How can I install Ubuntu on my Acer 6860 E1-472p?
How do I install Ubuntu alongside a pre-installed Windows with UEFI?
Should black holes be considered two-dimensional or three-dimensional? When using zero as a quantifier, is it correct to use the singular form on the object of the quantifier, or the plural form?  It sounds confusing when I put it that way, but what I mean is: Which is correct?   Your password expires in 0 days. Your password expires in 0 day.   Essentially I suppose I'm asking, does "singular" mean "one" or "the opposite of more than one", as zero is not "plural" in the traditional "more than one" sense?  I'm pretty sure "days" sounds correct, but I can't be sure. Local BLAST Copy Number per Hit I generated a series of local BLAST databases using makeblastdb of metagenomic data and am searching for the presence of a particular gene. While I can do the normal BLAST analysis looking at e-values, % identity, etc. I haven't been able to figure out how to determine how many times a given sequence exists in the database.  So: given a BLAST database of metagenomic data where a particular gene may exist multiple times, how do I determine the copy number of a given BLAST hit?  Thanks in advance. rails - outputting user input with Paragraph Returns? Solving the recurrence $T(n) = \sqrt n T(\sqrt{n}) + \sqrt{n}$ A former student of mine was TA-ing an algorithms class last quarter and asked students to solve this famous recurrence relation:  $$T(n) = \sqrt n T(\sqrt{n}) + n$$  There are  and this question has already been asked here.  I was talking to my student about this problem and mentioned that it's quite hard to solve and is pretty dependent on the particular choices being made. As an example, I suggested this variant of a recurrence relation that was less simple to solve:  $$T(n) = \sqrt n T(\sqrt{n}) + \sqrt{n}$$  The TA and I tried solving this recurrence for about an hour and a half without making any progress. Here are a few things we tried:   My approach to solving the initial recurrence relation was to draw out a recursion tree and notice that each level of the tree contributes $n$ to the total and that there are $O(\log \log n)$ layers, so the recurrence solves to $O(n \log \log n)$. When I tried doing this here, I noticed that the work per layer was no longer constant; instead, the top layer sums to $\sqrt{n}$, the second layer to $n^{3/4}$, the third to $n^{7/8}$, etc. We got stuck working with the summation $n^{1/2} + n^{3/4} + n^{7/8} + ...$. We tried using the iteration method to unroll the recurrence. With the original recurrence relation, this works out nicely; here, we got stuck at the same summation given above.   I'm completely stuck trying to figure out how to solve this recurrence relation. There's nothing riding on it per se - I don't need to solve it for any particular reason - but the fact that we arrived at it by a straightforward modification of a common algorithms problem set question makes it all the more enticing.  Any idea how to solve this recurrence?  Thanks! I have enum and use the variables like myEnum.SomeNameA, myEnum.SomeNameB, etc.  When I return one of these variables from a function, can I print their names (such as myEnum.SomeNameA) instead of the value they returned? Resolving ambiguous overload on function pointer and std::function for a lambda using + How do I execute the following shell command using the Python  module?  echo "input data" | awk -f script.awk | sort &gt; outfile.txt   The input data will come from a string, so I don't actually need echo.  I've got this far, can anyone explain how I get it to pipe through sort too?  p_awk = subprocess.Popen(["awk","-f","script.awk"],                           stdin=subprocess.PIPE,                           stdout=file("outfile.txt", "w")) p_awk.communicate( "input data" )   UPDATE:  Note that while the accepted answer below doesn't actually answer the question as asked, I believe S.Lott is right and it's better to avoid having to solve that problem in the first place! Ring of fractions of $R$ is integral over $R$ Let $R$ be a commutative ring and let $S\subseteq R$ be a multiplicative closed subset containing 1.     I need to show that if $S^{-1}R$ is integral over $R$, then $S^{-1}R=R$.   I don't see how to get started. Controls in container form come over child form? Family name pluralization I know that Wubi is slower in disk IO, but I can't find anything about it online other than vague statements such as "slightly slower disk seeks".  Has anyone done any benchmarks to quantify the speed difference? Matrices $B$ that commute with every matrix commuting with $A$ There have been many questions in the vein of this one, but I can't find one that answers it specifically.     Suppose $A,B\in M_n(\mathbb C)$ are two matrices such that, for any other matrix $C\in M_n(\mathbb C)$, $$AC=CA\implies BC=CB.$$ Prove that $B=p(A)$ for a polynomial $p\in\mathbb C[t]$.    Nothing is assumed about $A$ being diagonalisable or having distinct eigenvalues or being invertible.  