Omit page number in list of symbos using nomencl I'm typesetting a thesis using a custom class based on book class and to meet the university requirements, all pages in frontmatter must not show the page numbers but must be counted, starting at the cover. I was able to change everything, except for the list of symbols and list of abbreviations and acronyms, that were made with the nomencl package.  I tried the solution in this post , but using that one, the entry in the toc does not show any page number. I need just to hide the page number of the lists. How can this be accomplished?
suppress page numbers in document's front matter I have a problem and I want that someone tell me how I can resolve that.  I want that the number of page in my report start in the page of introduction but in reality it starts in the didecas and liste of contents. I want that these pages have not number. This is my code  \documentclass[12pt,a4paper]{report} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[french]{babel} \usepackage{color} \usepackage{xcolor} %les couleurs \definecolor{myblue}{rgb}{0.2,0.4,1} \usepackage{layout} \usepackage[top=2 cm, bottom=2 cm, left=2 cm, right=2 cm]{geometry} \usepackage{setspace} \usepackage[pdftex]{graphicx} \usepackage{array} %\documentclass[landscape]{scrartcl} \usepackage{geometry} \usepackage{multirow}  \usepackage[justification=centering]{caption}  \usepackage{pgfplots}   \usepackage{fancyhdr} \fancyhf{} % clear all header and footers   \begin{document} \pagestyle{empty} \include{title-content} \color{black} \doublespacing \include{dedicaces-content} %\include{remerciement}   \tableofcontents  \listoffigures \listoftables     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \pagestyle{plain}  \pagenumbering{arabic} \chapter{Introduction générale}
Correct way of mounting a Windows share I mounted a Windows share to my Ubuntu 11.10 system via adding the following line to the /etc/fstab:  //serveraddr/sharedfolder /mnt/foldername/ smbfs credentials=/home/myname/.smbpasswd 0 0  using the .  But there's a problem with the permissions. The owner of the mount folder is root. I can not create any folders or files in the mounted point is I don't become the root. Even I can not change the ownership to my default user via   sudo chown -R myname .  I googled and saw that it might stem from the fact that Windows file formats (fat32/ntfs) can not save ownership. But when I connect to the same location via samba as   smb://serveraddr/sharedfolder  by providing my credentials, I have the privileges to write without being the root.  How can I make it to let me write to the mounted folder? Why does an open interval NOT have measure zero? Assume  $$G_1=\mathbb Z_5 \times \mathbb Z_{5^2}\times \mathbb Z_{5^3}\times \mathbb Z_{5^4} \times\ldots$$ $$G_2= \mathbb Z_{5^2}\times \mathbb Z_{5^3}\times \mathbb Z_{5^4} \times \ldots$$ How do I prove $G_1$ and $G_2$ aren't isomorphic? I asked  this question here  and I received tree answer, but the answer of Ludolila will be complete when  $G_1$ and $G_2$ are not isomorphic.    Thanks in advance Power Supply Capacity Formula Given that I am building a home computer from scratch and I have several components (MB, CPU, HDD, Video, Sound, etc) which have specific power requirements, is there a formula for me to calculate the size/capacity of the Power Supply I will need? I want to be able to expand the setup later on also... The question says to solve this equation: $(z+1)^5 = z^5$  I did. Just want to find out if I did it properly and if my run-around logic makes sense.  First I begin my writing the equations as:  $$ (z+1)^5 = z^5$$ $$ \mathbf{e}^{5 \mathbf{Log}(z+1)} = \mathbf{e}^{5 \mathbf{Log}(z)} $$ So $$  \mathbf{Log}(z+1) = \ln|z+1| + \mathbf{Arg}(z+1)i $$   $$  \mathbf{Log}(z) = \ln|z| + \mathbf{Arg}(z)i $$   Now, because the natural logarithm is one-to-one, I write:  $$ \ln |z| = \ln |z+1| \Rightarrow |z| = |z+1|$$   So assign $ z = a +bi$   So that $|z| = \sqrt{a^2 +b^2} = |z+1| = \sqrt{(a+1)^2 +b^2} \Rightarrow a^2 +b^2 = (a+1)^2 +b^2 \Rightarrow a^2 = (a+1)^2 \Rightarrow a = -\frac12 $  So, $z = -\frac12 + bi$ and $z+1 = \frac12 + bi$ for some $b \in \mathbb R$  Now to find $b$  $$ \mathbf{Arg}(z+1) = \mathbf{Arg}(z)$$ $$ \tan^{-1} \frac{b}{\frac12} = \pi - \tan^{-1}\frac{b}{-\frac12}$$  I have a feeling this last part isn't quite right, so I just want to find out if I'm approaching this question properly?  Ultimately, I get $ z = -\frac12$ which upon inspection...is wrong... How can I use pathauto to generate URLs for deeply nested vocabularies? I am generating breadcrumbs from URL arguments as described .  