File Protocol Ajax Request
Local html file AJAX Call and jQuery Woes
Is there a way to "hide" a normal windowed application in Windows?  I'm looking for something that removes it from the taskbar and possibly puts an icon in the system tray instead. Printing a .NET DataGridView I am fairly new to .NET and C#, but I have a DataGridView that I would like to print.  What would be the best way to go about doing so? Assignment statement used in conditional operators How do I make the player name show up on a /tellraw command? So I am currently making a minecraft custom map (for 1.9). I want to make the player "say" something using /tellraw commands. I did figure out how to make the player name come up using /say commands, but still have no idea how to do it using the /tellraw command.  How do I do that? How many points in RSA, such that $m^e = m \bmod n$ For every RSA-cryptosystem, there exist some messages $m$, for which it holds that $m^e \equiv m \pmod n$ As to the question, how many such messages exist, this question has already been asked and answered here:  However, I do only understand the answer up to a certain point Suppose that $n$ is only 1 prime. Then it holds that the number of unconcealable messages is $\gcd(e-1, p-1)+1$ This is because in $m^e \equiv m \pmod  p$ , either $m = 0$ or $m \not= 0$. In the latter case $m$ can be expressed as $g^t$, where $g$ is a Generator in $\mathbb{Z_p}$ $g^{t*e} \equiv g^t \pmod p$ $g^{t(e-1) \bmod p -1 } \equiv g^{0 \bmod p -1} \pmod p$ $t(e-1) \equiv 0 \pmod {p-1} $ has exactly $\gcd(e-1, p-1)$  solutions, so the total number of unconcealable messages is, together with $m=0$ , $\gcd(e-1, p-1)+1$ When we now look at real RSA, which has two prime factors in its modulus, it holds that the number of unconcealable messages is $(\gcd(e-1, p-1)+1)\cdot(\gcd(e-1, q-1)+1)$ Why is that so? I suppose it has to do with CRT, but I just cant understand why. Old phone cannot connect to computer, the "computer connection" icon appears on phone I would like to get some pictures and audio out of an AT&amp;T SGH-a687 phone onto my Mac computer. I first tried connecting it to a New PC Studio on a PC, but there was a problem after transfer 2 photos, so I tried again. But the PC could no longer detect the phone, even though the "connected to computer" icon appears on the phone. I tried a different cable, different USB ports. No use.  So I downloaded the Samsung Kies on my Mac. Same thing, it tries to search for the phone once I connect the phone, but the computer doesn't detect anything. The phone still shows the "connected to computer" icon.  There is also Bluetooth (no internet on the phone) on the phone, but it failed to send the picture each time I tried.  What can I do at this point? Please also suggest data extraction software for either Mac or PC, if that's the easier route. The work done against gravity is $mgh$, well at least that's what my textbook says. I have a question: I can apply a force say 50N, so total work done = $mgh + mah$. Where $ma$ = Force. But the truth is irrespective of the force applied, the work done against gravity is always $mgh$. Why?  For example, when I move an object with a force, the work done is more, so work depends on the Force. But in case of gravity it always depends upon the weight I would like to know how to get the version information of an installed program that doesn't have the      --version or -version   method implemented. How do I parse command line arguments in Java? What is a good way of parsing command line arguments in Java? What's wrong with this query: INSERT INTO Users( weight, desiredWeight ) VALUES ( 160, 145 ) WHERE id = 1;  It works without the WHERE clause. I've seemed to have forgot my SQL. Experimental observation of matter/antimatter in the universe Ordinary matter and antimatter have the same physical properties when it comes to, for example, spectroscopy. Hydrogen and antihydrogen atoms produce the same spectroscopy when excited, and adsorb the same frequencies. The charge does not make a difference in the potential (regardless if it's generated by a proton or an antiproton) nor in how the positron behaves in this potential (being its mass equal to the mass of an electron)  How can astronomy evaluate if a far galaxy is made of matter or antimatter, given that from the spectroscopy point of view, they behave in the same way? In other words, how do we know that an asymmetry exists between matter and antimatter in the universe? mv * folder (avoiding 'cannot move' error) A coin is tossed three times: The probability of zero heads is 1/8 and the probability of zero tails is 1/8.  