Let me raise an example to be clear: say I have a 2*2 contingency table as follows:  5 10  10000 10000  I assume it satisfies the assumptions of chi-square test. But the proportion ~ 0.05% is very small. Will it cause a problem? E.g. the effect size can be small I suspect.
I applied a survey consisting of 12 questions to 120 people and each questions include 4 nominal categories; I want to make comparison of people's answers according to their socio-demographic characteristics such as their educational level or socioeconomic status. All of my comparison criteria consist of nominal categories and number of categories change between 3 to 6. I cannot drop or combine comparisons; categories in other words number of categories are fixed.    My question is when I compare the frequency distribution of people's answers based on their education levels (for example) through chi-squared test, I got a warning that says (for instance) 7 cells have less than count five; minimum expected count is; 43.  I got this warning for almost my all questions - demographic characteristics comparisons.    Shall I underestimate this warning and use my test results or is there a different test I should use? If I should use a different test, which one?
Prove that for every odd integer n, there is a group... Prove that for every odd integer n, there is a group with exactly n elements of order 2.  I am new to abstract algebra and proofs in general. Proving existence is difficult for me, but I will show what I have.  $$n = 2k + 1$$ $$k \in \mathbb{Z^+}$$  $$\exists G \supset H = (a_1 , a_2 , ..., a_n, e)$$  $$s.t. |a_1| = |a_2| = ... = |a_n| = 2$$  My instinct is to use Lagrange's Theorem, although I would like to try another method to prove this as we have not gone over cosets in class yet. If someone could just give me a hint of how to start, it would be very beneficial. Thank you. I have a problem with using "too" in a sentence. In what situation would you use "too" instead of "so" or "very"? Please include an example to illustrate the difference. How can I reference the script element that loaded the javascript that is currently running?  Here's the situation. I have a "master" script being loaded high in the page, first thing under the HEAD tag.  &lt;!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"&gt; &lt;html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"&gt; &lt;head&gt; &lt;script type="text/javascript" src="scripts.js"&gt;&lt;/script&gt;   There is a script in "scripts.js" which needs to be able to do on-demand loading of other scripts. The normal method doesn't quite work for me because I need to add new scripts without referencing the HEAD tag, because the HEAD element hasn't finished rendering:  document.getElementsByTagName('head')[0].appendChild(v);   What I want to do is reference the script element that loaded the current script so that I can then append my new dynamically loaded script tags into the DOM after it.  &lt;script type="text/javascript" src="scripts.js"&gt;&lt;/script&gt; loaded by scripts.js--&gt;&lt;script type="text/javascript" src="new_script1.js"&gt;&lt;/script&gt; loaded by scripts.js --&gt;&lt;script type="text/javascript" src="new_script2.js"&gt;&lt;/script&gt; It seems that most people who use the ignored tags feature end up with a fairly long list of those tags.  Although , it is sometimes necessary to add new ignored tags manually (typically when adding some-tag*).  It would be a lot easier if the add box was above the long list of ignored tags rather than below. I suppose the same would apply to the favorites tags for consistency's sake. What is the reason for the observation that across the board fields in physics are generally governed by second order (partial) differential equations?    If someone on the street would flat out ask me that question, then I'd probably mumble something about physicists wanting to be able to use the Lagrangian approach. And to allow a positive rotation and translation invariant energy term, which allows for local propagation, you need something like $-\phi\Delta\phi$.  I assume the answer goes in this direction, but I can't really justify why more complex terms in the Lagrangian are not allowed or why higher orders are a physical problem. Even if these require more initial data, I don't see the a priori problem.  Furthermore you could come up with quantities in the spirit of $F\wedge F$ and $F \wedge *F$ and okay yes... maybe any made up scalar just doesn't describe physics or misses valuable symmetries. On there other hand in the whole renormalization business, they seem to be allowed to use lots and lots of terms in their Lagrangians. And if I understand correctly, supersymmetry theory is basically a method of introducing new Lagrangian densities too.   