Finding smallest range to which at least one number belong from every list
How to pick one number from each sorted list so that the difference between the minimum and maximum is as small as possible?
Simplify $\sum (x_i- \mu)^2$ How can I simplify this relation? $$\sum_{i = 1}^m(X_i-\mu)^2=\sum_{i = 1}^m(X_i-\bar{X}_m)^2+m(\bar{X}_m-\mu)^2$$ I know how to set up: $$\sum_{i = 1}^m(X_i-\mu)^2=\sum_{i = 1}^m[(X_i-\bar{X}_m)+(\bar{X}_m-\mu)]^2$$ but when I open the bracket, how do I handle the mixed term? $$2\sum_{i = 1}^m(X_i-\bar{X}_m)(\bar{X}_m-\mu)$$  Thank you! I'm a little confused as to whether I should install Ubuntu on its own partition on my hard drive, use VirtualBox or another virtualization package to install it, or use Wubi to install it directly on top of my current OS (Win 7). I definitely want to learn and use Ubuntu, so this is not just for playing around with it.   Also, if I choose to partition, should I partition the hard drive myself or should I let the Ubuntu installation menu do it for me? I understand that I am going to need a main partition, for Ubuntu's core components, and also a swap partition. Then there is the option to add a partition for "home"- I don't understand what combination of these partitioning options I should choose, or whether it is better to partition in Windows before I install Ubuntu or just partition my hard drive when I install Ubuntu itself I have got a specific folder located in "/home/.../reboot/".  I have a series of users that require read-only access to the ever changing files within.  My problem is that the same users are not authorized to view any of the parent directory files.  Right now, the only thing I have in place is that their starting folder is at "/home/.../reboot/" by changing the etc/passwd file; and I need to block all access to other folders.  How could I only give them access to this directory? This is a  about IPv6 Subnetting.      Related:               I know a lot about , and as I prepare to (deploy|work on) an IPv6 network I need to know how much of this knowledge is transferable and what I still need to learn. IPv6 seems at first glance to be much more complex than IPv4. So I would like to know:   IPv6 is 128 bits, so why is /64 the smallest recommended subnet for hosts? Related to this:   Why is it recommended to use /127 for point to point links between routers, and why was it recommended against in the past? Should I change existing router links to use /127? Why would virtual machines be provisioned with less than a /64 worth of addresses? Are there other situations in which I would use a subnet smaller than /64?  Can I map directly from IPv4 subnets to IPv6 subnets? For instance, does an IPv4 /24 correspond directly to an IPv6 /56 or /120? My interfaces have several IPv6 addresses. Must the subnet be the same for all of them? Why do I sometimes see a % rather than a / in an IPv6 address and what does it mean? Am I wasting too many subnets? Aren't we just going to run out again? In what other major ways is IPv6 subnetting different from IPv4 subnetting? Where is Ubuntu terminal located? I am trying to add terminal as external tool to compile javascript with node.js. I am following this  which is based on windows , and i am stuck at figuring out location of terminal application.  Where is the terminal located in Ubuntu? Do you have any useful awk and grep scripts for parsing apache logs? I can use log analyzers, but often I need to parse recent web logs to see what's happening at the moment.   I sometimes do things like to figure out top 10 ips that request a certain file  cat foo.log | grep request_to_file_foo | awk '{print $1}' |  sort -n | uniq -c | sort -rn | head   What do you have in your toolbox? Are there critical hits for spells? Uniform continuity of $\sin(x\sin x )$ for all $x\in (0,\infty)$ Typesetting UML class diagrams What's the best way to draw UML class diagrams in LaTeX? Chrome unity shortcut acting up Whenever I click on my chrome shortcut I get a new window (as expected) but I also get a new Unity entry with a different icon (one from an iconset I had been using before upgrading to 14.04 - don't really remember the name). The new unity entry does not seem to originate from a correct .desktop entry because all the submenus are missing from it (New incognito window, New Window etc.) and it has a permanent tag '(1) Facebook' no matter the window title (see attached picture).  I have uninstalled and reinstalled chrome in hope that it would help which it did until restart but now I get the same behaviour.  Note that I haven't altered any .desktop files regarding chrome and I can only find one file:  /usr/share/applications/google-chrome.desktop which seems to be correct (with submenus etc.) Bug in Mathematica TeXForm generating \left\left I want to better understand why the construction something + [adjective] makes grammatical sense. Background: I was recently reading a Washington Post article and came across the following sentence.  So if you fall in that category, [the movie] feels more like magic than repetitive.  (Note: I've copied the relevant spoiler-free sentence above; following the link to the article will open you up to movie spoilers! ) The phrase more like magic than repetitive struck me as odd, presumably because it was non-parallel (more like [noun] than [adjective]). I would have expected &quot;...more like magic than something repetitive.&quot; (Actually, that needs to be rewritten to flow better, but the point is that at least it appears to be grammatical.) Thus, one would have the noun phrase &quot;something repetitive,&quot; of the form &quot;[noun] [adjective].