Name for a singular noun that is used like a plural?
Are collective nouns always plural, or are certain ones singular?
Bonferroni adjustment in SPSS - what does it do? I'm probably being a bit simplistic here, but from what I understand the Bonferroni post hoc test is simply when we adjust the alpha to account for inflated error when several post hoc tests are conducted. You divide the alpha by number of tests. So what exactly does SPSS do when we click the button for Bonferroni?   Shouldn't the dividing of alpha be done by us in interpreting the result, e.g. if we do 10 post hoc tests our alpha criterion should be .005 so if the p value is .012 then it is not significant, but SPSS hasn't done anything there, we have just changed our interpretation. So what exactly is it adjusting? List of Figures + Beamer? I want to have a List of Figures in my Beamer presentation.  I spent some time searching the web for a solution and only found desperate people telling others that it's not possible, i.e.  or .  I also made some experiments on my own, i.e. "abuse" tocloft to do the job. However, even \usepackage{tocloft} causes pdflatex to fail with ! LaTeX Error: \l@section undefined. (and lying to tocloft by defining the command as empty beforehand doesn't to any good either).  I can't really see why it should be that difficult to make a List of Figures in beamer, since \tableofcontents works flawlessly.   I'm not afraid of doing some coding on my own, but I'm not deep enough into TeX to really understand the magic behind the "List of Figures", "Table of Contents" and alike.  Is anyone out there having suggestions or links that might help me figure this out?  \documentclass{beamer}  % uncomment the following line % to get the error message % `! LaTeX Error: \l@section undefined.`  %\usepackage{tocloft}   % uncomment the following lines INSTEAD % to lie to tocloft % and get the next error message  %\makeatletter %\newcommand\l@section{} %\makeatother %\usepackage{tocloft}  \begin{document}  \begin{frame}{A frame} \begin{figure} \mbox{A} \caption{An 'A'} \end{figure} \end{frame}  \begin{frame}{List of Figures} \listoffigures \end{frame}  \end{document}   EDIT: Solution  Starting from Marco Daniels answer to my question, I modified his code to reflect my needs. Here is what I came up with. Please feel free to comment, improve and reuse this.  \RequirePackage{ifthen}  \makeatletter  \AtEndDocument{%   \clearpage   \beamer@tempcount=\c@page\advance\beamer@tempcount by -1%   \if@filesw   \newwrite\tf@lof   \immediate\openout\tf@lof\jobname.lof\relax   \newwrite\tf@lot   \immediate\openout\tf@lot\jobname.lot\relax   \fi }  \long\def\beamer@makecaption#1#2#3#4{%   \def\insertcaptionname{\csname#1name\endcsname}%   \def\insertcaptionnumber{\csname the#1\endcsname}%   \edef\insertframenumber{\theframenumber}%   \ifthenelse{\equal{#3}{\empty}}{%       \def\insertlistcaption{#2}%   }{%     \def\insertlistcaption{#3}%   }   \def\insertsource{#4}%   \def\insertcaption{#2}%   \ifthenelse{\equal{#1}{figure}}{%       \addtocontents{lof}{\protect\listoffigureformat{\insertcaptionnumber}{\insertlistcaption}{\insertframenumber}{\insertsource}}{}{}%   }{}   \ifthenelse{\equal{#1}{table}}{%       \addtocontents{lot}{\protect\listoftableformat{\insertcaptionnumber}{\insertlistcaption}{\insertframenumber}}{}{}%   }{}   \nobreak\vskip\abovecaptionskip\nobreak   \sbox\@tempboxa{\usebeamertemplate**{caption}}%   \ifdim \wd\@tempboxa &gt;\hsize   \usebeamertemplate**{caption}\par   \else   \global \@minipagefalse   \hb@xt@\hsize{\hfil\box\@tempboxa\hfil}%   \fi   \nobreak\vskip\belowcaptionskip\nobreak% }  \def\listoffigureformat#1#2#3#4{%   \makebox[2ex][r]{#1}%   \hspace{1ex}%   {\usebeamercolor[fg]{bibliography entry author} #2}%   \ifthenelse{\equal{#4}{\empty}}{}{ -- #4}%   \dotfill%   \makebox[2ex][r]{#3}\par% } \def\listoffigures{%   \setlength{\leftskip}{3ex}   \setlength{\parindent}{-3ex}   \@starttoc{lof}% } \def\listoftableformat#1#2#3{\makebox[2ex][r]{#1}\hspace{1ex}#2\dotfill\makebox[2ex][r]{#3}\par} \def\listoftables{%   \setlength{\leftskip}{3ex}   \setlength{\parindent}{-3ex}   \@starttoc{lot}% }  \long\def\@caption#1[#2]#3{   \par\nobreak   \begingroup     \@parboxrestore     \if@minipage       \@setminipage     \fi     \beamer@makecaption{#1}{\ignorespaces #3}{\ignorespaces #3}{\ignorespaces #2}\par\nobreak     \endgroup}  \makeatother A Basic Question