Please someone tells me why, after installing Ubuntu 12.04, that my clock displays the incorrect time. My operating system time is delayed several hours, and I don't know how to fix it. Thank you.
Dual boot system Windows XP Pro and Ubuntu 12.04.  I have the bios set for the correct time and Ubuntu set for US Eastern time.  Ubuntu will boot up and the time will be off by -4 hours.  If I correct the time in Ubuntu then when I boot up in Windows XP the time will be off by + 4 hours.  This is a fresh install of 12.04. I did not have this problem before reinstalling Ubuntu.
Exponent and abelian groups What exponent $e$ can guarantee the group to be abelian? Are there any known results except the case $e = 2$? Thanks for help. How does the momentum operator act on state kets? I have been going through some problems in Sakurai's Modern QM and at one point have to calculate $\langle \alpha|\hat{p}|\alpha\rangle$ where all we know about the state $|\alpha\rangle$ is that $$\langle x|\alpha\rangle=f(x)$$ for some known function $f$. ($|\alpha\rangle$ is a Gaussian wave packet.) Sakurai says that this is given by:  $$\langle p\rangle = \int\limits_{-\infty}^{+\infty}\langle\alpha|x\rangle\left(-i\hbar\frac{\partial}{\partial x}\right)\langle x|\alpha\rangle dx.$$  I am wondering how we get to this expression. I know that we can express   $$|\alpha\rangle =\int dx|x\rangle\langle x|\alpha\rangle$$  and  $$\langle\alpha|=\int dx\langle\alpha|x\rangle\langle x|,$$   so my thinking is that we have:  $$\langle\alpha|\hat{p}|\alpha\rangle =\iint dx dx'\langle\alpha|x\rangle\langle x|\hat{p}|x'\rangle \langle x'|\alpha\rangle, $$  and if we can 'commute' $|x\rangle$ and $\hat{p}$ this would become: $$\iint dxdx'\langle\alpha |x\rangle\hat{p}\langle x|x'\rangle \langle x'|\alpha\rangle,$$  which is the desired result as $$\langle x|x'\rangle=\delta(x-x').$$ Is this approach valid?  I think my question boils down to: Does the operator $\hat{p}$ act on the basis kets $|x\rangle$ or on their coefficients? In the latter case, if we had some state $|\psi\rangle = |x_0\rangle$ for some position $x_0$, then would we say that for this state $$\langle p\rangle =\langle x_0|\left(-i\hbar\frac{\partial}{\partial x}\right)|x_0\rangle = 0$$ as the single coefficient is $1$ and the derivative of $1$ is $0$? Why are we using the squared residuals instead of the absolute residuals in OLS estimation?  My idea was that we use the square of the error values, so that residuals below the fitted line (which are then negative), would still have to be able to be added up to the positive errors. Otherwise, we could have an error of 0 simply because a huge positive error could cancel with a huge negative error.  So why do we square it, instead of just taking the absolute value? Is that because of the extra penalty for higher errors (instead of 2 being 2 times the error of 1, it is 4 times the error of 1 when we square it). I have a realy strange problem in Ubuntu 11.10. I have by accident my password set to empty. If I wan't to install software now the system asks for a password, but I haven't one anymore. If I enter my old password it sais it's wrong and if I leave it empty it also does. Now I wan't to set my password back, but the system asks for the old password before I can change it. That password doesn't exist, so I can't do anything anymore! Can someone help me please? Use induction to prove the following: $1! + 2! + .... + n! &lt; (n + 1)!$  Base case:  $n = 1$  $1! &lt; 2!$ true   Inductive step: Assume that $1! + 2! + .... + k! \le  (k + 1)!$ is true   let $n = k + 1$  $1! + 2! + .... + (k + 1)! &lt; (k + 2)!$  $1! + 2! + .... + k! + (k + 1)! &lt; (k + 2)!$  $1! + 2! + .... + k! + (k + 1)! &lt; (k + 2)!$  $(k + 1)! + (k + 1)! &lt; (k + 2)!$  $2(k + 1)! &lt; (k + 2)!$  $2(k + 1)(k)(k - 1).. &lt; (k + 2)(k + 1)(k)(k - 1)....