&quot; This got me to thinking about the use of adjectives after nouns. There is a great Stack Exchange article  that involves the characteristics of the adjective that indicate it should come after the noun. But I haven't been able to find examples of how the word &quot;something&quot; requires a postponed adjective. Consider the following sentences:  I want to buy something red. I am looking for something sweet to eat.  You cannot say:  *I want to buy red something. *I am looking for sweet something to eat.  But you can say:  I want to buy something (that is) red. I am looking for something (that is) sweet to eat.  So, my question:  Do the above examples mean that the &quot;something [adjective]&quot; construction is an example of the ellipsis of &quot;that is&quot;?  Is there something special about the word &quot;something&quot; that requires the postponed adjective? Note that I am NOT looking for details about adjectives that are postponed; I'm interested in the word that precedes the adjective (either &quot;something&quot; or similar words).   By the way, I am not interested in the following constructions, which are different:  I want to paint the house red. I tried to build the brick house taller.  Thanks in advance! What does  signify when using Vim LatexSuite? I have a LaTeX editing environment that does some assistive stuff such as auto-indenting, closing parentheses, etc.  However, one of the things that it does is unknown to me, and is difficult to search for. If I type an empty set of braces, it replaces them with  {}&lt;++&gt;   If I type two empty sets, for example when I'm setting up a fraction \frac{}{} what I actually get is  \frac\left\{ } \right\}&lt;++&gt;}&lt;++&gt;   Is this a problem? What does &lt;++&gt; do?  Edit: This is using Is it possible to prove a null hypothesis? Essentially, I wonder that if X is a plural noun, should we use a plural or singular verb in a sentence such as "X are a subject that [verb] ..."? I believe that we should always use the singular verb form there (for example, "Eatable flowers are a subject that comes up ..."). However, I was surprised to find this sentence on the  page:     Eatable flowers are a subject that come up in almost all cooking classes eventually and I am a big fan of them.   as well as this on Google Books:     Tools are a subject that come up at our house from time to time.   I think they are typos, but I would like a double check to be sure. Fraleigh(7ed) Theorem33.12. Let $p$ be a prime and let $n\in\mathbb{Z}^+$. If $E$ and $E&#39;$ are fields of order $p^n$, then $E \simeq E&#39;$.   Proof in the text: Both $E$ and $E&#39;$ have $\mathbb{Z}_p$ as prime field, up to isomorphism. By Corollary 33.6(A finite extension $E$ of a finite field $F$ is a simple extension of $F$), $E$ is a simple extension of $\mathbb{Z}_p$ of degree $n$, so there exists an irreducible polynomial $f(x)$ of degree $n$ in $\mathbb{Z}_p[x]$ such that $E\simeq \mathbb{Z}_p[x]/ \langle f(x) \rangle$. Because the elements of $E$ are zeros of $x^{p^n}-x$, *we see that $f(x)$ is a factor of $x^{p^n}-x$ in $\mathbb{Z}_p[x]$*. Because $E&#39;$ also consists of zeros of  $x^{p^n}-x$, we see that $E&#39;$ also contains zeros of irreducible $f(x)$ in $\mathbb{Z}_p[x]$. Thus, because $E&#39;$ also contains exactly $p^n$ elements, $E&#39;$ is also isomorphic to $E\simeq \mathbb{Z}_p[x]/\langle f(x) \rangle$.  I don't know why $f(x)$ divides $x^{p^n}-x$. $E$ has a zero $\alpha$ of $f(x)$, but it doesn't need to have all the zeros of $f(x)$. So $f(x)=(x-\alpha)g(x)$ in $E[x]$ and $g(x)$ need not be splitted into linear factors. How can $f(x)$ divide $x^{p^n}-x$? If-else working, switch not Convergence of the series $\sum_n \frac{n^{\log(n)}}{\log(n)^n}$ I have a rough outline of a proof. Is this correct?  $$\sum_{n \geq 2} \frac{n^{\log(n)}}{\log(n)^n} \to a \in \mathbb{R} \iff  \sum_{n \geq 2} 2^n \frac{(2^n)^{\log(2^n)}}{\log(2^n)^{2^n}} \to b \in \mathbb{R}$$ via dyadic criterion (or Cauchy condensation test).  Analyzing the right-hand side:  $$\sum_{n \geq 2} 2^n \frac{(2^n)^{\log(2^n)}}{\log(2^n)^{2^n}} = \sum_{n \geq 2} 2^n \frac{(2^n)^{n\log2}}{(n\log2)^{2^n}} = \sum_{n \geq 2} \frac{(2^n)^{n\log2 + 1}}{(n\log2)^{2^n}} = \sum_{n \geq 2} \frac{2^{n^2\log2 + n}}{(n\log2)^{2^n}} = (\ast). $$  We know for some sufficiently large $N$, $n \log 2 \ge 2$ for all $n\ge N$, so we can bound the series above by:  $$(\ast) \leq C_N + \sum_{n \geq N} \frac{2^{n^2\log2 + n}}{2^{2^n}} = \sum_{n \geq N} 2^{n^2\log2 + n - 2^n} = (\ast\ast),$$  where $C_N$ is the finite sum of the original series until $N$. And once again, for sufficiently large $M$, $m^2\log2+m - 2^m \leq -m^2$ for all $m \ge M$. Hence  $$ (\ast\ast) \leq C_{\max(N,M)} + \sum_{n \geq \max(N,M)} 2^{-n^2} &lt; \infty.$$  We then use the $p$-test to conclude that the sequence does converge. How to add border to listview and different row colors at the same time Didn't understand relationship 'AccountContactRelation' in FROM part of query call I'm getting the following error message:     System.QueryException: Didn't understand relationship 'AccountContactRelation' in FROM part of query call. If you are attempting to use a custom relationship, be sure to append the '__r' after the custom relationship name. Please reference your WSDL or the describe call for the appropriate names.   Select Email, Id, (Select ContactId, AccountId From AccountContactRelation) From Contact  In AccountContactRelation object there are two lookup fields for Account and for Contact.  Please advise how to avoid such error message?