I am banned from asking questions due to having no rating on three questions, this doesn't really seem necessary.  My questions were valid, except no one up voted them, so this leads me to question ban?  If anyone can help me not do this in the future and help me out right now, please do!  Thanks!
If you are reading this, you may have been affected by one of our post quality bans. If so, it is important that you  to understand what has happened and how to regain the permissions that you lost. While trying to ask a question, one could see:  We are no longer accepting questions from this account. See  to learn more.  Likewise, for answers:  We are no longer accepting answers from this account. See  to learn more.   Why am I getting this message? Are deleted posts taken into account too? Is a question/answer ban the same as a suspension? How do I avoid getting a question ban? How long do I have to wait before I can post again? What can I do to release the ban? How can I reactivate my account? Can I simply create a new account? I'll just ask somewhere else on the SE network, and they'll migrate my question to the correct site! Will a ban on one Stack Exchange site affect my standing on other sites in the network? Does this apply to meta sites too? My account is in good standing. Why am I still blocked?
GAP: how to work with elements of the group ring of the symmetric group $\mathbb{C}[S_k]$? Wikipedia entry on likelihood seems ambiguous I'm trying to get apache to redirect from http to https, however I want them both on the same port (20100, but I doubt that will matter). Basically what's going on  except I'm not doing this for webmin. Currently I have it configured to serve HTTPS, and when I access with HTTP I get:  Your browser sent a request that this server could not understand. Reason: You're speaking plain HTTP to an SSL-enabled server port. Instead use the HTTPS scheme to access this URL, please.   There seem to be a lot of somewhat similar questions, however I am unable to find one that actually answers my question. Communities list is counting my reputation wrong on some sites I have 101 rep on  and am not suspended there. But the site switcher thinks I only have 1 rep there:     I noticed this after I .   To test this, I just added Expats and Reverse Engineering to my site switcher list. On both of those sites I have 101 rep and am not suspended. Results:    Huh? It's a well-known theorem (Corollary 8.10 in Lee Smooth) that given a smooth map of manifolds  $\phi:M\rightarrow N$ and a regular value $p\in N$ of $\phi$, the level set $\phi^{-1}(p)\subset M$ is a closed embedded submanifold. Is the converse true? That is, given an embedded submanifold $S\subset M$, is there necessarily a manifold $N$, smooth map $\phi:M\rightarrow N$, and regular value $p\in N$ of $\phi$ such that $S=\phi^{-1}(p)$?   Prop. 8.12 in Lee Smooth shows that this is true locally; specifically,     Let $S$ be a subset of a smooth $n$-manifold $M$. Then $S$ is an embedded $k$-submanifold of $M$ if and only if every point $p\in S$ has a neighborhood $U\subset M$ such that $U\cap S$ is a level set of a submersion $\phi:U\rightarrow\mathbb{R}^{n-k}$.   (and any level set of a submersion is of course the level set of a regular value). I feel like this is the kind of question where, if there is a counterexample, it probably is very simple, but I wasn't able to come up with one. Stripboard/veroboard/matrix board design software I'm prototyping a simple layout for a small project on matrix board, but figured there must be a better way than drawing it all out on paper. Are there any aimed at matrix board?  I've found the following software for stripboard, but nothing explicitly for matrix board:     Stripboard Magic (seems to be abandonware, but there's page about it )  Stripboard Editor Limited free version, designed to integrate well with     Any more I've missed? Any recommendations? How is it, that $\sqrt{x^2}$ is not $ x$, but $|x|$? As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.  But then I remember that $\sqrt(x)$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$.   So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.  But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.  Where does the absolute value in this derivation of the calculation come from?  I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there. What does linear time series refer to? Compare two dates with JavaScript Can someone suggest a way to compare the values of two dates greater than, less than, and not in the past using JavaScript? The values will be coming from text boxes. java.lang.ClassNotFoundException: org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter when starting Simple Struts2 Application The choice of measurement basis on one half of an entangled state affects the other half. Can this be used to communicate faster than light? It is often stated, particularly in popular physics articles and videos, that if one measures a particle A that is entangled with some other particle B, then this measurement will immediately affect the state of the entangled partner.  