Do we know the limit for making up these objects? What is the fundamental justification for order two? The set of all finite subsets of the natural numbers is countable I am looking for a closed form (ideally expressed as elementary functions) of the function $\exp_n(x) = \sum_{k=0}^n x^k / k!$.  I am already aware of expressing it in terms of the gamma function. Background / Motivation When counting combinations of objects with generating functions, it is useful to be able to express the partial sum $1 + x + \cdots + x^n$ as $\frac{1-x^{n+1}}{1-x}$. For example, to count the number of ways to pick 5 marbles from a bag of blue, red, and green marbles where we pick at most 3 blue marbles and at most 2 red marbles, we can consider the generating function $f(x) = (1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)$. By using the partial sum identity, we can express it as $f(x) = \left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^3}{1-x}\right)\left(\frac{1}{1-x}\right)$.  Simplify, express as simpler product of series, and find the coefficient of the $x^5$ term. I want to be able to do the same for a generating function in the form $g(x) = \exp_{n_1}(x)^{p_1} \exp_{n_2}(x)^{p_2} \cdots \exp_{n_j}(x)^{p_j}$ The easiest way to extract the coefficient of a given term $x^p / p!$ would be to use a similar closed form expression for $\exp_n(x)$ and a similar technique to $f$. Attempted Solutions Differential equation Recall that the way to prove the identity $1+x+x^2+\cdots+x^n = \frac{1-x^{n+1}}{1-x}$ is to define $S = 1 + x + x^2 + \cdots + x^n$ and notice that: $S - Sx = 1 - x^{n+1}$. Likewise, notice that $y(x) = \exp_n(x)$ satisfies $y - y' = x^n/n!$.  Via SAGE, the solution is $y(x) = \frac{c+\Gamma(n+1,x)}{n!}e^x$.  Our initial condition $y(0) = 0$ so $c=0$.  By , $\Gamma(n+1,x) = n! e^{-x} \exp_n(x)$ so we have $y(x) = \exp_n(x)$. Recurrence Relation Notice that $\exp_n(x) = \exp_{n-1}(x) + x^n/n!$. Using the unilateral Z-Transform and related properties, we find that $\mathcal{Z}[\exp_n(x)] = (z e^{x/z})/(z-1)$. Therefore, $\exp_n(x) = \mathcal{Z}^{-1}\left[(z e^{x/z})/(z-1)\right] = \frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz$. $(z^n e^{x/z})/(z-1)$ has two singularities: $z = 1$ and $z = 0$.  The point $z = 1$ is a pole of order one with residue $e^x$.  To find the residue at $z = 0$ consider the product $z^n e^{x/z} (-1/(1-z)) = -z^n \left( \sum_{m=0}^\infty x^m z^{-m} / m! \right) \left( \sum_{j=0}^\infty x^j \right)$.  The coefficient of the $z^{-1}$ term is given when $n - m + j = -1$. The residue of the point $z=0$ is then $-\sum_{m,j} x^m / m! = -\sum_{m=n+1}^\infty x^m / m!$. Let $C$ by the positively oriented unit circle centered at the origin.  By Cauchy's Residue Theorem, $\frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz = \frac{1}{2 \pi i} 2 \pi i \left(e^x - \sum_{m=n+1}^\infty x^m / m!\right) = \exp_n(x)$. Finite Calculus I've tried to evaluate the sum using finite calculus, but can't seem to make much progress. Draw a bivariate normal distribution in TikZ For a small class I want to draw a bivariate normal distribution and show where the means from the two variables meet in the space. Has anyone done this in TikZ? Divergence of $\frac{ \hat {\bf r}}{r^2} \equiv \frac{{\bf r}}{r^3}$, what is the 'paradox'? Get selected element's outer HTML Is an empty set equal to another empty set? I have a definition that claims that two sets are equal A = B, if and only if:  $\forall x ( x \in A \leftrightarrow x \in B)$  An empty set contains no elements. If I define the sets:  A = $\emptyset$  B = $\emptyset$  then can I conclude that A $\neq$ B? Since they contain no elements? Or would I have to show that an elements exists in one but not the other? To me it seems that these are not equal, but I'm struggling to wrap my head around this.  Thanks, Paul Zip folder without including path to folder and require only one argument I'm trying to make a zip function that'll do all the steps I typically do when zipping a file or folder. Say my current directory is ~, and I want to zip a folder named Folder1, whose file path is ~/Folder2/Folder1.  Issuing  cd ~/Folder2 &amp;&amp; zip -r Folder1 Folder1 &amp;&amp; cd -   does exactly what I want. Namely, it zips so that the output Folder1.zip does not contain folders that lead up to the folder I'm zipping (namely, /home/user/Folder2), it zips recursively, and it takes in the first argument of zip with the exact name as the file or folder I'm zipping (Folder1.zip).  