I've had strange output on answer submit :)
Has Stack Overflow been haiku-hacked?
What type of buildings offer what level of jobs? There are three tiers of workers (as is everything else); tier 1, 2 and 3. What I am having a hard time with is figuring out what I can build to target workers from each of the tiers. I gather what I am sure everyone else has, low tech industrial (regardless of density really) provides mostly tier 1 jobs.  But what are my other building options for tier 1 jobs? If I want to target tier 2 or 3 what types of buildings should I consider? I need a non trivial  examples of group isomorphism that maps $G\to H$ such that   $G/H \cong G$. Existence of a simple closed curve which is not null-homotopic Problem. Assume that $U$ is an open and connected subset of $\mathbb R^2$, and $\gamma :[0,1]\to U$ is a closed curve, which is not null-homotopic in $U$ and not necessarily simple closed. Show that there exists a simple closed curve, which is not null-homotopic in $U$.  I am sure that this is a standard problem, and I am looking for an elegant, if possible, solution or a reference. I have in mind the following sketch of proof:  a. Construct a polygonal closed curve $\tilde\gamma$ which is homotopic with $\gamma$ in $U$.  b. Construct $\tilde\gamma$ so that it intersects itself in finitely many points, and not whole segments.  c. Use induction on the number of self–intersections of $\tilde\gamma$, in order to show that $\tilde\gamma$ is written as a product (in the fundamental group's sense) of simple closed polygonal curves. jQuery animate backgroundColor I am trying to animate a change in backgroundColor using jQuery on mouseover.  I have checked some example and I seem to have it right, it works with other properties like fontSize, but with backgroundColor I get and "Invalid Property" js error. The element I am working with is a div.  $(".usercontent").mouseover(function() {     $(this).animate({ backgroundColor: "olive" }, "slow"); });   Any ideas? $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})=\mathbb{Q}(\sqrt{p}\sqrt[3]{q})$ for distinct prime $p,q$ How do you find the Lie algebra of a Lie group (in practice)? Given a Lie group, how are you meant to find its Lie algebra?  The Lie algebra of a Lie group is the set of all the left invariant vector fields, but how would you determine them?  My group is the set of all affine maps $x \rightarrow A.x+v$ from $R^n$ to $R^n$ under function composition. Divergence of $\sum_{n=1}^{\infty} \frac{1}{2+i^n}$? I notice that,  $i^n = i, \quad if \text{ } n \equiv \text{ } 1 \text{ } mod \text{ } 4$  $i^n = -1, \quad if \text{ }n \equiv \text{ } 2 \text{ } mod \text{ } 4$  $i^n = -i, \quad if \text{ }n \equiv \text{ } 3 \text{ } mod \text{ } 4$  $i^n = 1, \quad if \text{ } n \equiv \text{ } 0 \text{ } mod \text{ } 4$  so the series $\sum_{n=1}^{\infty} \frac{1}{2+i^n}$ seems to be an alternating series that does not converge to any point in the complex plane. Is there a rigorous way to prove this divergence? How would you write a proof to this divergence or convergence (if I am wrong)?  Thanks! Is it possible to upgrade a BGA 1224 CPU? I want to upgrade the CPU of my T430s laptop from i7-3520M to i7-3632QM. I'm sure i7-3632QM is supported (same socket (BGA1224), same TDP (35W) and microcode found in BIOS (306A9.PAT)).  So my question is: 1) is it possible to de-solder the current CPU without damaging it, and to solder the new CPU without damaging it and the motherboard ? 2) if so, what tools are necessary? who can do that at a reasonnable cost?  Until now, I was unable to find anyone able (or willing) to do that (I'm located in France, and I have tried a local fablab, electronics shops, TV repair shops, ...). Any hint would be appreciated. Move iTunes folder outside Music folder iTunes by default stores its settings, the iTunes Library.itl file, and the music library audio tracks, inside the \Users\username\Music\iTunes folder.  I don't want my Music folder polluted with program settings, especially because I'm synchronizing that folder on the cloud. Via the settings panel I can only change the path of the audio tracks, but the settings remain in the default position.  Is there a way to completely move the iTunes folder to another path?  PS: I've looked at the answers to , but the focus there is on moving the media library only, not the whole iTunes directory. How to resume baking fluid simulation? It happens sometimes that after finishing bake (e.g., for fluid simulation) we wish we could continue the simulation for two more seconds :(    How to use the cached bake to continue the simulation?  