Is the term PoWs (POWs) inconsistent with typical acronym pluralization? Prompted by recent remembrance day celebrations, it seems to me that the term PoWs is inconsistent with pluralization conventions.  Runs Batted In --&gt; RBI Prisoners of War--&gt; PoW (shouldn't this be the plural?)  I read that laser appears to have become a noun unto itself and is thus pluralized as . Is PoWs an accepted noun or is it an oddly pluralized acronym?   Other acronym pluralization questions that don't address PoW(s) directly:
Pluralizing abbreviations where the noun is not the last word For abbreviations - both acronyms and initialisms - where the last word in the abbreviation is the noun, I'm accustomed to adding an 's' with no apostrophe (e.g. ATMs), as described in the answer to this question:   But how do you pluralize abbreviations where the noun is not the last word? E.g. 'Point of Presence' (POP)  I can see how 'POPs' could read 'Point of Presences', when the intention is 'Points of Presence.'
How to prove that $\hat\sigma^2$ has $\chi^2_{n-p}$ distribution (linear regression) Consider the linear regression model:   $$Y_i=r(x_i)+\varepsilon_i\equiv\sum_{j = 1}^p x_{ij} \beta _j + \varepsilon _i,\quad i=1,\ldots,n.$$  where $x_1,\ldots,x_n\in \mathbb{R}^p$ are fixed, $E(\varepsilon_i)=0$,  $\operatorname{Var}(\varepsilon_i)=\sigma^2$. Denote $Y=(Y_1,\ldots,Y_n)^T$, $\beta=(\beta_1,\ldots,\beta_p)^T$, $X=(x_{ij})_{n\times p}$. As is known, $\hat \beta  = \arg\min\limits_{\beta  \in \mathbb{R}^p} (Y - X\beta)^T (Y - X\beta)=(X^TX)^{-1}X^TY$ if the matrix $X^TX$ is invertible, and so an estimator of $r(x)$ at $x=(x_1,\ldots,x_p)\in\mathbb{R}^p$ is given by $\hat r_n(x)=x^T\hat\beta$.  Let $$\hat{\sigma}^2 = \frac{1}{\sigma^2}\sum_{i = 1}^n (Y_i - {\hat r}_n (x_i))^2.$$ I am stucking the problem: $\hat\sigma^2$ has $\chi^2_{n-p}$ distribution. How to prove the statement? Java has a convenient split method:  String str = "The quick brown fox"; String[] results = str.split(" ");   Is there an easy way to do this in C++? Windows update doesn't progress beyond 0 KB 0% I have Windows&nbsp;7 64-bit and the  isn't working properly. What can I do? Is there any way to call the parent version of an overridden method? (C# .NET) In the code below I tried in two ways to access the parent version of methodTwo, but the result was always 2. Is there any way to get the 1 result from a ChildClass instance without modifying these two classes?  class ParentClass {     public int methodOne()     {         return methodTwo();     }      virtual public int methodTwo()     {         return 1;     } }  class ChildClass : ParentClass {     override public int methodTwo()     {         return 2;     } }  class Program {     static void Main(string[] args)     {         var a = new ChildClass();         Console.WriteLine("a.methodOne(): " + a.methodOne());         Console.WriteLine("a.methodTwo(): " + a.methodTwo());         Console.WriteLine("((ParentClass)a).methodTwo(): "          + ((ParentClass)a).methodTwo());         Console.ReadLine();     } }   Update ChrisW posted this:     From outside the class, I don't know   any easy way; but, perhaps, I don't   know what happens if you try   reflection: use the Type.GetMethod   method to find the MethodInfo   associated with the method in the   ParentClass, and then call   MethodInfo.Invoke   That answer was deleted. I'm wondering if that hack could work, just for curiosity. find the number of injections/surjections I've been banging my head on this problem for some time now, and could really use help. Bear in mind I'm not very good at this sort of thing and am struggling to get by in class.  Problem:     Given $ \alpha : A \to B       \;\;\;\;\;   |A|=p \;\;\;\;|B|=q $      Find:         Number of Injections: $A \to B$   Number of Surjections: $ A \to B$    Prove that ∃ a bijection $A \to B$       $\Rightarrow$ p = q   find number of bijections: $A \to B$         I apologize for dropping such a big problem on you all, but I just can't get anywhere with this. Help?  EDIT: These answers are really helpful and I believe I've got down the number of Injections, but  we haven't gone through permutations or combinatorics in class so my understanding of it is really hamstrung. I'm unclear on the Stirling theorem as discussed here: , specifically the part where inclusion-exclusion comes into play. I'm not sure how to apply that, since right now I'm letting p = 4 and q = 3, so I don't know where inclusion-exclusion would be used (if at all!) since I don't know how elements would be missing. Ubuntu Detects 2 monitors when I only have one Is "a total of 10 payments" singular or plural? A total of 10 payments were made.   