evidence for evaluating each feature. A theoretical justification of how informative truncation helps improving truncation bias can be found in Lemma 2 (Section 4). This feature-specific gravity attunes to the sparsity structure incurred at each burst without ad-hoc adjustment. It also avoids data pre-processing for locating sparse entries, which can be computationally expensive and compromises the advantage of online computation. In comparison to the truncated gradient algorithm in Langford et al. (2009) that quickly shrinks many features to zero indiscriminately, informative truncation keeps sparse features until enough evaluation is conducted. In doing so, sparse yet important features will be retained. The proposed approach also reduce the variability in the resulted sparse weight vector during the training process. Duchi et al. (2011) uses a similar strategy that allows the learning algorithm to adaptively adjust its learning rates for different features based on cumulative update history. They use the norm of accumulated gradients to regulate the learning rate. By adapting the gravity with the counter within each burst, our proposed strategy here can be viewed as applying the norm to the accumulated gradients that is refreshed every steps. ### 3.2 Stability Selection Despite of its scalability, subgradient-based online learning algorithms commonly suffer from instability. It has been shown both theoretically and empirically that stochastic gradient descent algorithms are sensitive to random perturbations in training data as well as specifications of learning rate (Toulis et al., 2015; Hardt et al., 2015). This instability is particularly pronounced in sparse online learning with sparse data, as discussed in Section 1. Under an online learning setting, using random ordering of the training sample as inputs, the algorithm would produce distinct weight vectors and unstable memberships of the final active feature set. Moreover, there has been a lot of discussion, in the literature, on the link between the instability of an learning algorithm and its deteriorated generalizability (Bousquet and Elisseeff, 2002; Kutin and Niyogi, 2002; Rakhlin et al., 2005; Shalev-Shwartz et al., 2010). To tackle this instability issue, in the proposed algorithm, we exploit the method of stability selection to improve its robustness to random perturbation in the training data. Stability selection (Meinshausen and Bühlmann, 2010) does not launch a new feature selection method. Rather, its aim is to enhance and improve a sparse learning method via subsampling. The key idea of stability selection is similar to the generic bootstrap (Meinshausen and Bühlmann, 2010). It feeds the base feature selection procedure with multiple random subsamples to derive an empirical selection probability. Based on aggregated results from subsamples, a subset of features is selected with low variability across different subsamples. With proven consistency in variable selection, stability selection helps remove noisy irrelevant features and thus reduce the variability in learning a sparse weight vector. Incorporating stability selection into our proposed framework, each truncated burst with gravity parameter is treated as an individual sparse learning engine. It takes random samples and carries out a feature selection to obtain a sparse weight vector. In the following, we define first the notion of selection probability for the stability selection step in our proposed algorithm. ###### Definition 1 (selection probability). Let be a random subsample of of size , drawn without placement. Parametrized by the gravity parameter , the probability of the feature being in the active set of a truncated burst that returns is Πgj=P∗(j∈^Sg(^w;XK))=ED[1(\|^wj\|>0)], where the probability is with respect to the random subsampling of . Let . For simplicity, we drop the superscript of in later discussions. For the rest of the paper, the selection probability always refers to that corresponds to weight vector with gravity parameter . Under unknown data distribution, the selection probabilities cannot be computed explicitly. Instead, they are estimated empirically. Since each truncation burst performs a screening on all features, the frequency of each feature being selected by a sequence of bursts can be used to derive an estimator of the selection probability. We denote a sequence of truncated bursts as a stage. A preliminary empirical estimate of the selection probability is given by ^Πj=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩∑τ:~kj,τ>01(\|^wj,τ\|>0)nk∑τ=11(~kj,τ>0),for j s.t.nK∑τ=1~kj,τ>01,otherwise, (7) where are the counters of informative updates for burst , . Different from the conventional stability selection setting, ’s are obtained sequentially and thus are dependent with each other. When is small, different subsamples produce selection probability estimates using (7) exhibit high variability, even when initialized with the same weight vector at . On the other hand, a large value of requires a prohibitively large number of iterations for convergence. To resolve the issues of estimating selection probability using a single sequence of SGD updates, we introduce a multi-thread framework of updating paths. Multiple threads of sequential SGD updates are executed in a distributed fashion, which readily utilizes modern multi-core computer architecture. With processors, we initialize the algorithm on each path of SGD updates with a random permutation of the training data, , denoted as . Then independently, stages of bursts run in parallel along paths, which return with , , . The joint estimate of selection probability with gravity is obtained as ^Πj=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩M∑m=1∑τ:~k(m)j,τ>01(\|^w(m)j,τ\|>0)M∑m=1nk∑τ=11(~k(m)j,τ>0),for j s.t.M∑m=1nK∑τ=1~k(m)j,τ>01,otherwise. (8) When more processors are available, a smaller is required for the algorithm to obtain a stable estimate of selection probability. The dependence among ’s is also attenuated when random subsets of samples are used for the estimation. This strategy falls under parallelized stochastic gradient descent methods, which is discussed in detail by Zinkevich et al. (2010). Under the framework of stability selection, each stage on every path uses a random subsample. The estimated selection probability quantifies the chance that a feature is found to have high relevance to class differences given a random subsample. At the end of each stage, stable features are identified as those that belong to a large fraction of active sets incurred during this stage of bursts. ###### Definition 2 (Stable Features). For a purging threshold , the set of stable features with gravity parameter is defined as ^Ωg={j:Πgj≥π0}. (9) For simplicity, we write the stable set as when there is no ambiguity. Stability selection retains features that have high selection probabilities and discard those with low selection probabilities. At the end of a stage of paths, we purge the features that are not in the set of stable features by permanently setting their corresponding weights to zero, and remove them from subsequent updates. We define the stabilized weight vector as ~w=^w⋅1^Ω. (10) As discussed above, due to the nature of online learning with sparse data, there are two undesirable learning setbacks in a single truncated burst. The first occurs when an important feature has its weight stuck at zero due to inadequate information in the subsample used, while the second case is when a noise feature’s weight gets sporadic large updates by chance. Using informative bursts, we can avert the first type of setbacks and using selection probability based on multiple bursts, we can spot noisy features more easily. In the presence of a large number of noisy features, the learned weights for important features suffer from high variance. Via stability selection, we systematically remove noisy features permanently from the feature pool. Furthermore, the choice of a proper regularization parameter is crucial yet known to be difficult for sparse learning, especially due to the unknown noise level. Applying stability selection renders the algorithm less sensitive to choice of the base gravity parameter in learning a sparse weight vector via truncated gradient. As we will show using results from our numerical experiments, this purging by stability selection leads to a notable reduction in the estimation variance of the weight vector. Here, is a tuning parameter in practice. We have found that the learning results in the numerical experiments are not sensitive to different values of within a reasonable range. Under mild assumptions discussed in Section 4, we derive a lower bound of the expected improvement in convergence by employing stability selection in the learning process in Lemma 1. ### 3.3 Adaptive Gravity with Annealed Rejection Rate The truncated SGD (Langford et al., 2009) adopts a universal and fixed base gravity parameter at all truncations. As pointed out in Langford et al. (2009) , a large value of the base gravity achieves more sparsity but the accuracy is compromised, while a small value of leads to less sparse weight vector yet attaining better performance. In other words, different extents of shrinkage serve different purposes of a learning algorithm. The needs for shrinkage also changes as the weight vector and the stable set evolves. Intuitively, the truncation is expected to be greedy at the beginning so that the number of nonzero feature can be quickly reduced for better computational efficiency and learning performance. As the algorithm proceeds, fewer features remain in the stable set. We should then be careful not to shrink important features with a truncation that is too harsh. A large base gravity is effective in inducing sparsity at the beginning of the algorithm when the weight vector is dense. As the algorithm proceeds, the same value of gravity is likely to impose too much shrinkage when the learned weight vector becomes very sparse, exposing some truly important features at the risk of being purged. On the other hand, a small fixed gravity is over-conservative so that the algorithm will not shrink irrelevant features effectively, leading to slow convergence and a dense weight vector overridden by noise. Tuning a reasonable fixed base gravity parameter for a particular data set does not only creates additional computational burden, but also inadequate in addressing different learning needs during different stages of the algorithm. As the role of gravity in a learning algorithm is to induce sparse estimates, in this paper, we propose an adaptive gravity scheme that delivers the right amount of shrinkage at each stage of the algorithm towards a desirable level of sparsity for the learned weight vector. We propose to control sparsity by a target rejection rate , that is, the proportion of updates that are expected to be truncated. Guided by this target rejection rate, we derive the necessary shrinkage amount and the corresponding gravity. As we discussed in Section 3.1, a base gravity is used in our learning algorithms to create gravity values for individual features that are attuned to their data sparsity levels. Therefore our adaptive gravity scheme is carried out by adjusting . At the beginning of a particular stage, we examine the truncations carried out during the previous stage. The base gravity is then adjusted to project the target rejection rate during the current stage. Specifically, at stage , we look at the pooled set of non-truncated weight vectors and informative truncation counters from all the bursts conducted in the previous stages on multiple threads. The adaptive base gravity for a target rejection rate is then obtained as g0,s(βs)≜sup{g0≥0:^ps(g0)≤βs}. (11) Here is the empirical probability, i.e., ^ps(g0)≜M∑m=1nk∑τ=1∑{j:j∈^Ωs,~k(m)j,τ>0}1(∣∣ ∣∣Δw(m)j,τ~k(m)j,τ∣∣ ∣∣>g0)M∑m=1nk∑τ=1∑j∈^Ωs1(~k(m)j,τ>0), where is the amount of updates on feature during the burst. We initialize the algorithm with a high rejection rate so that a large proportion of the weight vector can be reduced to zero at the end of each burst during the early stage of the algorithm. It allows the algorithm to explore as many sparse combination of features as possible at the early stage of the learning process. Along with the stability selection, the set of stable features can be quickly reduced to a manageable size by removing the majority of noises. When the weight vector becomes sparse, we decrease the rejection rate proportionally. With a lower rejection rate, and consequently a lower gravity, the algorithm can better exploit the subsequent standard SGD updates for a more accurate estimate of the true weight vector. As the rejection rate decreases to 0, the algorithm converges to the standard stochastic gradient descent algorithm on a small subset of stable features. To achieve the balance between exploration and exploitation, we construct an annealing function for the rejection rate that decreases monotonically as the level of sparsity decreases. Let be the maximum rejection rate at initialization and let be the annealing rate. The annealing function for the rejection rate at stage is given by βs+1 =ϕ(ds;β0,γ) =⎧⎨⎩β0[exp(−γds)−dse−γ]γ≥0β0log(1−γ(1−ds))log(1−γ),γ<0, (12) where is the level of weight vector sparsity at the end of stage . The greater the value is, the faster the rejection rate is annealed to zero as the number of stable features decreases. A positive, zero and negative value of corresponds to exponential decay, linear decay and logarithmic decay of the rejection rate, respectively. Figure 2 presents examples of the rejection rate anneal function with , and respectively. By using adaptive gravity (11) with annealed rejection rate (12), the amount of shrinkage is adjusted to the current level of sparsity of the weight vector quantified by the size of the stable set or the norm of the purged . Instead of tuning a fixed gravity parameter as in Langford et al. (2009), for our proposed algorithm, we tune the annealing rate and the maximum rejection rate . Here balances the trade-off between exploration and exploitation and determines the initial intensity of truncation. It enables the tuning process to be tailored to the data at hand as well as being comparable across different datasets. In Section 6, the tuning results instantiate that a negative annealing rate is preferred for highly sparse data, such as the RCV1 dataset, since the a high rejection rate needs to be maintained longer allowing sufficient information of sparse features to be evaluated by the learning process. On the other hand, a positive annealing rate is chosen for relatively dense data, such as the Arcene dataset, where the high frequency of nonzero values permit fast reduction of the active set. The complete algorithm of the stabilized truncated stochastic gradient descent algorithm is summarized in Algorithm 4. ## 4 Properties of the Stabilized Truncated Stochastic Gradient Descent Algorithm The learning goal of sparse online learning is to achieve a low regret as defined in (3). In this section, we analyze the online regret bound of the proposed stabilized truncated SGD algorithm in Algorithm 4 with convex loss. For simplicity, the effect of adaptive gravity with annealed rejection rate is not considered here. To achieve viable result, we make the following assumptions. ###### Assumption 1. The absolute values of the weight vector are bounded above, that is, for some , . ###### Assumption 2. The loss function is convex in , and there exist non-negative constants and such that, for all and , . For linear prediction problems, the class of loss function that satisfies Assumption 2 includes common loss functions used in machine learning problems, such as the loss, the hinge loss and the logistic loss, with the condition that for some constant . ###### Assumption 3. Assume that the set for identifying has the following properties: 1. is the parameter space for weight vectors that is subject to a sparsity constraint \|\|w\|\|0=d∗. For the optimal weight vector , we denote and . We further assume that is sufficiently small and the gravity parameter associated with is reasonably large so that, for , the average number of active selected features from each truncated bursts, , given a gravity , is greater than or equal to . 