I login Stack Overflow with a Google Account, but I can not find a button or link to let me log out. The only way I got so far is clear my browser's cookie...  Do any guys know about it? Thanks.
Today I open my SO account like I always do. But the UI of menus has changed. I am not able to see the chat tab, the logout link or the privileges link. Are they missing or moved somewhere else?  Changed UI:
Process descendants If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$ Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is  Try: From $$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ $$=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})+2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$ $$=2-2(a_{1}+a_{3})(a_{2}+a_{4})=2+2(a_{1}+a_{3})^2\geq 2$$ and equality hold when $\displaystyle a_{1}=-a_{3}$ and $a_{2}=-a_{4}$ Could some help me how to find its upper bound, Thanks What noun or adjective would one give to a person that is intolerant of stupidity? A person that becomes annoyed, angry, or of very short temper when conversing or interacting with people of a lower intelligence?  For example:       The man is very xyzxyz; his son can't do multiplication very well.     or     She is a(n) xyzxyz. She gets very frustrated when people forget things.   I'm looking almost for a scientific/medical/psychological term. One that would fall into the same kind of group as words like haemophiliac, masochist, nihilist, etc. I am having some issues when rendering an image, my rendering screen is turning into grey even when making a full render. What am I doing wrong? I've just downloaded a CD image of Ubuntu. I know the installer has a check CD option, but how can I check that the image is good before burning it to CD? How can I tell if two Minecraft mods are compatible? If I want to use two mods together, e.g.  and  (which don't state compatibility with one another), how can I tell if they will work together (and not break Minecraft - forcing me to re-download a fresh copy)? Should reputation from SO carry over to Meta? Representing several points with same coordinates on map using ArcGIS Desktop? I'm working on Arcgis 10.  I have a point shapefile representing events with several points having coinciding coordinates (They have the same coordinates). For representing them in a dynamic way, I found a nice way - .  Problem is: I was asked to produce a static (paper) map of the events. Is there a tool that can take these events and shift them apart a bit so all will be seen?  (I know that spreading them about will alter their coordinates, but I am prepared to do so on a duplicate shapefile for the display on the map). Hide specifics messages in chat What is the "Bootable flag" option when installing a distro? Is the "bootable flag" needed in today's distributions? If not, then why is it still in the installers? What is it exactly? Does a size large or larger creature killed by Negative Energy Flood rise as a medium sized zombie? I usually compress my tex PDFs with ghostscript:  pdflatex -shell-escape file.tex  gs -sDEVICE=pdfwrite        \    -dCompatibilityLevel=1.4   \    -dPDFSETTINGS=/prepress      \    -dNOPAUSE -dQUIET -dBATCH      \    -sOutputFile=file_compressed.pdf \     file.pdf   Normally that works well, but if I have relative links to local files, the links don't work afterwards.  Minimal example:  \documentclass{beamer} \usepackage{hyperref}  \begin{document}     \begin{frame}         \href{run:./audio.mp3}{test}     \end{frame} \end{document}   The link works in file.pdf but not in file_compressed.pdf. Is there any pdflatex or ghostscript setting that retains the link in the compressed version?  I use Acrobate Reader 10.1.14 for viewing, Ghostscript 9.10 and pdfTeX 3.14159265-2.6-1.40.15 (TeX Live 2014) on OS X 10.9.5.    Update: The breakage of hyperlinks had nothing to do with compressing the PDFs. The breakage happened for all re-processing of PDFs containing hyperlinks in its input PDF. I am looking to identify a story that I read around 2001 (it is probably much older).  