OR     A total of 10 payments was made.   Which is correct? Or can both be correct? $\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx$ Trying to compute   I was facing: \begin{align}J=\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx\end{align}  I want to prove that $\displaystyle J=0$, or equivalently, that,  \begin{align}\int_0^1 \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx=-\dfrac{1}{2}\text{G}\end{align}  $\text{G}$ being the Catalan constant.  Read carefully please.   I know, using so-called Feynman's trick, how to prove this. I would like to obtain a proof, using only integration by parts and change of variable in simple integrals (that is, no multiple integrals) I don't know, if, under these restrictions, such computation is possible.  NB: You're probably wondering what is the link between : $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$ and $J$.  The link is, for $x\in\left[0;\frac{\pi}{2}\right]$, $(\sin{x}+\cos{x}+\sqrt{\sin{2x}})(\sin{x}+\cos{x}-\sqrt{\sin{2x}})=1$ and $\sin(2x)=2\sin x\cos x$ Can an electron be in two places at the same time? So I've been reading a bit and watching some videos about the double slit experiment, and therefore the wave particle duality; I've also read this "implies" that a particle can be in two places at the same time. Can you clarify this, and how could that be possible? I'm looking for the title of an anthology of children's science fiction stories from the 1950s. The best of the bunch involved a Martian who visited earth, and realized we'd blown ourselves up in a nuclear war and had fallen into savagery. The Martian is a vegetarian who subsists on biscuits made by a machine he brought with him, and communicates by using one of his tentacles to feed messages direction to/from the brain of the person he's talking to.  Inadvertently left behind by his expedition, he initially befriended some children, and eventually managed to rebuild civilization. Presently he dies, but when another expedition comes a generation later, they find the civilization he rebuilt.  there were other stories, but that was the best one.   If I can't find the name of the book it was in, I'd be happy to settle for the name of the author and the title of the story. iptables drop all incoming ICMP requests except from one IP Currently, I have something like:  iptables -A INPUT -p ICMP --icmp-type 8 -j DROP iptables -A INPUT -s x.x.x.x -p ICMP --icmp-type 8 -j ACCEPT   However, when I run the second command, it looks as if iptables just stops.  I have to break out of it to get back to terminal.  Perhaps I am doing it all wrong, but some insight would be helpful. Citing a range of papers using numeric keys as in \cite{a, b, c} -> [1-3] How can I cite a range of papers, the output is a range of numbers, instead of a list of numbers? In other words, when I type  ... some dummy text here is due to me \cite{me1, me2, me3, me4} ...   I want the output to be     ... some dummy text here is due to me [3-6] ...   instead of     ... some dummy text here is due to me [3.4.5.6] ...   (I use LaTeX, and am currently using the amsart class and whatever it includes. But I am open to other suggestions.) How can I see flags I cast from all sites? This morning I flagged something on one of the sites where I have fairly low rep. Now I can't remember which site it was. I've checked my flag history on about half of them, even editing the URL for the site where I have no helpful flags, and I can't find this flag. I've been to my network profile and looked at the Activity tab, but it only shows public things like asking, answering, commenting, editing, and reviewing.  Is there a way to see my flagging history across sites? Can we have such a feature, if it's not possible? Is there a way to forget node names between TikZ pictures? I have a document with several TikZ pictures in it.  I generally use simple names for the nodes such as:  \node (A) at .....  \node (B) at .....   and so forth.  Because of this I have name clashes between the various pictures.  That's generally not a problem.  However I find myself in a position where I may or may not have defined a certain node within a given picture.  So I use the following snippet to test whether the node has been defined:  \makeatletter \long\def\ifnodedefined#1#2#3{%%   \@ifundefined{pgf@sh@ns@#1}{#3}{#2}} \makeatother   Unfortunately this fails if I've already created a node by the same name in another picture.  