There's a curious bug that keeps tripping me up on the Android app: votes sometimes appear to revert in the client, though they stick successfully server side. I'm using app version 1.0.54 on Android 4.4.3.   I've found a couple of ways this happens:   I'll give a post an up/downvote, the vote arrow will light up and the score will adjust. About a second later, enough time for a server ping and reply, the score will revert and the vote arrow will become unlit. This one happens sometimes. It's been happening for a while, but I'm not sure what version it started. I'll vote, and then leave a comment. When I leave the comment, my vote will revert: arrow becomes unlit, score reverts. This one happens every time. I am pretty sure it emerged just now in this version, 1.0.54.   In either case, if I attempt to re-cast my vote, the app will give me an error saying I've already voted on this answer and can't vote again. If I refresh, I see my vote has stuck, with the arrow lit and score adjusted accordingly. So the votes definitely get cast and stick server side, but the client side goes back to a state before they happened.  This one trips me up all the time: I'll vote, take some action (comment or read or scroll), then think "I thought I voted for that", try again, and get the error.  I suspect it might have something to do with reloading the posts in question from some kind of internal cache after a server reply, and the cached version doesn't have my vote recorded.
Related:   I am using 1.0.71 on KitKat.  Immediately after voting, the arrow is displayed in red and the vote count is correct. However, when I scroll to read answers and then back to the top, the arrow is grey again and the vote count is reset to the previous value.  A reload of the question (pulling the finger down) is needed to show the correct situation, which is kept from then on.  This case was on security.stackexchange.com but I observed the same behavior on other sites too.  Screenshots   original state:  upvoted:  scrolled down and back up  after a reload
I'm trying to solve the well known Coupon Collector's Problem by explicitly finding the probability distribution (so far all the methods I read involve using some sort of trick). However, I'm not having much luck getting anywhere as combinatorics is not something I'm particularly good at.   The Coupon Collector's Problem is stated as:  There are $m$ different kinds of coupons to be collected from boxes. Assuming each type of coupon is equally likely to be found per box, what's the expected amount of boxes one has to buy to collect all types of coupons?  What I'm attempting:  Let $N$ be the random variable associated with the number of boxes one has to buy to find all coupons. Then $P(N=n)=\frac{|A_n|}{|\Omega _n|}$, where $A_n$ is the set of all outcomes such that all types of coupons are observed in $n$ buys, and $\Omega _n$ is the set of all the possible outcomes in $n$ buys. I think $|\Omega _n| = m^n$, but I'm not even sure about that anymore, as all my attempts so far led to garbage probabilities that either diverged or didn't sum up to 1. I have been trying to install ubuntu (I tried 12.04LTS, 12.10, 12.10-secure-remix) using the wubi installer on Windows 8 and all the three times ran into the same problem showing some cannot find /ubuntu/winboot/wubildr when I tried to start ubuntu.  I tried to change the boot settings from UEFI to Legacy and then my laptop restarted and showed that "No operating System Found".  I then tried to install Ubuntu using a cd and after a lot of effort and using boot repair,etc. I landed into a new trouble. When I select UEFI on, the laptop loads Windows and windows too does not detect ubuntu. When I select Legacy mode, it starts ubuntu, shows me an option to start windows 8 but gives an error showing windows failed to load.  So in legacy mode, i am able to use only ubuntu and in UEFI mode, I am able to use only Win8. Please explain any process of starting windows in legacy mode (which I tried without any success) or starting ubuntu in UEFI mode. The result set has no current row while using JDBC I am using the following code in selenium, but I am not able to retrieve data from database. I am getting the errors "The result set has no current row"  My code is   public class Auto {      public static void main(String[] args) throws InterruptedException, SQLException, ClassNotFoundException, InstantiationException, IllegalAccessException     {           Class.forName("com.microsoft.sqlserver.jdbc.SQLServerDriver");         System.out.println("Driver loaded");           Connection con=DriverManager.getConnection("jdbc:sqlserver://192.168.