When updating packages like docker in Ubuntu using apt the systemd file for docker is reset to it’s default. I have a modification in the systemd file for docker which points the volume to another drive. When I update docker it resets the systemd file to default. Is there a way when a package is update via apt to keep the modified systemd file the same?
Many sysv init scripts used a corresponding file in /etc/default to allow the administrator to configure it. Upstart jobs can be modified using .override files. How do I override or configure systemd units, now that systemd is the default in Ubuntu?
Firstly, I apologise that this question is specific to two polynomials. I understand that this post will not help a lot of people, and for that I am sorry. I will make it up to you all by anwsering lots more questions in the future! However, as my lecturers enjoy time off on the weekend, I am unable to ask this to anyone else who would know what the heck I'm talking about. Sorry if my notation is not widely used, I can elaborate more if required. Please excuse my poor Latex too.  Let:  $B = Y^2 - X^3 + X$, and   $F = (X^2 + Y^2)^3 - 4X^2Y^2$.  I wish to find the intersection number of $B$ and $F$ at $P = (0,0)$  i.e $I(P,F \cap B). $  However, I want to use the following proposition to do this:  If $P$ is simple on $B$ then $I(P,F \cap B) = ord_P^B(F) $.  I will omit proving that $P$ is simple on $B$ as I hope it is clear to see it is.  I know that $I(P,F \cap B) &gt; M_p(F) * M_p(B) = 1 * 4 $, where $M_P(G)$ is the multiplicity of $P$ on $G$.  Now, consider the Image of $X$ and $Y$:  $K[X,Y] \longrightarrow \frac{K[X,Y]}{Y^2 - X^3 + X} \longrightarrow O_P(Y^2 - X^3 + X)$  $X \longrightarrow x$  $Y \longrightarrow y$  Where lower case denotes the image.  Then $ord_P^{Y^2 - X^3 + X}((X^2 + Y^2)^3 - 4X^2Y^2)$  Can be found by substituting $y^2 = x^3 - x$ into $(x^2 + y^2)^3 - 4x^2y^2$  Which gives:  $ord_P^{Y^2 - X^3 + X}(x^9+3x^8 -5x^6 -4x^5 +3x^4 +3x^3)=3$  i.e $I(P,F \cap B) = 3$, a contradiction to  $I(P,F \cap B) &gt; 4$ as above.  The correct answer I know is 6, so where have I gone wrong? Upgraded to 13.10 now clock settings are all disabled and clock does not display I've upgraded to 13.10 and now I don't have the standard menu clock, which I need for work.  I checked and 'indicator-datetime' is installed.  I even uninstalled/reinstalled that applet with no luck.  Also my "clock" settings under System Preferences are all disabled.  I can't change anything.  Does anyone know how I can get the old menu clock back?  Alternatively is there another menu clock app I can download?  Edit 1  Thanks for the responses    I've restarted several times and that didn't fix the issue. I just ran the 13.10 update today.  But I ran it again a few minutes ago.  I got   about 200KB in random updates.  The issue is still present after a reboot. # apt-cache policy indicator-datetime indicator-datetime: Installed: 13.10.0+13.10.20131016.2-0ubuntu1 Candidate: 13.10.0+13.10.20131016.2-0ubuntu1 Version table: *** 13.10.0+13.10.20131016.2-0ubuntu1 0     500 http://us.archive.ubuntu.com/ubuntu/ saucy/main amd64 Packages     100 /var/lib/dpkg/status   Edit 2  I believe the issue is related to 'rarings' changing the systray somehow...   Or , although I haven't got that solution to work yet.  Edit 3  Because of a suggestion from RolandiXor I've check the binaries.  My system is missing 'libical.so.0' but it has 'libical.so.1'.    ldd /usr/lib/x86_64-linux-gnu/indicator-datetime-service libical.so.0 =&gt; not found   Does anyone know how to get the libical.so.0 binary that the datetime service needs but doesn't install? How can I encourage Stack Overflow to rein in the 'subjective' vigilantes? I am trying to understand   How can a set contain itself? Can you show example of set which is not a set of all sets and it contains itself. Draw a vertical line over the entries of a column in an array I'd like to draw vertical/horizontal lines over the entries of a column/row in an array. I find the method to draw horizontal lines at . But when I used the same way to draw the vertical lines. It didn't look well. My codes are:  \usepackage{tikz} \usetikzlibrary{calc} \newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};} \newcommand{\DrawLine}[3][]{% \begin{tikzpicture}[overlay,remember picture]     \draw [#1] ($(#2)+(0,0.6ex)$) -- ($(#3)+(0,0.6ex)$); \end{tikzpicture}% }%  \begin{displaymath} \left[ \begin{array}{cccc} \tikzmark{11}{15} &amp; 0 &amp; \tikzmark{13}{0} &amp;0\tikzmark{14} \\ 0 &amp; 50 &amp; 20 &amp; 25 \\ 35 &amp; 5 &amp; 0 &amp; 10\\ \tikzmark{41}{0} &amp; 65 &amp; \tikzmark{43}{50} &amp; 65 \end{array} \right]. \DrawLine[black,thick]{11}{14} \DrawLine[black,thick]{11}{41} \DrawLine[black,thick]{13}{43} \end{displaymath}   The result looks like     But the result I want is     Does anyone know how to fix this? Thank you. To hyphenate or not? As a non-native speaker of English and an engineer by training, I always get confused about hyphenation and almost always end up referring to Google every time I need to make that decision.  Does anybody know of a concise, comprehensive style guide to hyphenation that explains this rather complex issue, once and for all? Unmangling the result of std::type_info::name Is it appropriate to put a comma before "which"? On Server Fault, we were discussing  in chat. There are non-native speakers in the chat and somehow we got on the topic of grammar and proper use of commas.   The sentence of interest:     I have just got a new Ubuntu LAMP VPS server, which I am trying to configure.   Some people thought that the comma was superfluous. Others did not.   When is it appropriate for a comma to preceed the word which, and is the example sentence above correct?    Also, before you vote to close this question as a dupe of , please consider the differences between the two questions and ignore the similarity of the titles. The OP's example sentence was restructured in that question instead of addressing the proper use of commas. Prove that for any piecewise smooth curve it is possible to find the parametrisation $\phi$ that is consistent with its length, ie.  length of a curve segment between $\phi(a)$ and $\phi(b)$ is equal to $|b-a|$.  It seems intuitive, but I don't have any idea how can I show this in the formal way. Quick Look support for non-native video files (.avi, .mkv) on Mavericks In the past, we've been able to get codecs for DivX/XviD/MPEG4 and support for containers like AVI, MKV in Mac OS X's Quick Look feature using  (as per ).  As noted on the Perian site, development has ceased, and future versions may or may not continue to work. It seems with the upgrade to OS X 10.9 "Mavericks", the OS or Quick Look API has changed sufficiently that Perian no longer works.  Are there any alternatives? Ideally a one-app-does-all like Perian, but even a list of individual codecs and/or plug-ins for video container files would be useful.  NB: This question isn't about video players—there are many that support more than those playable via the built-in QuickTime player. This question is about codecs/plug-ins for QuickTime on OS X to allow it and Quick Look (and apps which rely on the OS to play videos) to open/play video files/formats lacking native support. SQL JOIN - WHERE clause vs. ON clause Concecutive last zeroes in expansion of $100!$ Adding shapefile or feature class as layer in ArcGIS Desktop using Python/ArcPy? biblatex: Tighter integration of \textcite "in the flow of text" This question led to a new feature in a package:      Many biblatex citation styles feature the \textcite command which is "intended for use in the flow of text, replacing the subject of a sentence" (manual, section 3.6.2). At the moment, this works well for citations of single works -- e.g. (assuming the authoryear style), one may write in the source  The reference work is \textcite{knuth:ct:a}.   which will be typeset as     The reference work is Knuth (1984).   Citing multiple works, on the other hand, is somewhat tedious. If one simply writes (1):  Important works are \textcite{knuth:ct:a,knuth:ct:b,markey}.   the output is rather a list than a correct sentence (2):     Important works are Knuth (1984); Knuth (1986); Markey (2005).   One has to write (3):  Important works are \textcite{knuth:ct:a}, \textcite{knuth:ct:b}, and \textcite{markey}.   to get (4):     Important works are Knuth (1984), Knuth (1986), and Markey (2005).   Can \textcite be modified so that input (1) will yield output (4), hereby achieving a tighter integration "in the flow of text"?  \documentclass{article}  \usepackage[style=authoryear]{biblatex}  \addbibresource{biblatex-examples.bib}  \begin{document}  This is how \texttt{\textbackslash textcite} currently works:\\ Important works are \textcite{knuth:ct:a,knuth:ct:b,markey}.  \bigskip  But I'd like \texttt{\textbackslash textcite} to output the following:\\ Important works are \textcite{knuth:ct:a}, \textcite{knuth:ct:b}, and \textcite{markey}.  \printbibliography  \end{document} When I parent an object to follow a path and animate it, I only get a linear mode animation and changing that in Grapth Editor to bezier doesn't work! How to do that? How to fix "error: unknown filesystem. grub rescue> I get this error:  Error: unknown filesystem. grub rescue&gt;   I'm not exactly sure what I did. I believe I installed Ubuntu over Windows on my Acer netbook. I then tried to install , but it wouldn't start installing.  I turned the netbook off and back on. Now I get the error.  I have read a lot of other questions like this, but in my case I cannot boot a CD. If I put a Ubuntu CD or a Linux Mint CD into my external CD/DVD drive and change my BIOS to boot the CD-ROM first, it just gives me the same error screen.  Update copied from comment dated 2012-05-26 02:54:29Z  Here are some results from my commands:   grub rescue&gt;ls (hd0) (hd0,msdos5) (hd0,msdos1)  grub rescue&gt;ls (hd0,msdos5) unknown filesystem  grub rescue&gt;ls (hd0,msdos1) unknown filesystem  grub rescue&gt;ls (hd0) unknown filesystem I just installed my first Ubuntu Server (12.04) instance and immediately ran into the following problems:   Running apt-get install or update fails (see below) Running ifconfig produces peculiar IP addresses starting with 10.x.x.x and my router does not see the server at all in its "Attached Devices" page (its a consumer grade Netgear)   When I run apt-get I get 50+ of the following errors:  Failed to fetch &lt;some-http-url&gt;. Temporary failure resolving &lt;blah&gt;.ubuntu.com   I read up on this error for 12.04, and found some similar posts on this forum and on server fault that recommended going into /etc/resolv.conf and adding an entry:  nameserver 8.8.8.8   To my surprise, I found that /etc/resolv.conf was a symlink pointing to somewhere else  and I had to delete it first, then create a new one with the entry indicated above.  I restarted the server and tried running apt-get again, same results.  How do I start diagnosing the problem (I'm a relatively new Linux/Ubuntu user)?  Additional details that may help:   This is actually a virtual machine running as a VirtualBox guest OS The physical host that is running this VM is my laptop which has a wireless connection; I'm wondering if (somehow) the laptop is getting the network wirelessly, but perhaps either the VBox VM and/or the Ubuntu server OS isn't configured for wireless and so nothing is "getting through"?   Thanks in advance for any pointers!  Edit:  Here is a snapshot of my VM's Network settings (only Adapter 1 has any information populating its fields, so I assume my VM has only 1 adapter):    Here is the output when I run ifconfig from the terminal:    And here is the output when I run route from the terminal:    Running ping google.com: Where does the term $|\psi\rangle\langle\psi|$ come from while calculating the expectation value? Proving that ${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $ How can I prove that $${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $$  I tried the following:  We use the :  $$y^{\underline k}=\underbrace{y(y-1)(y-2)\ldots(y-k+1)}_{k\text{ factors}},$$ so that $\binom{y}k=\frac{y^{\underline k}}{k!} .$   Then  $${x +y+n- 1 \choose n} = \frac{(x +y+n- 1)!}{n! ((x +y+n- 1) - n)!} =  \frac{1}{n!}. (x +y+n \color{#f00}{-1})^{\underline n} $$  And  $$ {x+n-k-1 \choose n-k}{y+k-1 \choose k}$$ $$\frac{1}{(n-k)!}.(x+n-k-1)^{\underline{n-k}}.\frac{1}{k!}.(y+k-1)^{\underline{k}}$$ $$\frac{1}{k!.(n-k)!}.(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$ $$\sum_{k=0}^n{n \choose k}(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$  According to the Binomial-coefficients:  $$ ((x+n-k-1) + (y+k-1))^{\underline{n}}$$ $$ (x+y+n\color{#f00}{- 2})^{\underline{n}}$$  What is wrong ? und How can I continue?  :/ What does this triangle/diamond schematic symbol represent? Could someone please tell me what the symbol ringed in red represents?     I know it has ON Switch pointing to it, but the symbol has other lables in different parts of the document.