Japanese Stack Overflow is almost as old as Jesus
What's with the Japanese SO's stats?
Embedding $S_n$ into $A_{n+2}$ I am trying to prove that for all $n$, $S_n$ is isomorphic to a subgroup of $A_{n+2}$.  Say $S_n$ acts on $\{\alpha_1,...,\alpha_n\}$ and $A_{n+2}$ acts on $\{\alpha_1,...,\alpha_n,\alpha_{n+1},\alpha_{n+2}\}$.  Let $\sigma \in S_n$ and let $\phi: S_n \to A_{n+2}$ be given by $\phi(\sigma)=\sigma$ if $\sigma$ is an even permutation, and $\phi(\sigma)=\sigma(\alpha_{n+1}\alpha_{n+2})$ if $\sigma$ is an odd permutation.  This way $\phi(\sigma) \in A_{n+2}$.  Now, let $\sigma, \rho \in S_n$.  Then $\phi(\sigma \rho)=\sigma \rho (\alpha_{n+1}\alpha_{n+2})$.  Also, $\phi(\sigma)\phi(\rho)=\sigma(\alpha_{n+1}\alpha_{n+2})\rho(\alpha_{n+1}\alpha_{n+2})=\sigma\rho$.  So, $\phi(\sigma \rho)$ does not seem to equal $\phi(\sigma)\phi(\rho)$.  But the idea is that we want $\phi(S_n) \cong S_n$, and $(\alpha_{n+1}\alpha_{n+2})$ does not permute any members of $\{\alpha_1,...,\alpha_n\}$, so $\phi(\sigma \rho)$ and $\phi(\sigma)\phi(\rho)$ are equal as permutations on this (sub)set.  This feels a little subtle to me and I'm not sure if my idea is valid or not.  If it is not, can this proof still be saved?  I appreciate any thoughts on this.  Thanks. I am faced with the following problem: Servers get saturated since current load balancing strategy is based on client IP. Some corporate clients access our servers from behind large proxies so all clients appear with same IP to our load balancer. I think we are using some hardware load balancing device (can investigate further if necessary). We need to maintain session affinity (site is constructed in ASP), so all requests with same IP get routed to the same node.  Since all the communication goes over the HTTPS, no request data (like session Id) is available to balancer as a client discriminator. Is there a way to use some other data besides the IP to distinguish between clients and route the clients even when coming from same IP to different nodes?  Note: I need to maintain the traffic between the balancer and nodes safe (encrypted). Taken from :  Show that it cannot be the case that $U=X+Y$ where $U$ is uniformly distributed on [0,1] and $X$ and $Y$ are independent and identically distributed. You should not assume that X and Y are continuous variables.  A simple proof by contradiction suffices for the case where $X$, $Y$ are assumed discrete by arguing that it always possible to find a $u$ and $u'$ such that $P(U\leq u+u') \geq P(U\leq u)$ while $P(X+Y \leq u) = P(X+Y \leq u+u')$.  However this proof does not extend to $X,Y$ being absolutely continuous or singular continuous. Hints/Comments/Critique? Downloading all my questions with the answerer names I am writing several papers in which I use answers which I received in various StackExchange websites. Naturally I would like to acknowlege and thank the SE members that gave the useful answers. But there are many such answers, and each answer is relevant to a different paper.   So, I would like to download all the questions that I asked in the various SE websites, including the answerer names, to an Excel worksheet in which I can organize the questions by subject.  Is this possible? I used to be able to color code each folder in Finder.   I know I can use tags in Mavericks, but is there a way to restore the ability to actually color code folders?  Thanks. How to determine an object's class? If class B and class C extend class A and I have an object of type B or C, how can I determine of which type it is an instance? jQuery Event Handler created in loop Restore Python installation to default plus apt package dependencies Using SQLite3 with Python 2.5, I'm trying to iterate through a list and pull the weight of an item from the database based on the item's name.  I tried using the "?" parameter substitution suggested to prevent SQL injections but it doesn't work. For example, when I use:  for item in self.inventory_names:     self.cursor.execute("SELECT weight FROM Equipment WHERE name = ?", item)     self.cursor.close()   I get the error:     sqlite3.ProgrammingError: Incorrect number of bindings supplied. The current statement uses 1, and there are 8 supplied.   I believe this is somehow caused by the initial creation of the database; the module I made that actually creates the DB does have 8 bindings.  