When I vote to close a question and I open back up the dialogue window it will say      You voted to close this question {time} ago   which is fair enough.    However, if I flag a post I will get told      You have already flagged this post for moderator attention    only after I try and submit a second flag. I think it would be helpful to display      You flagged this post as {reason}, {time} ago   if I open the dialogue box again.
I flagged a post today, typed out a fairly thorough explanation, clicked the button... and got told that I'd already flagged this post (for something else, which I'd forgotten about) so this flag wasn't allowed.  (This happens if the first flag is still active.)  The dialogue with my explanation was gone at this point, so if I wanted to communicate with the moderators about that, I was going to have to start over.  (In case you're wondering how this could happen, one flag was related to the post itself and one was related to the user -- there being no way to flag a user directly, you have to hang it off of a post.  But I've also seen this with cases where the first flag hangs around unhandled for a week or two and I've forgotten about it.)  Since the system is going to detect that I've already flagged a post and reject a second flag, can it do that after I click the "flag" link instead of waiting until the end of the process?  (Or gray out the link so I can't click it in the first place?)  2015-12-18: someone suggested to me in chat today that this had been implemented, so I tested again.  The behavior has not changed: I flagged a post using "other", then flagged it again (also "other") before the first flag was handled, and the flag was rejected only after I'd typed the message.  (I did this on a site where I'm a moderator so I could then clean up after myself.)
Expression Functions for Lightning Web Components Can't connect to wireless 13.10 I have ubuntu installed as the sole OS on a T420. Before this I had installed the #! distro (and Xubuntu before that) and wireless worked seamlessly on either. My home hotspot shows up but when I click on it tries for a while before disconnecting.  Before going into the problem further, I wanted to double check that I entered the wifi password for my home network correctly. I went to Edit Connections >> Wi-Fi Security and changed the key, but it is still not able to connect. So I think the password is definitely correct. I don't understand why it can't connect now. I'm having difficulty with exercise 1.43 of Lang's Algebra. The question states     Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroup that is isomorphic to $G/H$.   Thinking about this for a bit, the only reasonable approach I could think of was to construct some surjective homomorphism $\phi\colon G\to K$ for $K\leq G$, and $\ker\phi=H$, and then just use the isomorphism theorems to get the result.  After a while of trying, I've failed to come up with a good map, since $H$ seems so arbitrary. I'm curious, how can one construct the desired homomorphism? This is just the approach I thought of, if there's a better one, I wouldn't mind seeing that either/instead. Thank you. Last night I took some night sky shots, and I've noticed that with long exposure times (15s, 30s) the time it took to store the images in the camera memory card (jpeg) took roughly as long?  The actual file size is fairly small at 1 to 3MB per file.  Any idea why this is? Is it safe to format msftres, msftdata and hidden partitions? FrameLabel and OverTilde /usr/bin/sudo must be owned by uid 0 and have the setuid bit set I have run following command accidentally   sudo chown [username] -hR /   Now sudo su getting error:  sudo: /usr/bin/sudo must be owned by uid 0 and have the setuid bit set   How to Solve This? $\lim_{n\to\infty}a_nb_n=? $ given that $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}b_n=\infty$ A. $\lim_{n\to\infty}a_nb_n=?$ given that $\lim_{n\to\infty}a_n=0$ and $ \lim_{n\to\infty}b_n=\infty$  B. $\lim_{n\to\infty}{a_n \over b_n}=?$ given that $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}b_n=0$  C. $\lim_{n\to\infty}{a_n \over b_n}=?$ given that $\lim_{n\to\infty}a_n=\infty$ and $\lim_{n\to\infty}b_n= \infty$    These seem to be difficult situations I come across when I have to evaluate limits.   