Someone showed me the story on pages printed from the internet, and it was attributed to Isaac Asimov (although I cannot state that is the true author).    The story is told in glimpses at different points of time in the future.  In each glimpse, humans are asking a more and more "intelligent" computer how to prevent the universe from collapsing into a singularity in the future, and the computer states it needs more computational power and resources to solve this problem.  At some point, man becomes one with the computer, and the computer exists in hyperspace (or subspace?).    In the final glimpse, it is the end of time and the universe is collapsing into a singularity.  At the instant the universe reaches the singularity, the computer figures out how to prevent it and says "Let there be light" and the universe begins to expand again.  Is anyone familiar with this story? Proving $\sqrt2$ is irrational I used the method of contradiction by assuming that $\sqrt 2$ is a rational number. Then, by the definition of rational number, there exist two integers $p$ and $q$ whose ratio equals $\sqrt 2$. Thus, $$\frac pq = \sqrt2\tag{x}$$  Squaring both sides, $$p^2/q^2 = 2\tag{a}$$  or $$p^2 = 2q^2\tag{b}$$ This means that $p^2$ is an even number, implying that $p$ is even.  Now, any even integer can be written as $2^kf$ where $f$ is any odd integer and $k$ is some positive integer (the minimum value of $k$ and $f$ is $1$ since even numbers start from $2 = 2^1\cdot1$). For odd numbers, $k=0$.  For example, $8 = 2^3 \cdot 1$, $4= 2^2 \cdot 1$, $18= 2^1 \cdot 9$, $24 = 2^3 \cdot 3$, $-12= 2^2 \cdot(-3)$, etc.  Now, from $(b)$, $q^2$ can be even or odd.  Case 1: $q^2$ is even (thus meaning $q$ is even). Then $p= 2^{k_1} \cdot f_1$ (say) and $q = 2^{k_2} * f_2$ (say). Note here in this case, both conditions $k_1=k_2$ and $f_1=f_2$ can't hold simultaneously since that would mean $p/q =1$ and here $p/q =\sqrt2$ (from $(x)$)). Let's consider the condition when $k_1=k_2=k$ but $f_1\ne f_2$. Then $$\frac{p^2}{q^2}=\frac{(2^kf_1)^2}{(2^kf_2)^2} = \frac{f_1^2}{f_2^2} =2$$ (from $(a)$), i.e.: odd/odd ($f_1$ and $f_2$ are odd) can never equal $2$.  Now lets consider $f_1=f_2=f$ but $k_1\ne k_2$, thus $(2^{k_1}f)^2/(2^{k_2}f)^2= 2^{2(k_1-k_2)}= 2$ meaning $k_1 - k_2 = 0.5$ but $k_1,k_2$ are integers, so their difference can't be $0.5$. (Also here, $k_1-k_2$ must be greater or equal to $1$ since $2^{k_1-k_2}=\sqrt 2$, $k_1-k_2$ can't be negative since $\sqrt 2&gt;1$, but $k_1-k_2\ge1$ can't satisfy $(a)$ since the minimum value of $p/q$ in this case will be $2$ which is surely greater than $\sqrt2$.)  Now lets consider $f_1\ne f_2$ and $k_1\ne k_2$, thus $\frac{(2^{k_1} f_1)^2}{(2^{k_2} f_2)^2}= 2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal $2$ since $\frac{f_1^2}{f_2^2}$ is either odd or "odd/odd" ie: it doesn't contain 2 as a factor and $2^{2(k_1-k_2)}$ is a power of $2$. So in $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$, there's no chance of cancellation of the odd factor $\frac{f_1^2}{f_2^2}$.  Case 2: $q^2$ is odd (thus meaning $q$ is odd). $q=f'$ (say, here $k=0$ for $q$ but not for $p$), thus from $(a)$ $2^{2k}\frac{f^2}{f'^2} = 2$ but this isn't possible since a power of $2$ multiplied with odd factor can't equal $2$.  Thus, both case 1 and 2 suggest that for any possible combination of $k_1,k_2,f_1$ and $f_2$, $p/q\ne\sqrt2$, i.e.: for no value of integers $p$ and $q$, $p/q = \sqrt2$. Thus, this contradicts our assumption that $\sqrt2$ is rational. Therefore it must be irrational.  PLS NOTE that i would like to clarify that $f_1^2$/$f_2^2$ is either an odd integer or a fraction of odd/odd or 1/odd form, thus not containing  2 as a factor,for any $2^n$, if it has to be reduced to 2 , must be multiplied with $1/ 2^(n-1)$ but $f_1^2$/$f_2^2$ can't be of $1/2^[n-1]$ form since for odd nos, in $2^k$*f notation, k is 0 , so 2 vanishes as $2^k$ becomes 1 in this case thus f1/f2 can't be represented as  1/ $2^(n-1)$ in which n has to be existent.  