Why do we need to use the definite article when we are speaking about inventions like in the sentence below:     Alexander Bell invented the telephone.   Thank you in advance!
I am making a test where I am asked to add the missing words.     [...] chair is [...] useful piece of furniture.    I would have said that a is the missing word in both the cases, but the test says the words I should have added are the and a respectively.     [...] last time I saw [...] bear, I was travelling in [...] Europe.   I would have added: nothing, a, nothing. The correct answer is: the, a, nothing.  Why is the the correct article to use in both the sentences?
Looking to draw this block diagram in Tikz I'm looking to create something like this in Tikz environment. Any tips? help? How to mount an internal HDD on start-up? I have a 500GB internal HDD where I store all my music and photos but it is annoying to manually mount it after a restart and I was wondering how to mount it automatically and make it available to Rhythimbox for playback.  Thanks! Preview on the right please, optionally? Securely erase hard drive using the Disk Utility How do we solve (i.e. get the closed form of) $\int \sqrt{1 + t^2} dt$ ?   shows the closed form of it but not the steps in solving it.  I think I need some algebraic trick..   I thought of some hyperbolic functions and their derivatives and equalities:  If I make $ t = \sinh x$, then I can reduce the integral to  $$ \int \cosh ^2 x dx $$ though I'm not sure if this is valid. But from here, I cannot go on evaluating the integral... Is there a way to take a screenshot using Java and save it to some sort of image? Simple as the title states: Can you use only Java commands to take a screenshot and save it?  Or, do I need to use an OS specific program to take the screenshot and then grab it off the clipboard? This code in JS gives me a popup saying "i think null is a number", which I find slightly disturbing. What am I missing?    if (isNaN(null)) {   alert("null is not a number"); } else {   alert("i think null is a number"); }     I'm using Firefox 3. Is that a browser bug?  Other tests:    console.log(null == NaN);   // false console.log(isNaN("text")); // true console.log(NaN == "text"); // false     So, the problem seems not to be an exact comparison with NaN?  Edit: Now the question has been answered, I have cleaned up my post to have a better version for the archive. However, this renders some comments and even some answers a little incomprehensible. Don't blame their authors. Among the things I changed was:   Removed a note saying that I had screwed up the headline in the first place by reverting its meaning Earlier answers showed that I didn't state clearly enough why I thought the behaviour was weird, so I added the examples that check a string and do a manual comparison. Non-static variable cannot be referenced from a static context Prove $[A,B^n] = nB^{n-1}[A,B]$ I am trying to show that $[A,B^n] = nB^{n-1}[A,B]$ where A and B are two Hermitian operators that commute with their commutator. However, I am running into a little problem and would like a hint of how to proceed.  If A and B commute then $[A,B] = ABA^{-1}B^{-1} = e$ where e is the identity element of the group.  $$\therefore AB=BA$$  $$n=1; [A,B^1] = (1)B^0[A,B] = e$$ This statement is certainly true. however moving on to $n = 2$ I find...  $$[A,B^2] = AB^2A^{-1}B^{-2} = ABBA^{-1}B^{-1}B^{-1} = BBAA^{-1}B^{-1}B^{-1}$$  Where in the last step I have used the fact that A and B commute to rearange the terms. However, it is plain to see that this last term simply reduces to the identity as well and for the n = 2 case we have:  $$[A,B^2] = e \ne 2B[A,B] = 2Be = 2B$$  Clearly I have assumed something I shouldn't have. The fact that there is a multiplicative factor of n implies I should be adding things, but I thought if I kept it as general as possible, the answer should just fall out naturally. I don't want an answer please, only guidance. Enumerate behaves differently with beamer 3.42 The following is produced by beamer version 3.41 with enumerate:  \documentclass{beamer} \mode&lt;presentation&gt;     {\usetheme{Rochester}} \begin{document} \begin{frame}   \begin{enumerate}[(a)]     \item Foo     \item Bar   \end{enumerate}   \begin{enumerate}[1]     \item Foo     \item Bar   \end{enumerate} \end{frame} \end{document}     And with beamer version 3.42 the following:    The latter output is so, well, ugly, that it's probably a bug. While waiting for a bug fix (?), how can I temporarily fix this so that I get the output produced by beamer 3.41? We're trying to debug some applications performing broadcast.  What is the difference between the broadcast address 255.255.255.255 and as e.g. reported by ifconfig,  Bcast:192.168.1.255 What can I do about uncounterable spells? I cast , sacrificing my only creature to kill my opponent.  My opponent responds with , which will win the game the other way.  Last Word says "This spell can't be countered."  What else can I do about it besides countering it to prevent my plans from being foiled?  (I'd really like to have the last word here.)    Historic Note: before of the April 2018 rules change, uncounterable spells with targets used to actually have different wording (still visible on the printed cards) implying that they still could be countered.  For example, Last Word said:     Last Word can't be countered by spells or abilities.   Meanwhile Akroma, Angel of Fury said:     Akroma, Angel of Fury can't be countered. Login screen loops unless you login as Guest I am running a VMWare Player with a Ubuntu 12.04 LTS Precise Pangolin as Guest on my Windows 7. Sometimes I get the shutdown blue screen error in Windows, this time it happened when I was running the Player. When I restarted everything Ubuntu gave me the &quot;Login Loop&quot; on adminstrator login. I log in and there's this black screen where I can only read:  piix4...smbus:0.0.0.07.3 Host Smbus controller not enabled .    