about Gravity, Inertia or Momentum or something along those lines $$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$  Now, $\log(n!) = \Theta (n\log(n))$ so I think we could write,   $$\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) $$  $$\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) $$  $$\log\left(2n\right) - \log\left(n\right) = \log(2)$$  Is this legitimate? I feel this might be wrong since, the $\Theta$ notation conceals a constant factor. Prove that $\phi(n) \geq \sqrt{n}/2$ I've already done a lot of searching (in particular: ) to try to prove this statement:      $f: A \to B$ is injective if and only if it has a left inverse. [I'm going to also assume $A$ and $B$ are nonempty.]   My attempt so far:     $\Leftarrow$ Suppose $f$ has a left inverse and $f(a_1) = f(a_2)$ for    $a_1, a_2 \in A$. Then there is a function $g: B \to A$ such that   $g\left(f(a_1)\right) = a_1$ and $g\left(f(a_2)\right) = a_2$. Since   $g$ is well-defined, it follows that $g\left(f(a_1)\right) =  g\left(f(a_2)\right)$, and hence $a_1 = a_2$.         $\Rightarrow$Now   suppose $f$ is injective. Then $f(a_1) = f(a_2) \implies a_1 = a_2$.   Furthermore, since $A$ is nonempty, there is an element $x \in A$.   Define $g: B \to A$ by  \begin{equation*} g(b) = \begin{cases}  f^{-1}(b), &amp; b \in \text{Im}(f) \\ x, &amp; b \in A \setminus  \text{Im}(f)\text{.} \end{cases} \end{equation*} This mapping is   well-defined, since if $f$ is injective, $f^{-1}(b)$ is unique.   My main question: does this imply that if $f: A \to B$ is injective that any mapping $g: \text{Im}(f) \to A$ is bijective? Otherwise, I don't understand why we can even use $f^{-1}$ on an element. [Please read the link above for more details - in proof 1.] Pgfplots: how to fill bounded area under a curve using addplot and fill? How to evaluate environment variables into a string in Python? Retrieve the position (X,Y) of an HTML element relative to the browser window Two column float with [t] using the cas-dc template I am using the cas-dc template for Elsevier. When trying to use \begin{figure*}[t], except I get the error 'The module 'cas/fig' does not have a key called 'cas/fig/t'. Other templates I use work fine. I have included \usepackage{stfloats}. I don't see anything mentioned that strictly forbids it, unless I missed this. Anyone know why? Using processing.runandload in QGIS 3 Get attention to a new answer to an old question It happened to me that I found a solution to an older question. This question was commented by many people with comments like "very interesting, important, ..." such that I thought that my answer has also some value. My answer was the only one. It was upvoted and accepted by the asker. However, I assume, that very few of the "commentators" saw my answer, because the system does not inform automatically about new answers.   I think it would have been helpful for me and the commentators when they would get informed. I could achieve this by adding my own comment to their comments, because this triggers a system message. My question is: would this action be considered admissible or not? Are there some other possibilities? putExtra treeMap returns HashMap cannot be cast to TreeMap android Proof conclusion to: Let m,n $\in$ $\mathbb{N}$ $\backslash$ $\{0\}$ . Prove that $s |\tau (m^{s}) - \tau(n^{s})$, where $\tau$ is a divisor function.   Hello, so I have a tiny question about the summarization of my proof.  I got to the point where I defined my $\tau$ divisor function and got them both $\tau(m^s)$= $\prod$ something +1 and $\tau(n^{s})$ = $\prod$ something +1.    I've noticed that the last terms of the polynomial are both one. Why is it with the +1 part, I can conclude with that:  s| $\tau(m^s)$ - $\tau(n^s)$.   The line for why that is, isn't quite clear. I've got a small piece of code that is parsing an index value to determine a cell input into Excel.  It's got me thinking...  What's the difference between   xlsSheet.Write("C" + rowIndex.ToString(), null, title);   and  xlsSheet.Write(string.Format("C{0}", rowIndex), null, title);   Is one "better" than the other? And why? Progress bar with dynamic text & text color update A question I'm struggling with is to prove or disprove that the automorphism group of a finite cyclic group must be cyclic. I think the statement is false but have not been able to come up with a counterexample.  