Kill all descendant processes
Process descendants
Solve the following equation in positive integers $x$ and $y$ What's the difference between stochastic and random? What does [param: NotNull] mean in C#? Should I include a link to my Stack Overflow profile on my résumé? Usage of the word "meet" in the context of a phone call/video conference Are there massless bosons at scales above electroweak scale? Spontaneous electroweak symmetry breaking (i.e. $SU(2)\times U(1)\to U(1)_{em}$ ) is at scale about 100 Gev. So, for , gauge bosons $Z$ &amp; $W$ have masses about 100 GeV. But before this spontaneous symmetry breaking ( i.e. Energy > 100 GeV) the symmetry $SU(2)\times U(1)$ is not broken, and therefore gauge bosons are massless.  The same thing happens when we go around energy about $10^{16}$ GeV, where we have the Grand Unification between electroweak and strong interactions, in some bigger group ($SU(5)$, $SO(10)$ or others). So theoretically we should find gauge bosons $X$ and $Y$ with masses about $10^{16}$ GeV after GUT symmetry breaks into the Standard Model gauge group $SU(3)\times SU(2)\times U(1)$, and we should find massless X and Y bosons at bigger energies (where GUT isn't broken).  So this is what happened in the early universe: when temperature decreased, spontaneous symmetry breaking happened and firstly $X$ &amp; $Y$ gauge bosons obtained mass and finally $Z$ &amp; $W$ bosons obtained mass.   Now, I ask: have I understood this correctly? In other words, if we make experiments at energy above the electroweak scale (100 GeV) we are where $SU(2)\times U(1)$ isn't broken and then we should (experimentally) find $SU(2)$ and $U(1)$ massless gauge bosons, i.e. $W^1$, $W^2$, $W^3$ and $B$ with zero mass? But this is strange, because if I remember well in LHC we have just make experiments at energy about 1 TeV, but we haven't discovered any massless gauge bosons. 2016: a year in closing Error: the question could not be loaded When I was trying to open  I just added using the iOS app I received this error:     Error The question could not be loaded   Reproduction scenario:   add a question; in my case it didn't open it directly; it doesn't show up in the list yet; I opened my profile and clicked the question; bang! I have seen this construction quite often:      Online ads have been around since the dawn of the Web, but only in   recent years have they become the rapturous life dream of Silicon   Valley.   What is the rule there?. When your sentence doesn't start with pronoun + verb, invert them as verb + pronoun?. I know it sounds awkward but is it possible (grammatically correct) to use something similar to:      Online ads have been around since the dawn of the Web, but only in   recent years they have become...   And in any case, does this only work with have (or has)? Maybe it works fine with 'had' but I can't think of an example right now. I have posted a proof below, and would appreciate it if someone could review it for accuracy. Thanks!   Problem:   Let n $\in$ $\mathbb{Z}$ with $n$ $\ge$ 3. Prove the following:   (a) Z(D$_{2n}$) = 1 if $n$ is odd.   (b) Z(D$_{2n}$) = {1, r$^k$} if $n$ = $2k$.   Note that $r$ and $s$ generate D$_{2n}$ with the group presentation { $r$,$s$ | $r$$^n$ = $s$$^2$ = 1, $rs$ = $sr$$^{-1}$ }   Proof:  part (a)   Let $n$ $\ge$ 3 where $n$ $\in$ $\mathbb{Z}$ and $n$ is odd.   For any x $\in$ Z(D$_{2n}$), x must commute with both $s$ and $r$, since both are in $D_{2n}$. Note that if $x$ commutes with both $s$ an $r$, then $x$ commutes with any element of $D_{2n}$ since $r$ and $s$ generate D$_{2n}$.    Then for any x$\in$ Z(D$_{2n}$), we have $xr$ = $rx$, where x is of the form $x$ = s$^j$r$^w$ (any such element can be arranged into this form via the relation $rs$ = $sr$$^{-1}$ and the fact that $r$ and $s$ generate D$_{2n}$). Note that since r and s have finite order, the exponents are modulo n and modulo 2 for $r$ and $s$ respectively.   Then we have  s$^j$r$^w$$r$ = $r$s$^j$r$^w$.   Which implies s$^j$r$^{w+1}$ = s$^j$r$^{w +- 1}$  In the case where the power of r on the RHS is $w$ - 1, it is clear we cannot have equality unless |r| = 1, which it is not. Note that if $j$ is even then we have only the $w$ + 1 case. Hence $x$ may be of the form $1$$r$$^w$ = $r$$^w$.   Also we have that x must commute with $s$, hence $s$$x$ = $x$$s$.   Then $s$$r$$^w$ = $r$$^w$$s$.  Which implies $s$$r$$^w$ = $s$$r$$^{-w}$.   Applying s$^{-1}$ to both sides we arrive at the task of finding when r$^w$ = r$^{-w}$ i.e when $w$ = $-w$. But since r has finite order we have:   $$ \bar{w} = (n-1)*w $$   where $\bar{w}$ is the residue class of w modulo n. Hence for some a $\in$ $\mathbb{Z}$$_+$   $$ w + an = wn - w $$  $$ 2w = wn - an $$ $$ 2w = n(w-a) $$   Now since 0 $\lt$ w $\le$ n we have that the LHS is greater than 0. Also since n $\ge$ w then (w-a) $\le$ 2, since otherwise the RHS would be greater than the LHS. So since n is odd we must have that (w-a) is even and hence (w-a) = 2, implying that 2w = 2n and hence n = w.   