The sign-up process for alternative SE sites should be in the background It's a bit of a hurdle to participate on a Stack Exchange site you've never used before. After reading something interesting, you have to stop and log in before being able to comment, vote, etc. Wouldn't it be better to be able to participate straight away without logging in?   I know with the current set-up, being automatically logged in to every site would mean that hundreds of sites would be displayed in your profile, but perhaps that can be changed so that a site would only display on your profile after the first activity is logged.  The way it is set up at the moment disincentivizes me from going through the sign up process. If my first vote is the difference between one click and three (with the screen changing pages), I'm less likely to contribute my first vote on that site, which in the grand scheme of things, means fewer votes on the network.
Why separate Stack Exchange accounts? I understand the reason why Stack Exchange has different sites to separate  from  or . I understand too why the reputation is associated on each site.   Why do I have to create a new  for each web site? I know that when I create a new account, it is automatically associated to my other accounts. But, is there a reason why, when I originally create a user on the first site, I can't go on another site with this user?
I have sucessfully installed AniMove by activating experimental plugins (yay! thanks gis.stackexchange). But I can't find it in Qgis! Yes I have activated it. Where is it? Is this a common bug? I searched for it the toolbox, I looked up the names of the several single tools based on this  and only found "Kernel Density", which is filed under SAGA. But I don't have SAGA installed (and don't know if I need to?) Removing lines from text file that contain string at beginning and string at end - and keep one It took me a while to word this best.  Problem 1:  Is there a way to find lines in a text file that have the same string at the beginning and the same strings at the end?   Beginning and end strings not necessarily equal to each other.  Text between the two strings is not consistant in number of characters or number of hyphens.  A hyphen is in between notable strings (which I added to my large text file).  For example:  File:    AAAAA-tjhut-ghyut-FILE1.txt   BBB-fhuj-dnbhg-frt76-FILE2.txt   AAAAA-ghtujukr-hgute7gr-xhyutt-FILE1.txt   AAAAA-VGYT-4d7tt-xhyutt-FILE1.txt   CCCC-fjuk-dryy-fhuj-FILE3.txt   Lines to detect:    AAAAA-tjhut-ghyut-FILE1.txt   AAAAA-ghtujukr-hgute7gr-xhyutt-FILE1.txt   AAAAA-VGYT-4d7tt-xhyutt-FILE1.txt   Problem 2:  Is there a way to remove all but one of the lines containing the duplicate strings at the beginning and end from above?  Before:    AAAAA-tjhut-ghyut-FILE1.txt   AAAAA-ghtujukr-hgute7gr-xhyutt-FILE1.txt   AAAAA-VGYT-4d7tt-xhyutt-FILE1.txt   After:    AAAAA-tjhut-ghyut-FILE1.txt   Big thanks for any advice.  Note: I think using the occurances of the first hyphen and last hyphen can be used to locate the duplicate strings. What is the idiom, proverb for "Little problems often become big problems if no one takes the initiative to correct them" What is the idiom, proverb for "Little problems often become big problems if no one takes the initiative to correct them"  Which means in an example that If the employees don’t bother to report a malfunctioning machine or a slip-and-fall hazard, serious injuries could occur – injuries that could have been preventable. Ignoring any hazard can be detrimental to one's business and the safety of their employees. Create WiFi hotspot on ubuntu Although many questions have already been asked regarding WiFi hotspot, I didn't get any of the solutions working for me.  I connect to the Internet using a 3G Dongle of Huawei 3772 (Vodafone). The connection type is Mobile Broadband connection. I have to share this Internet connection with my other devices like mobile phones and laptops.  How can I create a WiFi hotspot connection for this connection.  I am using Ubuntu 12.04 LTS.  I would like to note that I have already tried some help links from Ubuntu but, they are so confusing that I could not understand them. Prove that $\langle a^n \rangle \bigcap \langle a^k \rangle = \langle a^{lcm (n,k)} \rangle$ How to change IP address to point to localhost? Scan vs. Map vs. Apply How to boot from iso with Grub2/Burg boot loader This is a hypothetical question about how a CPU operates. If I purchase two identical CPUs, and use one long term (say one year), will it be identical in speed to the unused CPU? Will the number of clock cycles, latency of requests, etc on the used CPU be less than that of the unused CPU?  A supporting argument may be that mechanical devices degrade over time, While a CPU has no moving parts (other than the external fan), it does have circuits that can be damaged by heat, and voltage spikes. Lets say that after a year of intensive use, the circuits degrade and fewer electrons can pass since the pathway is narrower, etc.   