Some examples,    $\lim_{n\to\infty}n^a\sin(1/n^b)$ with $a,b&gt;1$   $\lim_{n\to\infty}{n^a\over \sin(1/n^b)}$ with $a,b&gt;1$   $\lim_{n\to\infty}{1/n^a\over \sin(1/n^b)}$ with $a,b&gt;1$   $\lim_{n\to\infty} nf(x_n)$ [where we only know that $\lim \ f(x_n) = 0]$    My first question is : Are the two situations  A and B  equivalent ? I feel that they are equivalent as $\lim_{n\to\infty}b_n=\infty \iff \lim_{n\to\infty}{1\over b_n}=0$, The problem is that I'm not sure if we can apply the product rule for limits to take the limit inside.  My second question is : What are some ways to solve such problems ? I only know that we can apply stolz- cesaro in some cases. I have searched in %appdata%, pulling up the list of files there, but there is no file labelled .minecraft. Searching %appdata%\.minecraft returns "no results found".   I also opened Minecraft to click on the resources button and could not locate that either. Prime dividing the binomial coefficients It is quite easy to show that for every prime $p$ and $0&lt;i&lt;p$ we have that $p$ divides the binomial coefficient $\large p\choose i$; one simply notes that in $\large \frac{p!}{i!(p-i)!}$ the numerator is divisible by $p$ whereas the denominator is not (since it is a product of numbers smaller than $p$ and $p$ is prime).  My problem is with generalizing this argument for $q=p^n$. I'm looking for the most elegant and simple way to prove that $p$ still divides $\large q\choose i$. Opening PDF in Illustrator breaks up text objects I recently had to convert some PDF files to SVG, and did this by opening the PDF in Illustrator, and saving out to SVG.    The issue was that when Illustrator opened the pdf, many (but not all) text objects would be broken up into different text objects.  For instance, the word "policy" wouldn't be one text object, but rather several text objects, "po", "l", "icy".  There didn't seem to be any rhyme or reason to it.    How can I prevent this from happening? How can I distribute a native executable for a Python program? Lately all of my work has been in Python (and Ruby), and I'm in love with those two languages. There are many libraries for game development on Python but the problem I see is that many of those frameworks don't cover the "distribution" aspect of game creation. I'd like to code my game on Python (and any other pure Python extra libraries I need) and be able to distribute an executable for the main PC platforms (Windows, Mac and Linux).  Any options out there or what are the best combinations for this? (maybe pyglet + cx_freeze) Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$ Prove by induction the summation of $\frac1{2^n}$ is greater than or equal to $1+\frac{n}2$.  We start with $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$$ for all positive integers.   I have resolved that the following attempt to prove this inequality is false, but I will leave it here to show you my progress. In my proof, I need to define P(n), work out the base case for n=1, and then follow through with the induction step. Strong mathematical induction may be used.  This is equivalent to $$\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n}2\;.$$  Let $P(n)$ be summation shown above.   Base case for $n=1$, the first positive integer,   $$\sum_{k=0}^1\frac1{2^k}=\frac1{2^0}+\frac1{2^1}=1+\frac12=\frac32\ge 1+\frac12=\frac32\;,$$  so base case is true.  Induction step: Assume $P(n)$ is true and implies $P(n+1)$. Thus   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge\frac1{2^{n+1}}+\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n+1}2\;.$$  This can be written as   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge \frac1{2^{n+1}}+1+\frac{n}2\ge 1+\frac{n+1}2\;.