I have html code with radio buttons and some common button. After click common button I copy this code with .clone and insert after with .after. In first block I have radio buttons with name like 'name', in second I change this name like '1name', in third -'2name' and etc. Moreover I have js code      $('input[type=radio]').change(function(){             var index= $(this).attr('rbgindex');             var name = $(this).attr('name');             $('.hid-'+name).remove();             $(this).parent().after('&lt;input class=hid-'+name+' type=hidden name='+name+index+' value='+index+' checked&gt;');     });   It should create hidden filed after change radio button. But I have problem with it. If I change radio button in firsy work, it work, but in another it doesn't work. What's wrong?
I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.
Can ANOVA be significant when none of the pairwise t-tests is? Is it possible for one-way (with $N&gt;2$ groups, or "levels") ANOVA to report a significant difference when none of the $N(N-1)/2$ pairwise t-tests does?   In  @whuber wrote:      It is well known that a global ANOVA F test can detect a difference of means even in cases where no individual [unadjusted pairwise] t-test of any of the pairs of means will yield a significant result.   so apparently it is possible, but I do not understand how. When does it happen and what the intuition behind such a case would be? Maybe somebody can provide a simple toy example of such a situation?  Some further remarks:   The opposite is clearly possible: overall ANOVA can be non-significant while some of the pairwise t-tests erroneously report significant differences (i.e. those would be false positives). My question is about standard, non-adjusted for multiple comparisons t-tests. If adjusted tests are used (like e.g. Tukey's HSD procedure), then it is possible that none of them turns out to be significant even though the overall ANOVA is. This is covered here in several questions, e.g.  and . Update. My question originally referred to the usual two-sample pairwise t-tests. However, as @whuber pointed out in the comments, in the ANOVA context, t-tests are usually understood as post hoc contrasts using the ANOVA estimate of the within-group variance, pooled across all groups (which is not what happens in a two-sample t-test). So there are actually two different versions of my question, and the answer to both of them turns out to be positive. See below. I was trying to dual boot elementary os with windows on my laptop (Asus Zenbook UX305). To do that I shrank the windows partition using the windows disk manager, and then used the Elementary installer's "install alongside windows" option to set up the linux partitions automatically. When I rebooted after installing I was put into the grub console. I was able to boot linux manually with the commands  grub&gt; set root=(hd0,gpt4) grub&gt; linux /boot/vmlinuz-3.19.0-26-generic root=/dev/sda4 grub&gt; initrd /boot/initrd.img-3.19.0-26-generic grub&gt; boot     and I found that running the command  grub&gt; configfile (hd0,gpt4)/boot/grub/grub.cfg   would load the normal grub menu as is supposed to happen - so there don't seem to be any problems with the config file itself. I also found that the root and prefix are on (hd0,gpt1), so running  grub&gt; set root=(hd0,gpt4) grub&gt; set prefix=(hd0,gpt4)/boot/grub grub&gt; insmod normal grub&gt; normal   also loaded the normal grub menu. So possibly the problem is that grub is looking on the wrong partition?  In my BIOS, I have a choice between using grub and the windows bootloader to boot. Selecting the windows bootloader boots me directly into windows without any problems.  However, I can't find any way to get any fixes to stick after a reboot. I've tried rerunning grub-install /dev/sda, recreating the config file with grub-mkconfig, and using the Ubuntu boot repair gui tool - none of these have worked. Is there any way to fix this short of reinstalling either or both OSs? I'd rather not have to reinstall windows if possible. Sign of Work and potential energy in electrostatics Conceptual question:  Suppose we have a configuration of point charges. If the potential of the energy of the system is negative, this means work is positive.   I'm kind of rusty with my mechanics, does the fact that the work is positive mean that to arrange the configuration you have to do work against the electric field? I.e. supply energy via some external force?  