Hence x = $r$$^w$ = $r$$^n$ = 1. So Z(D$_{2n}$) = 1.   part (b)   From the argument above any x $\in$ Z(D$_{2n}$) must be of the form $s$$^j$$r$$^w$ for j even and j $\in$ $\mathbb{Z}$. Then x is of the form $r$$^w$ for $0$ $\lt$ w $\le$ $n$. Since $n$ is even we have the equation below is satisfied when $2w$ = $n$ or $w$ = $n$, since (w-a) $\le$ 2 but also (w-a) $\ge$ 0 for (w-a) $\in$ $\mathbb{Z}$$_+$.   $$ 2w = n(w - a) $$   Hence $r$$^w$ $\in$ Z(D$_{2n}$) when $2w$ = $n$, as desired. Nonlinear function continuous but not bounded I would like an example of a map $f:H\rightarrow R$, where $H$ is a (infinite dimensional) Hilbert space, and $R$ is the real numbers, such that $f$ is continuous, but $f$ is not bounded on the close unit ball $\{ x\in H : \|x\| \leq 1\}$.  Actually, $H$ could be replaced by any Banach space (but not just a normed space-- that's too easy).  My motivation is that if $f$ is linear, this is impossible; but I have next to no intuition about non-linear functions.  Edit: Here's an example for $c_0$ which is even differentiable (disclaimer: I found it here: ).  Define $f:c_0\rightarrow F$ (where F is your field, real or complex) by $$ f(x) = \sum_{n=1}^\infty x_n^n \qquad (x=(x_n)). $$  You can estimate the sum by a geometric progression, so it does converge.  A bit of checking shows that f is Frechet differentible (so certainly continuous).  But $f(1,1,\cdots,1,0,\cdots)=n$ (if there are $n$ ones) so $f$ is not bounded on the closed unit ball.  What I don't immediately see is how to adapt this to $\ell^2$, say. Generator of rational functions unchanged under $\sigma(X) = X + 1$ Let $L = K(X)$ be the field of rational functions over a field $K$ with characteristic $p &gt; 0$, and let $\sigma \in \operatorname{Aut}_K(L)$ with $\sigma(X) = X + 1$.   Show that $G = \left&lt;\sigma\right&gt;$ is finite, and determine a generator of $L^G = \{ f \in L: \rho(f) = f$ for all $\rho \in G \}$ over $K$.    My solution so far: $G = \left&lt;\sigma\right&gt;$ is finite because char $(K) = p &gt; 0$ and therefore $\sigma^p (X) = X + p = X $.  A rational function unchanged under $\sigma$ will also be unchanged under any $\sigma^n$, so we don't have to worry about powers of $\sigma$ I think.   Any ideas for finding a generator? Thanks. What's wrong with this .desktop file? prevent the deletion of account records if they have oppurtunity trigger preventAccountDeletionifopp on Account (before delete) {     for(account accdel:trigger.old){        Opportunity oppprevent =[select Name,(select Name,Owner from Opportunitys)from ACCOUNT =: accdel.id ];         if(oppprevent.size() &gt; 0){           accdel.adderror('Opportunity record is refering this record.So, you cannot delete it...!');         }    } } Has a movie from a movie series ever been removed/retconned due to fan/general outcry? A petition to strike the events of  from the main Star Wars timeline that is going around online has made me wonder if there has ever been a retcon/removal of a movie from a movie series or just renunciation of a movie by those with creative control due to backlash, as opposed to the director(s) just deciding to go a new direction/starting fresh. Google just produces said petition when searching (please note this is not for or against Star Wars).  Addendum, by removal I am referring to the events in the movie no longer being recognized in the universe. Post a Form to Popup window not to the parent With English not being my first language, I usually need support for at least two languages when using features like auto-correct and dictation.  Is it possible to toggle the language which is used for dictation via a shortcut?  As of now, I always have to manually navigate through the menus in the System Preferences which if far from convenient. I've tried to see if there are any pre-defined actions in Automator, but found none. Let $f$ be a continuous function on $[0,1]$ such that $f([0,1])=[0,1]\times[0,1].$ Then show that $f$ is not one-one.  Hints will be appreciated. Why do people sometimes use the past perfect form of a verb when the past would suffice? (e.g. "you had mentioned") Is there any way to run file explorer as administrator under Windows 10 After install Windows 10, it seems that I lost the ability to run explorer.exe as administrator with a user in administrator group.   The reason I want to do that is, in our team's development environment, we are using a C# exe to do some environment configuration which will start command prompt under administrator mode and subst a drive. As a result those substed drives are not visible within the file explorer since it is run not as administrator. That will be kind of inconvenience and sometimes make mistakes.   I was able to run explorer.exe from task manager with the option "create with privileges" checked and then I can see all the drives in explorer. But now this is not working anymore.   I knew there are other options to workaround this workflow, but just want to make sure that if it is now totally impossible under Windows 10?  Any comment is appreciated.