And my question is: What is the probability that all three tosses result in the same outcome?   So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  If so, we could use the Addition Rule which is $P(A \cup B) = P(A) + P(B)$  where A and B are disjoint events, i.e. $A \cap B = \emptyset$, A is the event of tossing all heads and B is the event of tossing all tails.  I'm not sure how to continue after that... would the complement be used? Research statement ideas got used by interviewing committee Let $f$ be increasing on $[a,b]$ and $a < x_1 < \dotsb < x_n < b$. Show that $\sum_{k = 1}^n [f(x_k^+) - f(x_k^-)]\leq f(b^-) - f(a^+)$. The youngest person on Earth in a society of immortals The protagonist is the youngest person on Earth -- the only one to be born in many centuries. Terrestrial society is small but every individual is unique and multitalented, as they are all immortal and have lived for centuries. They live in flying habitats.  No one knows how to raise children anymore; the protagonist wears diapers into adulthood because he was never potty-trained, and his education was haphazard. Consequently, he is quite lost in this advanced society, and only tolerated by the sophisticates around him.   When the Earth is invaded, the immortals put up only a show of a fight, arriving at the scene of battle in a colorful flotilla of unarmed flying habitats. To the horror of the protagonist, they are wiped out by the invaders. He realizes that the immortals were weary of life and welcomed this end.   This is probably a novel, though it might be a novella. It was written in English and I read it probably in the 1970s or 1980s. Can Xdebug be triggered by Drush commands? I'm running into an issue with my Features modules and installation profile when installing the site with drush si. There are a number of probably related issues in the Features issue queue, but none with good solutions. The interesting part (mentioned in some of the issues) is that it works fine when installing the site through the browser, but fails when installing with drush si.  I have an idea of what's wrong, but need to confirm it. The only way I can think of confirming it is to step through the code with a debugger (Xdebug is what I have set up). So, is it possible to trigger Xdebug from a command run with Drush? Specifically, can I get Xdebug to kick in when I run drush si? FPDF error: Some data has already been output, can't send PDF I have a couple of illustrations done in Illustrator and I am planing to animate it for a website that I am working on, I've heard of  toolkit with flash, but is it the best way to go or is there another "better" way of doing it?  Here's an example of the kind of animations I am aiming for: I'm a mathematician interested in abstract QFT.  I'm trying to undersand why, under certain (all?) circumstances, we must have $T^2 = -1$ rather than $T^2 = +1$, where $T$ is the time reversal operator.  I understand from the Wikipedia article that requiring that energy stay positive forces $T$ to be represented by an anti-unitary operator.  But I don't see how this forces $T^2=-1$.  (Or maybe it doesn't force it, it merely allows it?)  Here's another version of my question.  There are two distinct double covers of the Lie group $O(n)$ which restrict to the familiar $Spin(n)\to SO(n)$ cover on $SO(n)$; they are called $Pin_+(n)$ and $Pin_-(n)$.  If $R\in O(n)$ is a reflection and $\tilde{R}\in Pin_\pm(n)$ covers $R$, then $\tilde{R}^2 = \pm 1$.  So saying that $T^2=-1$ means we are in $Pin_-$ rather than $Pin_+$.  (I'm assuming Euclidean signature here.)  My question (version 2): Under what circumstances are we forced to use $Pin_-$ rather than $Pin_+$ here?  (I posted a  on physics.stackexchange.com last week, but there were no replies.)    EDIT: Thanks to the half-integer spin hint in the comments below, I was able to do a more effective web search.  If I understand correctly, Kramer's theorem says that for even-dimensional (half integer spin) representations of the Spin group, $T$ must satisfy $T^2=-1$, while for the odd-dimensional representations (integer spin), we have $T^2=1$.  I guess at this point it becomes a straightforward question in representation theory: Given an irreducible representation of $Spin(n)$, we can ask whether it is possible to extend it to $Pin_-(n)$ (or $Pin_+(n)$) so that the lifted reflections $\tilde R$ (e.g. $T$) act as an anti-unitary operator.