Is there a way to scope the definitions of nodes or is there a way to undefine nodes once you no longer need them?  P.S.  I'm aware that I could embed each picture in a standalone file and import it using \includegraphics.  Unfortunately, this approach will not work for the document I'm creating.  MWE  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc}  \makeatletter \long\def\ifnodedefined#1#2#3{%%   \@ifundefined{pgf@sh@ns@#1}{#3}{#2}} \makeatother  \setlength{\parindent}{0pt} \pagestyle{empty} \begin{document}   There's no node \textbf{D} here, so I draw a triangle:\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1); %  \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}     There is a node \textbf{D} here, so I draw a square:\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1);   \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}      There isn't supposed to be a node \textbf{D} here.  I'm expecting a triangle, but get a square.\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1); %  \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}  \end{document}   produces: "INSERT IGNORE" vs "INSERT ... ON DUPLICATE KEY UPDATE" While executing an INSERT statement with many rows, I want to skip duplicate entries that would otherwise cause failure. After some research, my options appear to be the use of either:  ON DUPLICATE KEY UPDATE which implies an unnecessary update at some cost, or INSERT IGNORE implies an invitation for other kinds of failure to slip in unannounced.  Am I right in these assumptions? What's the best way to simply skip the rows that might cause duplicates and just continue on to the other rows? Why is $\infty \cdot 0$ not clearly equal to $0$? I did a bit of math at school and it seems like an easy one - what am I missing?  $$n\times m = \underbrace{n+n+\cdots +n}_{m\text{ times}}$$  $$\quad n\times 0 = \underbrace{0 + 0 + \cdots+ 0}_{n\text{ times}} = 0$$  (i.e add $0$ to $0$ as many times as you like, result is $0$)  So I thought an infinite number of $0$'s cannot be anything but $0$? But someone claims different but couldn't offer a reasonable explanation why. Google results seemed a bit iffy on the subject - hopefully this question will change that. Let $G$ be  a group and $H$ be a subgroup of $G$ of index $2$ .Then show that if $a,b\notin H$ then $ab\in H$.   $a,b\notin H\implies aH,bH\neq H\implies $ $aH=bH$  since $H$ has only two distinct left cosets in $G$.  $\implies a^{-1}b\in H$  How to show that $ab\in H$ from here? Many-worlds: how often is the split how many are the universes? (And how do you model this mathematically.) One problem I have found when learning calculus is that there are many different ways to denote the derivative. If $y=f(x)=x^2$, then we could write \begin{align} f'(x)&amp;=2x \\ y'&amp;=2x \\ \frac{df}{dx}(x)&amp;=2x \\ \frac{df(x)}{dx}&amp;=2x \\ \frac{d}{dx}f(x)&amp;=2x \\ \frac{dy}{dx}&amp;=2x \end{align} And this is just Lagrange and Leibniz's notations alone. What I find troubling is that they all seem to be suggesting subtly different things about what the derivative actually is. Is it a function, a limit of a quotient, or both? In the interests of keeping my post brief, I'll focus my attention on $f'(x)=2x$ and $\frac{dy}{dx}=2x$, as these seem to be the most common notations.  $$ f'(x)=2x $$  It does make sense to think of the derivative as the gradient function: $$ f'\colon x\mapsto\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} $$ In this case the limit expression is equal to $2x$, and so we can write $$ f' \colon x \mapsto 2x $$ However, this notation seems a little counter-intuitive when we consider what it means to differentiate a function with respect to a variable other than $x$. If I ask what is the derivative of $f(x)$ with respect to $\frac{x}{2}$, does this question make sense? Is it simply $f'(\frac{x}{2})$? Or do we have to express $x^2$ in terms of $\frac{x}{2}$? And how can we can express this derivative using Lagrange's notation?  $$ \frac{dy}{dx}=2x $$  There are many things which are nice about Leibniz's notation, including the fact that it is explicit which variable you are differentiating with respect to. However, in this case, it is unclear whether we are talking about a function, or something else entirely. There are other issues. Some people say they dislike the Leibniz formulation of the chain rule $$ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} $$ saying that they find it to be inaccurate. I don't really understand why this is the case. Could someone please elaborate? Is JavaScript a pass-by-reference or pass-by-value language?