$  $2 &lt; (k + 2)$   the right is always bigger for k are non negative integer. order of a sylow p-subgroup What do you call an 'unselfish' action made with a selfish reason? There are many examples of this, and I'd like to give a few:   A person who puts a lot of effort to help the community and earns reputation points. But that reputation is the motivation behind helping the community, not because they really want to. A student approaches a professor, and helps them in their teaching and doing research in order to have a good LOR (letter of recommendation). To a SEO, the content of the website must be good (well, this one is a little tricky, but you get the spirit.)   I'm not saying that the reputation/LOR/high-ranked is not deserved. It is. I just want to focus more on the selfish reason(s) behind that.  Any kind of answer can be accepted: single word, phrase, proverb or idiom. I have the following code for my masterpage:  &lt;%@ Master Language="C#" AutoEventWireup="true" CodeFile="MasterPage.master.cs" Inherits="MasterPage" %&gt; &lt;?xml version="1.0" encoding="UTF-8"?&gt; &lt;!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"&gt;  &lt;html xmlns="http://www.w3.org/1999/xhtml"&gt; &lt;head runat="server" style="position:relative"&gt;     &lt;title&gt;Masterpage&lt;/title&gt;     &lt;asp:ContentPlaceHolder id="head" runat="server"&gt;     &lt;/asp:ContentPlaceHolder&gt; &lt;/head&gt; &lt;body&gt;     &lt;form id="main" runat="server" style="position:relative"&gt;     &lt;div&gt;         &lt;asp:ContentPlaceHolder id="MainContent" runat="server"&gt;         &lt;/asp:ContentPlaceHolder&gt;     &lt;/div&gt;     &lt;/form&gt;     &lt;br style="clear:both" /&gt;     &lt;form id="footer" style="position:relative"&gt;     &lt;div&gt;         &lt;center style="font-size:small"&gt;Footer content&lt;/center&gt;     &lt;/div&gt;         &lt;/form&gt; &lt;/body&gt; &lt;/html&gt;   My problem is that when the MainContent isn't filling up the page the footer is not at the bottom of the page. Is there a way to force the footer to stay on the bottom of the page if the main content isn't pushing it all the way down? When the main content is "big" enough it pushes the footer down with the content and displays on the bottom when scrolling down, but not if it is to small.  Anybody know how to fix this? How can you sync Blender configs between multiple machines? Where is this picture from? How do I use Reverse Image Search to find the source of an anime/manga image? So I see an image that I believe is from either an anime or a manga. What are some ways that I can go about finding what series it's from? Try these options:   (browser built-in support and plugins)        Note: Please note that as of May 9, 2016, ALL  questions are now . Therefore, you can't ask an identification request question on this site. However, such questions are still welcome  for users with at least  (globally on StackExchange).    Protip: If you are using Google Image Search in a non-English language,  or just use  for better, more accurate results. I mean to ask  that $7$ is an arithmetic  expression  but it can also  be  written  as $7x^0$.  which is a constant polynomial  expression.  Every polynomial  expression  is an algebraic  expression  so with  this logic is $7$ an algebraic  expression  or an arithmetic  expression.  Further $7=5+2$, is it an arithmetic  equation  or algebraic  equation  because  it can also be written  in  $7x^0 = 5x^0 + 2x^0$ form. please help.  Thank you. Change how questions marked as duplicates by the asker are represented Situation: when a user self-marks their post as a duplicate, it shows as closed by "community" rather than the original poster.  Problems: this is not clear at all to most users of SE what has happened. It looks like the question was "mod closed" as a duplicate against the asker's wishes.    Suggestion: instead of the community user "closing" the post, have the final close user be the asker (or make it "Community &amp; Asker"). This will help clear up confusion and streamline the process for understanding what happens. Initiate an oAuth flow via an Auth Provider I have a directory which contains numbered image files, something like this:  01.png 02.png 03.png 03.svg 04.png 05.png 06.jpg 07.png 08.png 09.png 09.svg 10.png   Sometimes there may be multiple versions of a file in different formats (eg. a png and svg version of the 03 and 09 files above) but the numbers are otherwise consecutive.  