what happens with commanders and o ring effects?
How does the Commander interact with Banishing Light or Oblivion Ring?
I can enter U+0008 (backspace character) into a chat message I typed . I believe I tab-completed @rumtscho and fully typed @MarthaF (since she doesn't tab complete). However, other people tell me that a Unicode character was inserted into the message; I can't see it:  Originally, I abandoned this bug report for a lack of reproducibility. But on another machine, I can see the character and managed to reproduce it: press tab, then press backspace. It works anywhere in a message, not just in a @-ping. (Actually, it seems that you just need to un-focus the input box; clicking elsewhere then hitting backspace works, too)  First machine is Firefox 12 (Iceweasel) on Debian GNU/Linux AMD64, testing+unstable. No user scripts. Greasemonkey is installed, but disabled in the extension manager Second machine is Firefox 10 (Iceweasel) on Debian GNU/Linux AMD64, testing+unstable. No user scripts, Greasemonkey not even installed Third machine (just tested) is Firefox 11 (Iceweasel) on Debian GNU/Linux AMD64. Greasemonkey installed and enabled, and there is a global userscript. Does not happen on Chromium 18.0.1025.151, running on machine #3.  So far, that's all the machines I've tested on, and it happens on all of them. It only happens in chat. Easiest way to produce it is to click on the page background, somewhere outside the text box, to un-focus the text box. Then press backspace. Note that not all browsers will display the character, even if its there (check  to see if yours does) Occasionally, I get part of an application lingering behind on my desktop:    This is part of a context menu for an application. It is always top-most, meaning that it is always on the screen, and I cannot click on it (it just clicks through it). It has happened before, sometimes with whole windows. It remains on screen even if the application it originated from is no longer running.  How do I get rid of it without rebooting? Why is it called an “Indian file”? Using Illustrator CS6 to create shape out of multiple paths I am very new to Illustrator, in fact I am only experimenting with it in an attempt to help my friend develop a logo for a new business venture. I managed to draw out the shell of the logo using the line tool and modifying the lines with the anchor tool to get the shape I wanted.  However, I would like to somehow take all of these paths and create one object, or symbol which I can then apply a drop shadow, fill and various other effects too. Currently when I try to do this I get unwanted effects...  I'd appreciate any help on the subject. I had one batch class in which i am passing some data required for creating object record.But when i call that batch class from parent class variables are initialized but start method are not called as there is no record to query i also tried with iterable object but not able to do that.My code is as below  global class testBatchClass implements Database.Batchable&lt;SObject&gt;,Database.Stateful{     public List&lt;sObject&gt; sObjectList;     public string objectName;     public String dataList;      public testBatchClass(String objectName,String inputdata){          String listType = 'List&lt;'+objectName+'&gt;';         system.debug('list type value is'+listType);         List&lt;SObject&gt; castRecords = (List&lt;SObject&gt;)Type.forName(listType).newInstance();         this.sObjectList = castRecords;         this.objectName = objectName;         this.dataList = inputdata;      }     global Iterable&lt;SObject&gt; start(Database.BatchableContext batchableContext) {         return new sObjectList{};     }     global void execute(Database.BatchableContext batchableContext, List&lt;sobject&gt; scope){          List&lt;OpportunityRenewal__mdt&gt; mdtList = new List&lt;OpportunityRenewal__mdt&gt;([Select id,Api_Name_of_Fields__c,Sobject_Name__c from TestFileUploadCustomSetting1__mdt where Sobject_Name__c=:objectName]);         List&lt;String&gt; fieldName = new List&lt;String&gt;();         for(OpportunityRenewal__mdt md : mdtList){             if(md.Sobject_Name__c == objectName){                 fieldName = md.Api_Name_of_Fields__c.Split(',');                 break;             }         }         List&lt;String&gt; dataRowsSplit= dataList.split(',');         dataRowsSplit.remove(0);         for(String