therefore $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal 2 as to further explain the 3rd condition of case 1  Difficulty: is my approach correct? This proof which I thought is different from proofs found on the internet or books since I have used $p$ and $q$ as any integers, which may have a common factor. So I am not sure if I am on right track. Will someone please check this out and make kind comments? Where do I place my own .sty or .cls files, to make them available to all my .tex files? I know how to define a package or a class, but how do I make it available (to myself) at all times?  Say I have the package file commonstuff.sty (or myprettyclass.cls), that I want to be able to include in any .tex file I create on my computer. Where do I put it? Do I need to run some installer on it, or is it enough to keep it in a certain folder?  Note: I know I can just place it in the same folder as the .tex file, but that's not what I'm after here. I'm looking for a way to "install" the package centrally on my computer. Poisson distribution (solution check) Is there a word in English to qualify the highest-ranked student in a class in a year?  We are talking here about higher education, especially in the context of curricula that keep a given set of students over the course of several years, being ranked at the end of each year, possibly with competitive admission for the following year. The ranking is a weighted average of all marks obtained over the year.  I am looking for a single word, or expression, to quickly designate the one student with the highest ranking. If there is no such thing in English, then let's try to get the most concise and idiomatic definition.  In all cases, it is not emotionally connoted, i.e. not pejorative nor bragging.  For French speakers: I am looking for the translation of “major de promotion”.    EDIT: additional, not as much of a pure “word request” question: is this anything you should mention on a curriculum vitæ in the English-speaking world? If so, please mention whether the given word is appropriate for such a use&nbsp;&nbsp;:) Should I be offered the option to upgrade to 18.04.1? I understand that the first point release of Ubuntu 18.04, namely 18.04.1 is now available and that my system should now offer it as an update.  My current OS version is 16.04.5 LTS. In System Settings/Software and Updates I have set 'Notify me of a new Ubuntu version' to 'For long term support versions'. I have tried both the main server and the server for United Kingdom, neither offers an update when Software Updater is clicked or via the Ubuntu Software icon.  Can anyone please tell me why I don't get the option of updating to 18.04?  Edit:  I don't remember the details exactly, but I think that my last LTS point release update was offered automatically a lot sooner after the published release date than where we are now with 18.04.1.  Edit 10th August  It appears that there has been some problem with the update process and that it should be fixed next week, see:    Edit 14th August 18.04.1 LTS now available from UK server, Most games build on other games' successes and many are downright clones. Where is the limit of imitation before legal issues come into play? Is it down to literary or graphic work like characters and storyline that cause legal problems, or can someone actually claim to own gameplay mechanics?  There are so many similar clone games out there that the rules are probably very slack or nonexistent, but I'd like to hear the views of more experienced developers / designers. When to use a hyphen to coin a new word and when to omit a hyphen? Someone has asked for answer to these topics. However, I still want someone to provide me with simple and universal answers. I recently read a sentence from the English-speaking person. It is in an article. It is, ' The available options for mitigating pad cratering are not universally effective (see link below), but they may be helpful on a case by case basis until more crack resistant laminates are developed and become available.  '  If I am to write this, I will use 'a case-by-case basis until more crack-resistant laminates'   Can anyone tell me when to use a hyphen to coin a new word?