I log in to stackoverflow using Google account. I then logout by clicking logout link. It then takes me to a page where below message appears:     Clicking Log Out will clear our cookies and log you out on all   devices.   And I click on Logout button provided. After I press that button I am redirected to stackoverflow home page.  All works well till here. But if I open the Google page in the same window by entering URL www.google.co.in, I see that I am still logged into the Google account.  Why is it happening? Why aint I completely logged out after clicking on Logout button?  I am using Mozilla Firefox 23.0.
Why does logging in to Stack Overflow keep me signed in to my Gmail email after logging out and regardless of the Stay signed in checkbox?  Steps to reproduce:   Browse to  and confirm you are logged out Browse to  and confirm you are logged out Click login Click Google Enter Email/Password and confirm Stay signed in is unchecked Click Sign in Click logout (up top) Click Logout (down below) Browse to  and confirm you are still logged in Click Sign out &lt;- Shouldn't clicking Logout on Stack Overflow be doing the same thing as this? Browse to  and confirm you are logged out     Answers are making sense but my frustration is that clicking Logout (twice) on StackOverflow essentially does nothing. If I proceed to click login and Gmail I am immediately authenticated because I still have an active Gmail cookie. In order to effectively logout I need to click Logout twice on Stack Overflow and then manually browse to  and Sign out.
How can I understand and prove the "sum and difference formulas" in trigonometry? The "" formulas often come in handy, but it's not immediately obvious that they would be true.  \begin{align} \sin(\alpha \pm \beta) &amp;= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\ \cos(\alpha \pm \beta) &amp;= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \end{align}  So what I want to know is,    How can I prove that these formulas are correct? More importantly, how can I understand these formulas intuitively?   Ideally, I'm looking for answers that make no reference to Calculus, or to , although such answers are still encouraged, for completeness. Let $\theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+\theta$ in $\mathbb{Q(\theta)}$.  So problems like this really annoy me but I did crappy on the last homework after making a lot of arithmetic mistakes so I want to run everything by you guys. This one isn't actually for homework it's just a suggested practice problem but okay lets go!!!  So I'm not sure if the method I used to do this was standard or not, I don't remember my professor showing me, but what I did first was used the canonical euclidean algorithm in $\mathbb{Q[x]}$ and wrote:  $x^3+9x+6$  $=(1+x)(x^2-x+10-\frac{4}{x+1})$  $=(1+x)(x^2-x+10)-4$  the reduced modulo  the minimal polynomial of $\theta$ i.e. $x^3+9x+6$ and so  $4=(1+x)(x^2-x+10)$  $\rightarrow$  $1= \frac{1}{4}(1+x)(x^2-x+10)$ and sooo ah $(\theta^2-\theta+10)\frac{1}{4}$ is what I believe is the inverse of $(1+\theta)$ in the quotient field... Like i said I hate these problems but ehh is this correct? On a lighter note it's finally getting cool enough for me to go on proper runs here and so that's pretty bomb! Place questions with X net downvotes on hold automatically To resolve the frequent issue of poor question unfairly gaining answers, and thus paving the way for more poor questions to be asked  I Propose  EDIT: From feedback, have changed net vote threshold to 5 (to at least match that of the current 5 users at 3k rep) END EDIT     If a question has 5 net downvotes, the question is put   on hold automatically, and no answers are allowed until the question   is resolved/has been edited and net votes are > -5.   The net vote threshold could be perhaps different, to avoid grey area questions being on hold unfairly, but this also needs to be high enough to allow for early action.  On hold note, something such as:     This question has been put on hold via community voting on the   question, and cannot receive answers. To resolve this, edit and   improve the question.   This proposal would assist in stopping poor questions from receiving answers, as currently questioners not wanting to put in any effort know they will likely get an answer eventually.  This could also help improve site quality by making people improve their questions before they can have answers, so answers are only given to decent questions. Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing.  $$\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$$  Perhaps someone can help me out I don't understand how to move forward from here: $$\sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n+1}}\ge \sqrt{n+1}}$$ proof and explanation would greatly be appreciated :) Thanks :)  EDIT sorry meant GE not = fixed :) Does iTunes Home Sharing transfer in-app purchases? My wife recently got an iPhone and, being that I'm an iPhone user already, I set up iTunes Home Sharing (under my Apple ID) so that I could share some of my purchased apps with her.  While paid apps transferred to her phone without a problem, in-app purchases that I had made under my account didn't transfer over.  