*****\\SQLEXPRESS/*****_DW/EPS:****;user=******;password=******; ");          System.out.println("Driver loadedd");       java.sql.Statement stmt= con.createStatement();     ResultSet rs=stmt.executeQuery("select * from [*****_DW].[dbo].dim_grantee ");      System.out.println("Query executed");      while (rs.next());     {         String Name=rs.getString("NAME");          System.out.println("Name is");     }   } Here are two examples of what I mean:   I am sure you all too have experienced this frustration before. A dialog box opens that displays only part of a long string (e.g. a file path), but won't allow you to select the text so you can read the full string. Neither text container will resize upon window resizing. (By the way, you can force windows —specially dialog boxes— to resize with tools such as , but helps little in this case, since controls are fixed-width.) The text is hidden, period. Copying a very long drop-down list control contents to a text file.   Do you know any way to solve these problems? Let $p$ be prime and $(\frac{-3}p)=1$. Prove that $p$ is of the form $p=a^2+3b^2$ Let $p$ be prime and $(\frac{-3}p)=1$, where $(\frac{-3}p)$ is Legendre symbol. Prove that $p$ is of the form $p=a^2+3b^2$.   My progress:  $(\frac{-3}p)=1 \Rightarrow$ $(\frac{-3}p)=(\frac{-1}p)(\frac{3}p)=(-1)^{\frac{p-1}2}(-1)^{\lfloor\frac{p+1}6\rfloor}=1 \Rightarrow$ $\frac{p-1}2+\lfloor\frac{p+1}6\rfloor=2k$  I'm stuck here. This is probably not the way to prove that.  Also tried this way:  $(\frac{-3}p)=1$, thus $-3\equiv x^2\pmod{p} \Rightarrow$ $p|x^2+3 \Rightarrow$ $x^2+3=p\cdot k$ stuck here too.  Any help would be appreciated. Windows vs Linux Local Time? Ever since I started using Linux in dualboot with Windows I noticed that when I rebooted the machine, the time on the other OS was wrong, e.g. it was 12:00 on Linux, then I rebooted and on Windows it was 18:00, and when I checked the BIOS, the RTC was set to 18:00. Since my timezone is -06:00 CST I assume Linux just sets the RTC’s time to UTC and re-calculates the time based on the timezone, whereas Windows just sets the RTC to the local time. Both use the same NTP server to sync the time.  The question is, which one is doing the right thing? On which one should I set the timezone setting to UTC in order to have the right time on both? Rendered image looks pixelated When I hit F12 to render an image and zoom in it is quickly pixelated. How can I render a higher res display? I have looked through the settings and couldn't find a solution. How do I completely uninstall Google Chrome and Chromium? I'm having a problem with Chrome and one of its extensions, so I want to completely uninstall everything. I've uninstalled Chrome through the Software Center, but when I reinstall it, all of my extensions and preferences are still there.   If I want to start over from scratch, what do I need to do? How do you find a question's view counter? How do you see how many views a question has had when you're not in the newest category?  I've tried a word-for-word search for the question, but still can't see the view counter. What is postselection in quantum computing? A quantum computer can efficiently solve problems lying in the complexity class . I have seen a claim the one can (potentially, because we don't know whether BQP is a proper subset or equal to PP) increase the efficiency of a quantum computer by applying postselection and that the class of efficiently solvable problems becomes now postBQP = .  What does postselection mean here? Error mounting NTFS partition after hybrid shutdown with Windows 8 I've got this error and I can't access my files located in a different partition when I'm using Ubuntu, but when I'm using Windows 8 my files are accessible. I'm using Ubuntu as my main OS. I tried shutting down Windows 8 properly but still nothing happened. Is there any way where this can be fixed?  