We played with the settings, but with no success. What is wrong with this deduction to show that within a multiplet different state has different square angular momentum? I am watching Susskinds's lecture , in which he introduced square angular momentum operator $L^2$, which can be written as $$ L^2=L_z^2+L_-L_++L_z $$ where $$ L_\pm=L_x\pm iL_y $$  Besides, he argued that the spectrum of $L$ can only be $m,m-1,\cdots,-m+1,-m$ in which $m$ is an integer or half-integer. If we denote $|m\rangle$ the eigenvector corresponds to eigenvalue $m$, we have $$ L_\pm|l\rangle=|l\pm 1\rangle\;\text{or}\;0 $$  In the lecture he showed that $L^2|m\rangle=m(m+1)|m\rangle$ for $m$ the highest eigenvalue, by $$ \begin{align*} L^2|m\rangle&amp;=L_z^2|m\rangle+L_-L_+|m\rangle+L_z|m\rangle\\ &amp;=m^2|m\rangle+0+m|m\rangle\\ &amp;=m(m+1)|m\rangle \end{align*} $$  Then he left an exercise to show that for any $-m\leqslant l&lt; m$, we also have $|l\rangle$ is an eigenvector of $L^2$ with eigenvalue $m(m+1)$.  In the lecture, he hinted by showing that it suffices to show that $L^2$ commutes with $L_i$ for $i=x,y,z$, which gives us $$ L^2|m-1\rangle=L^2L_-|m\rangle=L_-L^2|m\rangle=L_-m(m+1)|m\rangle=m(m+1)L_-|m\rangle=m(m-1)|m-1\rangle $$  I believe it is true since I find the   online.   However, my question is that, why the following proof is wrong.  I tried to mimic the proof for $|m\rangle$ to have that  $$ \begin{align*} L^2|k\rangle&amp;=L_z^2|k\rangle+L_-L_+|k\rangle+L_z|k\rangle\\ &amp;=k^2|k\rangle+L_-L_+|k+1\rangle+k|k\rangle\\ &amp;=k^2|k\rangle+|k\rangle+k|k\rangle\\ &amp;=(k^2+k+1)|k\rangle \end{align*} $$  which shows that $|k\rangle$ does not have eigenvalue $m(m+1)$ but $k^2+k+1$ for $k&lt;m$. As the title asks, why didn't the basilisk bite received in the Chamber of Secrets destroy the Horcrux inside the bitten person?   Yes, I know Fawkes quite quickly healed the bite wound with his phoenix tears, but when Harry used the basilisk fang to destroy Tom Riddle's diary the destruction of that Horcrux was immediate. When Ron hit the locket with the Sword of Gryffindor, it was destroyed immediately. Although the destruction of the ring occurred off-page, Dumbledore used the Sword of Gryffindor to destroy it, so presumably, like the locket, it was gone right away. Nagini died straightaway when Neville decapitated her. The destruction of the Hufflepuff cup happens off-page. The diadem seems to have been destroyed by Fiendfyre immediately as well, as when Harry examined it outside the Room of Requirement it was already leaking a blood-like substance, and then it cracked in half. When Voldemort hit Harry with Avada Kedavra in the forest, Harry woke up in limbo with the last fragment of Voldemort's soul - that awful flayed baby-like being - separated from him immediately.   Is there a logical/canon-based explanation for why the fragment of Voldemort's soul residing in Harry wasn't destroyed upon Harry being bitten by the basilisk?  NOTE: I've read the HP Wiki's entry on Horcruxes; however, I find the HP Wiki to be inconsistent in its accuracy, so I'm hesitant to accept the Wiki's explanation on the matter. I'd prefer an explanation constructed from information in the books or a statement by JKR, or the like. Let $1, \omega, \dots, \omega^{n-1}$ be the roots of the equation $z^n-1=0$, so that the roots form a regular $n$-gon in the complex plane. I would like to calculate $$ \prod_{j \ne k} (\omega^j - \omega^k)$$  where the product runs over all $j \ne k$ with $0 \le j,k &lt; n$.    My attempt so far  Noting that if $k-j = d$ then $\omega^j - \omega^k = \omega^j(1-\omega^d)$, I can re-write the product as $$ \prod_{d=1}^{\lfloor n/2 \rfloor} \omega^{n(n-1)/2}(1-\omega^d)^n$$  I thought this would be useful but it hasn't led me anywhere.  Alternatively I could exploit the symmetry $\overline{1-\omega^d} = 1-\omega^{n-d}$ somehow, so that the terms in the product are of the form $|1-\omega^d|^2$. I tried this and ended up with a product which looked like $$\prod_{j=0}^{n-1} |1 - \omega^j|^n $$  (with awkward multiplicative powers of $-1$ left out). This appears to be useful, but calculating it explicitly is proving harder than I'd have thought.  The answer I'm expecting to find is something like $n^n$.    My motivation for this comes from Galois theory. I'm trying to calculate the discriminant of the polynomial $X^n+pX+q$. I know that it must be of the form $ap^n+bq^{n-1}$ for some $a,b \in \mathbb{Z}$, and putting $p=0,q=-1$, the polynomial becomes $X^n-1$. This has roots $1, \omega, \dots, \omega^{n-1}$, so that $(-1)^{n-1}b$ is (a multiple of) the product you see above. An expression for $a$ can be found similarly by setting $p=-1,q=0$. Is it possible to add new predictions/autofill in Overleaf? For example, when I use the command (in Overleaf/ShareLaTex):  \begin{figure}[H]     \centering     \includegraphics{some_graphic}     \caption{some_caption}     \label{fig:some_label} \end{figure}   the \includegraphics{} comes without the [] parameter to change the size of the image, and \begin{figure} without the [H] option by default.   