eachRow: dataRowsSplit){             List&lt;String&gt; splitString = eachRow.Split(',');             sObject sobj = Schema.getGlobalDescribe().get(objectName).newSObject();             for(Integer i=0; i &lt;= splitString.Size(); i++){                 sobj.put(fieldName[i],splitString[i]);             }             scope.add(sobj);         }         if(scope != Null &amp;&amp; scope.Size() &gt; 0){             upsert scope;         }     }      global void finish(Database.BatchableContext batchableContext) {} }   Thanks Chat related privileges are not site specific Let $A$ be an $n\times n$ complex nilpotent matrix. Then we know that because all eigenvalues of $A$ must be $0$, it follows that $\text{tr}(A^n)=0$ for all positive integers $n$.  What I would like to show is the converse, that is,      if $\text{tr}(A^n)=0$ for all positive integers $n$, then $A$ is nilpotent.   I tried to show that $0$ must be an eigenvalue of $A$, then try to show that all other eigenvalues must be equal to 0. However, I am stuck at the point where I need to show that $\det(A)=0$.  May I know of the approach to show that $A$ is nilpotent? I have been told [Warning: I leave this because it's what I asked and allows to understand the dialogues in the comments, but Azad, whom I thank, has pointed that the formula does not hold in general in the form it is expressed] that the angular momentum of and rigid body with respect to any point $P$ can always be expressed as $$\mathbf{L}_{P}=\mathbf{r}_{cm}\times M\mathbf{v}_{cm}+\big(\sum_im_iR_i^2\big)\boldsymbol{\omega}$$ where $\mathbf{r}_{cm}$ is the position of the centre of mass with respect to $P$, $M$ the mass of the body, $R_i$ the distance of the $i$-th point, having mass $m_i$, composing the body, and $\sum_im_iR_i^2=I$ its moment of inertia with respect to the instantaneous axis of the rotation around the centre of mass of angular velocity $\boldsymbol{\omega}$.  I know that the velocity $\mathbf{v}_i$ of each point $P_i$, having mass $m_i$, of a rigid body of mass $M$ can be see as the sum of a translation velocity of one of its points $C$ plus a rotation velocity around that point: $\mathbf{v}_i=\mathbf{v}_{C}+\boldsymbol{\omega}\times\overrightarrow{CP_i}$. If we chose $C$ as the centre of mass I see that $$\mathbf{L}_{cm}=\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_{i}=\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_{cm}+\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})$$$$=\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i}) $$because, if I am not wrong, $\sum_i \overrightarrow{CP_i}\times m_i\mathbf{v}_C=(\sum_i m_i\overrightarrow{CP_i})\times\mathbf{v}_C=\mathbf{0}$ since $\sum_i m_i\overrightarrow{CP_i}$ is the position of the centre of mass with respect to itself, which is $\mathbf{0}$.  How can it be proved that $\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})=(\sum_im_iR_i^2)\boldsymbol{\omega}$? I have searched a lot on the Internet and on books, but I find nothing. To give some background of mine, I have studied nothing of analytical mechanics. I find the formula very, very interesting both in itself and because, if the moment of inertia does not depend upon time, $\forall t\quad I(t)= I(t_0)$, the above expression can be differentiated to get the formula of the resultant torque with respect to the centre of mass $\sum\boldsymbol{\tau}_{cm}=\frac{d\mathbf{L}_{cm}}{dt}=I\boldsymbol{\alpha}_{cm}$ where $\boldsymbol{\alpha}$ is the angular acceleration around the centre of mass. I heartily thank you for any answer!    Some unfruitful trials: by using the "BAC CAB identity" as suggested by Azad, whom I heartily thank, $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=(\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c}$, I can see that$$\sum_i \overrightarrow{CP_i}\times m_i(\boldsymbol{\omega}\times\overrightarrow{CP_i})=\sum_im_i\|\overrightarrow{CP_i}\|^2\boldsymbol{\omega}-m_i(\overrightarrow{CP_i}\cdot\boldsymbol{\omega})\overrightarrow{CP_i}$$which, by decomposing every $\overrightarrow{CP_i}$ into an axial component $\mathbf{A}_i$ and a radial component $\mathbf{R}_i$, whose norms respectively are $A_i$ and $R_i$, with $R_i$ as the distance from $i$ to the axis of rotation, becomes $$\sum_im_iR_i^2\boldsymbol{\omega}+\sum_i m_i A_i^2\boldsymbol{\omega}-m_i(\mathbf{A}_i\cdot\boldsymbol{\omega})\overrightarrow{CP_i}$$but I cannot prove that $\sum_i m_i A_i^2\boldsymbol{\omega}-m_i(\mathbf{A}_i\cdot\boldsymbol{\omega})\overrightarrow{CP_i}=\mathbf{0}$. Different ways to prove there are infinitely many primes? VMware Linux Server -- how can you tell if you are a vm or real hardware? How many irreducible polynomials of degree $n$ exist over $\mathbb{F}_p$? I know that for every $n\in\mathbb{N}$, $n\ge 1$, there exists $p(x)\in\mathbb{F}_p[x]$ s.t. $\deg p(x)=n$ and $p(x)$ is irreducible over $\mathbb{F}_p$.     