So I want to make a function in my .zshrc something akin to  coolerzip() { cd /path/to/file &amp;&amp; zip -r file file &amp;&amp; cd - }   where it need only take in one argument, which in this case, it will be ~/Folder2/Folder1. Then Folder1 will be the name for both files in coolerzip, and I will cd right into ~/Folder2 so Folder1 lies in my current directory before zipping. Is there any way I can achieve this so that coolerzip only requires one argument rather than three? How to prove polynomials with degree $n$ does not form a vector space? This is one of my linear algebra problems:     Prove that polynomials of degree $n$ does not (The professor made these words bold intentionally) form a vector space.   From what I read, the set of polynomials of degree $n$ should be a vector space, because:   There is an "One" and a "Zero" in this set; We can find inverse for addition and multiplication from this set; It follows all the axioms of addition. It follows all the axioms of scalar multiplication.   Then can someone give me some hints to prove it does not form a vector space? Is there a problem with Beamer in the latest MiKTeX update? This is about MiKTeX Revision 2.9.6360.  Immediately after the latest MiKTeX update, my Beamer files stopped compiling.  When i tried to compile a file that had been working fine I got a Package installation popup saying:  tex\latex\beamer\beamerbasemodes.sty  Is missing.  It is part of the following package  beamer   So i clicked to have it installed.  All that achieved is that i got an error message  ! LaTeX error: File 'beamerbasemodes.sty' not found   I repeated this a few times with minor changes to the file, and then went to the MiKTeX Package Manager to see what parts of Beamer are installed.  It said none were.  Either the update had removed Beamer, or the attempt to install beamerbasemodes on the fly had done it.  So I reinstalled it by hand.  In fact i did this several times in various ways.  It made one difference.  Sometimes the problem package was identified as  translator.sty  In case it would help, I ran Update FNDB and restarted my machine.  Each time I installed Beamer, the Package Manager then reported that Beamer was installed.    But when I tried to compile the file, the same things happened all over.   And when i went back to the MiKTeX Package Manager it reported Beamer is not installed.   Is Beamer now out of date?  Or is this a problem on the new update?  Is there a simple fix i can use? Black screen after installation of Ubuntu 18.04 If $G$ is abelian, then the set of all $g \in G$ such that $g = g^{-1}$ is a subgroup of $G$ Prove that if $G$ is abelian then the set $H$ of all elements of $G$ that are their own inverses is a subgroup of $G$.   Naturally in an abelian group, $ab = ba$ for $a, b \in G$, however I'm not sure how to show the set elements that are their own inverses is a subgroup of $G$ using arbitrary elements. Distinct Sylow P subgroups I am trying to include long URLs in a bibliography, but those often overflow into the margin of the document.  I have seen a number of questions about breaking URLs in bibliography, but after trying every solution, I have not found anything that completely solves my problem. There are apparently a number of ways to allow LaTeX to break line at different places in URLs. But in my case, even though LaTeX breaks line at some places, I still get overflow for some URLs.  Here is the code I am using:  \usepackage[hyphens]{url} \usepackage{hyperref} \hypersetup{breaklinks=true}  \urlstyle{same} \usepackage{cite}   and with these bibliography entries:  @misc(webrtc,     organization={W3C WebRTC Working Group},     title={WebRTC 1.0: Real-time Communication Between Browsers},     howpublished={\url{http://www.w3.org/TR/2012/WD-webrtc-20120821/}} )  @misc(android_javascriptinterface,     organization={Android Open Source project},     title={Android WebView addJavascriptInterface reference},     howpublished={\url{http://developer.android.com/reference/android/webkit/WebView.html#addJavascriptInterface(java.lang.Object, java.lang.String)}} )   I'm getting this, for example:     The entry about Android overflows in the border: my column should end where you can read "browsers." on the first line or "http:" on the third line.  