Is it possible to renew the command to make it default? By that I mean that when I type the \begin{figure} command it adds the [H] and [h] as well as [width=\textwidth] in the \includegraphics command, rather than making a global change in the preamble e.g. I am using Ubuntu 12.10.  Please outline the different ways I can upgrade to 13.04. I'm searching how to remove the chapter+section number on the theorem numbering with a section counter.  With no section counter it's like:  \newtheorem{thm}{Title} Chapter1 ----Section 1 --------thm 1 --------thm 2 ----Section 2 --------thm 3 --------thm 4 Chapter2 ----Section 1 --------thm 5 --------thm 6 ----Section 2 --------thm 7 --------thm 8   And with the section counter:  \newtheorem{thm}{Title}[section] Chapter 1 ----Section 1 --------thm 1.1.1 --------thm 1.1.2 ----Section 2 --------thm 1.2.1 --------thm 1.2.2 Chapter 2 ----Section 1 --------thm 2.1.1 --------thm 2.1.2 ----Section 2 --------thm 2.2.1 --------thm 2.2.2   And I want numbering my theorems like this:  \newtheorem???????? Chapter 1 ----Section 1 --------thm 1 --------thm 2 ----Section 2 --------thm 1 --------thm 2 Chapter 2 ----Section 1 --------thm 1 --------thm 2 ----Section 2 --------thm 1 --------thm 2   I know it's always a pain to remove such things, thanks for whoever will help me ^^ Let $H$ be a subgroup of $G$ with $[G:H]=n$. Let $N$ be the kernel of the left-multiplication action on the cosets of $H$. Show $[G:N]$ divides $n!$ Let $H$ be a subgroup of $G$ with $[G:H]=n$ and $G$ finite.  Let $G$ act on left cosets of $H$ by left multiplication.  Let $N$ be the kernel of the action.  Show that $[G:N]$ divides $n!$.  Ok so I have a theorem that says that if $G$ is simple then it embeds into $S_n$.  So if $G$ is simple than I'm done because $G$ divides $n!$ and thus $N$ divides $n!$ and hence $[G:N]$ divides $n!$.  Now I also know that $N$ is the largest normal subgroup of $G$ contained in $H$.  Thus if $G$ isn't simple then it contains some non-trivial normal subgroup $K$ and thus $K\cap H$ is normal in $H$ and by the 2nd Iso Theorem $H/K\cap H\cong KH/H$...    As you can see I'm not sure where to go from here. I was hoping to prove that $K\cap H$ wasn't trivial and thus that $N$ couldn't be trivial since it must contain $K\cap H$, but I'm not sure I can, and even if I could I'm not sure that would help me in showing that $[G:N]$ is small enough to divide $n!$.  Can anyone give some guidance?  Thanks. I am planning to visit London for 6 days from the US. I have given my biometrics and when I was preparing to mail my documents to the UK visa processing center in the US, I used this link from the UK government website to cross-check the documents to be sent to the processing center.     (Guide to supporting documents).  Something that really surprised me was in Section 4 (documents that should NOT be sent along with the package) was hotel bookings and flights bookings.     Section 4: documents you should not send unless specifically requested      This page provides guidance on the types of documents that are not required to consider your application.         hotel bookings   flight bookings      Typically I have always been asked to provide my travel plans, accommodation plans and proof of them. This is the first time that travel plans and accommodation plans are not asked to be sent. What might be the reason behind it? This really confuses me. Any help in clearing my confusion would be greatly welcome. birthday problem - expected number of collisions There are many descriptions of the "birthday problem" on this site — the problem of finding the probability that in a group of $n$ people there will be any (= at least 2) sharing a birthday.  I am wondering how to find instead the expected number of people sharing a birthday in a group of $n$ people. I remember that expectation means the weighted sum of the probabilities of each outcome:  $$E[X]=\sum_{i=0}^{n-1}x_ip_i$$  And here $x$ must mean the number of collisions involving $i+1$ people, which is $n\choose i$. All $n$ people born on different days means no collisions, $i=0$; two people born on the same day means $n$ collisions, $i=1$; all $n$ people born on the same day means $n$ collisions, $i=n-1$.  Since the probabilities of three or more people with the same birthday are vanishingly small compared to two people with the same birthday, and decreases faster than $x$ increases, is it correct to say that this expectation can be approximated by  $$E[X]\approx {n\choose 0}p_{no\ collisions}+{n\choose 1}p_{one\ collision}$$  This doesn't look right to me and I'd appreciate some guidance.    Sorry - edited to change ${n\choose 1}$ to ${n\choose 0}$ in second equation. Sloppy of me. I want to clear all previous commands from the history of my server. I used history -c and it seems all things are cleared but when I ssh to the server, all the commands are still there.  How can I clear them permanently?