2. Assume features has various sparsity distribution that , where , for , and if , for . Assumption 3 posits that ’s parameter space of interest is substantially sparse, which is the main focus of sparse learning and of this paper. The theoretical analysis in the following concerns for an fixed optimal weight vector , if exists, under such a constraint. Nevertheless, this condition does not confine the applicable scenarios of the proposed method to a fixed subclass of problems. It suggests a balance between the model sparsity and the value of the gravity parameter that is implicitly embedded within the parameter tuning process. ###### Lemma 1. Let be a non-stabilized dense weight vector, i.e., an output weight vector from Algorithm 3. Let be the stabilized weight vector derived from , which is purged by the stability selection (10) with a set of stable features . Let be the average number of nonzero entries in ’s of the previous truncated bursts, i.e.,the number of selected features from truncated bursts, with gravity . Then, if Assumption 1 holds, there exists an with such that the bound on the expected difference between the distance from the non-stabilized weight vector to under Assumption 3 and the distance from the stabilized weight vector to is given by E(\|\|^w−w∗\|\|2−\|\|~w−w∗\|\|2) ≥ ε2(\|^Sε\|−\|^Ω\|)+2π0C2[(1−qg2π0p−p)qg−d∗] (13) When the purging threshold is sufficiently high such that , E(\|\|^w−w∗\|\|2−\|\|~w−w∗\|\|2)≥0. ###### Proof. : See Appendix A. ∎ Lemma 1 quantifies the gain of using stabilization when is highly sparse, where stability selection efficiently shrinks high variable estimates to zero. When the purging threshold is sufficiently high such that , the lower bound achieved by (13) is guaranteed to be positive. Furthermore, this result also indicates that the expected difference between distances from the non-stabilized and stabilized weight vector to the sparse depends on the differences between the sizes of the temporary nonzero set of features before purging, , and the size of the stable features after purging. In expectation, the stabilized weight vector is closer to the target sparse weight vector as the operation of purging efficiently reduces the size of stable features. This suggests a much faster convergence with stabilization. Lemma 1 also provides an insight on the benefit from using adaptive gravity with annealed rejection rate. At the beginning of the algorithm, the gap between the size of and the size of the set of stable features is large when aiming for extensive exploration of different sparse combination of features. Hence, the improvement brought by stabilization is more substantial during the early state of learning period. As the algorithm proceeds and the set of stable features becomes smaller and stabler, it dwindles the leeway that allows the aforementioned two sets to be different. Consequently, the proposed algorithm is gradually tuned toward the standard stochastic gradient descent algorithm to facilitate better convergence at the later period of the learning process. ###### Lemma 2. Let be the weight vector at initialization. After the first burst, let be the truncated weight vector using universal gravity as in Algorithm 1 and let be the truncated weight vector with informative truncation as in Algorithm 2. Then, under Assumption 3, E(\|\|¯w1−w∗\|\|2−\|\|^w1−w∗\|\|2)≥2g0K∑t=1∥ζtw∗∥1≥0 (14) where for and . ###### Proof. : See Appendix B. ∎ In Lemma 2, we compare the distances towards from 1) the weight vector with uniform gravity and 2) the weight vector with informative truncation that depends on the number of zero entries occurred in a burst. Such a gap suggests the effectiveness of informative truncation on sparse data in which feature sparsity is highly heterogeneous. In the scenarios where very few nonzero entries appear in a burst, the informative truncation imposes gravity that is proportional to the information presented in a burst. It is a fairer treatment than uniform truncation and leads to a large improvement in expectation. When features are all considerably dense in a burst, the informative truncation is equivalent to the uniform truncation. In short, Lemma 1 demonstrates the improvement in expected squared error due to stabilization on the weight vector. Lemma 2, on the other hand, quantifies the improvements in reduce truncation bias when implementing informative truncation on sparse features with heterogeneous sparsity levels. Given Lemma 1 and Lemma 2, we have the expected regret bound of the proposed Algorithm 4 in Theorem 1. ###### Theorem 1. Consider the updating rules for the weight vector in Algorithm 3. On an arbitrary path, with and , let be the resulted weight vector and be the gravity values applied to the weight vectors generated by Algorithm 4, along with the base gravity parameters . Set the purging threshold to be sufficiently large such that . If Assumption 1, 2, and 3 hold, then there exists a sequence of at each stability selection with the set of stable features such that the expectation of the regret defined in (3) is bounded above by E(T∑t=1[L(w––t,zt)+Kg–t\|\|w––t\|\|1]−T∑t=1[L(w∗,zt)+Kg–t\|\|w∗\|\|1]) ≤ ηA2−ηA(E[T∑t=1L(w∗,zt)+Kg–0,t(\|\|w∗\|\|1−\|\|w––t\|\|1)])+12−ηA(ηTB+1η\|\|w∗\|\|2) −12η−η2AT∑t=1ε2t1(tKnK∈Z)(\|^Sεt,t\|−\|^Ωt\|) (15) where and is the weight vector at time before stabilization. ###### Proof. : See Appendix C. ∎ In the result of Theorem 1, the first two parts of the right-hand-side of the expected regret bound (15) is similar to the bound obtained in Langford et al. (2009). It implies the trade-off between attained sparsity in the resulted weight vector and the regret performance. When the applied gravity is small under the joint effect of the base gravity and the size of each burst , the sparsity is less but the expected regret bound is lower. On the other hand, when the applied gravity is large, the resulted weight vector is more sparse but at the risk of higher regret. Based on Lemma 1, the proposed algorithm is guaranteed to achieve lower regret bound in expectation when the target sparse weight vector is highly sparse. As quantified in the third term of the right-hand-side of (15), the improvement comes from the reduction of the active set at each purging. By its virtue, noisy features are removed from the set of stable features and thus are absent in later SGD updates and truncations. Theorem 1 is stated with a constant learning rate . It is possible to obtain a lower regret bound in expectation with adaptive learning rate decaying with , such as , which is commonly used in the literature of online learning and stochastic optimization. However, the discussion of using an varying learning rate is not a main focus of this paper and adds extra complexity of the analysis. Without knowing in advance, this may lead to a no-regret bound as suggested in Langford et al. (2009). Instead, in Corollary 1, we show that the convergence rate of the proposed algorithm is with . ###### Corollary 1. Assume that all conditions of Theorem 1 are satisfied. Let the learning rate be . The upper bound of the expected regret is where . ###### Proof. By plugging in to the result from Theorem 1, we get E(T∑t=1[L(w––t,zt)+g–t\|\|w––t\|\|1]−T∑t=1[L(w∗,zt)+g–t\|\|w∗\|\|1]) ≤ A2√T−A(E[T∑t=1L(w∗,zt)+g–t(\|\|w∗\|\|1−\|\|w––t\|\|1]) +T2√T−A(ηTB+1η\|\|w∗\|\|2)−T2√T−AT∑t=1ε2t1(tKnK∈Z)(\|^Sε,t\|−\|^Ωt\|). The result is then straightforward. ∎ Assume that the input features have nonzero entries on average. With linear prediction model , the computational complexity at each iteration is . Leveraging the sparse structure, the informative truncation only requires an additional space for recording the counters. The purging process of stability selection consumes , , space for storing the generated intermediate weight vectors and computational complexity. Both storage and computational cost decrease when the set of stable features diminishes as the algorithm proceeds. Since the parameters , , and is normally set to be small values, the complexity mostly depends on . In summary, the proposed algorithm scales with the number of nonzero entries instead of the total dimensions, making it appealing to high-dimensional applications. ## 5 Practical Remarks When implementing Algorithm 4 in practice, the performance can be further improved in terms of both accuracy and computational efficiency by employing a couple of practical techniques. It includes applying informative purging and attenuating the truncation frequency to achieve more accurate sparse learning and steadier convergence. The first improvement can be implemented by better addressing the issue of scarcity of incoming samples. For computing selection probabilities, instead of using only information from the current stage, we can inherit information from previous stages for features that are too scarce to accumulate enough updates during one stage. Specifically, we introduce an accumulated counter at stage as the total number of times that a feature is updated within a burst during this stage: κj,s=M∑m=1nK∑τ=1~k(m)j,τ,j=1,…,p, which is essentially the denominator of the selection probability in (8). Similarly, we define an accumulated truncation indicator at stage as the total number of times that a feature is truncated to zero given valid update(s): bj,s=M∑m=1∑τ:~k(m)j,τ>01(\|^w(m)j,τ\|>0),j=1,…,p. A feature is then evaluated in the stability selection only if there are enough updates from the present stage and from any unused information carried over from previous stages. Given a threshold , let and . The selection probability is modified as ^Πj,s=⎧⎨⎩~bj,s~κs,for j s.t.~κs>δK1,otherwise, for j=1,…,p. (16) This strategy extends the key idea in Section 3.1 that, with sparse data, each decision need to be based on sufficient evidence. Using the “carried-over” information allows the algorithm to utilize information available in a sequence of SGD updates while attuned to the needs of features with different levels of sparsity. In practice, this modification facilitates faster convergence especially for ultra-sparse data. The second practical strategy is that the size of each burst, , can be adaptively adjusted in a similar fashion as the rejection rate in (12). At the end of each stage, the burst size is updated as Ks=⌈K0log(1αds−1)⌉, where is the initial burst size and, as in (12), . The tuning parameter adjusts the annealing rate of the truncation frequency. Although the result in Theorem 1 is based on a fixed , it can be easily shown that the same upper bound can also be attained with an increasing . By increasing in the later stage of the algorithm, when the majority of irrelevant features have been removed from the stable set, the chance of erroneous truncation is reduced. Such scheme further steers the algorithm from the mode of exploring potential sparse combination of features in the early stage toward the fine tuning of the weight vector by exploiting information from more samples in a sequence. It also facilitates faster convergence as the size of the stable set approaches to a sufficiently small number, as the algorithm converges to the standard stochastic gradient descent approximately. ## 6 Results In this section, we present experimental results evaluating the performance of the proposed stabilized truncated SGD algorithm in high-dimensional classification problems with sparsity regularization. In this paper, we focus on linear prediction model for binary classification where and with the observed class label . We consider two commonly used convex loss functions in machine learning tasks that both satisfy Assumption 1: • Hinge loss: • Logistic loss: Using five datasets from different domains, the performance of our algorithm and other algorithms for comparison are evaluated on classification performance and feature selection stability and sparsity. We first define measure of feature stability in Section 6.1. ### 6.1 Feature Selection Stability The goal of sparse learning is to select a subset of truly informative features with stabilized estimation variance as well as increased classification accuracy and model interpretability. Subgradient-based online learning methods depend heavily by the random ordering of samples on which they are fed to the algorithm. Such dependence leads to much deteriorated performance when it comes to high-dimensional sparse inputs. For a particular feature, the positions of its nonzero occurrences in a random ordering of samples greatly affect its learning outcome, in terms of learnt weight and membership in the set of selected features. Therefore, in addition to attaining a low generalization error, a desirable sparse online learning method should also produce an informative feature subset that is stable and robust to random permutations of input data. To evaluate feature selection stability of subgradient-based sparse learning methods, we define in the following a numerical measure of similarity between selected feature subsets resulted from different random permutations of data. Given an output weight vector from a subgradient-based algorithm with input data , similarly as in (6), we denote the selected feature subset as S(w––;D)={j:\|w––j\|>0,w––=Ψ(D}. Given two random permutations of the training data , and , the similarity between the two sets of selected feature subsets and is measured by the Cohen’s kappa coefficient (Cohen, 1960), κ(S1,S2)=qo−qe1−qe, where is the relative observed agreement between and	{"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": 0.9346431493759155, "perplexity": 632.4918669936706}, "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-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00545.warc.gz"}
http://math.stackexchange.com/questions/13562/an-application-of-the-optional-sampling-theorem	# An application of the Optional Sampling Theorem let $S(k), k\geq 0$ a discrete random process. Suppose $S(N)$ is with probability one either 100 or 0 and that $S(0)=50$. Suppose further there is at least a sixty percent probability that the price will at some point dip below 40 and then subsequently rise above 60 before time $N$. How do you prove that $S(k)$ cannot be a martingale?	{"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": 0.9163586497306824, "perplexity": 157.37901893299107}, "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/1438042988650.6/warc/CC-MAIN-20150728002308-00331-ip-10-236-191-2.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/interesting-analytical-chem-problem.634008/	# Interesting Analytical Chem Problem 1. Sep 6, 2012 ### ghelman Hi all, I need to solve this problem for my environmental chemistry class. It is supposed to be a "review" problem of concepts from analytical and gen. chem. but I certainly do not know how to do it. 1. The problem statement, all variables and given/known data If you work at the wastewater treatment plant, you are exposed to 1,4-dichlorobenzene, which is used in urinal deodorant cakes. a. If you work in a 500 m3 space with an exposed body of wastewater and can still smell the DCB (smell threshold 500 ppbv) what is the concentration of DCB in the water? b. In the winter the temperature of the water falls to 5C and you can no longer smell the DCB; what is the maximum concentration of DCB in the water at the new temperature? 3. The attempt at a solution V=500 m3 CDCB=500 parts per billion by volume P=1 atm (assumably) I assume that the wastewater plant operates at a temperature of 25o C. I am not sure if that is a good assumption or not. Anyone have any insight to that? I also assumed that the system is in equilibrium. I am not sure if I am allowed to assume that, but it seems reasonable. Doing so gives: Vapor pressure of DCB = 1.76 mm Hg At this point I do not know what to do. I have a concentration for DCB in air, but how do I relate that to the concentration of DCB in water at equilibrium? I know that the fugacities will be equal at equilibrium, but I am not sure whether that will be useful or if it is just a wild goose chase. Help! 2. Sep 7, 2012 ### AGNuke 1.) Assumption is justified. 25°C is totally justified. It is a standard Temperature. 2.) Henry's Law for conc. of DCB in water? 3. Sep 9, 2012 ### ghelman Hmmmm...but how would using Henry's Law make use of the given concentration in the air? 4. Sep 10, 2012 ### Staff: Mentor Unless I am missing something, there is not enough data to solve the problem. And no idea how you got 1.76 mmHg. 5. Sep 10, 2012 ### ghelman I got the vapor pressure from a research article. We are allowed to look up any values we think we may need. Given that, what will we need to know in order to solve the problem? 6. Sep 10, 2012 ### Staff: Mentor It is probably vapor pressure over a solid DCB at STP - so of no use here (unless it can be somehow related to solubility/pressure over solution). You need either a Henry's constant, or some kind of information that will let you calculate it. 7. Sep 10, 2012 ### ghelman I can probably find an article with a Henry's law constant, but if I solved it that way, how would that make use of the given concentration? Also, how would I solve part b? 8. Sep 10, 2012 ### Staff: Mentor Huh? Simple plug and chug. Or is your problem conversion between concentration and partial pressure? Assuming Henry's constant is not temperature dependent. Or using value for 5°C. It is almost exactly the same problem. 9. Sep 10, 2012 ### ghelman So would I assume that the system is in equilibrium and use the vapor pressure as the partial pressure in Henry's Law? If I did that then how would I use the given concentration of 500 ppbv? Or am I supposed to plug in the given concentration and solve for the partial pressure? But then what would I do with that? 10. Sep 10, 2012 ### AGNuke Convert 500ppbv into partial pressure. That's the amount of DCB in atmosphere, by volume, and by ideality (which we generally consider to make our life at least not miserable) 500ppb(pressure). 11. Sep 10, 2012 ### JohnRC (1) I suspect that the "given" 500 cubic metre of workspace is irrelevant except insofar as it suggests an environment where the vapour phase DCB will be in equilibrium with the liquid phase DCB -- that is, that the volume is fairly small, and suggests an enclosed, draught free workspace. (2) I also think that it is important to remember that you are dealing with inequalities here -- the conc of DCB is not = 500 ppbv but >= 500 ppbv (3) The equilibrium vapour pressure of the solid is relevant if you combine it with the (low) solubility of the solid in water, because the PCB activity in the solid is 1, and so the PCB concentration in the solution will therefore be equal to the same proportion of the saturation concentration as that of the vapour is to the saturated vapour in equilibrium with the solid (as expressed in the saturation vapour pressure). (4) Wastewater facilities are usually kept to around 5-10°C if the outdoor temperature is below freezing, but they are usually places where no other heating or cooling is applied. If a temperature is not specified, an assumption of 15°, 20° or 25 °C would be reasonable. 12. Sep 11, 2012 ### Staff: Mentor If you have air at 1 atm, and 21% of the air by volume is oxygen (in other words: concentration of oxygen is 21% v/v), what is the partial pressure of oxygen? This is an application of Avogadro's principle. 13. Sep 11, 2012 ### JohnRC I do not believe that an answer like this is really grasping the difficult part of this problem, which hinges on how to relate a concentration in air to a concentration in aqueous solution (Borek again): A datum like this can be brought into the solution of the problem, since the saturation vapour pressure is effectively the "solubility" of the solid in air, which can be matched to the actual solubility of the solid in water, and provide the information needed to match other airborne and aqueous concentrations. 14. Sep 11, 2012 ### Staff: Mentor My understanding of the situation is that OP has no idea how to convert 500 ppb to partial pressure (at least that's what I read between lines in earlier posts). Without doing this (simple) step, you can't proceed with the more difficult part. If I understand you correctly you want to assume activity of the DCP to be the same in the air and in the solution? Perhaps it will work, I have never seen it done this way (but I am far from claiming I have seen everything, quite the opposite). 15. Sep 11, 2012 ### JohnRC Not the activity, but the chemical potential -- the chemical potential must be equal in the solid, the saturated vapour, and the saturated solution. The last two can therefore be equated and scaled down to lower concentrations. This is really just the fugacity or Henry's law approach in another guise, and an accurate solubility figure for DCB can be fairly easily obtained. Similar Discussions: Interesting Analytical Chem Problem	{"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": 0.8564586639404297, "perplexity": 814.5510830859764}, "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-2018-05/segments/1516084889660.55/warc/CC-MAIN-20180120142458-20180120162458-00239.warc.gz"}