In terms of electrostatics, is there a general way of interpreting the sign of potential energy and work? How to get perfect UV sphere Mercator projection? I'm trying to unwrap a UV Sphere and I can't seem to get a perfect  projection. What process should I follow to get a perfect UV unwrap that looks like the black grid lines in this image?: Exclude menu item from the collapse of bootstrap 3 navbar Where is composition tab in QGIS Composer to change page size? I am trying to change the page size in the QGIS 3.0 Composer but I am not able to see the Composition tab.  I have right clicked an empty space in the toolbar but did not see anything about composition.  I am using Windows 7.  I tried looking at the QGIS docs but they are still in testing. I want to update ubuntu 11.04 throught apt-get. So I added the following lines in apt.conf  export http_proxy=http://username:password@proxyserver.net:port/ export ftp_proxy=http://username:password@proxyserver.netport/  export http_proxy=http://deepak:Deepak@123@12.1.1.1:3128   My question is: how do I insert special characters to a username or password?  For instance: my password for proxy is Deepak@123 and it is getting an error. I've been working through a bit of Ireland and Rosen's Number Theory for fun. Problem 4.11 of Ireland and Rosen asks     Prove that $1^k+2^k+\cdots+(p-1)^k\equiv 0\pmod{p}$ if $p-1\nmid k$, and $-1\pmod{p}$ if $p-1\mid k$.    My work is leading where I didn't expect to go. Here I assume $p$ id odd. First, if $p-1\mid k$, then we can wrike $k=(p-1)j$ for some $j$. But then $$ 1^k+\cdots+(p-1)^k\equiv 1^{(p-1)j}+\cdots+(p-1)^{(p-1)j}\equiv \underbrace{1+\cdots+1}_{p-1\text{ times}}=\frac{(p-1)p}{2}\equiv 0\pmod{p} $$ since $p-1$ is even.   Secondly, I suppose $p-1\nmid k$. WLOG, I can assume $0&lt;k&lt;p-1$, Since if $k=(p-1)j+r$ with $0&lt;r&lt;p-1$, then  $$ a^k\equiv a^{(p-1)j+r}\equiv a^r\pmod{p}. $$  Then if $g$ is a primitive root, $$ 1^k+\cdots+(p-1)^k=\sum_{i=1}^{p-1}(g^k)^i=\frac{g^k(1-g^{k(p-1)})}{1-g^k}. $$ Since $g$ is primitive, $p\nmid 1-g^k$, but $p\mid 1-g^{k(p-1)}$, so the above sum is again congruent to $0\pmod{p}$.  So in all cases, I get the sum is congruent to $0$ regardless. Have I messed up, or is there a typo? Thanks. Change Proxy Settings In Ubuntu I want to use the Ubuntu Software Center to install software, but I cannot download them.  How can I give my proxy settings in Ubuntu terminal (just like we use cmd in Windows to choose IE proxy settings)? Installing grounded light fixture on box with no ground conductor I'm installing a brand new motion activated security light to an existing box on my garage. After removing the old light, I notice it only has white/black wiring while the new light fixture has of course white/black/green (ground) wires to attach. What do I do if no ground wire is available with the old box? I am trying to evaluate the following:  $\displaystyle \int_{0}^{\infty} \frac{1}{1+x^a}dx$, where $a&gt;1$ and $a \in \mathbb{R}$  Any help will be much appreciated. If $R_{1}$ and $R_{2}$ are relations on a set S with $R_{1};R_{2}=R_{2};R_{1}$. Then $R_{1}$ and $R_{2}$ are bijective maps For the surjectivity part, I showed that if $(a,b)\in I$ then $(a,c)\in R_{1}$ and $(c,b)\in R_{1}$ and $(a,c)\in R_{2}$ and $(c,b)\in R_{2}$ for some $c\in S$. Now for arbituary $(x,y)\in R_{1}$ we have $(x,z)\in I$ which implies that $(y,z)\in R_{2}$. But also $(y,z)\in R_{1}$, so $R_{1}$ is surjective. Similar argument for $R_{2}$  And for Injectivity, I've tried to show that if $aR_{1};R_{2}c$ and $bR_{2};R_{1}c$, then $a=b$ but I am not entirely sure what $a=b$ means in this context. Do I have to show that there is an equivalence relation containing (a,b)?  edit: $R_{1};R_{2}$ is a composition of two relations and $I=R_{1};R_{2}$ Is there a way to get reverse numbering on the enumerate environment? To build a list of publications, I wish to show the most recent on top of the list, while keeping the numbering similar. This can be achieved manually by doing something like:  \begin{enumerate} \item[42] pub num 42 \item[41] pub num 41 \item[...] \item[1] pub num 1 \end{enumerate}   but is there a more principled approach? $$ A $$  Light is travelling away from the absolute center of the Schwarzchild black hole (i.e. none of that spinning/stretching field) at $\ C $, and so shouldn't $\ \Delta C$ need to be greater than $\ C $ in order for the light to collapse into the black hole, therefore violating the relativistic limit of $\ C $?    It seems like light would just 'hover' right in the center of a black hole then. Because if it goes away from it, the event horizon pulls it back, and if it goes into it, the velocity change must be greater than $\ C $ which violates general relativity; so technically there's no way for the light to move away, and so it would just be still.  