Removing artifact from PDF figure in LaTeX, compiled with pdflatex I am working on a document in LaTeX, and when inserting pdf figures I get a weird black line down the right side.  This is not present in the original pdfs (I've checked).  I thought it might be some scaling issue, since several of my figures are large and I have to scale them smaller than actual size in LaTeX, but it also appears in this figure, which is actual size. My code for the figure is:  \begin{figure}[t] \includegraphics[height = 9.6cm]{"testschema"} \caption{Database Schema for test case, simulating results from a test suite} \label{fig:schematest} \end{figure}   And the output after compiling with pdflatex is:
Suppress black box at end of line Under TeX, when I write a line too long, it prints a fat black box at the end of the line.  That is probably a feature, but I dislike it.  How to turn that OFF? Is there a command-line option or something else?   I usually use LaTeX, but I have a complicated file which happens to be plain TeX.
Writing to NTFS drive in Linux If I set multiple Alarms, can I tell them apart? Complete list of Lollipop system icons shown in notification bar? Is there a complete list of these? I've seen lots of Q&amp;A about "What does x icon mean" but it would be really useful if there were a complete list for reference! If you know it, also try to include the precise reason why the proof is fallacious. To start this off, let me post the one that most people know already:    Let $a = b$.  Then $a^2 = ab$  $a^2 - b^2 = ab - b^2$  Factor to $(a-b)(a+b) = b(a-b)$  Then divide out $(a-b)$ to get $a+b = b$  Since $a = b$, then $b+b = b$  Therefore $2b = b$  Reduce to $2 = 1$   As @jan-gorzny pointed out, in this case, line 5 is wrong since $a = b$ implies $a-b = 0$, and so you can't divide out $(a-b)$. Is this a plant or a fungus? It was soft and covered in flies. My guess is fungi.  Location: South Florida. Noticed in :         is 320, shown 322. Also locked in 2013.  And then :         has a score of 179, shown 180.  has a score of 123, shown 122.  Neither of these posts got any votes recently. "Compared with" vs "Compared to"—which is used when? Is only one of them correct? Are they used in different situations? Or are they interchangeable? How align 2 adjacent divs horizontally WITHOUT float? How to convert sshfs command to fstab entry? From another SO , I understand I should create an entry in fstab to permanently mount an access to a virtual machine.  I am not Linux expert. The magic command which allows me to perform this from a terminal is:  sudo sshfs -o idmap=user -o allow_other -o nonempty jverstrynge@devjverstrynge:/home/httpd /home/jverstrynge/httpd   When I check the above SO question, I see an entry looking like this:  /media/mybook/laptop_backup /export/laptop_backup none bind 0 0   Can someone explain how to transform the CLI command into an entry in fstab? Proving that any permutation in $S_n$ can be written as a product of disjoint cycles I have attempted a proof of this, but upon looking at my notes, I feel it might be incorrect: it is noticeably simpler than the one in my notes.  Proposition: any permutation in $S_n$ can be written as a product of disjoint cycles.  Proof (attempt): $s\in S_n$. Let $\langle s\rangle$ act on $X=\{1,\cdots,n\}$. Define $x\sim x'\iff x'\in\text{Orb}(x)$ for $x,x'\in X:$ this is clearly an equivalent relation, hence the orbits partition $X$. Each orbit corresponds a cycle in $s$, so we are done.  Does this hold, or have I missed something vital? I want to find the interval of convergence of $$\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(2x)^n$$ By the root test, $\sqrt[n]{|(\frac{n}{n+1})^{n^2}(2x)^n|}=|(\frac{n}{n+1})^{n}(2x)|$  And $(\frac{n}{n+1})^n=(1-\frac{1}{n+1})^n\to\frac1e$ as $n\to\infty$  So I've got an open interval(of convergence) s.t. $|\frac1e 2x|&lt;1\implies |x|&lt;\frac e2$.  Now I need to figure out whether the series is convergent or not at the end points, $x=\frac e2, -\frac e2$.  So there are two series I need to deal with, namely, $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}e^n$ and $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(-1)^ne^n$.  Suppose $(\frac{n}{n+1})^{n^2}$ is increasing. Then $$(\frac{n}{n+1})^{n^2}&lt;(\frac{n+1}{n+2})^{(n+1)^2}\iff 1&lt;(\frac{n+1}{n+2})^{(n+1)^2}(\frac{n+1}{n})^{n^2}$$ $$\iff 1&lt;(1-\frac{1}{n+2})^{2n+1}(1+\frac{1}{n^2+2n})^{n^2}$$  But $(1-\frac{1}{n+2})^{2n+1}\to\frac{1}{e^2}$ and $(1+\frac{1}{n^2+2n})^{n^2}&lt;(1+\frac{1}{n^2})^{n^2}\to e$  So I have $1&lt;\frac1e&lt;1$, which is a contradiction. Hence $(\frac{n}{n+1})^{n^2}$ is not increasing. But I don't know if it's decreasing. (is it true that $(1+\frac{1}{n^2+2n})^{n^2}\to e$?)  