I input a file inp.txt using its path relative to the directory containing the main.tex file and where I will run pdflatex:  \documentclass{article} \begin{document}     \input{../inputs/inp.txt} \end{document}   Is there a way to specify the absolute path of inp.txt, e.g. ~/inputs/inp.txt?  Since I am looking for a general solution to include \includegraphics, \input and \include files I try to use TEXINPUTS and modify my script:  \documentclass{article} \begin{document}     \input{inp.txt} \end{document}   When I run  TEXINPUTS=~/inputs pdflatex main.tex   it turns out I have changed my working directory producing the error message:  ! I can't find file `filepath.tex'.   Adding multiple paths TEXINPUTS=~/inputs:. produces other problems.  Is there a simple and concise solution?
Sometimes it is convenient to use an absolute path to include a preamble. To do so, I use the following "hack":  \documentclass[a4paper, 12pt]{article} \newcommand{\folder}{/path/to/folder} \input{\folder/preamble}  \begin{document} \end{document}   Now my question is, whether there is a more elegant way to do this, especially a way which is compatible with one of the flatten scripts from  ?
[See  for more on this topic, as well as better illustrations of the final product.]  I have a document with a lot of short excerpts from another book scattered through it.  Putting frames around them looks a little too 'clean'. How can I get a 'torn page' type effect, to make it look as if the excerpt has been physically torn out of a book?  (I've occasionally seen this done in printed books, although I can't find an example right now.)  I'm not wedded to a very specific look here --  gives plenty of examples. But just to give this some frame of reference, here are some of the nicer ones:          (The strong colouration in this second example is a probably a little too distracting for actual use, although it might be interesting as an option. It's more the edge effect I'm interested in.)    Edit (again): I tried typesetting my text according to the updated versions of the answers below. Here are the results:    Or with interpolated shading: Why does the checkbox stay checked when reloading the page? Closure of a Ball is Contained in Closed Ball (usual metric) Let $\overline{B_d (x, r)}$ be the closure of an open ball, and let $\hat B_d (x, r)$ be a closed ball. Show that in any metric space $(X, d)$, $\overline{B_d (x, r)} \subseteq \hat B_d (x, r)$ for all $x \in X$ and $r &gt; 0$.    I used the definition of closure: $\overline{B}$ = $B \bigcup B'$. If $x \in \overline{B}$, then $x \in B$ or $x \in \overline{B}$, or both. If $x \in B$, then clearly $x \in \hat B$ since $B \subseteq \hat B$. But, if $x \in B'$, then $x$ is an accumulation point. Since $\hat B$ is closed, it contains all accumulation points of $B$. Thus, $x \in \hat B$.  Is this approach correct, or at least in the right direction? Should I be using the closure definition, or try another one? Thank you. Use Results of Find Command to Call vi Is there a way that I can pipe the output of this command (assuming there is only one file found) to vi.  I want to find this command, and then call it up in vi.  Thanks.  find . -name 'id.properties' How to change LCD brightness from command line (or via script)? I lock my computer a lot, and was wondering if there was a way to get the lock screen when I hit Ctrl + Alt + L. Right now when I use that key combination it turns the screen black, and I have to hit a key or click to bring up the screen where it shows the option to enter your password or switch users, as well as displaying the desktop background behind it.   Is there anyway to avoid having to hit that extra key, to bring up the screen I want? And is there anyway to keep it that way? It's really a matter of personal preference, and I don't see any reason why not to have it stay that way as my monitor eventually is turned off and/or my computer suspended. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous and differentiable function in all $\mathbb{R}$. If $f(0)=0$ and $|f'(x)|\leq |f(x)|$ for all  $x\in\mathbb{R}$, then $f\equiv 0$.  I've been trying to prove this using the Mean Value Theorem, but I can't get to the result. Can someone help? Problems enabling Bitlocker on Surface Pro with Windows 8.1 Using a brand new Surface Pro (first generation) out of the box, installed current updates, then 8.1, and any further recommended updates after 8.1 was installed.   I've installed a two other things that don't seem connected to the issue: the Cisco Anywhere VPN client and Visual Studio 2013.  I'd like to enable Bitlocker, but during the verification step, the following error is displayed, with no options to continue:     This PC deosn't support entering a BitLocker recovery password during startup. Ask your   administrator to configure Windows Recovery Environment so that you can use BitLocker.   