Edit: when trying to create a minimal example for my problem, I noticed that this only occurs when using the IEICE template file. You can find it here: , but if you do not want to have to extract the archive and locate the correct file, I have put it there also:   Here is my minimal example:  .tex file:  \documentclass[paper]{ieice}  \usepackage[hyphens]{url} \usepackage[hidelinks]{hyperref} \hypersetup{breaklinks=true} \urlstyle{same} \usepackage{cite}  \title{Test document} \begin{document}  this\cite{upnp_spec} is a test\cite{android_javascriptinterface}  \bibliographystyle{plain} \bibliography{refs_min}  \end{document}   and refs_min.bib file:  @misc(android_javascriptinterface,     organization={Android Open Source project},     title={Android WebView addJavascriptInterface reference},     howpublished={\url{http://developer.android.com/reference/android/webkit/WebView.html#addJavascriptInterface(java.lang.Object, java.lang.String)}} )  @misc(upnp_spec,     organization = {UPnP Forum},     title = {UPnP Device Architecture 1.1},     howpublished = {\url{http://upnp.org/specs/arch/UPnP-arch-DeviceArchitecture-v1.1.pdf}} )   Is there any way I can fix this ? As a result, I am able to be understanding and accommodating to the people that I work with, including clients, which I believe is an important asset for a professional to have.   I wanted to make sure it is absolutely known that I mean to acknowledge clients as a class of people I can work with. Is this the right way to include it with commas? Why do we notate the greatest common divisor of $a$ and $b$ as $(a,b)$? In my textbook on elementary number theory from a class last year, as well as elsewhere through my academic experience and even posts here, I often see the greatest common divisor notated as $(a,b)$ (where this represents the greatest common divisor of $a,b$).  My question is, why is such an unusual notation used? I say it's unusual (if not a bit absurd), for a couple of reasons:   It is unintuitive. It does not grant the reader an immediate knowledge of what it means. In particular, I can easily see it being misinterpreted as an ordered pair of points, and not as a function (which it basically is and I touch on in a second) or representing a single value. Good notation would be immediately understood, no? One should not struggle with parsing through the context or meaning. A clearer notation is often used, and just three letters away. You could certainly think of $(a,b)$ as an ordered pair, if we consider it a function. I often see the alternate $gcd(a,b)$: a function, which takes in two inputs ($a$ and $b$), and outputs their greatest common divisor (or factor, if you prefer). Merely adding the three letters makes the notation clearer by clarifying it is  a function, and using a relatively well-accepted acronym. (Or, at least, I often hear it abbreviated "g.c.d." in school.)   I've mulled this over for a while and I can't think of much without borrowing inspiration from other "unusual" notations. For example:   Common understanding in context: consider the $\log(x)$ function. Depending on context, without a base stated as seen here, it can be presumed to mean whatever is most "appropriate." It could be base $e$ (common in mathematics), base $10$ (common in engineering), or base $2$ (common in computer science). The ambiguity is essentially resolved by context in other words - personally, not a fan of that (I like the explicit stating of bases for full clarity), but if such a motivation holds elsewhere, it might stand to reason that $(a,b)$ could just be taken as a shorthand and "understood" in the context of number theory to simply be the greatest common divisor. That it is in fact used despite $gcd(a,b)$ being clearer and somewhat common also does seem to suggest this playing a role. Perhaps it just started that way and it kept being used. For example,  (based on a ) regarding how logarithms, roots, and exponentiation can be unintuitive from a notation perspective, and proposes an alternative, easier to understand notation. I haven't been in middle school in nearly a decade, but guessing from discussions online, that obviously hasn't really picked up too much. In that sense, perhaps it's like that for $(a,b)$: like we still notate exponentiation, etc., the same way today despite how counterintuitive it is (at least for students), perhaps it just "stuck" and proliferated?   I wasn't able to easily find any reason why online, and it bugs me a bit, so I ask:     Why do we so unintuitively notate the greatest common divisor of two numbers $a,b$ as $(a,b)$?   I of course recognize that this is somewhat moot in light of how easily it's made clear - as established above, all I have to do is append the front of the parenthetical by $gcd$ or $gcf$ to make it way better. And of course, people do that. But I'm more curious in where this ambiguous former notation arose and, in light of a better notation, why it is still used today.