If $(1+\frac{1}{n^2+2n})^{n^2}\to e$ is true, then $(\frac{n}{n+1})^{n^2}e^n$ converges to some positive number, so $\sum(\frac{n}{n+1})^{n^2}e^n$ diverges. But I don't know if it's decreasing: can't apply the alternating series convergence test(Leibniz's criterion).  So, can you help me with the two end points? Greek numbering Route SSH to host through interface Is it possible to route all outgoing requests for SSH (port 22) to mydynamicdns.afraid.org via wlan0 instead of eth0 when both are connected?  Since my IP is dynamic, I would like to use a name instead of IP. How to use grep/awk/unix to match all lines from one file in another file, even if they are duplicates I'm trying to retrieve the the matching lines from a file, however the file that has the "index" or list of matches to retrieve also has duplicate entries and I want to print out the duplicates as well.  Example:  File 1 (index file or list of strings to match)  abc tgf abc ggg aaa fff   File 2 (what I want to retrieve for each of the matches in file 1)  Fred   1   3   abc Amy    2   4   ggg Dan    5   6   tgf Mike   9   1   fff Steve  2   1   aaa   My desired result would be (order is not important, just that I match every line):  Fred   1   3   abc Dan    5   6   tgf Fred   1   3   abc Amy    2   4   ggg Steve  2   1   aaa Mike   9   1   fff   I'm familiar and have tried grep as follows, but it does not print the duplicate entries.  grep -Fwf file1.txt file2.txt  Fred   1   3   abc Amy    2   4   ggg Dan    5   6   tgf Mike   9   1   fff Steve  2   1   aaa Study the following limit: $$ \lim\limits_{ (x,y,z) \to (0,0,0) } \frac{xy-z^2}{x^2+y^2+z^2} $$   I'm not sure if this way of proving that the limit does not exist is correct so I'm this is basically for a proof verification.  Approaching $(0,0,0)$ through the plane $z=0$, then the limit becomes  $$ \lim\limits_{ (x,y) \to (0,0) } \frac{xy}{x^2+y^2} $$  Now this limit does not exist, because approaching $(0,0)$ through the line $y=0$, the limit becomes  $$ \lim\limits_{ x \to 0 } \frac{0}{x^2} =0$$  But if we approach $(0,0)$ through the line $y=x$ we have that  $$ \lim\limits_{ (x,x) \to (0,0) } \frac{x^2}{2x^2}=\lim\limits_{ (x,x) \to (0,0) } \frac{1}{2} =\frac{1}{2}$$  So the limit $ \lim\limits_{ (x,y) \to (0,0) } \frac{xy}{x^2+y^2} $ doesn't exist. This means that the limit we began with doesn't exist, correct? There are many tutorials that use a pull-up or pull-down resistor in conjunction with a switch to avoid a floating ground, e.g.        Many of these projects use a 10K resistor, merely remarking that it is a good value.  Given a a particular circuit, how do I determine the appropriate value for a pull-down resistor?  Can it be calculated, or is it best determined by experimentation? What is the most bizarre thing you have seen done with TeX I think the questions says it all! :)  By TeX, I mean any derivative of TeX as long as the code is clearly a derivative of TeX, i.e. that it could belong on TeX.SX (and not pure Lua for example).  I'll give two answers myself so you see what I mean. It seems like maybe cancelled downvote might be less...weird? If $a^{n}-1$ is prime then $a=2$ and $n$ is prime? I was doing some basic Number Theory problems and came across this problem :     Show that if $a$ and $n$ are positive integers with $n\gt 1$ and $a^{n} - 1$ is prime, then $a = 2$ and $n$ is prime   My Solution : (Sloppy)        $a^{n}-1$ = $(a-1)$ . $(a^{n-1} + a^{n-2} + ... + a + 1)$   This means that $(a-1)$ | $(a^{n}-1)$    But $(a^{n}-1)$ is prime   So , $(a-1)$ = 1 $\Rightarrow$ $a = 2$   Now , let $n$ be composite   $n = kl$ , where $1 \lt k \lt n$ and $1 \lt l \lt n$   Now , $a^{kl}-1$ = $(a^{k}-1)$ . $(a^{k.{(l-1)}} + a^{k.{(l-2)}} + ... + a^k + 1)$   This means that $2^{n} -1$ is composite    Hence , we have achieved a contradiction      My Question : Am I correct ? Conditionally enabling systemd files through Debian packaging I've created a  package which installs a service.  On our embedded devices, I want this package to automatically enable the service.  On our developer workstations, I want the developers to systemctl start foo manually (it's a heavy service, and so it just consumes resources if run all of the time on a desktop environment).  How can I go about prompting the user for their decision during the apt-get step? Is that the best solution?  Note, I've created the package using dh_make and debhelper and enabled it with:  %:     dh $@ --with=systemd  override_dh_systemd_enable:     dh_systemd_enable --name=foo foo.service