I have a vocabulary that looks like this:  Arts-> Performing-> Dancing-> Ballet  I want my url to look like this: arts/performing/dancing/ballet  Using the pathauto module, I can do this for taxonomy terms:  [term:parent]/[term:name]   Which, on the Ballet page, gives me the following; not what I want:  dancing/ballet    If I use parents instead of parent,  ([term:parents]/[term:name]); this is close but not quite:  arts-performing-dancing/ballet   How can I get from arts-performing-dancing to arts/performing/dancing? Age of universe? Well relativity teaches us that time interval between two events is a frame dependent quantity, then how can we say that our universe is 13.8 billion years old? Should it not depend on who is asking this question? Line breaking with hyperref href with an underline in itemize I have a simple itemized list like the following MWE:  \documentclass[10pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{graphicx} \usepackage{hyperref} \usepackage{lmodern} \usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry} \usepackage{lipsum} \begin{document} \begin{itemize} \item \href{http://google.com}{[ID1]: \underline{\lipsum[1]}} \end{itemize} \end{document}   This unfortunately doesn't break the block of text as expected, as shown below:    How can I fix this item so it will line break properly? Lost supplementary user groups for GUI login session Ubuntu 16.04.5  Very weird situation - after Xorg crash, now every time when I login into Unity, my user account doesn't have it's system groups like adm, sudo, plugdev, fuse etc. But when I login on virtual console (Alt+F1) or over SSH - all my additional groups are there, as reported by id command.  Neither of /etc/group* or /etc/gshadow* files have changed, my user is still listed there in many system groups.  It looks like some systemd session parameters for GUI logins or lightdm cache files got screwed. But I have no idea where to start looking..  A lightdm log:  $ journalctl -au lightdm  -- Reboot -- systemd[1]: Starting Light Display Manager... systemd[1]: Started Light Display Manager. lightdm[3223]: pam_kwallet5(lightdm-greeter:setcred): (null): pam_sm_setcred lightdm[3223]: pam_unix(lightdm-greeter:session): session opened for user lightdm by (uid=0) lightdm[3345]: pam_succeed_if(lightdm:auth): requirement "user ingroup nopasswdlogin" not met by user "chm" lightdm[4867]: pam_ecryptfs: Passphrase file wrapped lightdm[3345]: pam_kwallet5(lightdm:auth): (null): pam_sm_authenticate lightdm[3345]: pam_kwallet5(lightdm:setcred): pam_kwallet5: pam_sm_setcred lightdm[3345]: pam_unix(lightdm:session): session opened for user chm by (uid=0) lightdm[3046]: Error writing to daemon: Bad file descriptor lightdm[3046]: Error writing to daemon: Bad file descriptor lightdm[3046]: Error writing to daemon: Bad file descriptor lightdm[3046]: Failed to write utmpx: Permission denied Why is propagation of uncertainties quadratic rather than linear? How to install with Wubi when it says "Could not retrieve the required disk image files"? Showing margins/frames in beamer slides Handle .bib in subfiles I have a problem very similar to the question being asked in  . My .bib is references in main.tex. When compiling main, everything works like a charm. Working on the subfiles, I have two issues though (which are most likely related):   I can't access the bib with auto-completion When compiling the subfile, it doesn't insert the citations.   I tried including  \bibliographystyle{plainnat} \bibliography{mybib}   in the subfiles, which also worked smoothly. Obviously, the bibs are included in every chapter of the document when compiling main.tex, then. I tried setting a global boolean in main.tex:  \newboolean{printBibInSubfiles} \setboolean{printBibInSubfiles}{false}    and in the subfiles:  \ifthenelse{\boolean{printBibInSubfiles}}    {}    {\bibliographystyle{plainnat}     \bibliography{../betterbib.bib}}    For some reason, then setting the boolean to false, the "bibliography"-header was printed in the subfiles as well as for the main file, but there was no output of citations under any of the "bibliography"-headers. Furthermore, compiling only the subfiles didn't work correctly.  My document structure is the following:  main.tex:  \begin[document] ... \subfile{sub/dataprocessing} ... \bibliographystyle{plainnat} \bibliography{mybib} \end{document}   the subfiles:  \documentclass[../main.tex]{subfiles} \begin{document} ... \cite{Random2014} ... \end{document}   After all this explanation: the only thing I want is to access the .bib from the subfiles without having to manually uncommenting something when compiling the main file. Any takers?  