https://hsm.stackexchange.com/questions/8050/what-geometric-results-were-first-proven-by-assuming-all-real-numbers-are-ration	# What geometric results were first proven by assuming all real numbers are rational? Pythagoras and his followers believed that all magnitudes are commensurable; that is, the ratio of two magnitudes of the same kind, like two lengths or two areas, is equal to the ratio of natural numbers. In modern language, this means that all real numbers are rational numbers. This allowed them to find out whether any two ratios of magnitude are equal: you just find out what ratio of natural numbers each one is equal to, and then compare those ratios to each other. But Hippassus of Metapontum discovered that the ratio of a diagonal of a square to its side is not equal to a ratio of natural numbers. So a new definition was needed for when two ratios of magnitudes are equal. This was done by Eudoxus of Cnidus. In modern language, his definition says that two real numbers are equal if the same rational numbers are less than, equal to, and greater than them (basically the idea behind Dedekind cuts). My question is, what results of geometry did the Pythagoreans prove using the incorrect assumption that all real numbers were rational? Long ago, I read a book, whose name I don't recall, which listed some results, and described how they had to be reproved with Eudoxus' new definition.	{"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": 0.8737915754318237, "perplexity": 327.37213732382605}, "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-2019-35/segments/1566027314852.37/warc/CC-MAIN-20190819160107-20190819182107-00355.warc.gz"}
http://slideplayer.com/slide/1420999/	# 15.1 Notes Numerical approach to evaluating limits as x approaches negative or positive infinity. ## Presentation on theme: "15.1 Notes Numerical approach to evaluating limits as x approaches negative or positive infinity."— Presentation transcript: 15.1 Notes Numerical approach to evaluating limits as x approaches negative or positive infinity 15.1 Notes In addition to limits where x approaches a number, there are limits where x approaches infinity (or negative infinity). A limit as x is approaching infinity (or negative infinity) is asking what f(x) is getting close to as x is getter larger and larger. 15.1 Notes In todays do now assignment you were actually evaluating a limit where x approaches infinity. As larger and larger values are x were evaluated in the function provided, the f(x) values got closer and closer to one half. Although f(x) will never reach one half, the limit is equal to one half. 15.1 Notes Just like a graphing calculators table feature was used to evaluate limits as x approached a number, it can be used to evaluate limits as x approaches infinity. However, rather than setting up the table to simulate x jumping by small increments to get close to some number, the table is set up to simulate x approaching infinity. 15.1 Notes The functions equation is typed in on the Y= screen as before. In the Table Setup menu we were starting the table (TblStart) at the value x is approaching and choosing a small number for the increment (ΔTbl) by which the x values changed. How should the menu be set up to simulate approaching infinity? By choosing relatively large numbers for both the TblStart and the ΔTbl. 15.1 Notes Consider todays do now assignment again: Go to the Y= screen and type in the function. 15.1 Notes Press 2 nd and WINDOW (TBL SET) to bring up the Table Setup Menu. Start the table at some large value, say 10, and set the increment ( Tbl) to some large value, say 50. 15.1 Notes Press 2 nd and GRAPH (TABLE) to view the table. 15.1 Notes Scrolling further down the table reveals that.5 appears in the y column. Realize however, that because the Y column only has five decimal places, the calculator is rounding. The f(x) values get very close to but never actually reach it. 15.1 Notes Another example… Type the equation on the Y= screen. 15.1 Notes Choose large numbers in the Table Setup Menu. This time 50 is chosen for TblStart, 100 for ΔTbl. 15.1 Notes View the t-table: As you scroll up the table (simulating x approaching negative infinity) the y values appear to be getting closer to -.5. 15.1 Notes Another example… Type the equation on the Y= screen. 15.1 Notes Make sure the calculator is in radian mode since the limit is of a trigonometric function. Choose large numbers in the Table Setup Menu. This time 100 is chosen for TblStart and ΔTbl. 15.1 Notes View the t-table: As you scroll down the table (simulating x approaching infinity) the y values appear to be getting closer to 0. 15.1 Notes One more example… Type the equation on the Y= screen. 15.1 Notes Setup the table. 100 is used for TblStart and ΔTbl. 15.1 Notes View the table: Notice as you look down the values in the y column that they are getting larger and larger rather than approaching some value. The y values are also approaching infinity. 15.1 Notes – Practice Problems: 1. 2. 3. 4. Download ppt "15.1 Notes Numerical approach to evaluating limits as x approaches negative or positive infinity." Similar presentations	{"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": 0.8178882002830505, "perplexity": 1381.2572609368224}, "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/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00272.warc.gz"}
https://math.stackexchange.com/questions/2584490/corollary-25-3-humphreys-linear-algebraic-groups	# Corollary 25.3, Humphrey's Linear Algebraic Groups I'm reading Humphrey's Linear Algebraic Groups (GTM 21). I don't understand the proof of Corollary 25.3(d). Corollary. Let $G$ be reductive, $rank_{ss}G = 1$, $T$ a maximal torus of $G$, $Z = Z(G)^\circ$. Then: (a) $(G,G)$ is semisimple, of dimension 3. (b) $G = (G,G)\cdot Z$, the intersection of $(G,G)$ with $Z$ being finite. (c) $C_G(T) = T$, and in particular, $Z(G) \subset T$. (d) If $\varphi : G \to \mathrm{PGL}(2,\mathsf{K})$ is an epimorphism, $\operatorname{Ker}\varphi = Z(G)$. You don't need to read all the proof. I can understand the proof of (a) (b) (c). I just don't understand the last sentence: "So (c) and (d) follows." Proof. We get from part (f) of the theorem an epimorphism $\varphi : G \to \mathrm{PGL}(2,\mathsf{K})$, with $(\operatorname{Ker}\varphi)^\circ = R(G)$. Since $G$ is reductive, $R(G) = Z$ and is a torus. On the other hand, $\mathrm{PGL}(2, \mathsf{K})$ is its own derived group: for example, otherwise the derived group would be solvable, contrary to the semisimplicity of $\mathrm{PGL}(2,\mathsf{K})$. It follows that $\varphi$ maps $(G,G)$ onto $\mathrm{PGL}(2,\mathsf{K})$. Combined with the fact that $(G,G)$ is connected and the fact that $(G,G)\cap Z$ is finite, this proves both (a) and (b). It is easy to see that a maximal torus in $\mathrm{PGL}(2,\mathsf{K})$ is its own centralizer, by computing in $\mathrm{SL}(2,\mathsf{K})$, so $C_G(T)$ and $T$ have the same image under $\varphi$, forcing $T \subset C_G(T) \subset T\cdot \operatorname{Ker}\varphi$. But $T$ has finite index in the right side, while $C_G(T)$ is connected. So (c) and (d) follows. It seems that the proof doesn't say anything about $\operatorname{Ker}\varphi$ except that $(\operatorname{Ker}\varphi)^\circ = Z$. we also know that $Z(G) \subset \operatorname{Ker}\varphi$, because the center of $\mathrm{PGL}(2,\mathsf{K})$ is trivial. So $Z(G)$ has finite index in $\operatorname{Ker}\varphi$. How can I prove that $\operatorname{Ker}\varphi \subset Z(G)$? Am i missing something? • I don't remember all these details when I learned linear algebraic groups. But I do remember constantly switching between Borel, Springer, and Humphreys when I was stuck on one book's explanation of something. – D_S Dec 31 '17 at 16:01 I'm not sure what Humphreys had in mind, but I think the following argument works. If $N$ is a finite normal subgroup of a connected group, then $N$ is contained in the center of that group (Exercise 7.11). We have $(\operatorname{Ker}\phi)^0 = Z(G)^0 = Z$, the radical of $G$. And we want to show that in fact $\operatorname{Ker}\phi = Z(G)$. In the quotient group $G/(\operatorname{Ker}\phi)^0$, the subgroup $\operatorname{Ker}\phi/(\operatorname{Ker}\phi)^0$ is finite, hence contained in the center of $G/(\operatorname{Ker}\phi)^0$. Let $x \in \operatorname{Ker}\phi$, so that $x(\operatorname{Ker}\phi)^0$ is in the center of $G/(\operatorname{Ker}\phi)^0$. Then for all $y \in G$, $xyx^{-1}y^{-1} \in (\operatorname{Ker}\phi)^0 = Z$. But also $xyxy^{-1}$ is in the derived group of $G$, which has finite intersection with $Z$ (Lemma 19.5). So the image of the morphism $$G \rightarrow G, y \mapsto xyx^{-1}y^{-1}$$ is finite and irreducible, hence a point. Therefore, $x \in Z(G)$. On the other hand, if $x \in Z(G)$, then as you say, the image of $x$ in $G/\operatorname{Ker}\phi \cong \operatorname{PGL}_2$ is trivial, because $\operatorname{PGL}_2$ has trivial center.	{"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": 0.9703934788703918, "perplexity": 61.20209396181475}, "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-2019-43/segments/1570987836368.96/warc/CC-MAIN-20191023225038-20191024012538-00373.warc.gz"}
https://brilliant.org/problems/two-squares-2/	# Two squares How many integers between 1 and 100 inclusive can be expressed as the sum of 2 perfect squares? ×	{"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": 0.8357451558113098, "perplexity": 167.5709465475675}, "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-2017-51/segments/1512948595858.79/warc/CC-MAIN-20171217113308-20171217135308-00510.warc.gz"}
http://physics.stackexchange.com/questions/57000/number-of-configurations-of-the-universe	# Number of configurations of the universe I have read that quantum mechanics says that the amount of possible particle configurations is $10^{10^{122}}$ to be exact in the universe. Do we know this figure to be exactly true to the exact figure? Wouldn't we need to know a true theory of quantum gravity to know the exact answer? Is the amount exactly that figure or just an estimate? - John, you have the power to delete question with no upvoted answers, but you should not remove the content from posts and leave it on the site. Posts here are not ephemeral, they are expected to last delete it if you must but do not erase it that way. – dmckee Mar 16 '13 at 0:18 I'm sorry I was embarrassed by my question in the end will not happen again – John rose Mar 16 '13 at 1:53 This is clearly an estimate, not an exact figure. It comes from ideas about quantum gravity (not proven but very strong conjecture) that say that the maximum entropy of a region is $$S = \frac{A}{4 \ell_P^2},$$ where $A$ is the area of a surface bounding the region and $\ell_P \approx 10^{-35}\ \mathrm{m}$ is the Planck length. Now entropy is a measure of the number of configurations available to a system (units where $k_B=1$): $$S = \ln \Omega.$$ Putting this together with the observed size of the universe for $A$ gives roughly the figure you mention. -	{"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": 0.8174467086791992, "perplexity": 241.89274932648792}, "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-2016-18/segments/1461860127407.71/warc/CC-MAIN-20160428161527-00077-ip-10-239-7-51.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/cylindrical-spherical-coordinates.667344/	# Cylindrical / Spherical Coordinates 1. Jan 27, 2013 ### eurekameh I'm trying to convert the below Cartesian coordinate system into cylindrical and spherical coordinate systems. For the cylindrical system, I had r,vector = er,hat + sint(e3,hat). While I do have a technically correct answer for the spherical coordinate system, I believe, I was wondering if there was a way to simplify the expression 1/cos[tan^-1*(sint)]. Thanks. 2. Jan 27, 2013 ### Staff: Mentor $r=\sqrt{x^2+y^2+z^2} = \sqrt{1+\sin^2(t)}$? 3. Jan 27, 2013 ### eurekameh Also, am I right in thinking r = rad(x^2 + y^2) = 1 for cylindrical coordinates? 4. Jan 27, 2013 ### Staff: Mentor Would be interesting to check this in an analytic way. If you use (r,theta,z) as coordinates: Right.	{"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": 0.9133432507514954, "perplexity": 2359.9271986509557}, "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/1510934806771.56/warc/CC-MAIN-20171123104442-20171123124442-00769.warc.gz"}
http://math.stackexchange.com/questions/227042/closed-subset-of-complete-metric-space-dont-understand-last-part-of-theorem?answertab=oldest	# Closed subset of complete metric space…don't understand last part of theorem. A closed subset of a complete metric space is a complete subspace. Proof. Let $S$ be a closed subspace of a complete metric space X. Let $(x_n)$ be a Cauchy sequence in $S$. Then $(x_n)$ is a Cauchy sequence in $X$ and hence it must converge to a point $x$ in $X$. But then $x \in \bar{S} = S$. Thus $S$ is complete. I have seen this theorem in several places but I never know why they are able to say "But then $x \in \bar{S} = S$"...Why is $x$ in $\bar{S}$? - Because S is a closed subset, hence $S = \bar{S}$ – amWhy Nov 1 '12 at 20:48 Do you know what $\bar S$ means? – Chris Eagle Nov 1 '12 at 20:48 Yes, its the closure of $S$, ie. the smallest closed subset containing $S$ or alternatively the set $S \cup$ the limit points of $S$. I don't see how they can say $x$ is in the closure of $S$ though. – sonicboom Nov 1 '12 at 20:53 Let $U$ be any open nbhd of $x$. There is an $r>0$ such that $B(x,r)\subseteq U$, where $B(x,r)$ is the open ball of radius $r$ centred at $x$. Since $\langle x_n:n\in\Bbb N\rangle\to x$, there is an $m\in\Bbb N$ such that $d(x_n,x)<r$ whenever $n\ge m$. In particular, $d(x_m,x)<r$, so $x_m\in B(x,r)\subseteq U$. Moreover, $x_m\in S$, so $x_m\in U\cap S$. This shows that $U\cap S\ne\varnothing$ for every open nbhd $U$ of $x$, and it follows at once that $x\in\operatorname{cl}S$. Added: Here in one place are some definitions that seem to be giving you a bit of trouble. Let $\langle X,d\rangle$ be a metric space, $S\subseteq X$ and $x\in X$. • $x\in\operatorname{cl}S$: for each $r>0$, $B(x,r)\cap S\ne\varnothing$. In words, every open ball at $x$ contains at least one point of $S$. • $x$ is a limit point (or cluster point) of $S$: for each $r>0$, $B(x,r)\cap(S\setminus\{x\})\ne\varnothing$. In words, every open ball at $x$ contains at least one point of $S$ other than $x$. • $x\in\operatorname{int}S$: there is an $r>0$ such that $B(x,r)\subseteq S$. - Cheers, I am going to have to write that out and draw some diagrams to make sure I understand whats going on. – sonicboom Nov 1 '12 at 21:00 When they say "Let $(x_n)$ be a Cauchy sequence in $S$" are they saying that all values of $x_n$ will be in $S$ and the only question is then where the limit of the sequence lies? – sonicboom Nov 1 '12 at 21:07 I get it now...no matter how small we make $r$ there will be an open ball with center $x$, radius $r$ that contains a point of $S$ other than $x$ itself, hence $x$ is a limit point. – sonicboom Nov 1 '12 at 21:14 @sonicboom: Actually, it’s possible that $x\in S$ and that the sequence is eventually constant at $x$, so $x$ might not be a limit point. But either it’s a limit point or it’s in $S$, and either way it’s in $\operatorname{cl}S$. – Brian M. Scott Nov 1 '12 at 23:23 @sonicboom: No, points of $S$ are not necessarily limit points of $S$: $x$ is a limit point of $S$ if every open neighborhood of $x$ contains a point of $S\setminus\{x\}$. Thus, $3$ is not a limit point of $S=[0,1]\cup\{3\}$: $(2,4)$ is an open nbhd of $3$ that contains no point of $S\setminus\{3\}$. If $x$ has an open nbhd that lies entirely within $S$, then $x$ is in the interior of $S$. – Brian M. Scott Nov 2 '12 at 11:11	{"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": 0.9833533763885498, "perplexity": 78.36144861652556}, "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-2015-27/segments/1435375096290.39/warc/CC-MAIN-20150627031816-00260-ip-10-179-60-89.ec2.internal.warc.gz"}
http://blog.jpolak.org/?p=977	# Highlights in Linear Algebraic Groups 13: Centralisers of Tori In Highlights 12, we used some of the equivalent conditions for a connected algebraic group $G$ over a field $k=\overline{k}$ to have semisimple rank 1 in the study of reductive groups (these are the groups whose unipotent radical $R(G)_u$ is trivial). Precisely, we showed that such a $G$ must have a semisimple commutator $[G,G]$ subgroup whose dimension is three, and that we can write $G = Z(G)^\circ\cdot [G,G]$ where the $-\cdot-$ denotes that this is an almost direct product: in other words, the multiplication map $Z(G)^\circ\times[G,G]\to G$ is surjective with finite kernel. Let $T\subseteq G$ be a maximal torus. We will show in this post that $C_G(T) = T$ for a connected reductive group $G$ of semisimple rank 1. This post shall conclude the detailed explanation of the concise proof of Corollary 25.3 in Humphrey’s book, which I encourage the reader to peruse for more information. ### Centralisers of Tori in $\mathrm{PGL}_2(k)$ The first step to show that $C_G(T) = T$ is to show that this is true for $G = \mathrm{PGL}_2(k)$. We have shown that $\mathrm{PGL}_2(k)$ is the image of a connected group in the previous post so it is connected. We have the quotient morphism $\mathrm{SL}_2(k)\to \mathrm{PGL}_2(k)$ of connected groups, and so each maximal torus of $\mathrm{PGL}_2(k)$ is the image of some maximal torus in $\mathrm{SL}_2(k)$. A couple lines of calculations will convince the reader that to show $C_G(T) = T$ for $G = \mathrm{PGL}_2(k)$, it is sufficient to show that $C_G(T) = T$ for $G = \mathrm{SL}_2(k)$ , which is also a short calculation. ### Centralisers of Tori in Reductive Groups of SS Rank 1 We can now proceed with the general $G$. Theorem. Let $G$ be a connected reductive group of semisimple rank $1$ (i.e. the group $G/R(G)$ has rank 1). If $T\subseteq G$ is a maximal torus then $C_G(T) = T$. Proof. Fix a morphism $\varphi:G\to\mathrm{PGL}_2(k)$ with $\ker\varphi = R(G) = Z(G)^\circ$. From the previous section, we see that the torus $T$ and its centraliser $C_G(T)$ have the same image under $\varphi$. Hence we have the inclusions $T\subseteq C_G(T) \subseteq T\ker\varphi = TR(G)$, since for any element $x\in C_G(T)$ there is a $t\in T$ such that $\varphi(y) = \varphi(t)$. However, $R(G)$ is a normal torus and hence contained in $T$ so $TR(G) = T$. QED Notice that since $Z(G)$ centralises any torus, $Z(G)$ is contained in every torus.	{"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": 0.9572075009346008, "perplexity": 96.93969282494417}, "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-2018-05/segments/1516084890874.84/warc/CC-MAIN-20180121195145-20180121215145-00798.warc.gz"}