Is this the correct behavior? Our phones are set up under different Apple IDs, but I was still able to transfer purchased (paid upfront) apps to her – which is completely allowed under the Terms of Use – so I assumed the same would apply to in-app purchases as well. Let $(B_t)_{t\geq 0}$ be a standard Brownian motion in $\mathbb{R}^d$. It is intuitive that, for fixed $s&lt;t&lt;u$  $$\mathbb{E}[B_t\mid \sigma(B_s,B_u)]=B_s+\frac{t-s}{u-s}(B_u-B_s).$$  However, I cannot think of a way to show this rigorously.   If first attempted to take $A\in\sigma(B_s,B_u)$ and show that $\mathbb{E}[1_A B_t]=\mathbb{E}[1_A(B_s+\frac{t-s}{u-s}(B_u-B_s))]$. But I cannot manage to show this equality.   I'd be very thankful for any ideas and suggestions on how to tackle this problem. I am rendering my scene with only one sample but all the renders of the same picture have a different seed value. How can I combine them correctly? I wrote a java-program that loads all the images and takes the average pixel color for each pixel of the images. Is that a correct way of combining the images? It doesn't reduce the noise very much so I am not happy with it. What can I do? Proving that free modules are flat (without appealing projective modules) Suppose $R\neq 0$ is a commutative ring with $1$. Let $M$ be a free $R$-module. I would like to prove that $M$ is a flat $R$-module. Everywhere I have looked (mostly online) this is proved by first proving that every free module is projective, and then proving that every projective module is flat. Unfortunately, Atiyah &amp; Macdonald's "Introduction to Commutative Algebra" (Chapter 2) does not discuss projective modules. But the result that every free module is flat comes very handy in the exercises.   So my question is,     Is it possible to prove that every free module is flat just by   definitions and without appealing to projective modules?   Thanks! The new 17.10 only has 64 bit ISOs. However 32 bit is still supported, there just are no live ISOs for it.  I have a 32 bit machine with WIFI, and I need to install from a USB disk. I tried the "mini.iso" but it doesn't have WIFI support.  What options do I have?  (PS I don't want another distro, and I can't use ethernet for installation, must be WIFI). I would like to declare functions in tikz for multiple uses of the same function within the code.  Sadly, it seems that the tikz "declare function" is not compatible with the french option in babel. Indeed, the following code (that I simplifed on purpose) runs fine :  \documentclass[english]{article} \usepackage{babel}                       \usepackage{tikz}  \begin{document} \begin{tikzpicture}     \tikzset{declare function={Carre(\t)=\t*\t;}}     \draw plot [domain=-1:1] (\x,{Carre(\x)}); \end{tikzpicture} \end{document}   But when I replace   \documentclass[english]{article}   with   \documentclass[english,french]{article}   compilation generates an error. This is what the log file says :     Missing character: There is no = in font nullfont!   Missing character: There is no @ in font nullfont!   Missing character: There is no @ in font nullfont!      Runaway argument?   -1:1] (\x ,{Carre(\x )}); \end {tikzpicture} \end {document}    ! Paragraph ended before \pgfmath@local@@functions was complete.                          \par    l.18       I suspect you've forgotten a `}', causing me to apply this   control sequence to too much text. How can we recover?   My plan is to forget the whole thing and hope for the best.   Before posting this, I updated all packages without success.  Any help welcome! I am seeking advice on which memory to use with my motherboard and CPU. Below are my components and some relevant information extracted from their official websites.   NOTE: Although this is a very specific case, a general answer is equally good for me — and also alot more useful for everyone else!  CPU: AMD Phenom II X4 810   ASUS M4A785TD-V EVO listed as one of many  motherboards.   Motherboard: ASUS M4A785TD-V EVO   4 x DIMM, Max. 16 GB, DDR3 1800(O.C.)*/1600(O.C.)/1333/1066/800 ECC,Non-ECC,Un-buffered Memory Dual Channel memory architecture (Complete list of specs: ).   NOTE II: I wouldn't want to do any overclocking as I have no experience with that. So I think the 1333 MHz is the most appropriate. Since the higher frequencies are marked with "O.C."  Questions:   How can I be certain that a specific RAM will work well (or not work) together with my CPU and motherboard? (The Qualified Vendor List is a good starting point, but it's a bit limited and sometimes a bit outdated). Is it true that choice of memory is not independent of the processor? How?    I'm worried I'll be missing something essential when buying my next RAM chips :) Google map API does not plot on different continents I have a simple cylinder which is mirrored on both the X and Y axis.  When I scale it along the X axis, the two halves of the cylinder overlap, rather than scale as it would before the mirror modifier was applied:    I haven't used Blender in a while and I know this is a simple fix. Given a logarithm is true, if and only if, $y = \log_b{x}$ and $b^y = x$ (and $x$ and $b$ are positive, and $b$ is not equal to $1$), are true, why aren't logarithms defined for negative numbers?  