I am interested in counting how many such $p(x)$ there exist (that is, given $n\in\mathbb{N}$, $n\ge 1$, how many irreducible polynomials of degree $n$ exist over $\mathbb{F}_p$).   I don't have a counting strategy and I don't expect a closed formula, but maybe we can find something like "there exist $X$ irreducible polynomials of degree $n$ where $X$ is the number of...".  What are your thoughts ? "compare with" or "compare to" Please check both sentences and let me know, which one is correct and why?   She compares me with her boyfriend.  She compares me to her boyfriend. Does $1^{\infty}=e$ or $1^{\infty}=1$? Setting Blend Width on Mosaic Dataset using ArcMap How can I remap a keyboard key? Specifying latest revision of a particular branch with bower Is there a test for the different styles within maths mode?  What I really want is to be able to save the current style to reinvoke it later on, but a \if along the lines of \ifmmode would work just as well.  In the specific example, I want to measure how big something will be when displayed in maths mode, so I do:  \settoheight{\mathheight}{some bit of maths}   except that it complains because the some bit of maths gets put in a box and measured, and the box doesn't know that it's maths.  So instead I put:  \settoheight{\mathheight}{\(some bit of maths\)}   except that then I get the wrong answer if this is used in displayed maths.  So ideally, I want:  \ifdisplaystyle  \let\currentstyle\displaystyle \else  \iftextstyle   \let\currentstyle\textstyle  ...  \fi \fi \settoheight{\mathheight}{\(\currentstyle some bit of maths\)}   and then I'll be happy.  Is there such a test?    Okay, so \mathchoice is clearly the way to go (thanks Martin!), but its behaviour is a little odd and since I want to use it to define a macro depending on which mode I'm in, this oddity is quite important.  Here's the code:  \newcommand{\mathtest}{% \gdef\name{a}% \mathchoice{% \name \gdef\name{b}% }{% \name \gdef\name{c}% }{% \name \gdef\name{d}% }{% \name \gdef\name{e}% }% \name}  \[ \mathtest \]   If \mathchoice behaved like \if, this ought to print ab.  But it prints ae.  All the options to \mathchoice get processed (which makes me wonder a bit how it works, but that's by-the-by).  So, is there an obvious way to make \mathchoice behave a little more like \if? Give moderators the option to select a deletion reason when removing posts What would be the Slater's determinant representation for an excited state? Setup  Introducing this spinorbital notation:  \begin{align} \Psi_1=\chi_{(r1)}\alpha_{(\omega1)}  = 1 \\ \Psi_1=\chi_{(r1)}\beta_{(\omega1)}  = \bar{1} \end{align}  and the Slater's determinant, for a system of two electrons, is:  \begin{align} |\Psi\rangle = \frac{1}{\sqrt{2}} \Big|  \begin{array}{cc} 1 &amp; \bar{1}  \\ 2 &amp; \bar{2}  \\ \end{array}  \Big| = \frac{|1\bar{2}\rangle-|\bar{1}2\rangle}{\sqrt2} \end{align}  Question  Based on the example, what would be the Slater's determinant representation for an excited state like $\Psi_a^r$ (i.e. with parallel spins like: $|\bar{1}\bar{2}\rangle$) and $\Psi_{ab}^{rs}$ (feel free to use any number) ? Suddenly Okular will not open any more I have made my laptop double boot a few months again, with Ubuntu 18.04.1 . I have been working with okular to open pdf files (for the commenting option) but suddenly okular doesn't work anymore. When trying to open okular I get the following error:     Unable to find the Okular component: Cannot load library /snap/okular/39/usr/lib/x86_64-linux-gnu/qt5/plugins/okularpart.so: (libKF5JSApi.so.5: cannot open shared object file: No such file or directory)   I don't understand how to solve this problem. Okular was working fine until today and I didn't change anything to my ubuntu system in the mean time. Perhaps the problem comes from the fact that I have been working on windows the last few days and also used the windows ubuntu app.   How can I make okular work again?