http://tex.stackexchange.com/questions/56093/pdfpages-addtolist-and-list-of-tables	# pdfpages, addtolist and list of tables These may be separate questions, but the following MWE: \usepackage{pdfpages} ... \includepdf[pages={-}, pagecommand={\label{tab:UserRoles}}, offset=0in -1in, addtolist={9, table, {The User Roles table}, tab:UserRoles}]{UserRolesTable} ... has several unexpected behaviors. 1. I get the following warning: Package pdfpages Warning: There's something wrong with the entries of addtolist'. Not all entries were processed. Check addtolist' [in the MWE]." 2. Probably due to the \label not being bound to the inserted pdf (in a table environment), my inserted PDF is numbered using the section, and not the expected "Table X" value. 3. The manually-inserted LoT line (which also uses the section number rather than the incremented table number) is not formatted like the other lines in the LoT - the number lines up, but the table caption itself is not aligned with the others in the list. The source PDF has no errors and displays correctly in the output PDF. PDFLaTeX -> BibTex MWE: \documentclass[11pt,letterpaper,twoside,final]{article} \usepackage{pdfpages} \begin{document} \listoftables \begin{center} \textbf{Table \ref{tab:UserRoles} - The User Roles Table} \addcontentsline{lot}{subsection}{\ref{tab:UserRoles} \textbf{User Roles}} % manual entry in ToC \includepdf[pages={-}, pagecommand={\label{tab:UserRoles}}, offset=0in -1in, addtolist={9, table, {The User Roles table}, tab:UserRoles, lot}]{UserRolesTable} \end{center} \end{document} - How many pages does the included PDF have? – egreg May 16 '12 at 16:01 5 pages in the included PDF. I tried specifying ('\includepdf{pages=1-5...' but that threw the same errors. – Yonah May 16 '12 at 16:45 Can you upload the pdf file? – Marco Daniel Jun 2 '12 at 22:07 Can't figure out how to attach to this page. The PDF is not hosted online (nor will it be), and I've had to strip out all identifying info. By way of update, the new PDF is only 2 pages, but that isn't an issue in the MWE - the "unspecified error" (#1 above) still shows. – Yonah Jun 4 '12 at 15:32 There are some errors in the MWE: • pagecommand={\label{tab:UserRoles}}: This is executed on each builded page. In case of more than one page the result are warnings because of multiply defined labels. Also the label setting is already done by your option setting for addtolist. (Edit: Fix of PolGab applied, thanks) • The argument list for addtolist expects four parameters, the page number, the type, the title and the label. However there is a fifth element lot in the MWE. The package pdfpages sets the list entry by calling \caption, where \@makecaption is a dummy. Thus the caption is not actually typeset, but the counter is incremented and the entry for the list is done. • With a working addtolist the manual \addcontentsline is not longer needed. Also it should be called at the right page to get the correct page number. The corrected MWE: \documentclass[11pt,letterpaper,twoside,final]{article} \usepackage{pdfpages} \begin{document} \listoftables \section{Table \ref{tab:UserRoles} - The User Roles Table} \includepdf[ pages={-}, offset=0in -1in, addtolist={3, table, {The User Roles table}, tab:UserRoles}, ]{test} \end{document} The result: The first page contains the list of tables and the "table section title". The third page of the included pages gets the entry in the list of tables, that is the fourth page. - Thanks so much for your help! – Yonah Jul 30 '12 at 18:54 Thanks so much for your help! The reason I used \pagecommand is that it adds my custom header/footer around the inserted PDF (which also explains the offsets). Your MWE does not insert a line to the LoT (I also tried changing it to \subsection to fit with my hierarchy, but no dice). I get a section titled "Table ?? - The User Roles Table" with nothing else on the page, then the PDF starts on the following page. Thanks, all, for your help so far. – Yonah Jul 30 '12 at 20:30 The entry in the LoT is present, see the result image. The entry is done by option addtolist alone and has nothing to do with the previous \section*. Ignore this line if you like; it's only there to show \ref. – Heiko Oberdiek Jul 30 '12 at 21:34 So now I have to wonder what else is causing TeX to ignore the label. I'm still not seeing (MWE below) an entry in the ToC and the table comes out a page too late (and dialing it back to p12 doesn't help). \includepdf[ pages={ - }, pagecommand={}, offset=0in -1in, addtolist={13, table, {The User Roles table}, tab:UserRoles}, ]{UserRolesTable} Is it possible that pdfpages conflicts with another package I use? – Yonah Jul 31 '12 at 13:34 First make sure that you are indeed including 13 or more pages, your addtolist specification fires on the 13th page. If it does not work, a complete MWE would make sense (it could be appended to the question). – Heiko Oberdiek Jul 31 '12 at 14:13 The addtolist option uses 4 (or any multiple of 4) arguments. For example, if your external PDF file (UserRolesTable.pdf) contains three pages, you can say: \documentclass[11pt,letterpaper,twoside,final]{article} \usepackage{pdfpages} \pagestyle{empty} \begin{document} \listoftables \includepdf[% pages={-}, offset=0in -1in,%	{"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": 0.86757892370224, "perplexity": 4111.498238288428}, "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-2014-52/segments/1418802773058.130/warc/CC-MAIN-20141217075253-00031-ip-10-231-17-201.ec2.internal.warc.gz"}
https://www.codecogs.com/library/engineering/fluid_mechanics/floating_bodies/the-oscillation-of-floating-bodies.php	• https://me.yahoo.com # The oscillation of Floating Bodies The oscillation of floating bodies including the angle of heel and the period of oscillations ## Introduction A single hulled ship subjected to any disturbing force will heel or roll. This page investigates the degree of roll and the periodic time of the movement. ## The Period Of Oscillation Of A Floating Body. NOTE: • 1) Torque = Moment of Inertia X Angular Acceleration. • 2) The Solution of $\frac{d^2y}{dx^2}\;+\;a^2y\;=\;0\;is\;&space;given\;&space;by\;y\;=\;A\,sin\,a\,x\;+\;B\,cos\,a\,x$ For small angles of heel, the Body can be regarded as Oscillating about it's Metacentre, in a manner similar to a Pendulum about it's point of suspension. If • W = The weight of the Ship • M = The Metacentric Height • k = The Radius of Gyration of the Ship about a horizontal Axis through the C.of.G. • $\inline&space;\theta$ = The Rotation after a time t. $Angular\;Acceleration&space;=&space;-\;\frac{d^2\theta&space;}{dt^2}\;(&space;Direction&space;\;towards\;&space;equilibrium\;&space;position)$ The Moment of Inertia of the ship about the C of G is $\inline&space;\frac{W}{g}k^2$ The Moment of Inertia of the ship about M is $\inline&space;&space;\frac{W}{g}\left(k^2+m^2&space;\right)$ Provided that m is small compared with k, these two can be regarded as equal $\therefore\;\;\;\;\;I_G\approx&space;I_M$ For small angles of heel , the Righting Couple = $\inline&space;W\;m\;\theta$ $\therefore\;\;\;\;\;W\;m\;\theta&space;\;=\;I_m\times&space;-\frac{d^2\theta&space;}{dt^2}\;\approx&space;I_G\times-&space;\frac{d^2\theta&space;}{dt^2}$ $=\;-\frac{W}{g}\;k^2&space;\frac{d^2\theta&space;}{dt^2}$ $\therefore\;\;\;\;\;\frac{d^2\theta&space;}{dt^2}\;+\;\frac{gm}{k^2\theta&space;}&space;=&space;0$ The Solution of this Equation is given by: $\theta&space;\;=\;A\;sin\sqrt{\frac{gm}{k^2}}\;t\;+\;b\;cos\sqrt{\frac{gm}{k^2}}\;t$ When t = 0 $\inline&space;\theta$= 0 and therefore B = 0 When $\inline&space;\theta$ = o and t = T/2 i.e. T = the time of a complete oscillation. $A\;sin\sqrt{\frac{gm}{k^2}}\times\frac{T}{2}&space;=&space;0$ $Since\;A\;\neq&space;0\;\;\;\;sin\sqrt{\frac{gm}{k^2}}\times\frac{T}{2}&space;=&space;0$ The simplest solution;_ $\sqrt{\frac{gm}{k^2}}\times\frac{T}{2}\;=\;\pi$ $\therefore\;\;\;\;\;The\;Periodic\;Time\;T\;=\;2\pi&space;\sqrt{\frac{k^2}{mg}}$ ### Example 1 A Solid Cylinder of Uniform material and with height equal to the Diameter is to float in water with it's axis vertical. Calculate the Metacentric hight and the specific gravity of the Cylinder so that it may have a Rolling Period of six seconds when the Diameter is 4 ft. The Rolling Period is given by $\inline&space;T\;=\;2\pi&space;\sqrt{\frac{k^2}{hg}}\;Secs.$ Where h is the Metacentric Height and k is the Radius of Gyration of the Body about it's C of G. $T\;=\;6Secs.\;=\;2\pi&space;\sqrt{\frac{k^2}{gh}}$ To Find H $k^2\;=\;\frac{d^2}{16}\;+\;\frac{l^2}{12}\;=\;16\left(\frac{1}{16}\;+\;\frac{1}{12}&space;\right)\;=\;\frac{7}{3}\;ft^2$ $h\;=\;\frac{4\pi&space;^2k^2}{gT^2}\;=\;\frac{4\pi&space;^2\times&space;7/3}{32.2\times&space;36}\;=\;0.0795\;ft.$ To Find The Specific Gravity $\frac{\pi&space;}{4}D^2\times&space;4\times&space;62.4\times&space;S\;=\;\frac{\pi&space;}{4}D^2x\times&space;62.4$ $\therefore\;\;\;\;\;\;\;s\;=\;\frac{x}{4}$ $OD\;=\;\frac{x}{2}\;\;\;\;and\;\;\;\;OG\;=\;2\;ft.$ $BM\;=\;\frac{I}{V}\;=\;\frac{\pi&space;\times&space;4^4}{64}\div&space;\frac{\pi&space;\times&space;h^2x}{4}\;=\;\frac{1}{x}$ MG (h) = BM + BO - BO - OG = $\inline&space;\frac{1}{x}\;+\;\frac{x}{2}\;-\;2\;=\;0.0795\;ft.$ $\therefore\;\;\;\;\;\;2\;+\;x^2\;-\;4.159\;=\;0$ Solving the quadratic X = 3.604 ft. From which the Specific Gravity S = 0.901 ## Worked Examples The workings associated with these examples have been hidden. They can be seen by clicking on the red butttons. ### Example 2 A model ship weighs 80 lb. and floats in a tank of water. A ball of weight 1.6 oz. traveling horizontally at 15 ft./sec is arranged to strike the model broadside at a point 2 ft. vertically above its centre of gravity. The ball after striking, moves with the model and the amplitude of the resulting roll is $\inline&space;2^0$. The natural period of roll of the system is 1.5 sec. Find the metacentric height of the system assuming it to be constant over this amplitude. ( Assume that the vessel oscillates about its centre of gravity. (B.Sc. Part 2) Let k be the radius of gyration of the model about the longitudinal axis through G and let the metacentric height which equals GM be h When oscillating, the unbalanced couple, when $\inline&space;\theta$ which acts on the model is:- $W\times&space;GM\times&space;\theta$ $\therefore&space;\;\;\;\;\;\;W\times&space;GM\times&space;\theta=W\times&space;h\times&space;\theta=\frac{W\times&space;k^2}{g}\;\;\;\frac{d^2\theta}{dt^2}$ Re-writing $\frac{d^2\theta}{dt^2}=\frac{g\times&space;h}{k^2}\times&space;\theta$ This is Simple Harmonic motion with a periodic time of:- $t=2\pi\sqrt{\frac{k^2}{g\times&space;h}}$ Note. For more details on Simple Harmonic Motion please see" Maths: Differential Equations: Equations of Motion" It can be assumed that the ball, on striking the model looses its momentum and produces an impulsive torque which starts the oscillations of the model. To consider the impact of the ball on the model, let:- • $\inline&space;W_1$ be the weight of the ball. (Note in the question, the weight is given in ounces and therefore must be divided by 16 to convert it to lb.) • $\inline&space;V_1$ be the velocity just before impact. • $\inline&space;\omega_1$ be the angular velocity of the model after impact. • "a" be the distance of the point of impact, measure from the centre of gravity (G) of the model. • A be the maximum angular velocity of the model during its Simple Harmonic Motion. • t be the periodic time. $\frac{a\times&space;W_1\times&space;V_1}{g}=\frac{W\times&space;k^2\times&space;\omega_1}{g}$ Substituting in given values:- $2\times&space;\frac{1.6}{16}\times&space;15=80\times&space;k^2\times&space;\omega_1$ $\therefore&space;\;\;\;\;\;\;k^2\times&space;\omega_1=\frac{3}{80}$ As the model starts at the centre of its harmonic motion, it can be seen that $\inline&space;omega_1$ is the maximum angular velocity. $\text{Then}\;\;\;\;\;\;\omega_1=2\pi\frac{A}{t}\;\;=\;\;2\pi\times&space;\frac{2}{57.3\times&space;1.5}=0.146\;rad./sec.$ From equations (26) and (27) $k^2=\frac{3}{80\times&space;0.146}$ Substituting in equation (23) $1.5=2\pi\sqrt{\frac{3}{80\times&space;0.146\times&space;32.2\times&space;h}}$ $\text{From&space;which\;\;\;\;\;\;h=0.14\;ft.}$	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 44, "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": 0.8570154309272766, "perplexity": 802.1913862104717}, "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/1652662550298.31/warc/CC-MAIN-20220522220714-20220523010714-00516.warc.gz"}
http://www.bluegartr.com/archive/index.php/t-90889.html	PDA View Full Version : Statistics halp Julian 2010-03-25, 22:13 This is a practice test, not homework (I already know the answer, but I want to know how to find it) An article in a journal reports that 34% of American fathers take no responsibility for child care. A researcher claims that the figure is higher for fathers in the town of Cheraw. A random sample of 225 fathers from Cheraw, yielded 97 who did not help with child care. Find the P-value for a test of the researcher's claim. I don't really know this section that well :( I kinda get the null/alternate hypothesis stuff, for example, H0 here is that mu (would it be mu in this problem?) = .34, and H1: mu(?) > .34 Yeah I'm kinda lost, and this book is a piece of shit that isn't really helping, any help would be appreciated, thanks! edit: Here's 2 more problems I don't really get: In a poll of 278 voters in a certain city, 67% said that they backed a bill. The margin of error in the poll was reported as 6 percentage points (w/ a 95% degree of confidence) Which statement is correct? A) The sample size is too small to achieve the stated margin of error B) For the given sample size, the margin of error should be larger than stated C) The reported margin of error is consistent with the sample size (correct answer) D) There is not enough info to determine whether the margin of error is consistent w/ the sample size E) For the given sample size, the margin of error should be smaller than stated. Also another problem, except it's a poll of 390, 77% backed it, ME is 5% w/ 95% degree of confidence, and the answer is "The stated margin of error could be achieved with a smaller sample size" Don't really get what's going on with those 2, any help would be appreciated :x thanks again Heavencloud 2010-03-25, 23:00 H0 = u=.34 Ha = u>.34 Is a significance level given? I'm trying to remember how to solve this but it doesn't feel like I have enough information =p Moss 2010-03-25, 23:48 Damn dude, I was doing these exact problems like 2 months ago. I'll look at my old stuff in class tomorrow. Weiing 2010-03-26, 00:12 Ok... Let me see if I can recall stats looong time ago. I don't know what formulae you have learned, but I was using this for my calculations: N= 4p(1-p)/m² Where N = sample size; p = percentage; m = margin of error In a poll of 278 voters in a certain city, 67% said that they backed a bill. The margin of error in the poll was reported as 6 percentage points (w/ a 95% degree of confidence) Which statement is correct? A) The sample size is too small to achieve the stated margin of error B) For the given sample size, the margin of error should be larger than stated C) The reported margin of error is consistent with the sample size (correct answer) D) There is not enough info to determine whether the margin of error is consistent w/ the sample size E) For the given sample size, the margin of error should be smaller than stated. key: p=.67 m=.06 sticking it into the formula above: N= (4)(.67)(1-.67)/(.06)² = (4)(.67)(.33)/(.0036) = .8844/.0036 = 245.667 thus, 245 is the minimum sample size needed for that margin of error. Since 278 is more than 245, then it is consistent (?????) edit: punched in to see the sample size needed for a 5% margin of error, and you would need 353.76. so 278 falls within the range of 245-353 Also another problem, except it's a poll of 390, 77% backed it, ME is 5% w/ 95% degree of confidence, and the answer is "The stated margin of error could be achieved with a smaller sample size key: p=.77 m=.05 N=(4)(.77)(1-.77)/(.05)² = (4)(.77)(.23)/(.0025) = .7084/.0025 = 283.36 Thus, the minimum sample size needed for that margin of error with that confidence interval is 283 people. 390 exceeds 283, so could have been done with fewer. HOWEVER, I haven't done stats in over 3 years now. So I may be totally wrong and pulling stuff out of my butt. I sat here for a while punching in random numbers to try to figure it out so... I tried! CDF 2010-03-26, 05:42 This is not a thorough explanation but may help: 1) Assuming you already covered the relevant material, the p-value to obtain is based off a test statistic that was already presented in lecture. Your textbook should have a section about "test for one proportion" or "test for a binomial proportion" where the normal approximation to the binomial distribution is applied and the corresponding test statistic given. The following is a brief summary of what is involved in the problem, implicitly or otherwise. Stating the null and alternative hypotheses for a test of a single proportion: http://latex.codecogs.com/gif.latex?H_0: \pi \leq .34 http://latex.codecogs.com/gif.latex?H_1: \pi > .34 Finding the sample proportion, which estimates the "true" proportion: http://latex.codecogs.com/gif.latex?