PS: Obviously, I also tried the solution provided in "" but couldn't get it to work. I don't know whether that would solve my problem anyway, since I don't have the \printbibliography-issue of the thread opener. Is trace invariant under cyclic permutation with rectangular matrices? I'm working with trace of matrices. Trace is defined for square matrix and there are some useful rule, i.e. $\text{tr}(AB) = \text{tr}(BA)$, with $A$ and $B$ square, and more in general trace is invariant under cyclic permutation. I was wondering if the formula $\text{tr}(AB) = \text{tr}(BA)$ holds even if $A$ and $B$ are rectangular, namely $A$ is $n$-by-$m$ and $B$ is $m$-by-$n$. I figured out that if one completes the involved matrices to be square by adding zero entries in the right places, then the formula still works... but I want to be sure about this! Can I enable the Windows 7 Start menu in Windows 10? I've upgraded from Windows 7 to Windows 10 and now I only find this new slower Start menu with all the tiles in it   (picturesource: )  Can I enable the older Windows 7 one again which suites my needs more? If you click on the Stack Exchange menu or the inbox menu or the achievements/reputation menu, you can then move the mouse between those three menus without any additional clicks.  This behaviour does not appear to be applying to the Winterbash menu (or the "help" menu for that matter, but that's on the other side of the bar).  Could we get this sorted, since otherwise the behaviour is a little inconsistent, with extra clicking on a menu bar where in all other circumstances clicking would close the menu! I'm reading Baby Rudin and exercise 28 of chapter 2 reads "Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable." I don't understand why the set must be closed in order for this to be true.  My proof is as follows, and I don't make any reference to the closedness of the set:  Let $E$ be a set in a separable metric space.  Let $B$ be a countable base of that metric space.  Let $P$ be the set of condensation points of $E$.  I claim that $P$ is perfect and $E-P$ is at most countable.  If $p$ is a limit point of $P$, then every neighborhood of $p$ includes at least one point $q$ of P.  For a neighborhood $N_{p}$ of $p$, choose a neighborhood $N_{q} \subset N_{p}$ of $q$.  Since $q$ is a condensation point of $E$, uncountably many points of $E$ are in $N_{q}$, and thus in $N_{p}$.  So $p\in P$ and $P$ is closed.  If $p \in P$, then $p$ is a condensation point of $E$, so every neighborhood $N_{p}$ of $p$ must include uncountably many points of $E$.  If $p$ is not a limit point of $P$, then there exists a neighborhood $N_{p}$ of $p$ such that $N_p$ includes no condensation points of $E$ except for $p$.  This means that every point $q\in N_{p}-\{p\}$ must have a neighborhood $N_{q}$ that does not have uncountably many points of $E$.  For each $b_{k}\in B \cap \mathscr{P}(N_{p})$, choose such a neighborhood $N_{q_{k}} \supset b_{k}$ of some point $q\in N_{p}-\{p\}$.  Now $\{N_{q_{k}}\}$ covers $N_{p}$, and $\bigcup_{k=1}^{\infty}N_{q_{k}}\supset N_{p}$ has at most countably many points of $E$, for it is a union of countably many sets that have at most countably many points of $E$.  This contradicts the assumption that $p\in P$.  So $P$ is perfect.  Let $S=E-P$. There are no condensation points in $S$.  So there exists an open cover $\{N_s\}$, where each $N_s$ is a neighborhood of $s \in S$ with at most countably many elements of $S$. For each $b_{k}\in B \cap \mathscr{P}(S)$, choose such a neighborhood $N_{s_{k}} \supset b_{k}$ of some point $s\in S$.  Then we have a countable subcover $\{N_{p_{k}}\}$ of $S$. As $N_{p_{k}} \cap S$ is at most countable for all $k$, $S$ must be at most countable.  So what am I doing wrong?  Or is Rudin's assumption of closedness extraneous? Norm is sup of inner products (proof). Let $V$ be a vector space with an inner product $\langle.,. \rangle$ and associated norm $|| . ||$ Then:    Could I have a proof of this fact? Date and time panel is greyed out, how do I restore it? Date and time is showing in the login screen (unity-greeter) but not when I login to my session.  When I go to date and time settings the panel is all grey :    Any idea how to recover from this ? How would I implement a binary search using just an array?