What does -a and -G mean? Thanks.  sudo usermod -a -G www-data username
I want to add the Apache user (www-data) to the audio group. I've read the man page for useradd, but I'm not having any luck. I'm running xubuntu 11.10. Here's what I'm doing:  $ sudo useradd -G audio www-data useradd: user 'www-data' already exists   If I leave out the -G option, bash, prints the help info for useradd:  $ sudo useradd  audio www-data Usage: useradd [options] LOGIN Options: -b, --base-dir BASE_DIR       base directory for the home directory...   It's not clear to me from the man page what options I should use to make this work.
Underfull \hbox in bibliography I am getting lots of "Underfull \hbox" warnings that point to the *.bbl file. It looks like these messages appear only when URLs are very long, like in the below example:  @techreport{nistguidesec,   author    = {Wayne Jansen and Timothy Grance},   title     = {Guidelines on Security and Privacy in Public Cloud Computing},   month    = {January},   note     = {Draft Special Publication 800-144. Available at \url{http://csrc.nist.gov/publications/drafts/800-144/Draft-SP-800-144_cloud-computing.pdf}}, }   In the produced PDF text is correctly hyphened and links are correctly split over several lines (I am using hyperref package)  How can I avoid annoying warning messages about "underfull \hbox" pointing to bibliography? Example of a group $G$ such that $G = N_1\cdots N_n$ and $N_i \cap N_j = \{e\}$ for all $i \neq j$ but $G$ is not the internal direct product of them. I was trying to find an example of a group $G$ and normal subgroups $N_1 , \ldots , N_n$ such that $G = N_1\cdots N_n$ and $N_i \cap N_j = \{e\}$ for all $i \neq j$ and yet $G$ is not the internal direct product of  $N_1, \ldots , N_n$ .  It is easy to show that G is the internal direct product of normal subgroups $N_1 , \ldots , N_n$ iff  (i) $G = N_1\cdots N_n$  (ii) $N_i\cap (N_1 \cdots N_{i-1}\cdots N_{i+1} \cdots N_n) = \{e\}$ for $i = 1, 2, \ldots, n$.  So I guess the question boils down to produce an example of a group $G$ and its normal subgroups  $N_1 , \ldots , N_n$ such that   (1) $G = N_1\cdots N_n$  (2) $N_i \cap N_j = \{e\}$ for all $i \neq j$   but     (3) $N_i \cap (N_1 \cdots N_{i-1} N_{i+1} \cdots N_n) \neq \{e\}$ for at least one $1 \le i \le n.$ Thanks in advance for help.... TV show that starts with debris encompassing Earth $$\lim_{x \to 0} \frac{f(x)}{|x|}=1$$ Then prove that $f$ is not differentiable at $0$. I was given this assignment and I'm not sure about my proof.. it feels like I'm assuming too much and not explaining with enough detail. Is there anything I need to add to make it sufficient? Assume toward contradiction that $f$ is differentiable at $0$. Then by definition we know that $f$ is continuous at $0$. Therefore $\lim _{x\to 0}f\left(x\right)=f\left(0\right)$ We know that $\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}=1$ Therefore we can conclude $\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \left|x\right|=1\cdot 0=0$ But we can also rewrite this as $\lim _{x\to 0}\left(f\left(x\right)\right)\cdot \frac{\left|x\right|}{\left|x\right|}=f\left(0\right)\cdot 1=0$ Therefore $f(0) = 0$. Since $f$ is differentiable at $0$ we have $\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0}=\lim _{x\to 0}\frac{\left(f\left(x\right)-0\right)}{x}=\lim \:_{x\to \:0}\frac{\left(f\left(x\right)\right)}{x}\cdot \frac{\left|x\right|}{\left|x\right|}=\lim \:\:_{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\frac{\left|x\right|}{x}$ Since the one sided limits of $\frac{f\left(x\right)}{\left|x\right|}$ exist, and are both equal to 1, we have $\lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^+}\frac{\left(f\left(x\right)\right)}{x}\cdot \:\:\:\frac{x}{x}=1\cdot 1=1$ $\lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^-}\frac{\left(f\left(x\right)\right)}{-x}\cdot \:\:\:\frac{-x}{x}=1\cdot -1=-1$ Since the limit doesn't exist, we have a contradiction to the fact $f$ is differentiable at $0$. Therefore $f$ is not differentiable at 0. How to Generate surface from points Hi Guys,  I have a set of points in 3D. How can I convert this into a surface ? I have a larger, external monitor connected to my MacBook Pro and have arranged it such that the dock is in the external monitor (via System Preferences > Displays > Arrangement, by dragging the white bar to the larger screen). This works fine, except for the fact that every now and then, the dock will suddenly shift back to my MacBook screen, but when I open up Displays to change this back, the white bar is still on the external monitor. So I have to drag it to the smaller screen, and then drag it back to the external monitor.   It's a minor issue, obviously, but I've noticed that it has been happening more and more frequently and it's getting quite annoying. Perhaps there's something I've been doing unintentionally for this to happen?   I'm using Mavericks, 10.9.5. Bash 'unbound variable' error when variables are referenced in multiline comment (set -u) I have an unexpected behaviour- following snippet fail with 'unbound variable' error:  #!