\hat{\pi} = 97/225 Obtaining the variance of the sample proportion given the null hypothesis (under which the proportion is .34): http://latex.codecogs.com/gif.latex?Var(\hat{\pi}) = \frac{\pi(1-\pi)}{n} You wouldn't use the standard error (based on 97/225) because the problem asks for a p-value, which is based on the null hypothesis. Obtaining the Z-statistic: http://latex.codecogs.com/gif.latex?Z = \frac{\hat{\pi} - \pi }{\sqrt{Var(\hat{\pi})}} A statement of the p-value (you may use a standard normal table in the back of the book to obtain the actual value): http://latex.codecogs.com/gif.latex?P\left ( Z > \frac{\hat{\pi} - \pi}{\sqrt{Var(\hat{\pi} )}} \right ) 2) You should have been given the sample size formula for polling ("yes/no" response) somewhere. The general formula is: http://latex.codecogs.com/gif.latex?n = \frac{Z^2\pi(1-\pi)}{(ME)^2} where Z is the Z-score corresponding to your chosen significance level. Z is 1.96 given a 5% significance level, so Z^2 is often approximated as 4. Since you usually don't have a good idea what the true proportion really is, the margin of error is usually based on pi = .5 because that gives the maximum n required to achieve a target margin of error (you want to be sure you achieve the desired M.E.), so n is often approximated as http://latex.codecogs.com/gif.latex?n \approx \frac{1}{(ME)^2} Use this formula to answer the question. 3) Refer to the general sample size formula: http://latex.codecogs.com/gif.latex?n = \frac{Z^2\pi(1-\pi)}{(ME)^2} Note that Z^2 is actually 3.841, not 4. Also, the true proportion could be less or greater than .5. Either or both could be invoked to reach the given conclusion. Julian 2010-03-26, 08:26 Thanks, much appreciated! Here's another one: Assume that a simple random sample has been selected from a normally distributed population. State the final conclusion. Test the claim that the mean age of the prison population in one city is less than 26 years. Sample data are summarized as n=25, xbar=24.4 years, and s=9.2 years. Use a significance level of alpha = 0.05. H0: mu=26 Ha: mu<26 (don't reject) This problem is different than all the other problems of this type, so I'm not really sure how to go about it >_>. All the other problems have been just look at the p-value and alpha, and make a conclusion about it, which is easy, but I guess I gotta calculate the p-value for this one, and I have no idea how lol. edit: If I did this right, I calculated t to be .8695, DF is 24, not sure how to go about calculating the p value.. :x Well, if I look at this crappy table I have, t for .1 is 1.318, so since it's less than that, it means the p-value is greater than .1 (so it's also greater than .05, the given alpha), so you don't reject? Did I do that right? D: Weiing 2010-03-26, 08:49 Yeah. I just calculated and got .8695 and DF of 24. Looking at my percentage points of t distribution, df of 24 with alpha of .05 shows 2.064. But since .8695 is less than that, can't reject the H0. So looks like you're right :D Julian 2010-03-26, 09:24 Here's another: In a survey of 5500 TV viewers, 20% said they watch network news programs. Find the standard error for the sample proportion. (.0171) I tried root(p(1-p)/n), which is (.2.8/5500)^.5, but that gives .00539 <_< EDIT: ok this one I have no clue how to do >_> Scores on a certain test are normally distributed with a variance of 14. A researcher wishes to estimate the mean score achieved by all adults on the test. Find the sample size needed to assure with 98% confidence that the sample mean will not differ from the population mean by more than 2 units. (20) There's also a similar question to the above one: The weekly earnings of students in one age group are normally distributed with a standard deviation of 81 dollars. A researcher wishes to estimate the mean weekly earnings of students in this age group. Find the sample size needed to assure with 98% confidence that the sample mean will not differ from the population mean by more than 5 dollars. (1425) These are the last questions in my practice exam, got everything else down, so this is it :D Edit: trying to solve the above 2.. I basically have 5 +/- z SE (or 2 for the first one) (z being 2.326 for 98% confidence) Using SE = S/(n^.5) for standard deviation, and S=(variance/n)^.5 for the variance, I get 19 and 1420, which are... close enough? Especially compared to the other choices given. So I guess it's kinda right? idk <_<; CDF 2010-03-26, 14:10 1) The reasoning on the lower bound of the p-value is fine. The t-statistic is actually -0.8695, though. Be careful if you actually have to report the p-value. To avoid making a "sign error," it could help to summarize what the p-value actually means (if this actually means anything to you): http://latex.codecogs.com/gif.latex?P\left ( \bar{x} \leq 24.4 \right \| H_0) = P(t\leq -0.8695 \| H_0) 2) For n = 550, the standard error is 0.0171. 3) 20 and 1425 are based on the Z-score (2.32635) being rounded to the nearest hundredths place (2.33)	{"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": 0.9182250499725342, "perplexity": 762.628155116776}, "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/1394021338216/warc/CC-MAIN-20140305120858-00004-ip-10-183-142-35.ec2.internal.warc.gz"}
http://mathhelpforum.com/advanced-algebra/221347-systems-linear-equations-problem.html	# Thread: Systems of Linear Equations Problem 1. ## Systems of Linear Equations Problem I'm having trouble solving a linear equations system with an unknown coefficient. I am trying to solve it using the reduce row echelon form, but to no avail. x + y + 7z = -7 2x + 3y + 17z = -16 x + 2y + (k2 + 1)z = 3k [ 1 1 7 \| -7 ] [ 2 3 17\| -16] [ 1 2 k2+1\| 3k ] Specifically, I am trying to simplify this equation to find: a) no solution "inconsistent" b) infinite number of solutions "consistent" c) one solution "unique" Right now I am just attempting a. I believe I have accidentally already solved c. So following the procedure... I subtract 2R1 from R2, giving in the second row: [ 0 1 3 \| -2 ] I then subtract R1 from R3 giving in the third row: [ 0 1 k2-6 \| 3k+7 ] Then subtracting R2 from R3 to give: [ 0 0 k2-9 \| 3k+9 ] From here I am lost. My intuition tells me that 3rd row should simplify to something like this (x representing a non-zero value): [ 0 0 0 \| x ] Any help would be greatly appreciated. 2. ## Re: Systems of Linear Equations Problem From your last line simply divide through by k^2-9: [ 0 0 1 \| (3k+9)/k^2-9] This tells you that z = (3k+9)/(k^2-9). As long as this value is defined you get a unique set of solutions for x, y, and z. For example if k=0 you get x=-1, y=1, and z = -1. If k= 4 you get x= -17, y= -11, z = 3. But notice that the (3k+0)/(k^2-9) is not defined for k= 3 or k=-3, so we need to see what happens for these two cases. For k=3 your last row becomes [0 0 0 \| 15] which has no solution and for k = -3 it's [0 0 0\| 0] which means the three original equations are not linearly indpendent and hence there are an infininite number of solutions. 3. ## Re: Systems of Linear Equations Problem Hello, I can help you with the first one. For a system of equations to be consistent (with either one or infinite solutions) it has to be satisfied that $\det\left(\bf A\right)\neq 0$ For this reason, calculating the determinant of the coefficient matrix, you can determine the values of k where the system is not defined: $\det\left(\bf A\right) = \left\|\begin{matrix}1 & 1 & 7 \\ 2 & 3 & 17 \\ 1 & 2 & k^2 + 1 \end{matrix}\right\|=k^2 -9=0$ Thus, $k=\pm 3$ makes the determinant to be equal to zero, and thus, the system is not defined. 4. ## Re: Systems of Linear Equations Problem Hello, brettm! $\begin{array}{ccc} x + y +\qquad 7z &=& \text{-}7 \\ 2x + 3y +\qquad 17z &=& \text{-}16 \\ x + 2y + (k^2 + 1)z &=& 3k\end{array}$ Specifically, I am trying to simplify this equation to find: . . (a) no solution, "inconsistent" . . (b) infinite number of solutions, "consistent" . . (c) one solution, "unique" $\text{We have: }\:\left[\begin{array}{ccc\|c} 1&1&7 & \text{-}7 \\ 2&3&17 & \text{-}16 \\ 1&2& k^2\!+\!1 & 3k \end{array}\right]$ $\begin{array}{c}\\ R_2-2R_1 \\ R_3-R_1 \end{array}\left[\begin{array}{ccc\|c}1&1&7 & -7 \\ 0&1&3&-2 \\ 0&1&k^2\!-\!6 & 3k\!+\!7 \end{array}\right]$ $\begin{array}{c}R_1-R_2 \\ \\ R_3-R_2 \end{array}\;\left[\begin{array}{ccc\|c}1&0&4&\text{-}5 \\ 0&1&3&\text{-}2 \\ 0&0&k^2\!-\!9 & 3k\!+\!9 \end{array}\right]$ $\text{If }k = 3\text{, we have: }\;\left[\begin{array}{ccc\|c}1&0&4&\text{-}5 \\ 0&1&3&\text{-}2 \\ 0&0&0&18 \end{array}\right] \;\;\text{ . . . no solution}$ $\text{If }k = \text{-}3\text{, we have: }\;\left[\begin{array}{ccc\|c}1&0&4&\text{-}5 \\ 0&1&3&\text{-}2 \\ 0&0&0&0 \end{array}\right]\;\;\text{ . . . infinite solutions}$ $\text{If }k \,\ne\,\pm3: \:$ $\begin{array}{c} \\ \\ R_3 \div (k^2-9)\end{array}\;\left[\begin{array}{ccc\|c} 1&0&4 & \text{-}5 \\ 0&1&3&\text{-}2 \\ 0&0&1&\frac{3}{k-3}\end{array}\right]$ . . . $\begin{array}{c}R_1-4R_3 \\ R_2-3R_3 \\ \\ \end{array} \left[\begin{array}{ccc\|c} 1&0&0 & \frac{\text{-}5k+3}{k-3} \\ 0&1&0& \frac{\text{-}2k-3}{k-3} \\ 0&0&1& \frac{3}{k-3} \end{array}\right]\;\;\text{ . . . one solution}$ 5. ## Re: Systems of Linear Equations Problem Thank you very much for your help everyone. I've been trying to simplify the unique solution to the equation(s), i.e.: $\frac{-5k+3}{k-3}$ But I'm beginning to think that I am meant to just leave this answer in terms of k, is this correct? I'm confused because the question states "Determine the value of the constant k such that the system..." Also, what is the justification for arriving at $k = \pm3$ for no and infinite solutions? I am basically stating that for a system to have no solution, the bottom row of the matrix (excluding the augmented end) must have values of 0, with the augmented column having a non-zero value, $0x + 0y + 0z = 18$ (not possible therefore there is no solution. So I determine the value of k when $k^2 - 9 = 0$; then substituting in both values into $3k + 9$ seeing when it is a non-zero value, to find the answer for part a), no solutions. My justification for part b) is basically the same, just substituting in a value determined from $k^2 - 9 = 0$ that will give $3k + 9 = 0$ in the augmented column. I am sure there has to be a better, more conventional way to explain this on my homework. 6. ## Re: Systems of Linear Equations Problem Originally Posted by berni1984 Hello, I can help you with the first one. For a system of equations to be consistent (with either one or infinite solutions) it has to be satisfied that $\det\left(\bf A\right)\neq 0$ For this reason, calculating the determinant of the coefficient matrix, you can determine the values of k where the system is not defined: $\det\left(\bf A\right) = \left\|\begin{matrix}1 & 1 & 7 \\ 2 & 3 & 17 \\ 1 & 2 & k^2 + 1 \end{matrix}\right\|=k^2 -9=0$ Thus, $k=\pm 3$ makes the determinant to be equal to zero, and thus, the system is not defined. This is a better solution, but we should note that a zero determinant could mean that the system either has no solutions or INFINITELY MANY solutions. So we can state that if \displaystyle \begin{align} k \neq \pm 3 \end{align} there is unique solution, and then the OP will need to substitute in each of k = 3 and k = -3 and reduce the system to echelon form to determine if there are no solutions or infinitely many solutions.	{"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": 23, "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": 0.9233173131942749, "perplexity": 364.35193551385385}, "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-2016-44/segments/1476988718034.77/warc/CC-MAIN-20161020183838-00048-ip-10-171-6-4.ec2.internal.warc.gz"}
http://www.physicsforums.com/showthread.php?s=6d80cb3f5caf01dd0b1c96705fafb29d&p=4542421	# Confinement potential by Hluf Tags: confinement, potential P: 20 When we derive the Bethe-Salpeter equation, we have the relations of mass of constitute m, transverse momentum, q-hat and energy ω for equal mass constitute as ω^{2}=m^{2}+(q-hat)^{2}. What is the relations of ω,m and q-hat for low energy or long distance or low momentum? Can we say q-hat≈m? Thank you for taking time to help me! Related Discussions High Energy, Nuclear, Particle Physics 39 High Energy, Nuclear, Particle Physics 2 High Energy, Nuclear, Particle Physics 4 Quantum Physics 3 High Energy, Nuclear, Particle Physics 17	{"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": 0.960420548915863, "perplexity": 4547.9936586393105}, "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-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00421-ip-10-147-4-33.ec2.internal.warc.gz"}
http://one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/forceandmotion/impulse.html	# Impulse and Impulsive Force Card 1: What is Impulse? ## Impulse Impulse is defined as the product of the force (F) acting on an object and the time of action (t). Impulse exerted on an object is equal to the momentum change of the object. Impulse is a vector quantity. Card 2: Formula of Impulse ## Formula of impulse Impulse is the product of force and time. Impulse = F × t Impulse = momentum change Impulse = mv - mu Example 1 A car of mass 600kg is moving with velocity of 30m/s. A net force of 200N is applied on the car for 15s. Find the impulse exerted on the car and hence determine the final velocity of the car. Impulse = F × t = (200) × (15) = 300oNs Impulse = mv - mu (3000) = 600v - 600(30) 600v = 3000 + 18000 v = 21000/600 = 35 m/s [500,000N] Card 3: What is impulsive force? ## Impulsive Force Impulsive force is defined as the rate of change of momentum in a reaction. It is a force which acts on an object for a very short interval during a collision or explosion. Example 2 A car of mass 1000kg is traveling with a velocity of 25 m/s. The car hits a street lamp and is stopped in0.05 seconds. What is the impulsive force acting on the car during the crash? \begin{gathered} {\text{Impulsive Force,}} \hfill \\ F = \frac{{m(v - u)}} {t} \hfill \\ F = \frac{{1000(0 - 25)}} {{0.05}} \hfill \\ F = -500,000N \hfill \\ \end{gathered} Card 4: Effects of Impulse vs Force ## Effects of impulse vs Force A force determines the acceleration (rate of velocity change) of an object. A greater force produces a higher acceleration. An impulse determines the velocity change of an object. A greater impulse yield a higher velocity change. Card 5: Examples Involving Impulsive Force ## Examples Involving Impulsive Force • Playing football • Playing tennis • Playing golf • Playing baseball Card 6: Effect of time on impulsive force 1 - Long Jump ## Long Jump • The long jump pit is filled with sand to increase the reaction time when atlete land on it. • This is to reduce the impulsive force acts on the leg of the atlete because impulsive force is inversely proportional to the reaction time. Card 7: Effect of time on impulsive force 1 - High Jump ## High Jump (This image is licenced under the GNU Free Document Licence. The original file is from the Wikipedia.org.) • During a high jump, a high jumper will land on a thick, soft mattress after the jump. • This is to increase the reaction time and hence reduces the impulsive force acting on the high jumper. Card 8: Effect of time on impulsive force 1 - Jumping ## Jumping A jumper bends his/her leg during landing. This is to increase the reaction time and hence reduce the impact of impulsive force acting on the leg of the jumper. Card 9: Empty Card Card 10: Empty Card Table of Content	{"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": 0.8458269834518433, "perplexity": 2116.513758042424}, "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-2019-13/segments/1552912202131.54/warc/CC-MAIN-20190319203912-20190319225912-00101.warc.gz"}