The "administrator" in this case is me, and I don't:   ... understand why this has happened ... know how to fix the issue, as if it's really necessary to make this change, what would I do? According to , $[0,1]$ cannot be written as union of countable disjoint closed sets, is the same true about (uncountable) family of disjoint closed intervals ? Hi I have a question regarding finding the values of limit for the following question.  Let $a, b \in \mathbb R$. Find the limit $$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$ Prove $\limsup_{n \rightarrow \infty} a_n\geq1$ if $\lim_{n \rightarrow \infty} a_na_{n+1}=1$ Let $\{ a_n \}_{n=1}^{\infty}$ be a bounded sequence s.t. $a_n&gt;0,\ \forall n \in \mathbb{N}$. Suppose $\lim_{n \rightarrow \infty} a_na_{n+1}=1$, prove $\limsup_{n \rightarrow \infty} a_n\geq1$.   I tried to prove by contradicition using subsequence, but couldn't work it out.  Any help appreciated. Recording claimable expenses in GnuCash Passing functions to setTimeout in a loop: always the last value? QGIS area calculation differs when on the fly CRS transformation enabled When I open QGIS, add the layer, and calculate the areas of the shapefile via field calculator I get a different area than when I open QGIS and check "Enable on the fly CRS transformation" and calculate the area. This is despite making sure that the project and the layer have the same Coordinate system (same EPSG number). What am I doing wrong?   I have a shapefile with area calculations made with ArcGIS (not be me, the data was handed to me and I have no clue for which CRS the area was calculated with ArcGIS). The shapefile layer CRS is EPSG: 21781 (Switzerland). In QGIS, if I do not change the OTF settings and leave the project CRS as EPSG:4326 (WGS84) I get the same value as the ArcGIS area value. However, if I change the OTF before adding the layer to EPSG: 21781 I get different area values. As I understand this suggests that ArcGIS Area was calculated with the CRS EPSG: 4326.  First workflow:   open QGIS  project CRS: EPSG 4326  add layer  project CRS adapts automatically and is EPSG 21781 now  calculate $area with field calculator    Second workflow:   open QGIS  project CRS: EPSG 4326  Turn on OTF, set project CRS to EPSG 21781  add layer  calculate $area with field calculator    Step 5 of the first and second workflow DO NOT produce the same area. Explanation of method for showing that $\frac{0}{0}$ is undefined How to know if Mac OS X machine is being remotely viewed? Currently I have no way to know if my machine is being remotely viewed by our tech support, and would prefer to know, given the massive privacy breach that would entail should this be happening. I am running Mac OS X Snow Leopard.  Running  ps -A | grep Remote   I know this service is running:  /System/Library/CoreServices/RemoteManagement/ARDAgent.app/Contents/MacOS/ARDAgent   My knowledge of administration of Macs on a network is limited, so I don't know what tools allow this. Is there a way in photoshop to convert k values to alpha values? Basically, I would like to make it so that the lighter a pixel is (the more white it has) the more transparent it becomes. Black pixels would be opaque, white pixels would be transparent.  I currently use the slider in blending options on the first tab to get rid of the white and make it transparent but find it works slightly better if I use the magic wand tool with delete.  I figure that there is some blending option out there but there are so many and I'm not sure which to use to achieve my goal.  (Using Photoshop CS6) I am trying to make a simple pie chart with pgf-pie, but for a reason unknown to me the labels I use don't display correctly.   Here is the chunk of code I've copied from the manual into my .tex file :  \begin{tikzpicture}     \pie[color ={ cyan!10 , cyan!40, cyan!70,  cyan!100}, explode=0.1] {10/A, 20/B, 30/C, 40/D} \end{tikzpicture}   Here is a screenshot of what I get :    And here is a picture of what I'd like to obtain :   Does anybody knows why I get these numbers and how to get rid of them ?  edit : Here is a minimal working example   \documentclass{article} \usepackage[frenchb]{babel} \usepackage{pgf-pie}  \begin{document} \begin{tikzpicture}     \pie[color ={ cyan!10 , cyan!40, cyan!70,  cyan!100}, explode=0.1]{10/A, 20/B, 30/C, 40/D} \end{tikzpicture} \end{document} What is the "Install this third-party software" box in the installer and what third-party software is installed? If $P(A \cup B \cup C) = 1$, $P(B) = 2P(A) $, $P(C) = 3P(A) $, $P(A \cap B) = P(A \cap C) = P(B \cap C) $, then $P(A) \le \frac14$ We have ($P$ is probability): $P(A \cup B \cup C) = 1$ ; $P(B) = 2P(A) $ ; $P(C) = 3P(A) $ and $P(A \cap B) = P(A \cap C) = P(B \cap C) $. Prove that $P(A) \le \frac{1}{4} $. Well, I tried with the fact that $ 1 = P(A \cup B \cup C) = 6P(A) - 3P(A \cap B) + P(A \cap B \cap C) $ but I got stuck... Could anyone help me, please?