When I go to the prompt in root mode it fails to update and only upgraded, specially some plugins ( I think graphic plugins) which also appear in one an error message after quitting the prompt, but they´are successfully installed. They are not the error message. After that I have been working with the Fail/safe Mode recovery panel. When I try to update via Root I get errors like this: W:failed to fetch http://extras.ubuntu.com/ubuntu/dists/precise/release.gpg could not        resolve 'extras/ubuntu.com  There are 8 more like this referring to areas like: -archive/canonical.com -ppa.Launchpad.net -security.Ubuntu.com -Us.archive.ubuntu.com - release.gpg                         precise-updates/release.gpg                         precise_backport/release.gpg   Final Message: some index files failed to download.....they have been ignored or old files are used. The black screens most of the time pass by too fast for me to pick up any information. But in general I think I have done everything I was able to in the recovery panel including updating network and graphic packages and recovering filesystem packages and the basic stuff ( I am a beginner regarding Linux ) in the root prompt. Now I am stuck in this screen with graphic options: - Run in low-graphics mode just for one session - Reconfigure Graphics - Troubleshoot the error - Exit to console login  I am trying to choose to reconfigure graphics but the mouse disappears in the virtual machine screen and sometimes when options change it´s only the first and last option. But this happens from the blue without messages. This particular option menu is in the regular GUI style against a black screen in Terminal style. Really strange. What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$ I would like to evaluate the sum  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$  Here is my attempt:    Letting    $$f(z)=\frac{1}{az^2+bz+c}$$  The poles of $f(z)$ are located at   $$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$  and  $$z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$  Then  $$ b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)= \lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}= \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b} $$  Using L'Hopital's rule. Continuing, we have the limit is  $$ \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}= \frac{\pi\cot (\pi z_0)}{2az_0+b} $$  For $z_0 \ne 0$  Similarly, we find  $$b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}$$  Then   $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\ -\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)=  -\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)= \frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}= \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}} $$  Then we have  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}$$    Is this correct?  I feel like I made a mistake somewhere.  Could someone correct me?  Is there an easier way to evaluate this sum? How do I install LibreOffice 4? I want to download and install .  How can I do this? A general question asked from time to time on stack exchange is question of trigger bulkification.  Therefore I have 2 questions:   Which best practices are you using? What is the best way to handle updates of more than 200 records? (In case of more than 200 records the trigger is executed more than once.) Prove that if $n \in \mathbb{N}$, then $\sum_{d|n}{(d(n))^3}=(\sum_{d|n}{d(n)})^2$ where $d(n)$ is the divisor function. Here is my candidate for  the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists. I would be interested in seeing others.  $***$ Added after some comments: I prove here by very elementary means that the limit exists. Calling the limit "$e$" names it. $***$  It only needs Bernoulli's inequality (BI) in the form $(1+x)^n \ge 1+nx$ for $x &gt; -1$ and $n$ a positive integer, with equality only if $x = 0$ or $n = 1$. This is easily proved by induction: It is true for $n=1$, and $(1+x)^{n+1} = (1+x)(1+x)^n \ge (1+x)(1+nx) = 1+(n+1)x+nx^2 \ge 1+(n+1)x $. (If $-1 &lt; x &lt; 0$, if $1+nx \ge 0$, the above proof goes through, and if $1+nx &lt; 0$, $1+mx &lt; 0$ for all $m \ge n$ so certainly $(1+x)^m &gt; 1+mx$.)  This proof originally appeared in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563 and uses the arithmetic-geometric mean inequality (AGMI) in the form $\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$  (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.  Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$.  We will prove that $a_n$ is an increasing sequence  and $b_n$ is an decreasing sequence.  Since $a_n &lt; b_n$, this implies,  for any positive integers $n$ and $m$ with $m &lt; n$  that $a_m &lt; a_n &lt; b_n &lt; b_m$.  For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$.  Since their sum is $n+2$ and their product is $(1+1/n)^n$, by the AGMI,  $((n+2)/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $(1+1/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $a_{n+1} &gt; a_n$.  For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$.  Since their sum is $n$ and their product is $(1-1/n)^n$, by the AGMI,  $(n/(n+1))^{n+1} &gt; (1-1/n)^n$  or $(1+1/n)^{n+1} &lt; (1+1/(n-1))^n$,  or $b_n &gt; b_{n+1}$.  Since $b_n-a_n  = (1+1/n)^{n+1} - (1+1/n)^n = (1+1/n)^n(1/n) =a_n/n $ and every $a_n$ is less than any $b_n$ and $b_3 = (1+1/3)^4 = 256/81 &lt; 4$, $b_n-a_n &lt; 4/n$, so $b_n$ and $a_n$ converge to a common limit.  These proofs do not seem to be really elementary,  since they use the AGMI.  However, they use a special form of the AGMI,  where all but one of the values are the same,  and this will now be shown to be implied by BI,  and thus be truly elementary.   Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive.  The AGMI for these values is  $(((n-1)u+v)/n)^n \ge u^{n-1}v$  with equality if and only if $u = v$.  We will now show that this is implied by BI: $(((n-1)u+v)/n)^n \ge u^{n-1}v$  is the same as  $(u+(v-u)/n)^n \ge u^n(v/u)$.  Dividing by $u^n$,  this is equivalent to  $(1+(v/u-1)/n)^n \ge v/u$.  By BI,  since $(v/u-1)/n &gt; -1/n &gt; -1$, $(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$  with equality only if $n=1$ or $v/u-1 = 0$. Thus BI implies this version of the AGMI. How can I do the following in Git?  My current branch is branch1 and I have made some local changes. However, I now realize that I actually meant to be applying these changes to branch2. Is there a way to apply/merge these changes so that they become local changes on branch2 without committing them on branch1? Android, how do I stop a pointer appearing below an EditText