I just tried to log out of chat.SE (I'm an active user in GameDev and sometimes use other chat rooms), but I can't seem to. Could this possibly be tied to the new global-auth update? I have tried hitting &quot;log out&quot; in the dropdown menu, which just redirects me to . That link also does not work, just seemingly redirecting me to  and showing my active rooms (aka the site rooms for GameDev.SE). I can log out through the main GD.SE site proper, which also logs me out of chat, but upon clicking &quot;log in&quot; again on the chat site, I get: . Logging in on the site proper seems to again, work as expected. Some system information: Windows 8.1 Browser: Firefox 39.0
I attempted to log out with the following actions:    This accomplished absolutely nothing since I remained logged in after clicking "log out" on the confirmation page. I then tried logging out of Stack Exchange from one of the network sites and repeated the steps above. The results were the same. I remained logged in no matter which combination of steps I performed.  In fact, the only way I could log out at all was to clear all of the cookies that were set.
Probability of student passing an exam There're $25$ tests, each with $2$ questions. There're total of $50$ different questions (the questions do not repeat). The student knows the answer of $44$ questions. To pass the exam, the student needs to correctly answer both questions on the first test he drew, or answer $1$ question from the test he drew first, and $1$ question from the test he drew second. (meaning he draws twice if he answers a question incorrectly on his first try)  What is the probability that the student will pass the exam?  I have never faced a problem like this one so any help is appreciated. I was thinking about approaching it "manually" with some applied combinatorics but I feel like there's a much easier way that I am not familiar with. Endofunctors of $\textsf{FinVec}$ which are not $C^n$ Does the faction of your armor/clothes affect character interactions in Skryim? Finding $\lim\limits_{x\to1^-}\Bigl(\prod\limits_{n=0}^{\infty}\Bigl(\frac{1+x^{n+1}}{1+x^n}\Bigr)^{x^n}\Bigr)$ Compute : $$L=\lim_{x\to 1^-} \left(\prod_{n=0}^{\infty} \left(\frac {1+x^{n+1}}{1+x^n}\right)^{x^n}\right)$$   My try:   Writing out the first few terms I noticed that the limit can be expressed as $$L=\lim_{x\to 1^-} \frac 12\left(\prod_{n=1}^{\infty} (1+x^n)^{x^{n-1} -x^n}\right)$$ $$L=\frac 12\lim_{x\to 1^-} \left(\prod_{n=1}^{\infty} (1+x^n)^{x^{n-1}}\right)^{1-x}$$  Converting to exponential form I get $$L=\frac 12 \exp {\left(\lim_{x\to 1^-} (1-x)\sum_{n=1}^{\infty} \left(x^{n-1}\ln(1+x^n)\right)\right)}$$  And got stuck here. I did try to use approximation $\ln x\sim x$ hereon to get final answer as $\frac {\sqrt e}{2}$ but I think it was improper to use approximation and hence I believe that the answer I got is flawed. I also noticed that the form I obtained is quite similar to be used for a Riemann Sum but I didn't get a way to tackle the problem using that way.  Can someone please help me with this problem.... Is it possible to add non-number characters on the left of a code block?  Specifically, I want to add ... to correspond with some source code (see image).    I am currently using commands to stop / start the line numbers. Here is the code:  \documentclass[12pt]{article}  \usepackage[a4paper, margin = 20mm, top = 30mm, bottom = 25mm]{geometry} \usepackage{xcolor, soul} \usepackage{listings}  \definecolor{codebackground}{rgb}{0.95, 0.95, 0.92}  \lstset {     frame            = tb, % hrule above and below     keepspaces       = true,     columns          = flexible,     basicstyle       = {\normalsize\ttfamily},     escapeinside     = {(*@}{@*)}, % for escaping     backgroundcolor  = \color{codebackground},     showstringspaces = false,     language         = C,     keywordstyle     = \color{blue},     stringstyle      = \color{red},     commentstyle     = \color{teal},     numbers          = left, % {none, left, right}     firstnumber      = 1,     numberstyle      = \scriptsize\color{black},     numbersep        = 5pt }  \let\origthelstnumber\thelstnumber \makeatletter % for stopping line numbers \newcommand*\stopnumber {     \lst@AddToHook{OnNewLine}     {         \let\thelstnumber\relax         \advance\c@lstnumber-\@ne\relax     } } % for starting line numbers \newcommand*\startnumber[1] {     \setcounter{lstnumber}{\numexpr#1-1\relax}     \lst@AddToHook{OnNewLine}     {         \let\thelstnumber\origthelstnumber         \refstepcounter{lstnumber}     } } \makeatother  \begin{document} \begin{lstlisting}[firstnumber = 456] int third_function(const int c) {     /* error checking */     if(c &lt; 0)     {         return;     } (*@ \stopnumber @*)      ...     (*@ \startnumber{512} @*)     /* other stuff */     int d = 4;     char* word = "third function"; (*@ \stopnumber @*)      ...     (*@ \startnumber{678} @*)     return (c * d); } \end{lstlisting} \end{document} Let $\mathbb F$ be a field and $\mathbb  K $ be an extension field of $\mathbb F$  such that $\mathbb K$ is algebraically closed.   Let $\mathbb L$ be the field of all elements of $\mathbb K$ which are algebraic over $\mathbb F$. Then $\mathbb L_{|\mathbb F}$ is an algebraic extension.   My question is : Is $\mathbb L$ algebraically closed ?  I am trying to prove the existence of algebraic closure, so please don't assume that every field has an algebraic closure. Nonconstant polynomials do not generate maximal ideals in $\mathbb Z[x]$ Radio antenna producing waves in the visible spectrum If a radio could produce waves in the visible light spectrum, what would the result be?  This is a thought experiment that I've pondered for a few years now.  I realize there are a few/many real-world constraints, but if we lifted these constraints for the sake of thought, what could we expect?  Personally, I don't see why we wouldn't observe visible light emitting from the antenna, disregarding any light from Blackbody Radiation. The following is an real analysis qualifying exam problem that I cannot solve:  Suppose $X$ is a Banach space and that $(x_n)$ converges weakly to $x$. Show that $\liminf ||x_n|| \geq ||x||$.  Using the Uniform Boundedness Principle I can show that $\sup_{n \in \mathbb{N}} ||x_n -x||$ is finite. Using Alaoglu's Theorem I can show that some subset of $(x_n)$ converges in norm to $x$. I feel like I am close with this but cannot seem to finish the problem. I normally dual-boot with 12.04 and Win XP pro sp3 on a Thinkpad R60.  I needed to reinstall windows because of problems piling up, so I ran the computer's built-in hard-disk recovery software.  I told it not to mess with the other partitions, but somehow it deleted the Linux partition.  That's what I was using to backup all of my windows files, so now everything I have is gone.  I just ran the USB setup with linux, and the partition is deleted - it just shows it as empty space.    Is there ANY way to get all of my files back? I have a scenario where I have multiple threads adding to a queue and multiple threads reading from the same queue. If the queue reaches a specific size all threads that are filling the queue will be blocked on add until an item is removed from the queue.  The solution below is what I am using right now and my question is: How can this be improved?  Is there an object that already enables this behavior in the BCL that I should be using?  internal class BlockingCollection&lt;T&gt; : CollectionBase, IEnumerable {     //todo: might be worth changing this into a proper QUEUE      private AutoResetEvent _FullEvent = new AutoResetEvent(false);      internal T this[int i]     {         get { return (T) List[i]; }     }      private int _MaxSize;     internal int MaxSize     {         get { return _MaxSize; }         set         {             _MaxSize = value;             checkSize();         }     }      internal BlockingCollection(int maxSize)     {         MaxSize = maxSize;     }      internal void Add(T item)     {         Trace.WriteLine(string.Format("BlockingCollection add waiting: {0}", Thread.CurrentThread.ManagedThreadId));          _FullEvent.WaitOne();          List.Add(item);          Trace.WriteLine(string.Format("BlockingCollection item added: {0}", Thread.CurrentThread.ManagedThreadId));          checkSize();     }      internal void Remove(T item)     {         lock (List)         {             List.Remove(item);         }          Trace.WriteLine(string.Format("BlockingCollection item removed: {0}", Thread.CurrentThread.ManagedThreadId));     }      protected override void OnRemoveComplete(int index, object value)     {         checkSize();         base.OnRemoveComplete(index, value);     }      internal new IEnumerator GetEnumerator()     {         return List.GetEnumerator();     }      private void checkSize()     {         if (Count &lt; MaxSize)         {             Trace.WriteLine(string.Format("BlockingCollection FullEvent set: {0}", Thread.CurrentThread.ManagedThreadId));             _FullEvent.Set();         }         else         {             Trace.WriteLine(string.Format("BlockingCollection FullEvent reset: {0}", Thread.CurrentThread.ManagedThreadId));             _FullEvent.Reset();         }     } } Round a double to 2 decimal places can't read from SQlite database Show that $$\int_{0}^{\infty} \frac{x - \sin(x)}{x^3} \, dx = \frac{\pi}{4}$$   My attempt is as follows: Let $$f(z) = \frac{z - i e^{iz}}{z^3}$$ and consider the contour on $[\epsilon, R]$ followed by a semicircular arc in the counter clockwise direction, then on $[-R, -\epsilon]$, then the semicircular clockwise contour avoiding the origin. We have, then, that  $$0 = \int_{\Gamma} f(z) dz = \int_{[\epsilon, R]} f(t) dt + \int_{C_R}f(Re^{it})Rie^{it}dt + \int_{[-R, -\epsilon]}f(t)dt + \int_{C_{\epsilon}}f(\epsilon e^{-it})\epsilon i e^{-it} dt$$  Then the first and third integrals ( $I_1$ and $I_3$) combine so that  $$I_1 + I_3 = 2\int_{\epsilon}^R \frac{t - \sin{t}}{t^3}\,dt$$  Further,   $$|I_{C_R}| \leq \int_0^\pi \left|\frac{Re^{it} - ie^{-R\sin{t}}e^{iRcos{t}}}{R^2 e^{2it}} \right|dt \rightarrow 0 \text{ as } R\rightarrow \infty$$  (I've omitted the details, it isn't too bad to bound)  However, I'm having trouble computing the limit  $$\lim_{\epsilon \rightarrow 0}\int_{C_{\epsilon}} f(\epsilon e^{-it})\epsilon i e^{-it} dt$$  No matter which way I look at it, it seems like this limit does not exist. Perhaps I'm seeing something wrong or have I chosen a bad $f(z)$? Prove that the average of iid Gaussian random variables is Gaussian Given $x_1, \ldots, x_N$, independent and all distributed as a   Gaussian with mean $\mu$ and variance $\sigma^2$. Then, the average   $$\bar{x} = \frac{1}{N}\sum_{i=1}^Nx_i$$ is distributed as a Gaussian   with mean $\mu$ and variance $\frac{\sigma^2}{N}.$   This is a very well-known result. Anyway, I'm looking around to find a proof for this and I'm not having luck. I shall come, not will Native speakers, how can you comment such a sentence:  I shall come, not will  Can I use it for  I will definitely come, not maybe    Adding some information other people asked me about.  How I came to such a question.  Here   I asked about "Shall I win?" pointing that it could have some variants of understanding:  1) Shall I win? = Should I win? (suggestion)  2) Shall I win? = Will I win? (archaic with 1st person)  3) Shall I win? = is it 100% that I will win? (predestination\inevitability)  The excellent guy Mixolydian said either variant was possible, that it didn't matter what person you used "shall" with, it could have this "predestination\inevitability" anyway. Out of this we're getting that nowadays we can use "will" for simple future actions and "shall" for almost predicted (almost 100% surety of happening) future actions.  If I want to say to someone:  Don't be afraid, I will come anyway.  Don't get panicked, I will definitely come  Be sure, I will come in any case  I can use "shall" for it meanining an action into which I will put 100% effort:  I shall come!  And what if my original sentence is  Don't doubt, I will definitely come, not maybe or probably  Can I use  I shall come, not will  pointing that "shall" guarantees me putting all my strengths into it while "will" doesn't give any assuarance that I will do everything to finally come.  As I understood, it wasn't very easy to ask such a question because English was different everywhere - US, UK, Australie, Ireland etc. I heard it was acceptabse to use even "thee\thou" in Ireland or somewhere like that while all other world has already forgotten such words...  P.S. What I got from the answers ias that it was better to forget about the existence of "shall" cause it created so many misunderstandings:D Let $\int_{- \infty}^{\infty} f(x) dx =1$. Then show that $ \int_{- \infty}^{\infty} \frac{1}{1+ f(x)} dx = \infty.$ How do I iterate over the words of a string? I'm trying to iterate over the words of a string.  The string can be assumed to be composed of words separated by whitespace.  Note that I'm not interested in C string functions or that kind of character manipulation/access. Also, please give precedence to elegance over efficiency in your answer.  The best solution I have right now is:  #include &lt;iostream&gt; #include &lt;sstream&gt; #include &lt;string&gt;  using namespace std;  int main() {     string s = "Somewhere down the road";     istringstream iss(s);      do     {         string subs;         iss &gt;&gt; subs;         cout &lt;&lt; "Substring: " &lt;&lt; subs &lt;&lt; endl;     } while (iss); }   Is there a more elegant way to do this? I would like to pause a render that I have started, with the intention of resuming it at a later time (but during the same Blender session).  How can I do this? What is the right way to indent your source code?