I have different users on the machine who use different keyboard layouts.  In 11.04, you could change the keyboard layout on the login screen (and it would remember it per user).  How can I do this in 11.10?  Or is it not possible?
I have a netbook with an AZERTY layout. I've changed it to "bépo" to try it, but now I've reset it to a french layout.  While all the apps have taken the change into account, Ubuntu's login manager still uses bépo, making entering passwords a challenge. I've enabled the virtual keyboard for the time being so that I could login again.  I don't remember what I had done in the first place to switch the login screen to bépo, but now I can't switch it back.    EDIT : In the keyboard settings page, if I click "use system defaults", then everything is in bépo ! Where is this different keyboard layout set ?  ANOTHER EDIT : Here's what did not work :   Changing the layout in settings using the "use system defaults" checkbox in settings sudo dpkg-reconfigure keyboard-configuration changing it in Xconf adding a line in /etc/lightdm/lightdm.conf  sudo dpkg-reconfigure locales slaming head against desk dpkg-reconfigure lightdm sudo gsettings set org.gnome.libgnomekbd.keyboard layouts "['fr']" combinations of the above     Is it possible to relaunch the keyboard configuration tool that's in ubuntu's setup ? The thing has automatic layout detection and it changes the setting everywhere.
The full prophecy by Trelawney, the part my question is about in bold:     "The one with the power to vanquish the Dark Lord approaches ... born to those who  have thrice defied him, born as the seventh month dies ... and the Dark Lord will mark him as his equal, but he will have power the Dark Lord knows not ... and either must die at the hand of the other for neither can live while the other survives ... the one with the power to vanquish the Dark Lord will be born as the seventh month dies ..."    – Sybill Trelawney (to Dumbledore)    I am interpreting      for neither can live while the other survives   as: Harry can't live while Voldemort survives and vice versa. But why? It clearly worked until the end of book 7. One could argue with Dumbledore's thery of the self-fulfilling prophecy. But clearly Voldemort was smart enough to see that there was no mechanism that would kill him just because Harry stayed alive.    and either must die at the hand of the other    Again: Why? If somebody else destroyed all Horcruxes and then killed Harry, Voldemort would be just as vulnerable and could be killed by anyone. On the other hand, Harry could have been killed by anyone - his mum's protection transferred to Voldemort by Harry's blood was only helping against Voldemort.  I guess that I am interpreting this part of the prophecy in the wrong way or maybe there is an alternate, less obvious and more tricky one. So my question is: What exactly does the highlighted part of Trelawney's prophecy state? What kind of material is the Robert Downey Jr. Iron Man's armor supposed to be made from? I know he crafted the prototype from the things he had on hand, but once he was able to build the next one(s) in his own lab, what hi-tech material did he use for the movie armor? How can I format a date as dd/mm/yyyy or mm/dd/yy ?   Like in VB format("dd/mm/yy",now)  How can I do this in C#? Fast way of copying IMAP to IMAP on CentOS I'm trying to copy of IMAP emails from one server to my new server but after Googling for hours and trying several tools, nothing is working. Is there a tool that actually works?  I am copying hundreds of thousands of emails so using a email client is out of the question, I tried it, and at the rate of a email per 2 seconds, it would take a few weeks or months!  www.linux-france.org/prj/imapsync/ is commonly mentioned, however, the tool is no longer available.  I tried www.athensfbc.com/imap_tools/ which just hangs and does nothing.  Any migration tools that work? It shows no question owner, not even a simple name (without link).   What happened? Is there a word for four times as much, analogous to once, twice, and thrice? Is there a word for 'four times as much', analogous to once, twice, and thrice? I am looking for help to point me out at a very old novel (likely from the '60s) about two empires at interstellar war. When it comes to starships, one is technologically very advanced, and the other has very low and inefficient technology.  The highly advanced empire has apparently strong chances to win the war, in particular considering a new technology they were developing. I remember the technology to work like this: consider the space as a flat surface, and on this surface, a disk around the spaceship. When the technology was turned on, this disk was stretched out, as made of rubber, so that it produced a cone, with the tip being the spaceship. The spaceship was therefore teleported 10 parsec away (if I remember correctly), but retained all the connections to the original position. If the spaceship moved, the cone moved as well. This allowed to move in position among enemy ships completely unnoticed, by stretching while afar, then moving in the proper position 10 parsecs away, then destretching.  Despite this technology, the empire lost the war because of delays. It's a perfect example of how having a disruptive technology does not guarantee to win against the competitors.  