Is this the nature of how a CPU operates, or is it simply working or broken, with no speed degradation in between? Not able to calculate sum of multiple column where file data in different format .csv I have a scenario where i want to calculate sum of column   SAL|CAL|TAG|VAL 12.11"|"1.1"|"2.1"|"1.1 13.11"|"1.1"|"2.2"|"2.2 14.11"|"1.1"|"2.4"|"2.4   My below code not giving output giving output as 0.00  I need output for this column CAL VAL  awk -F'|' '{T+=$2} END { printf "%.2f\n", T }' demo.txt   output :  6.7 5.7 Tikz: Arrowheads in the center Trying to purge MySQL fails I'm running in circles currently, trying to fix some mysql problem.  I couldn't use apt-get for a while now, because it always stuck at the line Setting up mysql-server-5.6 (5.6.27-0ubuntu0.15.04.1) ... and didn't let me do anything. So I tried to reinstall mysql, purging it. While I was able to remove mysql-server and mysql-client eventually, I'm still unable to purge mysql-common. Obviously, I managed to corrupt something, but how can I get my system back and running at this point?  ...:~# apt-get purge mysql-common Reading package lists... Done Building dependency tree        Reading state information... Done The following packages will be REMOVED:   mysql-common* 0 upgraded, 0 newly installed, 1 to remove and 5 not upgraded. After this operation, 0 B of additional disk space will be used. Do you want to continue? [Y/n] Y (Reading database ... 423600 files and directories currently installed.) Removing mysql-common (5.6.27-0ubuntu0.15.04.1) ... Purging configuration files for mysql-common (5.6.27-0ubuntu0.15.04.1) ... update-alternatives: error: /var/lib/dpkg/alternatives/my.cnf corrupt: invalid status dpkg: error processing package mysql-common (--purge):  subprocess installed post-removal script returned error exit status 2 Errors were encountered while processing:  mysql-common E: Sub-process /usr/bin/dpkg returned an error code (1)   -  ...# sudo dpkg -P --force-all mysql-common (Reading database ... 424733 files and directories currently installed.) Removing mysql-common (5.6.27-0ubuntu0.15.04.1) ... Purging configuration files for mysql-common (5.6.27-0ubuntu0.15.04.1) ... update-alternatives: error: no alternatives for my.cnf dpkg: error processing package mysql-common (--purge):  subprocess installed post-removal script returned error exit status 2 Errors were encountered while processing:  mysql-common   -  ...:~# lsb_release -a No LSB modules are available. Distributor ID: Ubuntu Description:    Ubuntu 15.04 Release:    15.04 Codename:   vivid   My goal is to just get it back and running again (mysql). What is the "correct" way of embedding text into math mode? What kind of behaviour causes an interrupted exception? I'm relatively new to Threading in Java and I've noticed that everytime I use Thread.sleep() I have to catch InterrupetdException.  What kind of behaviour causes this, and in simple applications where I have a monitor thread can I just Ignore the exception? Value of $\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks! I've already done a lot of searching (in particular: ) to try to prove this statement:      $f: A \to B$ is injective if and only if it has a left inverse. [I'm going to also assume $A$ and $B$ are nonempty.]   My attempt so far:     $\Leftarrow$ Suppose $f$ has a left inverse and $f(a_1) = f(a_2)$ for    $a_1, a_2 \in A$. Then there is a function $g: B \to A$ such that   $g\left(f(a_1)\right) = a_1$ and $g\left(f(a_2)\right) = a_2$. Since   $g$ is well-defined, it follows that $g\left(f(a_1)\right) =  g\left(f(a_2)\right)$, and hence $a_1 = a_2$.         $\Rightarrow$Now   suppose $f$ is injective. Then $f(a_1) = f(a_2) \implies a_1 = a_2$.   Furthermore, since $A$ is nonempty, there is an element $x \in A$.   Define $g: B \to A$ by  \begin{equation*} g(b) = \begin{cases}  f^{-1}(b), &amp; b \in \text{Im}(f) \\ x, &amp; b \in A \setminus  \text{Im}(f)\text{.} \end{cases} \end{equation*} This mapping is   well-defined, since if $f$ is injective, $f^{-1}(b)$ is unique.   My main question: does this imply that if $f: A \to B$ is injective that any mapping $g: \text{Im}(f) \to A$ is bijective? Otherwise, I don't understand why we can even use $f^{-1}$ on an element. [Please read the link above for more details - in proof 1.] Which words in a title should be capitalized? Are there any concrete rules that say which words (parts of speech) in a title should start with a capital letter? What would be a correct capitalization for the title of this question? Discovering new features A disabled module has a schema version of -1, how can I enable the module and then run DB updates using hook_update_N? How to prevent form from being submitted?