Why can't $b$ be negative? Take $(-2)^3 = -8$ for example. Turning that into a logarithm, we get $3 = \log_{(-2)}{(-8)}$ which is an invalid equation as $b$ and $x$ are not positive! Why is this? Change rep required to downvote on MSE To make your opinion known on MSE you need both up and downvotes. Given at the moment you need 125 rep to downvote on this site, the voting patterns will be skewed positive because not everyone is active on the site.   This is to be expected for regular Stack Exchange sites as you are expected to be involved in the community before you are deemed knowledgeable enough to be part of the community voice. However, MSE is different because you are voting for the entirety of the Stack Exchange network and you shouldn't need to be active on MSE itself to have a voice on changes to the entire network.  This is especially important for questions like  because it will effect every network to do with coding at all. This was brought up in .  I would suggest reducing the rep required to downvote on MSE down to 100 (or 101) rep. This is the amount you get for the site association bonus. This is given once you achieve 200 rep on any network site and shows that you are active in at least one community. I think this would be the perfect use of the association bonus in showing that you are a member of the Stack Exchange community. Sequential renaming of files I found an answer for  which works fine for renaming filenames as follows :  1.jpg &gt; 00001.jpg 2.jpg &gt; 00002.jpg . . . 9.jpg &gt; 00009.jpg   That uses the following command:  rename 's/\d+/sprintf("%05d", $&amp;)/e' *.jpg   But my filenames aren't numeric. They are an alphanumeric sequence, like: a.jpg, b.jpg, c.jpg...... m.jpg.  In a similar way to the original question, I want to rename these so they display in the correct sequence. How do I do this? Is this statement true?     You can not travel back in time. If I travel fast enough the clock will start to go backwards, but that does not mean I am traveling back in time. It would only mean that the time reference is producing a negative count. In this movie there are scientists who (spoilers ahead)       ...are creating a simulation. One scientist starts seeing odd looking characters but no one else sees them. He asks around and more and more characters are missing or don't know about the characters in the story he knows about.  He finds out that he is part of a simulation. He finds out that the characters only he remembers have been deleted. He also finds out that there are characters from his own simulation that have broken free of his simulation and are in his world.    Hints: I remember this being an older film (70's?); it might have been foreign film possibly with subtitles (not Hollywood); it might have been part of the Criterion collection; I'm pretty sure I streamed it from Hulu. Sorry, I can't remember more details. I have 2 monitors connected to a NVIDIA GeForce GTX 660:   Left is 1920x1080 via DVI (Samsung SyncMaster 2443BW) Right is 2560x1440 via DisplayPort and is set to Primary (Samsung SyncMaster SA850)   I often leave my computer running overnight (working on long render job, doing backups, etc) so I only power off my monitors manually. However, if I power off the primary monitor, it is disabled completely as if it were unplugged from the graphics card (i.e. icons are moved, open apps are moved and resized to fit on the new primary monitor). If I switch from DP to a DVI cable, this does not occur. It seems to be an issue when using DisplayPort.  Is there a way to disable the monitor from being disabled completely in Windows when I power it off physically?   I have found no relevant settings in either Windows Display settings, nor the NVIDIA control panel software.  Edit #1: I'm using an .  Edit #2: Here is a screenshot of my NVIDIA Control Panel v7.2.710.0:   Edit #3: Here are screenshots of my Monitor configuration:    Clicking on Advanced settings reveals a Generic PnP monitor running at 59 Hz? 60 Hz is the only other option available.    Here is my secondary monitor which is connected via DVI:    Edit #4: If I open the "Screen Resolution" configuration panel and power the monitor off (via the power switch), this is what is shown:    It looks like I disconnected the monitor from the video card. Folders Completely Hidden Except via Save As I'm using an audio program called Supercollider on Windows 10. I have saved some files in a folder called "tunes" from SC via Save As, but Windows Explorer and CMD both seem unable to recognise access or recognise the "tunes" folder.  C:\Users\HP&gt;cd "C:\Program Files (x86)\SuperCollider-3.7.2\supercollider-robin\tunes" The system cannot find the path specified.  C:\Program Files (x86)\SuperCollider-3.7.2\supercollider-robin&gt;dir/a doesn't show it either.  However, when saving from SC with Save As, it's there clear as day. I can even shift+right-click to get the path: "C:\Program Files (x86)\SuperCollider-3.7.2\supercollider-robin\tunes"  This makes no sense to me. Can anyone explain and tell me how to access my folder tunes folder please?