The thing is that the light must move at the speed of light or general relativity doesn't work. It's massless... But in the center of a black hole, if the above is true (it can't move or escape), then it is not travelling at the speed of light. So it cannot be still.  It cannot be spinning in a constant orbit either, because that would require velocity in an outward angle. And it cannot move in an outwardly fashion, only in an inwardly one.  Is it spinning in an infinitely decaying spiral towards the center?  If so what happens to it if and when it reaches the direct and complete center, i.e. The case of it being 'shot' from the center.    $$ B $$  If you shot light at an angle towards a black hole, but not directly to the center of it  then that light from the outside will eventually become the same as the light that I said was shot from the center to the outside; because in both cases, they end up in an infinitely decaying spiral.    If so it would end up exactly like the light in the above paragraph, so then we need to figure out what happens to that light to figure this one out, or vice versa.    $$ C $$  If you shot light directly towards the center of a black hole it would have no spiral to decay into, so therefore it must stop at some point in the center of a black hole. If it stops though, it loses it's momentum and violates general relativity again because it's energy "disappears".    If so then it violates conservation laws. So I'm assuming this simply is not true.    Is the light travelling into a black hole somehow still travelling at the speed of light, in every direction possible where that direction keeps it facing towards the center?  If so, why is there not a point at which the only thing left for the light to do is stay still, since all other paths would increase the distance between it and the black hole's center, and therefore become Case A. And if there isn't, what would happen if there was?  Also, if it's a time and observation answer where it supposedly takes and infinite amount of time for C to hit the center, wouldn't it be the case that the black hole is 'eating' all of the available light in the universe and keeping it in stasis?  Also, what happens in a Kerr black hole then? Does it just rip the field of time/gravity at an angle such that the light would never get closer than the center? Jackknife vs. LOOCV Is this a capacitor start motor and can I start it with a microwave capacitor I salvaged an old motor out of a air conditioner and I want to know if it's a capacitor start motor if I can start it with the capacitor I have and how to start it.  I included pictures of the motor and capacitor btw i have never worked with capacitive start motors so i need details on exactly how to wire it up with out ruining it "Newtonian limit" property of special relativity Books say that special relativity is indistinguishable from Newtonian mechanics when the speed of the primed frame ($v$) is small compared to the speed of light ($c$). This is what I mean by the "Newtonian limit" property of special relativity. (I don't know the correct name for it). But the transformation of the time coordinate  $$t' = \gamma \left(t + \frac{vx}{c^2}\right)$$  involves the spatial quantity $x$ which, if large enough, could balance out the smallness of $v/c^2$. So why doesn't this observation mess up the "Newtonian limit" property? Why are the lower 3 bits of curve25519/ed25519 secret keys cleared during creation? I am currently experimenting with ed25519 and I noticed that on secret key creation, bit 254 is always set and the lower 3 bits are always cleared. I found that bit 254 is always set to protect against timing attacks in this question:   But why are the lower 3 bits cleared. Obviously it has to do with the formula in the curve25519 paper:  The set of secret keys is defined to be $\{\underline{n} : n \in 2^{254} + 8\{0, 1, 2, 3,\ldots, 2^{251}-1\}\}$  It's because of the 8 in there, but why is that 8 there? I suspect it has something to do with theorem 2.1 in the curve25519 paper, but I am not sure, because I do not understand fully what is being proven there.   I am experimenting with ed25519 primitives in some cryptographic routines which need me to add scalars to the secret key (add secret keys). Even if I add two well formed ed25519 secret keys, the result will not always have bit 254 set and the lower 3 bits cleared. Is this a security problem? I understand the risk for bit 254 but not the lower 3. Can I gift a game via the PlayStation Network? Is it possible to "gift" a game (e.g., Castle Crashers) to someone through the PlayStation Network?  I don't have a PS3 but was considering buying Castle Crashers as a gift for a friend so that we could play co-op.  Is this possible? Where is a good place to learn classical optics for high school competitions?