https://www.physicsforums.com/threads/standard-candles.129119/	# Standard candles 1. Aug 17, 2006 ### wolram http://arxiv.org/abs/astro-ph/0608324 Cosmological Implications of the Second Parameter of Type Ia Supernovae Authors: Philipp Podsiadlowski (Oxford), Paolo A. Mazzali (MPA, Munich; Trieste), Pierre Lesaffre (Oxford, Cambridge, Paris), Christian Wolf (Oxford), Francisco Forster (Oxford) Theoretical models predict that the initial metallicity of the progenitor of a Type Ia supernova (SN Ia) affects the peak of the supernova light curve. This can cause a deviation from the standard light curve calibration employed when using SNe Ia as standardizable distance candles and, if there is a systematic evolution of the metallicity of SN Ia progenitors, could affect the determination of cosmological parameters. Here we show that this metallicity effect can be substantially larger than has been estimated previously, when the neutronisation in the immediate pre-explosion phase in the CO white dwarf is taken into account, and quantitatively assess the importance of metallicity evolution for determining cosmological parameters. We show that, in principle, a moderate and plausible amount of metallicity evolution could mimic a lambda-dominated, flat Universe in an open, lambda-free Universe. However, the effect of metallicity evolution appears not large enough to explain the high-z SN Ia data in a flat Universe, for which there is strong independent evidence, without a cosmological constant. We also estimate the systematic uncertainties introduced by metallicity evolution in a lambda-dominated, flat Universe. We find that metallicity evolution may limit the precision with which Omega_m and w can be measured and that it will be difficult to distinguish evolution of the equation of state of dark energy from metallicity evolution, at least from SN Ia data alone. May be not much, just another question mark. 2. Aug 18, 2006 ### wolram Double edged sword. http://arxiv.org/abs/astro-ph/0608351 The impact of neutrino masses on the determination of dark energy properties Authors: Axel De La Macorra, Alessandro Melchiorri, Paolo Serra, Rachel Bean Recently, the Heidelberg-Moscow double beta decay experiment has claimed a detection for a neutrino mass with high significance. Here we consider the impact of this measurement on the determination of the dark energy equation of state. By combining the Heidelberg-Moscow result with the WMAP 3-years data and other cosmological datasets we constrain the equation of state to -1.67< w <-1.05 at 95% c.l., ruling out a cosmological constant at more than 95% c.l.. Interestingly enough, coupled neutrino-dark energy models may be consistent with such equation of state. While future data are certainly needed for a confirmation of the controversial Heildelberg-Moscow claim, our result shows that future laboratory searches for neutrino masses may play a crucial role in the determination of the dark energy properties. 3. Aug 21, 2006 ### wolram No acceleration http://arxiv.org/abs/astro-ph/0608386 On the Absence of Cosmic Acceleration Authors: John Middleditch Report-no: LAUR 06-5685 Type Ia Supernovae (SNe) have been used by many to argue for an accelerated expansion of the universe. However, high velocity and polarized features in all nearby SNe Ia, show that the paradigm for Type Ia SNe is drastically and catastrophically invalid. By now it is also clear that an extreme version of the axisymmetry seen in SN 1987A is the correct paradigm for SNe Ia and Ic. A Ia/c is produced from the merger of two degenerate cores of common envelope WR stars, or of two CO white dwarfs. Its polar blowouts produce the observed high velocity and polarized spectral features in Ia's, and its equatorial bulge is much brighter in Ia's, due to the greater fraction of 56 Ni contained within it. These become classified as Ia's when viewed from the merger equator, and Ic's when viewed from the poles. Thus cosmology determined strictly from Ia's alone is flawed at its very foundation: the local sample is selectively biased. The problem arose with the more distant supernovae, when the high velocity polar blowout features, which initially obscure part of the Ia/c equatorial bulge, expose a greater fraction of it, particularly when viewed off the equator, during the interval when Delta m_15 is measured, leading to a smaller decrease in observed luminosity. The width-luminosity correction was thus too small, and the result was a distant SN Ia which appeared to be too faint for its redshift. When the errors introduced by this process and others are taken into account, there may be no cosmic acceleration effect in distant SNe. 4. Aug 21, 2006 ### Garth These papers may indicate the confidence placed in the standard $\Lambda$CDM model may be misplaced. Garth 5. Aug 21, 2006 ### wolram I am not going to say anything Garth, i will just post any thing that seems interesting in this thread, unless any one has objections. 6. Aug 21, 2006 ### Garth The point is: Are these 'SN Ia' standard candles? Such a lot in the standard model hangs on the assumption that they are. Yet the evidence that their luminosities are known depends only on those nearby, and therefore recent, ones that are able to be calibrated. As we go back into cosmological history metallicity may be expected to change, and as your links imply this may alter their expected luminosities and call into question the conclusions based on them. Garth 7. Aug 21, 2006 ### wolram May be i should ask you what is standard in cosmology, is there a standard galaxy ,star, planet, moon, if not it is difficult to understand how there can be a (standard) body in the U. 8. Aug 21, 2006 ### Garth The concept of a standard candle is simply a luminous object of which we think we know the Absolute Magnitude. (Such as the Cepheid Variables) The object's distance can then be calculated from the distance modulus. Garth 9. Aug 21, 2006 ### matt.o I think Wolram was asking a deeper question here, Garth. He is pointing out the differences in other observed phenomena such as galaxies and then questioning how there can be standard candles if things are so different. I would just like to say that as scientists, we generally like to have a few more cards up our sleeves, hence there are other tests which point to a dark energy density of ~.7 independant of the SN1a results. Such studies include the CMB, baryon acoustic fluctuations, galaxy cluster mass function etc. 10. Aug 21, 2006 ### SpaceTiger Staff Emeritus That hasn't been true for some time. Pretty much everything we infer from Type Ia SNe (so far) has been checked with other methods (e.g. CMB, LSS). In fact, it would be very strange if the results were found to be invalid after having already been found consistent with these more reliable measurements. 11. Aug 22, 2006 ### wolram Thankyou, Matt.o, i do not know about (baryon acoustic fluctuations) i will have to look up the subject. 12. Aug 22, 2006 ### Garth If SN Ia are not standard candles and the universe is not accelerating would that not alter the interpretation of the CMB and LSS data? Garth 13. Aug 22, 2006 ### matt.o Actually, I don't think baryon acoustic fluctuations have been used by themselves (yet) to constrain dark energy. The current galaxy redshift surveys don't cover enough volume. They have certainly been used in conjunction with WMAP data to constrain cosmological parameters http://arxiv.org/abs/astro-ph/0507583" [Broken] paper by Chris Blake and Karl Glazebrook. I do know of galaxy redshift surveys that are going to attempt to estimate the equation of state of the dark energy, but they are a little while off yet. There are also planned missions to survey for clusters and measure the dark energy equation of state using the cluster mass function. Last edited by a moderator: May 2, 2017 14. Aug 22, 2006 ### SpaceTiger Staff Emeritus The paper in the OP says the following: This doesn't suggest to me that the SN Ia results have been invalidated. As for the Middleditch paper, I took a quick look at it and it sounds very crankish. I doubt it will make any waves. If SNe Ia were not standard candles, why would you infer that the universe was not accelerating? The former statement would simply mean that they couldn't be used as distance indicators. 15. Aug 22, 2006 ### Tzemach There are still some big holes in our understanding, maybe if we collect enough details of anomolies we will be able to figure out just what is going on out there. Here are a couple for the collection' APM 8279+5255 the red shift of the quasar, a vibrant galaxy with a bright central region and massive central black hole, revealed that it contained much more iron than it should for its age. The amount of iron present is greater than the amount found in our own solar system. At a fraction of the age of the solar system this quasar should contain the lower percentage of iron. Fossil cluster RX J1416.4+2315 Current projections state that the fossil group should not have had enough time to form given the age of the universe. 16. Aug 22, 2006 ### Garth I was simply reverting back to the pre-'cosmic acceleration' understanding. The acceleration or non-acceleration of the universe determined from SNe Ia at varing high z is an important handle on the behaviour and therefore nature of DE. The consequences of them not being standard candles would therefore be rather significant. Tzemach has highlighted again the Age problem question, I think there is a degeneracy in the interpretation of the data that may be resolved by attention to these anomalies. Is there an age problem at high z? Are SNe Ia standard candles? Is there a genuine deficiency at low-l mode in the WMAP anisotropy power spectrum? Just a few questions to chew over.... Garth Last edited: Aug 22, 2006 17. Aug 22, 2006 ### SpaceTiger Staff Emeritus It's not clear to me that your above statements really make that case. It would have made a big difference historically, but with the high-precision CMB and LSS measurements we have now, I don't think you'd see many people jumping ship from standard cosmology. 18. Aug 22, 2006 ### Garth Well ST, surely a project such as Destiny depends on the assumption that SNe Ia are standard candles? Are you saying that if, for the sake of argument, they prove to be intrinsically fainter (say) at high z, then the interpretation of the high precision WMAP and LSS data would not be affected? Or do you think that WMAP, LSS, etc. have so proved cosmic acceleration independently of SNe Ia that the there is no question about it, and consequently SNe Ia have been verified to be standard candles? Garth Last edited: Aug 22, 2006 19. Aug 22, 2006 ### SpaceTiger Staff Emeritus Again, this depends crucially on the distinction between them not being standard candles and them being renormalized. If they were renormalized and subsequently showed no acceleration, then that would be a serious problem for mainstream cosmology in general. That alone likely wouldn't be enough to convince folks (after all, it would also demonstrate the unreliability of supernovae). However, if subsequent measurements with other distance indicators gave the same result, the mainstream model would have to be changed. If the supernovae were just shown to be unreliable as standard candles, then yes, I'm saying the interpretation would likely remain the same. Your choice of words in the above doesn't make it clear which case you're referring to. Those data sets do not directly prove acceleration, but they do suggest it. Believe it or not, $\Lambda CDM$ was under serious consideration even prior to the supernova results because of COBE CMB measurements. 20. Aug 22, 2006 ### matt.o Exactly. From COBE, it was known we live in an $$\Omega$$ = 1 Universe, but the observed matter density only made up ~.3 of the total density. People were considering a cosmological constant to make up the rest of the energy density and along came the SN 1a results! Similar Discussions: Standard candles	{"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": 0.860864520072937, "perplexity": 1685.5117784188426}, "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/1505818687837.85/warc/CC-MAIN-20170921191047-20170921211047-00246.warc.gz"}
http://www.talkstats.com/threads/differences-between-2-x1-and-x1-x2-under-normal-distribution.73857/	Differences between 2 X1 and X1 + X2 under normal distribution shawnteh1711 New Member The question is as below: 1. If X1 ~ N(μ, σ2) and X2 ~ N(μ, σ2) are normal random variables, then what is the difference between 2 X1 and X1 + X2? Discuss both their distributions and parameters. Assume that X1 + X2 have same μ =1, σ = 2 2 X1 = 2(1,22) = 2(1,4) = (2,8) X1 + X2 = (1 + 1, 22 + 22) = (2,8) spunky Doesn't actually exist You're almost there, but you're forgetting how variances/standard deviations change under linear transformations. Just go back to the definition of the variance and using the algebra of expectations see how Var(X1+X2) is different from Var (kX1) (where k is a constant) shawnteh1711 New Member You're almost there, but you're forgetting how variances/standard deviations change under linear transformations. Just go back to the definition of the variance and using the algebra of expectations see how Var(X1+X2) is different from Var (kX1) (where k is a constant) I look it up and got the following rule : 1) Var (kX1) = k ^2 Var (X) 2) If X and Y are independent, V(X + Y) = V(X) + V(Y) I try to apply to the question. Given that X1 ~ N(µ, σ2) and X2 ~ N(µ, σ2) Assume that X1 and X2 have same µ =1, σ =2 Var(2X1) = 2^2Var(1,2^2) = 4Var(1,4) = Var(4,16) Var(X1 + X2) = Var(X1) + Var (X2) = Var(1, 2^2) + Var(1, 2^2) = var (1,4) + var (1,4) = var (2,8) It means that Var (2X1) has 2 times more variance value than Var(X1+X2). Is it correct?	{"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": 0.8528398275375366, "perplexity": 3257.288977711554}, "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-2019-35/segments/1566027313803.9/warc/CC-MAIN-20190818104019-20190818130019-00197.warc.gz"}
http://math.stackexchange.com/users/36924/mike-flynn?tab=activity&sort=comments	# Mike Flynn less info reputation 29 bio website location age member for 2 years seen Aug 15 at 15:20 profile views 40 Mar3 comment How to properly take derivatives in calculus of variations (Euler-Lagrange formula) @SanathDevalapurkar It is not obvious to me why $df'/df$ is $0$. Aug14 comment P(tomorrow is the end of the world) =? "they both", do you mean "they all"? Aug13 comment Mean Distance on a 3-sphere? Can you show an attempt to solve the problem? Aug12 comment Linear programming simplex - can I have a constraint with a multiplication? so if your solution is $x = {x_1, x_2, \dots, x_n}$, you are saying all of $n_1, n_2, t_1, t_2$ are components of $x$? Aug9 comment Understanding detailed balance (crossposted from stats) @al-Hwarizmi Markov Chain Monte Carlo, a method of approximating a probability distribution with a random walk, often used in statistics. The second sentence should read "proportional to their ratio of probability densities". I've now edited it. Jul19 comment A method to test for uniform distribution over a convex polytope hit and run. Pick an initial point, choose a random direction, the next point is a random point between initial point and wall in that direction. Jul19 comment A method to test for uniform distribution over a convex polytope I have no way of knowing $\|P\|$, in fact, this method is how I would calculate $\|P\|$, provided the sampling is uniform. Jun24 comment What is the typical method for sampling uniformly in a convex polytope Not in practice actually, if you make the jump length long enough. And are you saying that there is no known way to do it quickly for 1000 dimensions? Because I was considering using CUDA to do it... Jun18 comment Changing a basis and satisfying inequalities Ah, I now realize you must be very careful with each step... Dec11 comment What is a covector and what is it used for? yes, so $dx([1,2])$ would be the function $dx$ acting on the vector $[1,2] \in R^2$. Dec4 comment How would you define a geodesic in $\operatorname{SO}(2n)$ and $\operatorname{SO}(2n+1)$? Thank you for the answer. In $R^{n^2}$, do we compute the Euclidean distance as if the matrix was a long vector, adding up the squares of the components and taking a square root? Dec3 comment What is the diameter of a manifold? @QiaochuYuan, it was a question unattached to a textbook, just "What are the intrinsic diameters of SO(2n) and SO(2n+1)?" The textbook we have been using is Frank Morgan's "Riemannian Geometry". My professor probably assumed that the choice of metric would be immediately obvious, but it is not, at least to me. Dec3 comment What is the diameter of a manifold? @QiaochuYuan does changing it to "intrinsic diameter" change anything? Nov19 comment What is a covector and what is it used for? Yes, I think most of my misunderstanding arose from the fact that I was interpreting $dx$ as a "length" as opposed to a function.	{"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": 0.9236931800842285, "perplexity": 521.4167242220439}, "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-2014-35/segments/1408500830746.39/warc/CC-MAIN-20140820021350-00273-ip-10-180-136-8.ec2.internal.warc.gz"}
http://link.springer.com/chapter/10.1007%2F978-3-642-22321-1_7	Chapter Developments in Language Theory Volume 6795 of the series Lecture Notes in Computer Science pp 70-81 # Avoiding Abelian Powers in Partial Words • Francine Blanchet-SadriAffiliated withDepartment of Computer Science, University of North Carolina • , Sean SimmonsAffiliated withDepartment of Mathematics, The University of Texas at Austin * Final gross prices may vary according to local VAT. ## Abstract We study abelian repetitions in partial words, or sequences that may contain some unknown positions or holes. First, we look at the avoidance of abelian pth powers in infinite partial words, where p > 2, extending recent results regarding the case where p = 2. We investigate, for a given p, the smallest alphabet size needed to construct an infinite partial word with finitely or infinitely many holes that avoids abelian pth powers. We construct in particular an infinite binary partial word with infinitely many holes that avoids 6th powers. Then we show, in a number of cases, that the number of abelian p-free partial words of length n with h holes over a given alphabet grows exponentially as n increases. Finally, we prove that we cannot avoid abelian pth powers under arbitrary insertion of holes in an infinite word.	{"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": 0.8809349536895752, "perplexity": 968.7997746928725}, "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-2016-44/segments/1476988718426.35/warc/CC-MAIN-20161020183838-00562-ip-10-171-6-4.ec2.internal.warc.gz"}
https://brilliant.org/problems/lets-do-some-calculus-28/	Let's do some calculus! (28) Calculus Level 4 $\large {\lim_{x \to 0}} \ \dfrac{\cos^2 x - \cos x - e^x \cos x + e^x - \dfrac{x^3}{2}}{x^n}$ Find the value of $$n$$ for which the above limit is finite and non-zero. Notations: $$e \approx 2.71828$$ is the Euler's number.	{"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": 0.9868950843811035, "perplexity": 574.6111969690453}, "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-2017-22/segments/1495463608107.28/warc/CC-MAIN-20170525155936-20170525175936-00032.warc.gz"}
http://nrich.maths.org/5433/index?nomenu=1	by Henry Kwok Rules of Pole Star Sudoku This is a variation of Sudoku on a standard 9x9 grid which contains a set of special clue-numbers. These are small numbers placed always on the border lines between selected pairs of neighbouring cells of the grid. Each clue-number is the difference between the two numbers that should be in the neighbouring cells just to the left and to the right from that clue-number. For example, a clue-number 7 on the border line means that one of the possible answers in the cell on the left can be 9, 2, 8 or 1. If we choose 9 for the answer in the cell, it means that the answer in the cell next to it on the right is 2. If we choose the answer 2, it means that the answer on the right is 9. Not much information is there? However, fortunately for the solver, you can use a starting digit (digit 9 in the last column) as the Pole Star to guide you out of the "wilderness" of the puzzle. Hence this Sudoku variant is named Pole Star Sudoku. The remaining rules are as in a standard Sudoku: the object of the puzzle is to fill in the whole 9x9 grid with numbers 1 through 9 (one number per cell) so that each horizontal line, each vertical line, and each of the nine 3x3 squares (outlined with the bold lines) must contain all the nine different numbers 1 through 9.	{"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": 0.8112038969993591, "perplexity": 360.85287596766324}, "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-2015-35/segments/1440645298065.68/warc/CC-MAIN-20150827031458-00326-ip-10-171-96-226.ec2.internal.warc.gz"}
