I appear to have the accepted answer to , and a rep audit on 7th December showed the +15 rep.  But an audit today no longer shows this acceptance, even though the green tick displays when you visit the question.  I can see that the OP account has been deleted, so presumably that means the acceptance rep, and any votes cast by the OP are lost?  Is there a definitive FAQ and explanation for how rep is re/de-allocated when old accounts are deleted or merged?
In my opinion the accepted answer should permanently stick as only the original question submitter has the ability to accept.
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I'd like to prove that any linear functional $\phi$ on $\ell^\infty$ that is positive, i.e. for any $(x_n) \in \ell^\infty$ such that $x_n \geq 0$ for all $n \in \mathbb{N}$ we have $\phi((x_n)) \geq 0$, is a bounded linear operator and has $\|\phi \| = \phi((1,1,...))$.  Nothing I have tried worked so far, so any hints would be appreciated, thanks! What is the Mediator in this Feynman Diagram $q\bar{q} \rightarrow gg$? Let's consider the t-channel version of the process $q\bar{q} \rightarrow gg$. What is the mediator in this case, a $q$ or a $\bar{q}$? Naively, I would have thought a $q$ (as to have one fermion line), but I am not $100 \%$ sure. I would like to know if anyone can help me with installing Ubuntu on VirtualBox If $g\circ f$ is injective and $f$ is surjective then $g$ is injective Why is the word 'number' abbreviated to 'No.' in UK English and '#' in American English? Why the disparity? And why use 'No.'? Is it from the French?  And the hash or pound sign seems a weird choice too, is there a history or any reason involved? Relation between confidence interval and testing statistical hypothesis for t-test It is well known that confidence intervals and testing statistical hypothesis are strongly related. My questions is focused on comparison of means for two groups based on a numerical variable. Let's assume that such hypothesis is tested using t-test. On the other side, one can compute confidence intervals for means of both groups. Is there any relation between overlapping of confidence intervals and the rejection of the null hypothesis that means are equal (in favor of the alternative that means differ - two-sided test)? For example, a test could reject the null hypothesis iff the confidence intervals do not overlap. Projection of a point onto a simple convex polytope I have a convex polytope described by the intersection of a hypercube and two parallel lower and upper hyperplanes cutting through the hypercube. I want to find the projection of a point onto the polytope. I read the question  and see that the general problem of finding a projection onto a convex polytope can be solved with quadratic programming. However, taken alone, projection onto a hypercube or a hyperplane is trivial. For the hypercube, each dimension is defined by a lower+upper bounds and we just pull each dimension within those bounds, which take $O(n)$ time in $n$ dimensions. For the hyperplane just find the projection onto it. So I'm wondering whether there is simpler method for finding such a projection, maybe somehow combining the two independent projections. Categorize my data into 3 groups CAML query daterange not working Here is a question as phrased in Advanced Linear Algebra (Steven Roman):     Show that a linear operator $\tau$ on a finite-dimensional complex inner product space $V$ is normal if and only if whenever $S$ is an invariant subspace under $\tau$, so is $S^\perp$.   I'm only interested in the $\Rightarrow$ direction. Wikipedia (and ProofWiki)  that uses some kind of computation with the trace. Linear Algebra Done Right chooses an orthonormal basis and then uses the fact that $\Vert \tau v \Vert = \Vert \tau^* v \Vert$.  However, the above  proofs seem a bit too computational to me. Can the following argument work?  (Use $\odot$ for orthogonal direct sum.) Suppose that $\tau$ is normal and write $$V=E_{\lambda_1} \odot \cdots \odot E_{\lambda_k}$$ where the $E_{\lambda_i}$ are the eigenspaces of $\tau$. If $S$ is a $\tau$-invariant subspace then $\tau|_S$ is diagonalizable, so $$S=\widetilde{E}_{\lambda_1} \odot\cdots\odot \widetilde{E}_{\lambda_k}$$ where the $\widetilde{E}_{\lambda_i} \le E_{\lambda_i}$ are the eigenspaces of $\tau|_S$ (which could be trivial). Then choose orthogonal complements $D_i$ such that $E_{\lambda_i} = \widetilde{E}_{\lambda_i} \odot D_i$, and then $$S^\perp = D_1 \odot\cdots\odot D_k.$$ Since each $D_i$ is a subspace of an eigenspace, they are all $\tau$-invariant. In an urn there are $n$ red balls and $m$ blue balls. I extract them without replacement. Let $X$=time of first blue. What is $E(X)$?  I found PMF of $X$ and it is, if $k &gt; n+1$  $$P(X=k) =\frac{n(n-1) \dots (n-k+1)m}{(m+n)(m+n-1) \dots (n+m-k+1)}$$  How could I evaluate $E(X)$?  Edit: I'm looking for an explicit form of $E(X)$ Are user accounts on data.stackexchange.com and Stack Exchange two different entities? I tried to search for similar questions, as I always do, but was not able to find one. I just created a new  on Data.Stack Exchange with my existing Google OpenID and it seems to be not associated with my regular Stack Exchange account. Is this by design? and if so, why? I am trying to define an alternative \item command that would allow me to use a different color for some of the items. My code has the following form:  \documentclass{article} \usepackage{color}  \definecolor{gray}{rgb}{0.7,0.7,0.7} \newcommand{\grayitem}[1]{{\color{gray} \item #1}}  \begin{document}  \begin{enumerate} \item First item. \grayitem Second item. \item Third item. \end{enumerate}  \end{document}   I was hoping that  \grayitem Second item.   would produce the same result as  {\color{gray} \item Second item.}   but for some reason, these are not equivalent. They produce the following results, respectively:  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;   Why are these two lines of code not equivalent? What am I doing wrong and how can I change the definition of \grayitem to produce the second result?  I have also tried removing a pair of braces, like this  \newcommand{\grayitem}[1]{\color{gray} \item #1}   but then the third item becomes gray as well.  The solution should also work for multi-paragraph items. Contextually grey out panel element in python 2.8 I have a panel with split column.  In the 2nd row is 3 boolean variables. I want to grey out the 2nd and 3rd option by default. I want the 2nd to be changeable if 1st is true, and 3rd to be changeable if both 1 and two are true. I've read something about "poll()" but not how to actually use it, or set conditions. Physics should only be enablable if singleMesh, and destructable only if physics enabled.      # Create two columns, by using a split layout.     split = layout.split()      # First column     col = split.column()     col.label(text="Building Size:")     col.prop(bpy.context.scene.user_vars, 'buildingHeight')     col.prop(bpy.context.scene.user_vars, 'buildingWidth')     col.prop(bpy.context.scene.user_vars, 'buildingDepth')      # Second column, aligned     col = split.column()     col.label(text="Dynamics:")     col.prop(bpy.context.scene.user_vars, 'singleMesh')     col.prop(bpy.context.scene.user_vars, 'enablePhysics')     col.prop(bpy.context.scene.user_vars, 'destructable_Building')  #EDIT This is what the specific solution to my example looks like. col = split.column() col.label(text="Dynamics:") col.prop(vars, 'singleMesh') sub = col.row() sub.enabled = vars.singleMesh sub.prop(vars, 'enablePhysics') sub = col.row() sub.enabled = vars.enablePhysics sub.prop(vars, 'destructable_Building') Is cross-posting a question on multiple Stack Exchange sites permitted if the question is on-topic for each site? 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