For example, if Alice and Bob share an entangled pair of electrons and Alice measures her spin in the $x$ direction, then Bob's spin will also end up spinning in that direction, and similarly if she measures in the $z$ direction. Moreover, the effect will be instantaneous, regardless of the spatial distance between the two particles, which seems at odds with special relativity.  Can I use a scheme like this to communicate faster than light? What's the name of each pseudo-rectangle in a spherical surface? How do I call one constructor from another in Java? Brace expansion not working in a script for i in {1..40} do     echo $i done   I got  {1..40}   and I would like to have something like  1 2 3 and so on   so I can use the variable i inside a command's parameter. "Error: Invalid signature" when configuring dual boot Ubuntu 14.04 and Windows 10 Here is a fun integral I am trying to evaluate:  $$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$  I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, using $\dfrac{e^{ix}-e^{-ix}}{2i}$ form in the binomial series.  But, I am having a rough time getting it set up correctly. Then, again, there is probably a better approach.   $$\frac{1}{(2n)!}\int_{0}^{\infty}\frac{1}{(2i)^{2n}}\sum_{k=0}^{n}(-1)^{2n+1-k}\binom{2n}{k}\frac{d^{2n}}{dx^{2n}}(e^{i(2k-2n-1)x})\frac{dx}{x^{1-2n}}$$  or something like that. I doubt if that is anywhere close, but is my initial idea of using the binomial series for sin valid or is there a better way?.  Thanks everyone. How do I go about porting Ubuntu Touch to different devices? I am following the first volume of the course of theoretical physics by Landau. So, whatever I say below mainly talks regarding the first 2 chapters of Landau and the approach of deriving Newton's laws from Lagrangian principle supposing Hamilton's principle of extremum action. Please keep this view in mind while reading and answering my queries and kindly neglect the systems to which Action Principle is not applicable:  If we use homogeneity of space in Euler-Lagrange equations, we obtain a remarkable result i.e. the conservation of momentum for a closed system.  Now, this result, using the form of Lagrange for a closed system of particles, transforms into $ \Sigma F = 0 $ . Now, how from this can we conclude that the internal forces that particles exert come in equal and opposite pairs?  Is it because for 2 particles this comes out as $ F_{1} + F_{2} = 0 $ and we take the forces exerted by particles on one other to be independent of other particles (i.e. Superposition Principle) as an experimental fact?  I doubt it as whole of Newtonian Mechanics is derivable from Lagrangian Mechanics and supposed Symmetries. So, according to me, a fact like Newton's Third Law should be derivable from it without using an additional experimental fact.  I have an idea to prove it rigorously. Consider two particles $i$ and $j$. Let the force on $i$ by $j$ be $F_{ij}$ and on $j$ by $i$ be $k_{ij}F_{ij}$. Now the condition becomes $\Sigma (1+k_{ij})F_{ij}=0$ where the terms to be included and rejected in summation understood. As this must be true for any value of $F_{ij}$, we get $k_{ij}=-1$. I don't know if this argument or refinement of such an argument holds or not. I can see many questions arising in this argument and it's not very convincing to me.  I would like to hear from you people as to if it is an experimental result used or not? If not, then is the method given above right or wrong? If wrong, how can we prove it?  Addendum  My method of proof uses the fact of superposition of forces itself, so it is flawed. I have assumed that the coefficients $k_{ij}$ are constants and don't change in the influence of all other particles which is exactly what superposition principle says.  As the superposition of Forces can be derived by superposition of potential energies at a point in space and potential energy being more fundamental in Lagrangian Mechanics, I restate my question as follows:  Is the principle of superposition of potential energies by different sources at a point in space derivable from the inside of Lagrangian Mechanics or is it an experimental fact used in Lagrangian Mechanics?  I, now, doubt this to be derivable as the fundamental assumption about potential energy is only that it is a function of coordinates of particles and this function may or may not respect superposition. I am going to Canada, I have a valid U.S. tourist visa (B1/B2). Can I transit via a U.S. airport on the way to Canada with this visa? Logically I know I can since I am allowed to enter the U.S. but I am just making sure. Why do I need 50 reputation to comment? What can I do instead? Users with less than 50 reputation points cannot write comments on questions and answers that they don't own.  Why does this limit exist? But I want to contribute now. What should I do instead? But I can't write a good answer without more information! Shouldn't I be allowed to post comments if I have reputation on a different site? But I saw a new user posting a comment! How did they do it? Why do I need 50 reputation to comment on my own post?   See also:    in the Help Center