Typically there are 40-80 such files in each directory.  The numbers correspond to the order these images appear in a manuscript (a Word document, but that's not important).  There is no other way to determine the order of the images.  If I add a new image to the manuscript I need to place a copy of the image in this directory with the correct numbering.  So if the new image is the fifth in the manuscript I need to rename the files in the directory to this in order to make room for it:  01.png 02.png 03.png 03.svg 04.png 06.png 07.jpg 08.png 09.png 10.png 10.svg 11.png   What is the most straightforward way from the command line, or from a script or macro to renumber all the files starting at a certain number?  I have a standard Fedora Linux install using bash. Show $x^p-t$ has no root in the field $\mathbb{F}_p(t)$ I don't think I fully understand.  Let's say there is a root $x_0 \in K=\mathbb{F}_p(t)$, where $p$ is a prime number.  Then $x_0 = \frac{P(t)}{Q(t)}$ for some polynomials $P,Q \in \mathbb{F}_p[t]$. We can assume $\gcd(P,Q)=1$  and $x_0^p-t= \frac{(P(t))^p}{(Q(t))^p}-t= \frac{P(t)^p-tQ(t)^p}{Q(t)^p} =0$, so coefficients of $P(t)^p, tQ(t)^p$ must be identical, which contradicts $\gcd(P,Q)=1$,   hence such $x_0$ does not exist and the polynomial has no root in $K$.   am I about right?    anyway, I would appreciate an explanation about $\mathbb{F}_p(t)$, what is $t$? what is the meaning of a variable which does not belong to any specific "world"? I cannot use $t$ as if it was a member of $\mathbb{F}_p$ and hence cannot assume $t^{p-1} = 1 \pmod p$... What's the correct way to do division using a backslash? If I write $a/bc + d$, will that be equal to $(a/(bc))+d$?   Basically, if I place a slash, will I divide by what's directly behind the slash $(b)$, the term that's behind the slash $(bc)$, or everything after the slash $(bc+d)$? How to install TeX on Android phones? I used  app in my Android phone. But this app does not support XeLaTeX and requires an internet connection to be used.  Is there any distribution of TeX for Android? How I can install TeX on my Android phone? How do the tens­es and as­pects in English cor­re­spond tem­po­ral­ly to one an­oth­er? Non-na­tive speak­ers of­ten get con­fused about what the var­i­ous tens­es  and as­pects mean in English. With in­put from some of the folk here I've put to­geth­er a di­a­gram that I hope will pro­vide some clar­i­ty on the mat­ter.  I of­fer it as the first an­swer to this ques­tion. Con­sid­er it a liv­ing doc­u­ment. In­put is wel­come, and good sug­ges­tions will be  in­cor­po­rat­ed in­to the di­a­gram.    No­ta bene: What this is not is a dis­cus­sion of whether there are  more than two tens­es in English. , to which this ques­tion is not in­tend­ed to sup­ply ar­gu­ments one way or the oth­er.  Here, the aim is to pro­vide an overview of what con­struc­tions English-speak­ing peo­ple use for con­vey­ing in­for­ma­tion about ac­tions re­fer­ring to past, present, and fu­ture, and to pro­vide it first and fore­most to pre­cise­ly the peo­ple who are like­ly to use "tense" as a catch-all term in their search, rather than to lin­guists who know bet­ter.  Break­ing News There is now an ex­cel­lent ELU blog ar­ti­cle ti­tled  . It is high­ly rec­om­mend­ed read­ing. Calculating Image boundary / footprint of satellite images using open source tools? Show that $\langle x,y\rangle_A = \langle Ax,Ay\rangle$ is an inner product on $\mathbb R^n$ Let $A$ be an $n \times n$ matrix with real enteries. Define $\langle x,y\rangle_A = \langle Ax,Ay\rangle, \quad x,y \in \mathbb R^n$ , where $\langle,\rangle$ is a standard inner product on $\mathbb R^n$. Then  $\langle x,y\rangle_A$ is an inner product iff  1)  $\ker A = \{0\}$.  2) $\operatorname{rank} A= n$.  3) all eigenvalues of $A$ are positive.  4) all eigen value of $A$ are non negative.  Clearly if $\ker A \neq \{0\}$, then there exist $x \neq 0$ such that $Ax =0$, then $\langle Ax, Ax\rangle_A = 0$, then $\langle ,\rangle_A$ is not inner product, so $\ker A = \{0\}$, thus rank of $A$ is $n$.  I am unable to check the 3) and 4) option. Please help to check the other option. Thank you