Edit: after further work, this might not actually be true. If it's not, a counterexample would be great. Question: If $(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\dotsm+\binom{n}{n}x^n$, prove that $\binom{n}{1}-2\binom{n}{2}+3\binom{n}{3}-\dotsm+(-1)^{n-1}n\binom{n}{n}=0$  My attempt:  I wrote the expression for $\binom{n}{k}$ for each term in the L.H.S and then tried to simplify but could not reach the R.H.S.  Please help. I have no idea on how to begin. I am not allowed to use calculus. Both options affect colors in a similar way, but I can't figure out the difference. There is no set of all sets.. So there are a few posts about this already, but they skip over the problem I have.  The proof of Cantor's theorem elegantly shows that if we consider $f:A\rightarrow \mathcal{P}(A)$, the set $B=\{x\in A : x\not\in f(x) \}$ by its definition cannot lie in the image of $f$, precluding it from being a bijection - hence power sets have strictly greater cardinality.  But now, I thought one (obvious) way to characterise the "set of all sets" would be $S=\mathcal{P}(S)$, as indeed, it contains all possible sets (perhaps here is the problem, but I don't immediately see how using this as a definition for the set of all sets is bad). Now consider simply taking the identity function $i:S\rightarrow S$. Then $B=\{x\in S : x\not\in x \}$ - the set of all sets which do not contain themselves. But this doesn't exist. It's exactly the set of Russell's paradox. If this isn't a set, and so is not in $S$, then the proof no longer offers an obstruction to $i$ being surjective.    Each other post I've seen on the matter uses Cantor's theorem to show such a set doesn't exist. My query is to find that something else that must be happening, since I think the proof has issues for this particular set. I'm expecting other issues to arise with the definition $S=\mathcal{P}(S)$... "Ten times fewer the number" versus "one-tenth the number" Lately I've been hearing and reading statistics that are communicated in wording that, frankly, confuses me.  Forgive me for not citing specific instances, but I can give a hypothetical statistic that exemplifies the kind I'm referring to:     Comparing the number of tax returns the IRS audited in 2002 and 2010, the number of unlucky folks who had their returns audited in 2002 was ten times fewer than the number in 2012.    There is something about that wording that bothers me, and I'm not sure why.    After a fruitless search on the internet using numerous combinations/permutations of words and phrases such as "X times fewer," "X times less," "mathematical expressions of 'times fewer' as opposed to 'a fraction fewer,' or 'a fraction less,'" and more, I came up empty. Perhaps this question is more appropriate on a math website, but in the off-chance members of EL&amp;U might give their imprimatur to this question, here goes:  Here's a second hypothetical example.  Is it grammatical—not to mention mathematical—to say the following?:     There are ten times fewer pollinating honeybees worldwide today than there were in 1912.  [Though this is a made-up statistic, there has been a dramatic reduction in the number honey bees in America of late!]   Should not the expression be:     There are one-tenth fewer honeybees worldwide today than there were in 1912.     On the surface, the "ten times fewer" locution seems to me to be a contradiction in terms.  How can something less be 10 times fewer?    I can understand readily how, for example, 2012's bee population of, say, 90 million can be one-tenth (point one) less than 1912's population of 100 million, but ten times fewer?   I'm confused.  Which expression is "more" correct? Automate Fixed Width Page, length > some minimum length, but only as long as needed I am trying to automate the solution proposed by egreg to  by hooking into \begin{document} and \end{document}.  So, egreg's solution works fine:  \documentclass{article} \def\MinimumPaperHeight{12cm} \usepackage{geometry} \geometry{paperwidth=\MinimumPaperHeight,paperheight=\maxdimen,margin=1cm} \usepackage{lipsum}  \begin{document} \setbox0=\vbox{ \lipsum[1-6] } \dimen0=\dp0 \pdfpageheight=\dimexpr\ht0+2cm\relax \ifdim\pdfpageheight&lt;\MinimumPaperHeight \pdfpageheight=\MinimumPaperHeight \fi \unvbox0\kern-\dimen0 \end{document}   But can't seem to get this to work:  \documentclass{article} \def\MinimumPaperHeight{12cm} \usepackage{geometry} \geometry{paperwidth=\MinimumPaperHeight,paperheight=\maxdimen,margin=1cm} \usepackage{lipsum} \usepackage{environ}  \NewEnviron{MyBox}{     \setbox0=\vbox{\BODY}% }{%     \dimen0=\dp0%     \pdfpageheight=\dimexpr\ht0+2cm\relax%     \ifdim\pdfpageheight&lt;\MinimumPaperHeight \pdfpageheight=\MinimumPaperHeight \fi%     \unvbox0\kern-\dimen0% }  \AtBeginDocument{\begin{MyBox}} \AtEndDocument{\end{MyBox}}  \begin{document} \lipsum[1-6] \end{document}   Hooking into \AtBeginDocument is per this solution on . Join matrix based on conditions How can I edit two materials simultaneously? Whenever I open a new shading window, whatever I do in one is just mirrored on the other. If I click a node in one window, the same node and material get selected in the other. I can't edit two shaders at once. How can I edit two shaders simultaneously?