Thank you for any assistance you can provide. Why is voltage added when batteries are connected in series but not in parallel? As I understand, the different voltage between two object is caused by the total different charged between two object. Thus, the higher extra electron between two opposite charged object, the higher voltage between them, isn't it? So if i have two battery or capacitor connect in parallel, the negative connect to negative, the positive connect to the positive. Caused the charge from each plate to be double. So that's as my expected, it must be double voltage, but ironically why it's out of my expected and the reality is vice-versa? I have this table definition as part of a invoice  \begin{tabularx}{\textwidth}[ht]{lXrr}%   It worked fine for years until I have a very short word in the X column for the first time. Now my table is very small in width and centered horizontally on the page.  In my expectation the X column should fill up all space until the width of the table reaches the width of the page except the margins.  I have read about workarounds specifying something like 1.2\textwidth, but I don't like that, because the table is a template included via \include, so it would affect all my invoices.  I also tried \linewidth, \hsize and \columnwidth, but everything results in the same layout.  Is there a way to make the only X column expand so that it uses all available space?    Complete included table definition (Rechnungstabelle.tex):  \newcommand{\Rechnungstabelle}[4]{% vielen Dank für Ihren Auftrag. \noindent \begin{tabularx}{\textwidth}[ht]{lXrr}% \toprule% \bf{Anzahl} &amp; \bf{Bezeichnung} &amp; \bf{Einzelpreis} &amp; \bf{Gesamtpreis} \\% \midrule% #1% \bottomrule% &amp; \bf{Rechnungsbetrag}  &amp; &amp; \bf{\EUR{#2}} \\% \ifthenelse{\equal{#3}{}}{}{&amp; Enthaltene MwSt. 19\% &amp; &amp; \EUR{#3} \\}% \ifthenelse{\equal{#4}{}}{}{&amp; * Enthaltene MwSt. 7\% &amp; &amp; \EUR{#4} \\}% \end{tabularx}% }   Changing the table header row to  \bf{Anzahl} &amp; \bf{Bezeichnung} \hspace{\textwidth} &amp; \bf{Einzelpreis} &amp; \bf{Gesamtpreis} \\%   is a workaround, but I don't think that would be good TeX style.  Minimum example for reproducing:  \documentclass[ fontsize=12pt,                  paper=a4,                 foldmarks=TBMPL,                 fromalign=center,                 ngerman                 ]{scrlttr2} \usepackage{babel} \usepackage[utf8]{inputenx} \usepackage[right]{eurosym} \usepackage{tabularx} \usepackage{ltablex} \usepackage{booktabs} \usepackage{ifthen}  \input{Rechnungstabelle} \begin{document} \begin{letter}  \Rechnungstabelle{     1,0 h &amp; Do something &amp; \EUR{99,00} &amp; \EUR{99,00} \\      }     {99,00}     {}{} \end{letter} \end{document} Is it possible to make signing up gracefully fail over to logging in if the SE OpenID account exists? For most OpenID options, it doesn't much matter if you pick the "Sign up" versus the "Log in" tab on the sign in page.  Whether you have an account on the site or not, it just does the right thing. Not so if you are using the Stack Exchange OpenID service:    (Bonus bug: the reCAPTCHA box is a bit too wide for the expanded "Sign up using Stack Exchange" container.)  Since we already have collected an email and password (well, two copies of the password), could we not check to see if they match and automagically log the user into their SE OpenID?  If not, the error needs to be reworded somehow to explain that while the email might have been used to create a Stack Exchange OpenID, that doesn't mean they can't log in with that OpenID to create an account.  Counterintuitively, they need to use the  even if they've never had an account there before.  The  link adds to the confusion since it sends an email that leads off with:     Someone requested an account recovery on The Workplace Stack Exchange for jericson@stackexchange.com, but we don't have an account on this site that matches this email address.   (For people with accounts on other sites and/or Stack Exchange OpenIDs, this paragraph is followed by other instructions that would probably work if followed. By that time I imagine many users are at wit's end.)  To simplify: If a user already has a Stack Exchange OpenID (likely because they have an account on another site) they should be able to use the  to sign up for an account or there should be clear instructions on what to do next if they fail.