Keep Column which is not in aggregate function in group by statement
Fetch the row which has the Max value for a column
How can I compile, run and decompile C# code in Ubuntu terminal? ToC numbering problem How can I figure out what's slowly eating my HD space? What should one look for when buying a used film camera? I'm planning to buy a used Nikon F5.   I have no experience with a film camera and I want to know what all should one be careful of before buying a used film camera and more specifically a Nikon F5. "Screaming" noise while steaming milk? List of Ubuntu Versions with Corresponding Linux Kernel Version Is there a list of Ubuntu versions with default corresponding Linux kernel versions somewhere?  I would specifically like to know the most recent version of Ubuntu that still used Linux Kernel 2.x. Example of $f,g: [0,1]\to[0,1]$ and Riemann-integrable, but $g\circ f$ is not? Sum of two squares modulo p Let $\mathbb{F}_{27}$ denote the finite field of size 27. For each $\alpha\in\mathbb{F}_{27}$ , define $ A_{\alpha}$ $=$ { $ 1, 1+\alpha , 1+\alpha+{\alpha}^2 , 1+\alpha+{\alpha}^2+{\alpha}^3 , .... $} . Then  prove that  the number of  $\alpha\in\mathbb{F}_{27}$ such that $|A_\alpha|$=26 equals 12.  $ 0\in A_\alpha $ if and only if $\alpha\neq 0$ For the first proof , the multiplicative group of the field $\mathbb{F}_{27}$ is cyclic and its order is 26  , so there are 12 generators in that group.   Now  I need  to  show  that  $ 1+\alpha+{\alpha}^2+.. {\alpha}^{25} $ =0  , where order of $\alpha$=26 . Since   $\alpha ^{26}$ =1 ($\alpha$-1)(1+$\alpha$+${\alpha}^2$+...+${\alpha}^{25}$)=0 , since $\alpha\neq 1$  , So $ 1+\alpha+...+{\alpha}^{25} $ =0 My question is , can I write these last two  lines?  I mean are those two lines alright to write? If those two lines are invalid then I need some help to show  $ 1+\alpha+{\alpha}^2+..+{\alpha}^{25} $=0  .Thanks can we have a parallel earth made of dark matter? Does 'rm .*' ever delete the parent directory? The expression .* is expanded by bash to include the current and parent directories:  $ ls -la total 2600 drwxrwxrwx   2 terdon terdon 2162688 Sep 10 16:22 . drwxr-xr-x 142 terdon terdon  491520 Sep 10 15:34 .. -rw-r--r--   1 terdon terdon       0 Sep 10 16:22 foo $ echo .* . ..   If I run rm -rf .* on my Debian using GNU bash, version 4.2.36(1)-release and rm from rm (GNU coreutils) 8.13, I get this message:  $ rm -rf .* rm: cannot remove directory: `.' rm: cannot remove directory: `..'   Is this a GNU thing or is it POSIX? Are there any *nix systems where the command above will silently delete . and ..?  Also, is this a safety feature of the shell or of the rm command itself? Package incompatibilites: etoolbox, hyperref, and bm, standalone? I am trying to transition to using the etoolbox package, but seem to have run into some difficulties and they appear related to the hyperref and bm packages which to me seem totally unrelated.  The MWE example compiles since \RemoveEToolbox is defined and shows that there is no apparent problem before etoolbox was used.    Commenting out the \def\RemoveEToolbox{} line produces two problems:   With both the \renewcommand, and the usepackage{hyperref} commented, this results in "Runaway argument? Paragraph ended before \bm@test@token was complete". Including the usepackage{hyperref} (with \renewcommand commented) results in "Undefined control  sequence. \Call@AtVeryEndDocument ...cumentHook \@undefined"   But if uncomment the \renewcommand{\bm}{#1} line (effectively doing without any \bm), then this seems to compile.  