I am trying to find out the author and name of this novel, hopefully with the date of publication and in which journal/magazine. I've discovered a rather weird behaviour of PdfLaTeX. In this example  \documentclass{article} \usepackage[breaklinks]{hyperref} \begin{document} \url{http://www.emarketer.com/blog/index.php/quick-stat-smartphone-users-account-38-mobile-phone-users/} \end{document}   PdfLaTeX won't break the URL and create a too long line. However, if I replace \url with \href like this:  \documentclass{article} \usepackage{hyperref} \begin{document} \href{http://www.emarketer.com/blog/index.php/quick-stat-smartphone-users-account-38-mobile-phone-users/}{http://www.emarketer.com/blog/index.php/quick-stat-smartphone-users-account-38-mobile-phone-users/} \end{document}   I get the line broken, but the font of the link is not right any more of course (normal instead of texttt). Why doesn't URL break the line at dashes and what could I do to get a nicely typeset line-broken link? Prove $3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$ if $x,y,z$ are positive real numbers,Prove:$$3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$$ Additional info:$\sum_{cyc}$ denotes sums over cyclic permutations of the symbols $x,y,z$. I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.  Things I have done: Writing Left hand side in expanded form:$$LHS=3(\sum_{cyc}x^3y^3+\sum_{cyc}xy^4z+3x^2y^2z^2)$$ Right now I don't have good idea were to start(Is expanding and looking for AM-GM is good?).So any hint is appreciated. Source : This inequality can be found as Problem 42, page 13, in the book &quot;Old and New inequalities&quot; by Titu Andreescu, where it is attributed to Manlio Marangelli. Principal component analysis "backwards": how much variance of the data is explained by a given linear combination of the variables? I have carried out a principal components analysis of six variables $A$, $B$, $C$, $D$, $E$ and $F$. If I understand correctly, unrotated PC1 tells me what linear combination of these variables describes/explains the most variance in the data and PC2 tells me what linear combination of these variables describes the next most variance in the data and so on.   I'm just curious -- is there any way of doing this "backwards"? Let's say I choose some linear combination of these variables -- e.g. $A+2B+5C$, could I work out how much variance in the data this describes? Refer to the "name" of an equation, while a List of Equations is generated using these names In what order do multiple readied actions with the same trigger resolve? Orange squares appearing on render? I'm trying to render a scene in Blender, and as of yesterday (it was not doing this when I was working on it 2-3 days ago), I am seeing orange squares dancing around the rendered image. Google tells me that these are render tiles (I think), but what are they, why are they there, and how do I get rid of them? WITH ROLLUP WHERE x IS NULL "PuTTY key format too new" when using ppk file for PuTTY SSH key authentication I wanted to connect to my Unix server using SSH keys. For that I had converted my id_rsa file from ./ssh directory to a ppk file using PuTTYgen, but after loading my .ppk key I get this warning:  PuTTY key format too new   If I use my default id_rsa file, then I get error:  OpenSSH SSH-2 private key (old PEM format)  Either way I am getting stuck. Can anyone please help me? Why must the eigenvalue of the number operator be an integer? My apologies if this question has already been answered. Using the notation of Sakurai, we have an energy eigenket N and an eigenvalue n.  $$N | n \rangle = n | n \rangle$$  For the number operator $N = a^\dagger a $. My question is precisely why does n have to be an integer. I do understand that it must be positive definite due to the norm condition on $ a | n \rangle $:  $$\langle n | a^\dagger a | n \rangle \geq 0$$  I can see that we can show:  $$ a | n \rangle = \sqrt{n-1} | n-1 \rangle $$ and similarly for $a^\dagger | n \rangle$  Sakurai argues that the sequential operation of $a$ operators leads to the form which he does not write, but I will write here for clarity:  $$ a^k | n \rangle = \left(\sqrt{n}\sqrt{n-1}...\sqrt{n-k+1}\right) | n - k + 1 \rangle $$   Since  $$ a|0 \rangle = 0 $$  the sequence must terminate, but can only terminate if n is an integer. How can we make that jump? Is there a formal mathematical proof that the sequence $\left(\sqrt{n}\sqrt{n-1}...\sqrt{n-k+1}\right)$ will terminate if and only if n is an integer? How can you change the href for a hyperlink using jQuery? is $y = \sqrt{x^2 + 1}− x$ a injective (one-to-one) function? I have a function  $y = \sqrt{x^2 + 1}− x$ and I need to prove if it's a Injective function (one-to-one).  The function f is injective if and only if for all a and b in A, if f(a) = f(b), then a = b  $\sqrt{a^2 + 1} − a = \sqrt{b^2 + 1} − b$  $\sqrt{a^2 + 1} + b = \sqrt{b^2 + 1} + a$  $(\sqrt{a^2 + 1} + b)^2 = (\sqrt{b^2 + 1} + a)^2$  $a^2 + 1 + b^2 + 2b\sqrt{a^2 + 1} = b^2 + 1 + a^2 + 2a\sqrt{b^2 + 1}$  $2b\sqrt{a^2 + 1} = 2a\sqrt{b^2 + 1}$  $b\sqrt{a^2 + 1} = a\sqrt{b^2 + 1}$  $(b\sqrt{a^2 + 1})^2 = (a\sqrt{b^2 + 1})^2$  $b^2(a^2 + 1) = a^2(b^2 + 1)$  $b^2a^2 + b^2 = a^2b^2 + a^2$  $b^2 = a^2$  $\sqrt{b^2} = \sqrt{a^2}$  $b = \pm a$  then the function is not injective because $b$ in not equal to $a$, but using online  the function is injective. Even when /tmp has no file called something, searching for it with find will return 0:    $ find /tmp -name something    $ echo $?   0   How can I get a non-zero exit status when find does not find anything? Another simple series convergence question: $\sum\limits_{n=3}^\infty \frac1{n (\ln n)\ln(\ln n)}$ I'm being asked to determine if $\displaystyle\sum\limits_{n=3}^\infty \frac1{n (\ln n)\ln(\ln n)}$ converges. So, using Cauchy's Condensation Test, I reduced the problem to one of determining the convergence of $\displaystyle\sum\limits_{n=3}^\infty\frac 1{n\ln (n\ln 2)}$. Am I on the right path, and how do I proceed from here?