I am using latex to write my thesis report, so i use the glossary package to generate the acronym list , but i get a small space between acronym and it description ,here is my minimal code:  \documentclass{report} \usepackage{graphicx} \usepackage[colorlinks]{hyperref} \usepackage[savewrites,nopostdot,toc,acronym,symbols,nogroupskip]{glossaries}  \makeglossaries   \newacronym{CICS}{CICS}{Customer Information Control System} \newacronym{EHDM}{EHDM}{Enhanced Hierarchical Development Methodology} \newacronym{ASF}{ASF}{Algebraic Specification Formalism} \newacronym{ProCos}{ProCos}{Provably Correct Systems} \newacronym{HOL}{HOL}{Higher Order Logic} \newacronym{LOTOS}{LOTOS}{Language Of Temporal Ordering Specification} \newacronym{CCS}{CCS}{Calculus of Communicating Systems} \newacronym{CSP}{CSP}{Communicating Sequential Processes} \newacronym{RAISE}{RAISE}{Rigorous Approach to Industrial Software Engineering} \newacronym{VDM}{VDM}{Vienna Development Method }  \begin{document}     \tableofcontents     \chapter{Sample}     This is a sample document that uses the dummy glossary entries     supplied with the glossaries bundle for testing.     \gls{ASF} \gls{CICS}     Here are all the entries (including acronyms):      \gls{ASF} \gls{CICS} \gls{CSP} \gls{ProCos}     \gls{VDM} \gls{RAISE} \gls{CCS} \gls{ASF}     \gls{HOL} \gls{EHDM} \printglossary[type=acronym,style=tree]  \end{document}    and I get this output:     How can I change the space between the acronym entry and it description?? Can you help me please??
How can I vertically align entries in my acronyms' list? I use the Glossaries package to generate it. Here is MWE:  % !TeX program = xelatex % !TeX encoding = utf-8 % !TeX spellcheck = ru_RU  \documentclass[12pt, a4paper]{article}  \usepackage{polyglossia} \setdefaultlanguage{russian} \setmainfont[Ligatures=TeX]{Times New Roman}  \usepackage[xindy={glsnumbers=false}, nonumberlist, nopostdot, nogroupskip]{glossaries} %\setglossarystyle{super} % uncomment for style "super" \makeglossaries  \begin{document}  \newacronym{bam}{ДАП}{Двунаправленная Ассоциативная Память} \newacronym{ai}{ИИ}{Искусственный Интеллект} \newacronym{am}{АП}{Ассоциативная Память} \newacronym{ann}{ИНС}{Искусственная Нейронная Сеть} \newacronym{an}{ИН}{Искусственный Нейрон} \newacronym{it}{ИТ}{Информационные Технологии} \glsaddall  \printglossaries  \end{document}   With the default style I have the following:  As you can see, the long names are not vertically aligned. What I want can be achieved with the    \setglossarystyle{super}   But it this case, short names lose the BOLDness, some small horizontal indent appears and the spacing between lines becomes smaller:  where RED is some extra indent, which I don't need. Thank you for your answers!
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I want to calculate the posterior distribution given a gamma-prior and a poisson likelihood. The task is from a textbook and I just have the solutions (without a walkthrough). Please find all given information in the following:  Given information:  $m$ countries each with a population of $n_i$; $y_i$ is the number of people having a disease; $y_i$ is a poisson random variable with mean $\frac{n_i*\lambda}{100}$  The prior distribution for $\lambda$: $\lambda \sim Gamma(a_0,b_0)$  Task:  Show that the posterior distribution is also a gamma distribution with $a_m = a_0 + \sum_{i = 1}^{m}y_i$ and $b_m = b_0 + \sum_{i = 1}^{m} \frac{n_i}{100}$  My solution:  $Posterior \propto Likelihood \times Prior$  $p(\theta|y) \propto \prod_{i=1}^{m}(\theta^{y_i}e^{-\theta})\times \theta^{a_0-1}e^{-b_0\theta}$  $p(\theta|y) \propto \theta^{\sum_{i = 1}^{m}y_i}e^{-m\theta}\times \theta^{a_0-1}e^{-b_0\theta}$  $p(\theta|y) \propto \theta^{\sum_{i = 1}^{m}y_i+a_0-1}e^{-\theta(m+b_0)}$  $a_m = a_0 + \sum_{i = 1}^{m}y_i$  $b_m = b_0 + m \neq b_0 + \sum_{i = 1}^{m} \frac{n_i}{100}$  I think I am doing a mistake, because I don't know how to incorporate the information that the mean of $y_i$ is given as $\frac{n_i*\lambda}{100}$. Or do I just have to transform $b_m$ somehow? Can someone may give me a hint?   Thank you already in advance!  Edit 1: Considering first advices  $Likelihood: p(y | \theta) \propto \prod_{i=1}^{m}(\frac{n_i\lambda}{100})^{y_i}e^{-\frac{n_i\lambda}{100}}$  $\propto \prod_{i = 1}^{m} ((\frac{n_i}{100})^{y_i})\lambda^{\sum_{i=1}^{m}y_i}e^{-\lambda\sum_{i=1}^{m}\frac{n_i}{100}}$  $\propto \lambda^{\sum_{i=1}^{m}y_i}e^{-\lambda\sum_{i=1}^{m}\frac{n_i}{100}}$  Is the transformation of $\prod_{i=1}^{m}$ I did from the first to the second step correct? In most sentences that use the word lunch, I found this word does rarely take the definite article the. For example:     I have lunch in three hours.      Shall we go out for lunch?      Let's have some lunch.      I'll take my lunch to school.   But now I wonder if this word really cannot take the definite article, and if it can, in what cases do I use the definite article and in what cases not?  