https://space.meta.stackexchange.com/questions/371/adding-a-hyphen-to-tagremotesensing/372	# Adding a hyphen to [tag:remotesensing] It seems that "remote sensing" is two separate words, so the tag should have a hyphen (i.e., be ). (I do not think this justifies having a synonym.) I am tagging this even though I do not think there would be much debate about whether "remote sensing" should be treated as two words. • On the other hand, there is earth-observation which is a subset of remote sensing. – gerrit Sep 24 '13 at 13:05	{"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": 0.8949571251869202, "perplexity": 976.2930626861611}, "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-2020-50/segments/1606141182794.28/warc/CC-MAIN-20201125125427-20201125155427-00595.warc.gz"}
https://intl.siyavula.com/read/science/grade-11/newtons-laws/02-newtons-laws-06	We think you are located in South Africa. Is this correct? # Don't get left behind Join thousands of learners improving their science marks online with Siyavula Practice. ## Forces and Newton's Laws Exercise 2.10 A force acts on an object. Name three effects that the force can have on the object. A force can change the shape of the object. A force can change the direction in which the object is moving. A force can accelerate or stop a body. Identify each of the following forces as contact or non-contact forces. The force between the north pole of a magnet and a paper clip. non-contact The force required to open the door of a taxi. contact The force required to stop a soccer ball. contact The force causing a ball, dropped from a height of $$\text{2}$$ $$\text{m}$$, to fall to the floor. non-contact A book of mass $$\text{2}$$ $$\text{kg}$$ is lying on a table. Draw a labelled force diagram indicating all the forces on the book. Where $$\vec{F}_{g}$$ is the force due to gravity and $$\vec{N}$$ is the normal force. A constant, resultant force acts on a body which can move freely in a straight line. Which physical quantity will remain constant? 1. acceleration 2. velocity 3. momentum 4. kinetic energy [SC 2003/11] acceleration Two forces, $$\text{10}$$ $$\text{N}$$ and $$\text{15}$$ $$\text{N}$$, act at an angle at the same point. Which of the following cannot be the resultant of these two forces? 1. $$\text{2}$$ $$\text{N}$$ 2. $$\text{5}$$ $$\text{N}$$ 3. $$\text{8}$$ $$\text{N}$$ 4. $$\text{20}$$ $$\text{N}$$ [SC 2005/11 SG1] $$\text{2}$$ $$\text{N}$$ A concrete block weighing $$\text{250}$$ $$\text{N}$$ is at rest on an inclined surface at an angle of $$\text{20}$$$$\text{°}$$. The magnitude of the normal force, in newtons, is 1. $$\text{250}$$ 2. $$250\cos\text{20}\text{°}$$ 3. $$\text{250}\sin\text{20}\text{°}$$ 4. $$\text{2 500}\cos\text{20}\text{°}$$ $$\text{250}$$ A $$\text{30}$$ $$\text{kg}$$ box sits on a flat frictionless surface. Two forces of $$\text{200}$$ $$\text{N}$$ each are applied to the box as shown in the diagram. Which statement best describes the motion of the box? 1. The box is lifted off the surface. 2. The box moves to the right. 3. The box does not move. 4. The box moves to the left. The box moves to the left. A concrete block weighing $$\text{200}$$ $$\text{N}$$ is at rest on an inclined surface at an angle of $$\text{20}$$$$\text{°}$$. The normal force, in newtons, is 1. $$\text{200}$$ 2. $$\text{200}\cos\text{20}\text{°}$$ 3. $$\text{200}\sin\text{20}\text{°}$$ 4. $$\text{2 000}\cos\text{20}\text{°}$$ $$\text{200}$$ A box, mass $$m$$, is at rest on a rough horizontal surface. A force of constant magnitude $$F$$ is then applied on the box at an angle of $$\text{60}$$$$\text{°}$$ to the horizontal, as shown. If the box has a uniform horizontal acceleration of magnitude, $$a$$, the frictional force acting on the box is... 1. $$F\cos\text{60}\text{°}-ma$$ in the direction of A 2. $$F\cos\text{60}\text{°}-ma$$ in the direction of B 3. $$F\sin\text{60}\text{°}-ma$$ in the direction of A 4. $$F\sin\text{60}\text{°}-ma$$ in the direction of B [SC 2003/11] $$F\cos\text{60}\text{°}-ma$$ in the direction of A Thabo stands in a train carriage which is moving eastwards. The train suddenly brakes. Thabo continues to move eastwards due to the effect of: 1. his inertia. 2. the inertia of the train. 3. the braking force on him. 4. a resultant force acting on him. [SC 2002/11 SG] his inertia A $$\text{100}$$ $$\text{kg}$$ crate is placed on a slope that makes an angle of $$\text{45}$$$$\text{°}$$ with the horizontal. The gravitational force on the box is $$\text{98}$$ $$\text{N}$$. The box does not slide down the slope. Calculate the magnitude and direction of the frictional force and the normal force present in this situation. Since the box does not move the frictional force acting on the box is equal to the component of the gravitational force parallel to the plane. \begin{align} F_{f} & = F_{gx} \\ & = F \cos \theta \\ & = (98) \cos(45) \\ & = \text{69,3}\text{ N} \end{align} This force acts at an an angle of $$\text{45}$$$$\text{°}$$ with the horizontal. The normal force acts in the opposite direction to the gravitational force and has a magnitude of $$\text{98}$$ $$\text{N}$$ at an angle of $$\text{45}$$$$\text{°}$$ with the horizontal. A body moving at a CONSTANT VELOCITY on a horizontal plane, has a number of unequal forces acting on it. Which one of the following statements is TRUE? 1. At least two of the forces must be acting in the same direction. 2. The resultant of the forces is zero. 3. Friction between the body and the plane causes a resultant force. 4. The vector sum of the forces causes a resultant force which acts in the direction of motion. [SC 2002/11 HG1] The resultant of the forces is zero. Two masses of $$m$$ and $$2m$$ respectively are connected by an elastic band on a frictionless surface. The masses are pulled in opposite directions by two forces each of magnitude $$F$$, stretching the elastic band and holding the masses stationary. Which of the following gives the magnitude of the tension in the elastic band? 1. zero 2. $$\frac{\text{1}}{\text{2}}F$$ 3. $$F$$ 4. $$2F$$ [IEB 2005/11 HG] $$F$$ A rocket takes off from its launching pad, accelerating up into the air. The rocket accelerates because the magnitude of the upward force, F is greater than the magnitude of the rocket's weight, W. Which of the following statements best describes how force F arises? 1. F is the force of the air acting on the base of the rocket. 2. F is the force of the rocket's gas jet pushing down on the air. 3. F is the force of the rocket's gas jet pushing down on the ground. 4. F is the reaction to the force that the rocket exerts on the gases which escape out through the tail nozzle. [IEB 2005/11 HG] F is the reaction to the force that the rocket exerts on the gases which escape out through the tail nozzle. A box of mass $$\text{20}$$ $$\text{kg}$$ rests on a smooth horizontal surface. What will happen to the box if two forces each of magnitude $$\text{200}$$ $$\text{N}$$ are applied simultaneously to the box as shown in the diagram. The box will: 1. be lifted off the surface. 2. move to the left. 3. move to the right. 4. remain at rest. [SC 2001/11 HG1] move to the left A $$\text{2}$$ $$\text{kg}$$ mass is suspended from spring balance X, while a $$\text{3}$$ $$\text{kg}$$ mass is suspended from spring balance Y. Balance X is in turn suspended from the $$\text{3}$$ $$\text{kg}$$ mass. Ignore the weights of the two spring balances. The readings (in N) on balances X and Y are as follows: X Y a) $$\text{19,6}$$ $$\text{29,4}$$ b) $$\text{19,6}$$ 49 c) $$\text{24,5}$$ $$\text{24,5}$$ d) 49 49 [SC 2001/11 HG1] b) $$\text{19,6}$$, 49 P and Q are two forces of equal magnitude applied simultaneously to a body at X. As the angle θ between the forces is decreased from $$\text{180}$$$$\text{°}$$ to $$\text{0}$$$$\text{°}$$, the magnitude of the resultant of the two forces will 1. initially increase and then decrease. 2. initially decrease and then increase. 3. increase only. 4. decrease only. [SC 2002/03 HG1] increase only The graph below shows the velocity-time graph for a moving object: Which of the following graphs could best represent the relationship between the resultant force applied to the object and time? (a) (b) (c) (d) [SC 2002/03 HG1] graph (b) Two blocks each of mass $$\text{8}$$ $$\text{kg}$$ are in contact with each other and are accelerated along a frictionless surface by a force of $$\text{80}$$ $$\text{N}$$ as shown in the diagram. The force which block Q will exert on block P is equal to ... 1. $$\text{0}$$ $$\text{N}$$ 2. $$\text{40}$$ $$\text{N}$$ 3. $$\text{60}$$ $$\text{N}$$ 4. $$\text{80}$$ $$\text{N}$$ [SC 2002/03 HG1] $$\text{40}$$ $$\text{N}$$ A $$\text{12}$$ $$\text{kg}$$ box is placed on a rough surface. A force of $$\text{20}$$ $$\text{N}$$ applied at an angle of $$\text{30}$$$$\text{°}$$ to the horizontal cannot move the box. Calculate the magnitude and direction of the normal and friction forces. The normal force is: \begin{align} N & = mg \\ & = (\text{12})(\text{9,8}) \\ & = \text{117,6}\text{ N} \end{align} This force points straight up from the surface. The friction force is: \begin{align} F_{f} & = F_{\text{app}} \cos 30° \\ & = (\text{20})\cos(\text{30}°) \\ & = \text{17,3}\text{ N} \end{align} The frictional force points in the opposite direction to the applied force. Three $$\text{1}$$ $$\text{kg}$$ mass pieces are placed on top of a $$\text{2}$$ $$\text{kg}$$ trolley. When a force of magnitude F is applied to the trolley, it experiences an acceleration a. If one of the $$\text{1}$$ $$\text{kg}$$ mass pieces falls off while F is still being applied, the trolley will accelerate at ... 1. $$\frac{\text{1}}{\text{5}}a$$ 2. $$\frac{\text{4}}{\text{5}}a$$ 3. $$\frac{\text{5}}{\text{4}}a$$ 4. $$5a$$ [SC 2002/03 HG1] $$\frac{\text{5}}{\text{4}}a$$ A car moves along a horizontal road at constant velocity. Which of the following statements is true? 1. The car is not in equilibrium. 2. There are no forces acting on the car. 3. There is zero resultant force. 4. There is no frictional force. [IEB 2004/11 HG1] There is zero resultant force. A crane lifts a load vertically upwards at constant speed. The upward force exerted on the load is F. Which of the following statements is correct? 1. The acceleration of the load is $$\text{9,8}$$ $$\text{m·s^{-1}}$$ downwards. 2. The resultant force on the load is F. 3. The load has a weight equal in magnitude to F. 4. The forces of the crane on the load, and the weight of the load, are an example of a Newton's third law 'action-reaction' pair. [IEB 2004/11 HG1] The load has a weight equal in magnitude to F. A body of mass $$M$$ is at rest on a smooth horizontal surface with two forces applied to it as in the diagram below. Force $${F}_{1}$$ is equal to $$Mg$$. The force $${F}_{1}$$ is applied to the right at an angle θ to the horizontal, and a force of $${F}_{2}$$ is applied horizontally to the left. How is the body affected when the angle θ is increased? 1. It remains at rest. 2. It lifts up off the surface, and accelerates towards the right. 3. It lifts up off the surface, and accelerates towards the left. 4. It accelerates to the left, moving along the smooth horizontal surface. [IEB 2004/11 HG1] It accelerates to the left, moving along the smooth horizontal surface. Which of the following statements correctly explains why a passenger in a car, who is not restrained by the seat belt, continues to move forward when the brakes are applied suddenly? 1. The braking force applied to the car exerts an equal and opposite force on the passenger. 2. A forward force (called inertia) acts on the passenger. 3. A resultant forward force acts on the passenger. 4. A zero resultant force acts on the passenger. [IEB 2003/11 HG1] A zero resultant force acts on the passenger. A rocket (mass $$\text{20 000}$$ $$\text{kg}$$) accelerates from rest to $$\text{40}$$ $$\text{m·s^{-1}}$$ in the first $$\text{1,6}$$ seconds of its journey upwards into space. The rocket's propulsion system consists of exhaust gases, which are pushed out of an outlet at its base. Explain, with reference to the appropriate law of Newton, how the escaping exhaust gases exert an upwards force (thrust) on the rocket. Newton's third law states that if body A exerts a force on body B, then body B exerts a force of equal magnitude on body A, but in the opposite direction. The exhaust gases exert a force on the rocket. This means that the rocket also exerts a force on the exhaust gases. Since the direction of the force from the exhaust is downwards the direction of the force from the rocket must be upwards. This is what propels the rocket upwards. What is the magnitude of the total thrust exerted on the rocket during the first $$\text{1,6}$$ $$\text{s}$$? We first need to find the acceleration. We can do this by using the equations of motion: \begin{align} v_{f} & = v_{i} + at \\ 40 & = 0 + (\text{1,6})a \\ a & = \text{25}\text{ m·s$^{-2}$} \end{align} Now we can find the magnitude of the total thrust, T: \begin{align} T & = W + ma \\ & = (\text{20 000})(\text{9,8}) + (\text{20 000})(\text{25}) \\ & = \text{696 000}\text{ N} \end{align} An astronaut of mass $$\text{80}$$ $$\text{kg}$$ is carried in the space capsule. Determine the resultant force acting on him during the first $$\text{1,6}$$ $$\text{s}$$. \begin{align} T & = W + ma \\ & = (\text{80})(\text{9,8}) + (\text{80})(\text{25}) \\ & = \text{2 784}\text{ N} \end{align} Explain why the astronaut, seated in his chair, feels “heavier” while the rocket is launched. Weight is measured through normal forces. When the rocket accelerates upwards he feels a greater normal force acting on him as the force required to accelerate him upwards in addition to balancing out the gravitational force. State Newton's second law of Motion. If a resultant force acts on a body, it will cause the body to accelerate in the direction of the resultant force. The acceleration of the body will be directly proportional to the resultant force and inversely proportional to the mass of the body. The mathematical representation is: $\vec{F}_{net} = m\vec{a}$ A sports car (mass $$\text{1 000}$$ $$\text{kg}$$) is able to accelerate uniformly from rest to $$\text{30}$$ $$\text{m·s^{-1}}$$ in a minimum time of $$\text{6}$$ $$\text{s}$$. Calculate the magnitude of the acceleration of the car. \begin{align} a & = \frac{\Delta v}{\Delta t} \\ & = \frac{30}{6} \\ & = \text{5}\text{ m·s$^{-2}$} \end{align} What is the magnitude of the resultant force acting on the car during these $$\text{6}$$ $$\text{s}$$? \begin{align} F & = ma \\ & = (\text{1 000})(5) \\ & = \text{5 000}\text{ N} \end{align} The magnitude of the force that the wheels of the vehicle exert on the road surface as it accelerates is $$\text{7 500}$$ $$\text{N}$$. What is the magnitude of the retarding forces acting on this car? The retarding force is the force that the cars wheels exert minus the resultant force: \begin{align} F & = F_{\text{app}} - F_{\text{res}} \\ & = \text{7 500} - \text{5 000} \\ & = \text{2 500}\text{ N} \end{align} By reference to a suitable Law of Motion, explain why a headrest is important in a car with such a rapid acceleration. Newton's first law states that an object continues in a state of rest or uniform motion (motion with a constant velocity) unless it is acted on by an unbalanced (net or resultant) force. There is a resultant force on the person's head and as the car accelerates their head is pulled back. The headrest stops the person's head from being pulled to far back and prevents whiplash. A child (mass $$\text{18}$$ $$\text{kg}$$) is strapped in his car seat as the car moves to the right at constant velocity along a straight level road. A tool box rests on the seat beside him. The driver brakes suddenly, bringing the car rapidly to a halt. There is negligible friction between the car seat and the box. Draw a labelled free-body diagram of the forces acting on the child during the time that the car is being braked. Draw a labelled free-body diagram of the forces acting on the box during the time that the car is being braked. Modern cars are designed with safety features (besides seat belts) to protect drivers and passengers during collisions e.g. the crumple zones on the car's body. Rather than remaining rigid during a collision, the crumple zones allow the car's body to collapse steadily. State Newton's second law of motion. If a resultant force acts on a body, it will cause the body to accelerate in the direction of the resultant force. The acceleration of the body will be directly proportional to the resultant force and inversely proportional to the mass of the body. The mathematical representation is:$\vec{F}_{net} = m\vec{a}$ The total mass of a lift together with its load is $$\text{1 200}$$ $$\text{kg}$$. It is moving downwards at a constant velocity of $$\text{9}$$ $$\text{m·s^{-1}}$$. What will be the magnitude of the force exerted by the cable on the lift while it is moving downwards at constant velocity? Give an explanation for your answer. We take upwards as the positive direction. The acceleration is $$\text{0}$$ since the lift is moving at constant velocity. \begin{align} \Sigma F & = T - W \\ (0)(m) & = T - W \\ \therefore T & = W \\ & = (\text{9,8})(\text{1 200}) \\ & = \text{11 760}\text{ N} \end{align} The lift is now uniformly brought to rest over a distance of $$\text{18}$$ $$\text{m}$$. Calculate the magnitude of the acceleration of the lift. \begin{align} v_{f}^{2} & = v_{i}^{2} + 2a \Delta x \\ 0 & = (9)^{2} + 2a(18) \\ 36a & = -81 \\ a & = \frac{-\text{81}}{\text{36}} \\ & = -\text{2,25}\text{ m·s$^{-2}$} \end{align} Calculate the magnitude of the force exerted by the cable while the lift is being brought to rest. \begin{align} T - W & = ma \\ T & = ma + mg \\ & = (\text{1 200})(-\text{2,25}) + (\text{9,8})(\text{1 200}) \\ & = \text{9 060}\text{ N} \end{align} A driving force of $$\text{800}$$ $$\text{N}$$ acts on a car of mass $$\text{600}$$ $$\text{kg}$$. Calculate the car's acceleration. \begin{align} F & = ma \\ a & = \frac{F}{m} \\ & = \frac{\text{800}}{\text{600}} \\ & = \text{1,33}\text{ m·s$^{-2}$} \end{align} Calculate the car's speed after $$\text{20}$$ $$\text{s}$$. \begin{align} v_{f} & = v_{i} + at \\ v_{f} & = 0 + (20)(\text{1,33}) \\ & = \text{26,7}\text{ m·s$^{-1}$} \end{align} Calculate the new acceleration if a frictional force of $$\text{50}$$ $$\text{N}$$ starts to act on the car after $$\text{20}$$ $$\text{s}$$. \begin{align} a_{2} & = a_{1} - \frac{F_{f}}{m} \\ & = \text{1,33} - \frac{\text{50}}{\text{600}} \\ & = \text{1,25}\text{ m·s$^{-2}$} \end{align} Calculate the speed of the car after another $$\text{20}$$ $$\text{s}$$ (i.e. a total of $$\text{40}$$ $$\text{s}$$ after the start). \begin{align} v_{f2} & = a_{1}t + a_{2}t \\ v_{f2} & = (20)(\text{1,33}) + (20)(\text{1,25}) \\ & = \text{51,6}\text{ m·s$^{-1}$} \end{align} A stationary block of mass $$\text{3}$$ $$\text{kg}$$ is on top of a plane inclined at $$\text{35}$$$$\text{°}$$ to the horizontal. Draw a force diagram (not to scale). Include the weight of the block as well as the components of the weight that are perpendicular and parallel to the inclined plane. Determine the values of the weight's perpendicular and parallel components. The perpendicular component is: \begin{align} W_{\perp} & = F \sin(\theta) \\ & = mg \sin (\theta) \\ & = (3)(9.8) \sin(35) \\ & = \text{16,86}\text{ N} \end{align} The parallel component is: \begin{align} W_{\|\|} & = F \cos(\theta) \\ & = mg \cos (\theta) \\ & = (3)(9.8) \cos(35) \\ & = \text{24,01}\text{ N} \end{align} A crate on an inclined plane Elephants are being moved from the Kruger National Park to the Eastern Cape. They are loaded into crates that are pulled up a ramp (an inclined plane) on frictionless rollers. The diagram shows a crate being held stationary on the ramp by means of a rope parallel to the ramp. The tension in the rope is $$\text{5 000}$$ $$\text{N}$$. Explain how one can deduce the following: “The forces acting on the crate are in equilibrium”. The sum of the forces is equal to the mass times the acceleration. Since the acceleration is $$\text{0}$$, the sum of the forces is $$\text{0}$$. This means that the forces acting on the crate are in equilibrium. Draw a labelled free-body diagram of the forces acting on the elephant. (Regard the crate and elephant as one object, and represent them as a dot. Also show the relevant angles between the forces.) The crate has a mass of $$\text{800}$$ $$\text{kg}$$. Determine the mass of the elephant. \begin{align} W_{E \|\|} & = W \sin \theta \\ \text{5 000} & = W \sin(15) \\ W & = \frac{\text{5 000}}{\sin (15)} \\ & = \text{19 318,52}\text{ N} \end{align} \begin{align} m_{E} & = \frac{W}{g} - m_{\text{crate}} \\ & = \frac{\text{19 318,52}}{\text{9,8}} - \text{800} \\ & = \text{1 171,28}\text{ kg} \end{align} The crate is now pulled up the ramp at a constant speed. How does the crate being pulled up the ramp at a constant speed affect the forces acting on the crate and elephant? Justify your answer, mentioning any law or principle that applies to this situation. There will be no changes to the forces since the acceleration is $$\text{0}$$. Car in Tow Car A is towing Car B with a light tow rope. The cars move along a straight, horizontal road. Write down a statement of Newton's second law of Motion (in words). If a resultant force acts on a body, it will cause the body to accelerate in the direction of the resultant force. The acceleration of the body will be directly proportional to the resultant force and inversely proportional to the mass of the body. As they start off, Car A exerts a forwards force of $$\text{600}$$ $$\text{N}$$ at its end of the tow rope. The force of friction on Car B when it starts to move is $$\text{200}$$ $$\text{N}$$. The mass of Car B is $$\text{1 200}$$ $$\text{kg}$$. Calculate the acceleration of Car B. The total force on car B is $$\text{600}\text{ N} - \text{200}\text{ N} = \text{400}\text{ N}$$. \begin{align} F & = ma \\ a & = \frac{\text{400}}{\text{1 200}} \\ & = \text{0,33}\text{ m·s$^{-2}$} \end{align} After a while, the cars travel at constant velocity. The force exerted on the tow rope is now $$\text{300}$$ $$\text{N}$$ while the force of friction on Car B increases. What is the magnitude and direction of the force of friction on Car B now? The frictional force is equal to the force exerted on the tow rope but in the opposite direction. $$F_{f} = \text{300}\text{ N}$$. Towing with a rope is very dangerous. A solid bar should be used in preference to a tow rope. This is especially true should Car A suddenly apply brakes. What would be the advantage of the solid bar over the tow rope in such a situation? Newton's first law states that an object will keep on moving in a straight line unless acted on by a force. A solid bar will stop car B from moving if car A breaks while a rope will not. The mass of Car A is also $$\text{1 200}$$ $$\text{kg}$$. Car A and Car B are now joined by a solid tow bar and the total braking force is $$\text{9 600}$$ $$\text{N}$$. Over what distance could the cars stop from a velocity of $$\text{20}$$ $$\text{m·s^{-1}}$$? The acceleration is: \begin{align} F & = ma \\ a & = \frac{\text{9 600}}{2(\text{1 200})} \\ & = \text{4} \end{align} So the stopping distance is: \begin{align} v_{f}^{2} & = v_{i}^{2} + 2a \Delta x \\ 0 & = (\text{20})^{2} + 2(-4)\Delta x \\ 8 \Delta x & = 400 \\ \Delta x & = \text{50}\text{ m} \end{align} Testing the Brakes of a Car A braking test is carried out on a car travelling at $$\text{20}$$ $$\text{m·s^{-1}}$$. A braking distance of $$\text{30}$$ $$\text{m}$$ is measured when a braking force of $$\text{6 000}$$ $$\text{N}$$ is applied to stop the car. Calculate the acceleration of the car when a braking force of $$\text{6 000}$$ $$\text{N}$$ is applied. \begin{align} v_{f}^{2} & = v_{i}^{2} + 2a \Delta x \\ 0 & = (\text{20})^{2} + 2(30)a \\ 60a & = -400 \\ a & = -\text{6,67}\text{ m·s$^{-2}$} \end{align} Show that the mass of this car is $$\text{900}$$ $$\text{kg}$$. \begin{align} F & = ma \\ \text{6 000} & = (\text{6,67})m \\ m & = \text{900}\text{ kg} \end{align} How long (in s) does it take for this car to stop from $$\text{20}$$ $$\text{m·s^{-1}}$$ under the braking action described above? \begin{align} v_{f} & = v_{i} + at \\ 0 & = \text{20} + (-\text{6,67})t \\ \text{6,67}t & = 20 \\ t & = \text{3,0}\text{ s} \end{align} A trailer of mass $$\text{600}$$ $$\text{kg}$$ is attached to the car and the braking test is repeated from $$\text{20}$$ $$\text{m·s^{-1}}$$ using the same braking force of $$\text{6 000}$$ $$\text{N}$$. How much longer will it take to stop the car with the trailer in tow? \begin{align} F & = ma \\ \text{6 000} & = (900 + 600)a \\ a & = \text{4} \end{align} \begin{align} t & = \frac{-\text{20}}{\text{4}} \\ t & = \text{5}\text{ s} \end{align} It will take $$\text{2}$$ $$\text{s}$$ longer. A box is held stationary on a smooth plane that is inclined at angle θ to the horizontal. $$F$$ is the force exerted by a rope on the box. $$w$$ is the weight of the box and $$N$$ is the normal force of the plane on the box. Which of the following statements is correct? 1. $$\tan\theta =\frac{F}{w}$$ 2. $$\tan\theta =\frac{F}{N}$$ 3. $$\cos\theta =\frac{F}{w}$$ 4. $$\sin\theta =\frac{N}{w}$$ [IEB 2005/11 HG] $$\sin\theta =\frac{N}{w}$$ As a result of three forces $${F}_{1}$$, $${F}_{2}$$ and $${F}_{3}$$ acting on it, an object at point P is in equilibrium. Which of the following statements is not true with reference to the three forces? 1. The resultant of forces $${F}_{1}$$, $${F}_{2}$$ and $${F}_{3}$$ is zero. 2. Force $${F}_{1}$$, $${F}_{2}$$ and $${F}_{3}$$ lie in the same plane. 3. Force $${F}_{3}$$ is the resultant of forces $${F}_{1}$$ and $${F}_{2}$$. 4. The sum of the components of all the forces in any chosen direction is zero. [SC 2001/11 HG1] Force $${F}_{3}$$ is the resultant of forces $${F}_{1}$$ and $${F}_{2}$$. A block of mass M is held stationary by a rope of negligible mass. The block rests on a frictionless plane which is inclined at $$\text{30}$$$$\text{°}$$ to the horizontal. Draw a labelled force diagram which shows all the forces acting on the block. Resolve the force due to gravity into components that are parallel and perpendicular to the plane. \begin{align} W & = Mg \\ W_{\|\|} & = Mg \sin \theta \\ W_{\perp} & = Mg \cos \theta \end{align} Calculate the weight of the block when the force in the rope is $$\text{8}$$ $$\text{N}$$. \begin{align} W_{\|\|} & = Mg \sin \theta \\ 8 & = M(\text{9,8}) \sin (\text{30}) \\ M & = \text{1,63}\text{ N} \end{align} A heavy box, mass m, is lifted by means of a rope R which passes over a pulley fixed to a pole. A second rope S, tied to rope R at point P, exerts a horizontal force and pulls the box to the right. After lifting the box to a certain height, the box is held stationary as shown in the sketch below. Ignore the masses of the ropes. The tension in rope R is $$\text{5 850}$$ $$\text{N}$$. Draw a diagram (with labels) of all the forces acting at the point P, when P is in equilibrium. By resolving the force exerted by rope R into components, calculate the... magnitude of the force exerted by rope S. \begin{align} S & = R \cos \theta \\ & = (\text{5 850}) \cos (\text{20}) \\ & = \text{5 497,2}\text{ N} \end{align} mass, m, of the box. \begin{align} mg & = R \sin \theta \\ \text{9,8}m& = (\text{5 850}) \sin (\text{20}) \\ m & = \text{204,17}\text{ kg} \end{align} A tow truck attempts to tow a broken down car of mass $$\text{400}$$ $$\text{kg}$$. The coefficient of static friction is $$\text{0,60}$$ and the coefficient of kinetic (dynamic) friction is $$\text{0,4}$$. A rope connects the tow truck to the car. Calculate the force required: to just move the car if the rope is parallel to the road. \begin{align} F_{s} & = \mu_{s}F \\ F_{s} & = \mu_{s} mg \\ & = (\text{0,6})(\text{400})(\text{9,8}) \\ F_{s} & = \text{2 352}\text{ N} \end{align} to keep the car moving at constant speed if the rope is parallel to the road. \begin{align} F_{f} & = \mu_{k}F \\ & = \mu_{k} mg \\ & = (\text{0,4})(\text{400})(\text{9,8}) \\ & = \text{1 568}\text{ N} \end{align} to just move the car if the rope makes an angle of $$\text{30}$$$$\text{°}$$ to the road. \begin{align} F_{s} \cos \theta & = \mu_{s}F \\ F_{s} & = \frac{\text{2 352}}{\cos\text{30}} \\ & = \text{2 715,9}\text{ N} \end{align} to keep the car moving at constant speed if the rope makes an angle of $$\text{30}$$$$\text{°}$$ to the road. \begin{align} F_{f} \cos \theta & = \mu_{f}F \\ F_{f} & = \frac{\text{1 568}}{\cos\text{30}} \\ & = \text{1 810,6}\text{ N} \end{align}	{"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": 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": 0.999161958694458, "perplexity": 555.082374877282}, "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/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00029.warc.gz"}
http://mathhelpforum.com/advanced-statistics/119491-one-2-tail-test-hypothesis-cobb-douglas-prod-function.html	# Thread: One or 2 Tail test hypothesis (cobb douglas prod. function) 1. ## One or 2 Tail test hypothesis (cobb douglas prod. function) Considering the Cobb Douglas production function as the initial function to start with. We regress that to Ln y = ln a + ln b1 k + ln b2 L Now we state in the hypothesis that ( h0; ) b1=b2=b3 = 0 is equal to zero And the alternative hypothesis is that it's not equal to zero. Considering the ' not equal to zero ' u might expect a two tail test because it can be either positive or negative. But since were' talking about Capital and Labor, we expect those 2 variables to have a positive! effect on output. Can we therefore say that it's better to use a 1 - tail test? 2. Originally Posted by Massachusetts Cowboys Considering the Cobb Douglas production function as the initial function to start with. We regress that to Ln y = ln a + ln b1 k + ln b2 L Now we state in the hypothesis that ( h0; ) b1=b2=b3 = 0 is equal to zero And the alternative hypothesis is that it's not equal to zero. Considering the ' not equal to zero ' u might expect a two tail test because it can be either positive or negative. But since were' talking about Capital and Labor, we expect those 2 variables to have a positive! effect on output. Can we therefore say that it's better to use a 1 - tail test? Negative Capital is rare occurences, but it happens sometimes, Goldman Sachs is one example. Negative Labor is an indication that the company needs to hire more labor to catch up with intensive production. There is nothing wrong in using two-tailed test in this case. If you prefer not to use two-tailed test, you must modify your null hypothesis. Hypotheses can always be tailored to you personal preference. You may set it up to accept or to reject or to postpone decisions.	{"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": 0.8473897576332092, "perplexity": 1392.3817630343383}, "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-2018-13/segments/1521257645824.5/warc/CC-MAIN-20180318145821-20180318165821-00043.warc.gz"}
https://math.meta.stackexchange.com/questions/12582/when-do-we-include-axiom-of-choice-as-a-tag	# When do we include "Axiom of Choice" as a tag? My question is the following: Suppose that the question is about a functional equation, where assuming the function we seek is continuous, we can show the existence of one particular type solution, whereas if continuity is not assumed, then the fact that $\mathbb R$ possesses a Hamel basis (as a linear space over $\mathbb Q$), guarantees the existence of discontinuous solutions of the functional equation. And my question is the following: It is non-appropriate to include the "Axiom of Choice" among the tags? It is important to say that many, if not most, of the subscribers who try to solve such functional equations are not familiar the Axiom of Choice/Zorn's Lemma, as such problems are popular in the IMOs and similar competitions. • – Asaf Karagila Mod Jan 25, 2014 at 20:30 I think the prevailing view is that since most mathematicians consider the Axiom of Choice a "fundamental truth" (or at least use it without much consideration), the tag should only in included when Choice is somehow of more central interest to the question itself. Examples of such question templates would be: • How do you prove that $\Phi$ is equivalent to the Axiom of Choice? • Can $\Psi$ be proven without the Axiom of Choice? • How much Choice is needed to prove $\Xi$? • Does $\Theta$ imply Choice? Just as I would not include the tag for most questions dealing with cardinal arithmetic, even in cases where $\mathsf{AC}$ and $\neg \mathsf{AC}$ may yield different answers (such as this recent example), I would not include it in questions from other mathematical areas where whether or not Choice holds could possibly change the answer (and the OP has not indicated interest in the connection between Choice and the particular problem). Of course, this doesn't preclude users from submitting answers which point out the connection between Choice and the given question. Such answers may be quite enlightening (though admittedly possibly not to the OP). • How about: "This functional equation has a discontinuous solution, but this requires Zorn's Lemma" - What happens then? Jan 25, 2014 at 21:03 • @Yiorgos: How about you just post the link to the question? – Asaf Karagila Mod Jan 25, 2014 at 21:10 • See for example: math.stackexchange.com/questions/641416/… Jan 25, 2014 at 21:13 • @Yiorgos: If your question was "Do discontinuous such functionals exist without assuming the Axiom of Choice," then, yes, I would include the axiom-of-choice tag. But I would rephrase your question to explicitly state that this is your central interest. As it stands, your Update seems to say that Choice yields a solution to your question, but this does not make your question about Choice. Jan 25, 2014 at 21:23 • Remark: much of the advice of this answer is independent of the axiom-of-choice tag. The phrase "when [blank] is somehow of more central interest to the question itself" is the general guideline for tagging. Another example: don't tag a linear PDE question with semigroups just because solutions can be expressed in terms of a continuous semigroup of mappings; use it when somehow the structure of a semigroup is critical to the question. Jan 27, 2014 at 8:35 I think an "axiom of choice equivalents" might be a reasonable tag. This includes gobs of important stuff. • I'm against this. I wrote on this in the thread linked on the comments of the question. – Asaf Karagila Mod Jan 27, 2014 at 1:16	{"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": 0.8542274832725525, "perplexity": 585.8116999429294}, "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-20221001132802-20221001162802-00564.warc.gz"}
http://mathhelpforum.com/advanced-algebra/144237-integral-domain.html	Math Help - Integral Domain 1. Integral Domain Hello I am just trying to work out some problems from herstein. Can you please tell me if this one is right. If D is an integral domian and if na=0 for some a not 0 in D and for some integer n not 0 , prove that D is of finite characterstic. so we have for some a not 0, na = 0 if b not 0 is another element in D then nab = 0b (ie) (na)b = 0 na is an element of D, an integral domain which has zero divisors, so(na)b=0 where na = 0 is justified. However nab=0 => (nb)a=0 Here nb is an element of D and a is nonzero, So nb = 0. Hence the proof. Am i right? 2. Originally Posted by poorna Hello I am just trying to work out some problems from herstein. Can you please tell me if this one is right. If D is an integral domian and if na=0 for some a not 0 in D and for some integer n not 0 , prove that D is of finite characterstic. so we have for some a not 0, na = 0 if b not 0 is another element in D then nab = 0b (ie) (na)b = 0 na is an element of D, an integral domain which has zero divisors, so(na)b=0 where na = 0 is justified. However nab=0 => (nb)a=0 Here nb is an element of D and a is nonzero, So nb = 0. Hence the proof. Am i right? That looks good to me.	{"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": 0.933155357837677, "perplexity": 1232.2448355359695}, "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-2014-41/segments/1412037663754.5/warc/CC-MAIN-20140930004103-00382-ip-10-234-18-248.ec2.internal.warc.gz"}

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.