$$  I work the math out but I get stuck contradicting my statement. Please show your steps hereafter so I can correct my mistakes. I tried to fix my hard drive using Ubuntu but I encountered an error message and didn't know how to proceed. I need to fix my bad sectors   fsck /dev/sdb   fsck from util-linux 2.20.1   e2fsck 1.42.5 (29-Jul-2012)   fsck.ext2: Permission denied while trying to open /dev/sdb   You must have r/w access to the filesystem or be root Application launchers in '/usr/share/applications' not working in Nautilus In Python, I can compile a regular expression to be case-insensitive using re.compile:  &gt;&gt;&gt; s = 'TeSt' &gt;&gt;&gt; casesensitive = re.compile('test') &gt;&gt;&gt; ignorecase = re.compile('test', re.IGNORECASE) &gt;&gt;&gt;  &gt;&gt;&gt; print casesensitive.match(s) None &gt;&gt;&gt; print ignorecase.match(s) &lt;_sre.SRE_Match object at 0x02F0B608&gt;   Is there a way to do the same, but without using re.compile. I can't find anything like Perl's i suffix (e.g. m/test/i) in the documentation. We are always introduced to magnets by pictures illustrating a parallepiped magnet, a horseshoe magnet, etc... Anyway the first time we actually calculate a magnetic field is for eelectromagnets. Indeed I've rarely seen explaining or calculating the magnetic field of a permanent magnet. I know formulas for the magnetic field of a moving charge, around a conductive wire, inside a solenoid, but I don't have idea on how to go about calcualting the filed of a simple permanent magnet. It's like that without the help of electricity I'm stuck. I'd guess it would be heavily dependent on the magnetic material, its shape and the medium in which is placed. I hope that someone would provide me some insights on why such formulas are  never introduced in most physics courses and lay down some examples of a permanent magnet field caluclation. How to smooth camera animation to have fluid feel? When dealing with camera motion, its very sensitive to any movement discontinuities. Is there a way to easily achieve natural looking motion when dealing with complicated motion paths and timings?  Follow path animation feels too robotic unless I adjust all the keyframe easings carefully and this has to be re-done everytime the animation changes (the automatic bezier type just doesn't work). A two column loop with one lead post I got a loop from internet (I forgot the site) for two column post view. It has a nice CSS for it too. But skipping that for the question now.  &lt;?php $counter = 1; ?&gt;                  &lt;?php query_posts( 'cat=3&amp;showposts=6' ); //Find whether the catID is 3 ?&gt;             &lt;?php while ( have_posts() ) : the_post() ?&gt;                   &lt;?php /* Two Column Support get class */ ?&gt;              &lt;?php $class = ( $counter % 2  ? ' one' : '' ) ?&gt;                  &lt;?php /* Two Column output open the div class */ ?&gt;              &lt;div class="column&lt;?php echo $class;// echo $first; ?&gt;"&gt;              &lt;?php news_format(); ?&gt;             &lt;/div&gt; &lt;!-- End Two Column Support close div class --&gt; &lt;?php $counter++; //Update counter ?&gt;   It printed my posts into 2 columns like:  ========= | 1 | 2 | --------- | 3 | 4 | --------- | 5 | 6 | =========  It's fine. But I want to print it like this:  ========= |   1   | --------- | 2 | 3 | --------- | 4 | 5 | --------- | 6 | 7 | =========  What the modification I need to do? I can't pass any custom limit into the WordPress' custom loop &lt;?php while ( have_posts() ) : the_post() ?&gt;. It's for a custom theme. I did it, but the problem is, it's showing like:  ========= |   1   | --------- | 1 | 2 | --------- | 3 | 4 | --------- | 5 | 6 | =========   is my whole bunch of code. The three files are read in this order?  .bash_profile .profile .bashrc   This is not happening when I first open a terminal.   I have trace statements in the files appending to the file init.log. Please take a look at the following. It begins after opening a terminal. I have placed a comment to show where the log picks up after the su command.  stephen@debian:~$ cat init.log reading .bashrc done reading .bashrc stephen@debian:~$ su - stephen Password:  stephen@debian:~$ cat init.log reading .bashrc done reading .bashrc # # after su # reading .bash_profile reading .profile reading .bashrc done reading .bashrc done reading .profile done reading .bash_profile stephen@debian:~$    So the su - login triggers the expected sequence however the initial login reads only the bashrc. This cannot be correct. Can someone explain under what conditions this would occur. I could modify the bashrc and profile files so that the initial read includes all expected files but I would rather get to the root of the problem and fix it there.