Error mounting /dev/sda4 at /media/kris/BackUp Files:  Command-line `mount -t "ntfs" -o "uhelper=udisks2,nodev,nosuid,uid=1000,gid=1000,dmask=0077,fmask=0177" "/dev/sda4" "/media/kris/BackUp Files"' exited with non-zero exit status 14:  The disk contains an unclean file system (0, 0). Metadata kept in Windows cache, refused to mount. Failed to mount '/dev/sda4': Operation not permitted The NTFS partition is in an unsafe state. Please resume and shutdown Windows fully (no hibernation or fast restarting), or mount the volume read-only with the 'ro' mount option. Cannot add a NOT NULL column with default value NULL in Sqlite3 What affects field upgrades in BF4? Figuring out algorithm/formula for item scaling in WoW Why are the Degrees of Freedom for multiple regression n - k - 1? For linear regression, why is it n - 2? I'm reading a textbook and I see this question:    So there are 200 women, and the DF is 196, implying that the equation for DF is $n - k - 1$. There are 3 variables: bp, age, and type so $k == 3$. What's the intuition behind this?  Also, why is the degrees of freedom for linear regression n - 2? Lithium-6 isotope has an approximate magnetic momentum of $0.88\ \mu_N$ in its fundamental nuclear state. I'm trying to find its angular momentum and parity.  I found in a standard table: $I=1^+$ and while I get the parity part, I don't understand the rest. What I have done is filling the shells separately for protons and neutrons and finding that both have their outer particle in the $1 P_{3/2}$ state. Sum of these gives possible total angular momenta to be either $0$, $1$, $2$ or $3$.  So my question is: how do you choose between these to find the real value $I=1$? Continuing enumerate Counters in Beamer I'm trying to continue the numbering of examples in enumerate environments in beamer. For example, I'd like one slide to feature   Foo Bar   and the next slide   Zip Yadda   My strategy so far has been to define a counter that saves the enumerate counter at the end of the first environment and can then be queried to set the enumerate counter at the start of the second enumerate environment. (This is essentially the proposal also found at )  That works wonderfully so long as I don't try to uncover elements of the slide sequentially. The problem, of course, is that as each of the pdf-pages that make up the slide is created, the counter is advanced. Here's a MWE.  \documentclass{beamer}  \setbeamercovered{highly dynamic}  \newcounter{saveenumi} \newcommand{\seti}{\setcounter{saveenumi}{\value{enumi}}} \newcommand{\conti}{\setcounter{enumi}{\value{saveenumi}}}   \begin{document}  \begin{frame}[&lt;+-&gt;]   \begin{enumerate}   \item foo   \item bar%     \seti   \end{enumerate} \end{frame}  \begin{frame}[&lt;+-&gt;]   \begin{enumerate}     \conti   \item zip   \item yadda%     \seti   \end{enumerate} \end{frame}  \end{document}   As the second slide is progressively uncovered (pages 3 and 4 of the resulting PDF), the counter at the beginning of the enumerate environment is advanced each time. So page 3 of the PDF has the right numbering, (3) and (4) respectively, while the fully revealed page has the wrong numbering. Pictures follow:   on one of the slides    on the following one Formatting Ticks and FrameTicks labels with a function Print out elements from an array with a comma between elements except the last word Evaluate   $$\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}.$$   $\underline{\bf{My\;\;Try}}::$ Let $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n} = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \left(\frac{(n+1)\cdot (n+2)\cdot (n+3)\cdots (n+n)}{(1)\cdot (2)\cdot (3)\cdots (n)}\right)$  $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \left\{\ln \left(\frac{n+1}{1}\right)+\ln \left(\frac{n+2}{2}\right)+\ln \left(\frac{n+3}{3}\right)+\cdots+\ln \left(\frac{n+n}{n}\right)\right\}$  $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{n+r}{r}\right) = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{1+\frac{r}{n}}{\frac{r}{n}}\right)$  Now Using Reinman Sum  $\displaystyle y = \frac{1}{2}\int_{0}^{1}\ln \left(\frac{x+1}{x}\right)dx = \frac{1}{2}\int_{0}^{1}\ln (x+1)dx-\frac{1}{2}\int_{0}^{1}\ln (x)dx =  \ln (2)$  My Question is , Is there is any method other then that like Striling Approximation OR Stolz–Cesàro theorem OR Ratio Test  If yes then please explain here  Thanks