And also, if the word lunch cannot take the definite article, why cannot it have it? If you provide 0 as the dayValue in Date.setFullYear you get the last day of the previous month:  d = new Date(); d.setFullYear(2008, 11, 0); //  Sun Nov 30 2008   There is reference to this behaviour at . Is this a reliable cross-browser feature or should I look at alternative methods? Every Christmas, Harry receives a mediocre gift from the Dursleys. In Harry Potter and the Chamber of Secrets he gets a toothpick.     At that moment, Hedwig swooped into the room, carrying a very small package in her beak.     “Hello,” said Harry happily as she landed on his bed. “Are you speaking to me again?”     She nibbled his ear in an affectionate sort of way, which was a far better present than the one that    she had brought him, which turned out to be from the Dursleys. They had sent Harry a toothpick    and a note telling him to find out whether he’d be able to stay at Hogwarts for the summer    vacation, too.    They hate owls, but obviously use them to contact Harry at school? Title is the question itself.  How can I show below without L’Hospital’s Rule  $$ \lim_{x\to \infty} ( 1 + x )^\frac{1}{x} = 1 $$ Find all functions $f(x)$ such that $f\left(x^2+f(y)\right)$=$y+(f(x))^2$ How to split larger files into smaller parts? Number of integer solutions of $x^2 + y^2 = k$ I'm looking for some help disproving an answer provided on a  I posted about computing the number of double square combinations for a given integer.    The original question is from the Facebook Hacker Cup       Source: Facebook Hacker Cup   Qualification Round 2011      A double-square number is an integer $X$   which can be expressed as the sum of   two perfect squares. For example, 10   is a double-square because $10 = 3^2 + 1^2$.   Given $X$, how can we determine the   number of ways in which it can be   written as the sum of two squares? For   example, $10$ can only be written as $3^2 + 1^2$   (we don't count $1^2 + 3^2$ as being different). On the other hand,   $25$ can be written as $5^2 + 0^2$ or as   $4^2 + 3^2$.      You need to solve this problem for $0 \leq X \leq 2,147,483,647$.      Examples:      $10 \Rightarrow 1$      $25 \Rightarrow 2$      $3 \Rightarrow 0$      $0 \Rightarrow 1$      $1 \Rightarrow 1$   In response to my original question about optimizing this for F#, I got the following response which I'm unable to confirm solves the given problem correctly.       Source:       Again, the number of integer solutions   of $x^2 + y^2 = k$ is four times the   number of prime divisors of $k$ which   are equal to $1 \bmod 4$.      Knowing this, writing a program which   gives the number of solutions is easy:   compute prime numbers up to $46341$ once   and for all.      Given $k$, compute the prime divisors of   $k$ by using the above list (test up to   $\sqrt{k}$). Count the ones which are   equal to $1 \bmod 4$, and sum. Multiply   answer by $4$.   When I go through this algorithm for $25$, I get $8$ which is not correct.          For each prime factor (pf) of $25$ ($5$, $5$ are prime factors of $25$)         if pf % $4$ = $1$ (true for both $5$'s), add $1$ to count      return $4$ * count (count would be $2$ here).          So for $25$, this would be $8$ At  I've read that being $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ by using  we get $\sin'(x)=\sum_{n=0}^{\infty}(2n+1) \frac{(-1)^n}{(2n+1)!}x^{2n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=\cos(x)$ but the page  says that if $f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$ then inside the radius of convergence of the power series we have $f'(x)=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1}$ so shouldn't it be $\sin'(x)=\sum_{n=1}^{\infty}(2n+1)\frac{(-1)^n}{(2n+1)!}x^{2n+1-1}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$?  Now I understand that this last result has a missing term due to the index of the series being $1$ and not $0$ but being the $1$ mandatory because of the above theorem and since trying to shift index doesn't help (we get $\sin'(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2}$) how can the result that ProofWiki shows be rigorously justified?   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How to avoid this ?  \documentclass[]{standalone} \usepackage{tikz,xparse} \usetikzlibrary{calc}  \tikzset{pics/.cd,     % Marque croix     Cross/.style args={#1 and #2}{%         code = {%         \draw[#2,rotate=45,scale=1.4] (0,#1 pt) -- (0,-#1 pt) ;         \draw[#2,rotate=-45,scale=1.4] (0,#1 pt) -- (0,-#1 pt) ;         }     },     Cross/.default={2 and black} }  \NewDocumentCommand{\Segment}{mO{}}{%     \begin{scope}[#2]     \path    (0,0) coordinate (sA) pic {Cross={1.5 and gray}}         -- (#1,0) coordinate (sB) pic {Cross={1.5 and gray}} ;     \draw (sA)--(sB) ; % Can't reuse (sB)     \end{scope} }   \begin{document}     \begin{tikzpicture}     \Segment{1.5}     \end{tikzpicture} \end{document} For $a,b,c&gt;0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$  how can I find the limit of sequence above?  Provide me a hint or full solution. thanks ^^