/bin/bash set -u &lt;&lt;EOF a=a b=$a EOF   Tested on:  GNU bash, version 4.4.12(1)-release  GNU bash, version 4.2.46(2)-release How to express a polar coordinate for a macro that only accepts space-separated values? \psElectricfield only accepts a coordinate in space-separated values as follows. For example, a charge of -0.1 unit at (-1,0) is written as [-0.1 -1 0]. I have attempted to write in PS notation but it still does not work.  \documentclass{article} \pagestyle{empty}  \usepackage{pst-electricfield,multido} \def\M{2}  \begin{document} \multido{\i=0+30}{12}{% \begin{pspicture*}(-\M,-\M)(\M,\M)     \psElectricfield     [         Q={[-0.1 -1 0][-0.1 1 0][0.1 !\i sin 2 mul \i cos 2 mul]}     ] \end{pspicture*}\newpage} \end{document}   How to express  a polar coordinate for a macro that only accepts space-separated values? Virtual functions and performance - C++ In my class design, I use abstract classes and virtual functions extensively. I had a  feeling that virtual functions affects the performance. Is this true? But I think this performance difference is not noticeable and looks like I am doing premature optimization. Right? We want to show that if $a,b\in G$ where $G$ is a finite Abelian group, we have  $\operatorname{LCM}(|a|,|b|) = |ab|$ given that $ab \neq e$.   How I approached this question was by saying let $\operatorname{LCM}(|a|,|b|) = L$. Then if we show that $L$ divides $|ab|$ and $|ab|$ divides $L$ then $|ab| = L$. Showing that $|ab|$ divides $L$ was fine. But then I am having trouble with the second part that is showing $L$ divides $|ab|$. For this, so far I have:  Consider $(ab)^{|ab|} = e = a^{|ab|}b^{|ab|} $  since we know $a \neq b^{-1}$ by assumption then we can conclude that $|a|$ divides $|ab|$ and $|b|$ divides $|ab|$. We know that $L = \frac{|a||b|}{\operatorname{gcd}(|a|,|b|)}$ from here can we conclude that $L$ divides $|ab|$? Community♦ user edit: How does it know what to edit? Show finite complement topology is, in fact, a topology XIV pages in page counts? Is it possible to make biblatex output page counts for both the regular pages and the roman-numbered pages, for example:     Cook, Theodore Andrea. The Curves of Life : Being an Account of Spiral Formations and Their Application to Growth in Nature, to Science and to Art; with Special Reference to the Manuscripts of Leonardo da Vinci. — London : Constable and Company, 1914. — xxx + 479 p.   Here there are "30 pages + 478 pages" in total. Thanks! On , the accepted answer doesn't float on top. Apart from that, the display is correct (i.e. the V mark and the green border)  If I sort by "vote", I've got the following answer order:   Jezell Marc Gravell chilltemp (Accepted) Shaun Bowe   By "oldest":   Marc Gravell Jezell chilltemp (Accepted) Shaun Bowe   and by "newest":   Shaun Bowe chilltemp (Accepted) Jezell Marc Gravell     I'm using Firefox V.3.5 : Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.9.1) Gecko/20090624 Firefox/3.5 (.NET CLR 3.5.30729)  Feel free to ask for more debug information if needed. How can I iterate, loop, process, or otherwise do something, with each element from a comma separated list of things?  This question       , and the are already lots of very good answers everywhere. However I think it would also be valuable to have a single reference point where all such alternative techniques are explained in detail, with their advantages and disadvantages discussed.  All of plain-TeX, LaTeX, LaTeX3, LuaTeX, Context, etoolbox, pgffor, etc. answers are invited. Please consider explaining, for example:   How does the command/method deal with whitespace around items? How does the command/method deal with empty/missing items? (e.g. a,b,,c) How does the command/method deal with a trailing comma? Can the command/method work of a list stored on a macro? Is each iteration evaluated globally or on a local group? (i.e., if a command is defined during an iteration, does the command survive the loop?) Do items get expanded or somehow mangled before being processed? What about lists where items are key=value or some other kind of pairs? Does the method work for lists separated with something other than a comma? Any other thing I should know before deciding to use this method? Story where the protagonist is turned off from an automated system of professional attribution and instant learning In WPF, what are the differences between the x:Name and Name attributes? Longest line using awk Simple Math Problem A bat and ball cost \$1.10.  The bat costs a dollar more then the ball.  How much does the ball cost?  If this is not the correct place to ask a question like this please tell me and I will remove it :)     The ball costs \$0.05, not \$0.10 I am trying to solve this exercise from Jech's book on set theory:  Ex. 3.6: There is a well ordering of the class of all finite sequences of ordinals such that for each $\alpha$, the set of all finite sequences in $\omega_{\alpha}$ is an initial segment and its order-type is $\omega_{\alpha}$. Definition: If $W$ is a well ordered set and $u \in W$, then the set $\{ x \in W \mid  x &lt; u \}$ is an initial segment in W  Any help?