cursor.execute("""CREATE TABLE Equipment      (id INTEGER PRIMARY KEY,      name TEXT,     price INTEGER,      weight REAL,      info TEXT,      ammo_cap INTEGER,      availability_west TEXT,     availability_east TEXT)""")   However, when I use the less-secure "%s" substitution for each item name, it works just fine. Like so:  for item in self.inventory_names:     self.cursor.execute("SELECT weight FROM Equipment WHERE name = '%s'" % item)     self.cursor.close()   I can't figure out why it thinks I have 8 bindins when I'm only calling one. How can I fix it? Suppose, I want to refer multiple equations (1), (2), (3), (4), (5) and (7) so that they appear as (1-5, 7):  \documentclass[fleqn,preprint,10pt]{article} \usepackage{amsmath} \usepackage{amssymb} \usepackage{amsfonts} \begin{document} \begin{align} \label{1}\psi([e_1,e_2]) = \psi(e_2) = e_2 = (e_2,2e_3) = (-2e_3,e_2) = (\psi(e_1), \psi(e_2))\\ \label{2}\psi([e_1,e_3]) = \psi(2e_3) = 2e_1 = (e_1,2e_3) = (-2e_3,e_1) = (\psi(e_1),\psi(e_3))\\ \label{3}\psi([e_2,e_3]) = 0 = (e_2,e_1) = (\psi(e_2),\psi(e_3)),\\ \label{4}\left\{e_4, e_1\right\},\,\,\,\left\{e_4, e_2\right\},\,\,\,\left\{e_4, e_3\right\},\,\,\,\left\{e_4, e_1+\epsilon\,e_2\right\},\\ \label{5}\left\{e_1, e_2\right\},\,\,\,\left\{e_3, e_1\right\},\,\,\,\left\{e_3, e_2\right\},\\ \label{6}\left\{e_4, e_1\right\},\,\,\,\left\{e_4, e_2\right\},\,\,\,\left\{e_4, e_3\right\},\,\,\,\left\{e_4, e_1+\epsilon\,e_2\right\},\\ \label{7}\text{Nor}_{L_4\oplus L_3}\left(\{e_2\}\right)/\{e_2\}=\,\{e_1, e_3, e_4\} \end{align} \end{document} $ \int_{0}^{ \infty} \int_{0}^{ \infty} \frac { e^{-(x+y)}}{x+y} dx dy $ Equivalence Relation problem Let $S$ be the relation on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b)S(c,d)$ if and only if $ad=bc$. Prove this is an equivalence relation on $\mathbb{N} \times \mathbb{N}$.  I think I've found this to be reflexive and symmetric, but I'm stuck on transitivity. Can someone check my work so far and assist with testing transitivity?  Reflexive: Let $(x,y)S(x,y)$. Then $xy = yx$. So $S$ is reflexive. Symmetric: Suppose $(a,b)S(c,d)$. Then, $ad = bc$. Therefore, $da = cb$ and $cb = da$. Therefore, $(c,d)S(a,b)$. Thus, $S$ is symmetric.  Transitive: Suppose $(a,b)S(b,c)$. Then $ac = bb$.   This is where I'm stuck. Any ideas? What do I do? I looked at a similar question to mine and there was an answer saying to make sure Java is allowed past the firewall. So I checked and it already is. I've tried uninstalling and installing both Minecraft and Java but nothing has changed.  says that all services are fine, so what's going on? I bought a 2nd account to use as an alternative in-game account and now I can't log in.  (I cleared the email and password boxes for obvious reasons in this screenshot: security purposes) Does anyone know the general strategy for summing a series of the form:  $$\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2},$$  where $a$ is a positive integer?  Any hints or ideas would be great! Can the expression $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m \in \mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $\sqrt{m} = a-\sqrt{n}, \ \ a \in \mathbb{Q}$, which I don't think is possible, but how to show it? Right identity and Right inverse implies a group Let $(G, *)$ be a semi-group. Suppose   $ \exists e \in G$ such that $\forall a \in G,\  ae = a$;  $\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$.    How can we prove that $(G,*)$ is a group? Upgrading from a modern smartphone camera without going all in on a DSLR kit Why None is the smallest in python? I went through a , but the answer was informing how to hide the monitor, but not how to disable it. I do not want any new windows to creep into the laptop display when the external display is connected. I could not find any in the Display Settings (under System Preferences). I don't want to close the lid because I think it impacts the air flow of the MacBook Pro. I don't want to close the lid because the sound quality is reduced. Deleting comment removes "Add comment" link on Mobile If I delete a comment on the mobile version of the site (at least if it's the only comment on an answer), the "Add comment" link will vanish along with the comment, so I have to reload to leave another comment.     but got declined as , which is why I'm reposting this (as it may well be a new bug with the same symptoms).  I'm using Firefox 32.0.3 on Android.