I have been using this for a while now and things appeared to have been working fine, but now that I attempt to use etoolbox I seem to run into problems.  \documentclass{standalone} \def\RemoveEToolbox{}% Shows that this works fine without {etoolbox}  \usepackage{amsmath} \usepackage{xcolor} \usepackage{bm}  \ifdefined\RemoveEToolbox     \newcommand*{\iftoggle}[3]{#2}% Default to true value of "if" \else     \usepackage{etoolbox}     \newtoggle{paper}     \toggletrue{paper}      %\renewcommand{\bm}[1]{#1}% Why can't I have a normal \bm with {etoolbox}? \fi  % \usepackage{hyperref}  \begin{document}  This document is intended for  \iftoggle{paper}{paper}{electronic} distribution.  \bm{\textcolor{blue}{Solve $x^2-1=0$}}  \begin{align*}     a &amp;= b\\     \iftoggle{paper}{         c &amp;= d\\     }{}     e &amp;= f\\ \end{align*} \end{document} System program problem detected? create mesh then add vertices to it in python using Ajax with 3 consecutive drop down lists depending on eachother Does linear ordering need the Axiom of Choice? Beamer: how to justify the whole text? How can I justify the text in the whole document in beamer without using \usepackage{ragged2e} and \justifying at the beginning of each frame?  \documentclass[8pt,handout,t]{beamer} \usepackage{pifont} %used inside itemize environment \item[\ding{226}] \usepackage{lipsum} \begin{document}  \begin{frame} \begin{columns}[T]  \begin{column}{0.3\textwidth} \includegraphics[width=\columnwidth]{example-image-1x1} \end{column} \begin{column}{0.65\textwidth} \begin{itemize} \item[\ding{226}] \lipsum[2] \item[\ding{226}] \lipsum[3] \end{itemize} \end{column}  \end{columns} \end{frame}  \begin{frame}  \lipsum[3]  \begin{itemize}    \item[\ding{226}] \lipsum[4]  \end{itemize} \end{frame} \begin{frame}   \begin{columns}[t] \begin{column}{0.4\textwidth} \lipsum[3] \end{column} \begin{column}{0.4\textwidth} \lipsum[5] \end{column} \end{columns} \end{frame} \end{document}     Edit: I had already checked  out but it does not work for both the columns and the itemize environment. Let $r$ be a relation on $A \times A$ such that $(a,b) r (c,d) \iff ad = bc.$ How can I show that this relation is transitive, ie. $(a, b)r(c,d)$ and $(c,d)r(e, f) \implies (a,b)r(e,f)$?  I tried to say that $(a,b) r (c,d)$ means that $c=ka$ and $d=kb,$ $k$ a coefficient, such that $ad = bc \iff akb = bka,$ and going on from there, but I'm not sure this is valid for all values of $a, b, c,$ and $d$. How can I show the transitivity of the relation? Find asymptotic of recurrence sequence Given a sequence $x_1=\frac{1}{2}$, $x_{n+1}=x_n-x_n^2$. It's easy to see that it limits to $0$.  The question is: is there exists an $\alpha$ such, that $\lim\limits_{n\to\infty}n^\alpha x_n\neq0$.  I tried to find explicit formula for $x_n$, but did not succeed. I stucked at that point: if $a_1=(\frac{1}{4})^2$, $a_n=(a_{n-1}+\frac{1}{4})^2$, then $x_n=-a_{n-2}+\frac{1}{4}$ for $n\ge3$.   Thanks for any ideas and help. How do I get the distance meter value between two geometries in PostGIS? I have a simple question about calculating distances in PostGIS.  I would like to get the distance between two geometries. I am using this sid : 4269 in meters what i am doing now is this : ST_Distance((a.geom,b.geom)) FROM ... but I am getting result in degrees. I think that I should work with geography but how can I cast a geom to a geography? I tried with (a.geom::geography) but i am getting an error.  What can I do to get my result in meters?   Thank You