jquery animation callback called once instead of for each element
Callback animation to begin only after ALL children have finished animating
iPhone, reproduce the magnifier effect Irreducible polynomial is always separable in char 0 field Render semi-transparent volume with cycles I want to render an image of a fire in cycles, but as a png with a transparent background. Without a transparent background, the output image looks like this:    But as soon as I tick the "Transparent" checkbox under Render->Film, all the semi-transparent parts of the image are not rendered and it looks like this    which is clearly not what I want. I've tried to add a holdout shader or a transparent shader to my smoke domain material nodes, but it didn't help. I've only found answered questions for semi-transparent glass, but none for volumes. My node setup looks like this:    but the transparent shader doesn't do anything.  Has anyone an idea how to fix this?  Thanks  Edit: I've just changed the flow type to "Fire and Smoke" instead of "Fire" and now the semi-transparent parts render too, but I don't want the smoke. How to tell a lambda function to capture a copy instead of a reference in C#? I've been learning C#, and I'm trying to understand lambdas.  In this sample below, it prints out 10 ten times.  class Program {     delegate void Action();     static void Main(string[] args)     {         List&lt;Action&gt; actions = new List&lt;Action&gt;();          for (int i = 0; i &lt; 10; ++i )             actions.Add(()=&gt;Console.WriteLine(i));          foreach (Action a in actions)             a();     } }   Obviously, the generated class behind the lambda is storing a reference or pointer to the int i variable, and is assigning a new value to the same reference every time the loop iterates.  Is there a way to force the lamda to grab a copy instead, like the C++0x syntax  [&amp;](){ ... } // Capture by reference   vs.  [=](){ ... } // Capture copies I want to run a ppp connection when my USB modem is connected, so I use this udev rule:  ACTION=="add", SUBSYSTEM=="tty", ATTRS{idVendor}=="16d8",\     RUN+="/usr/local/bin/newPPP.sh $env{DEVNAME}"  (My modem appears in /dev as ttyACM0)  newPPP.sh:  #!/bin/bash /usr/bin/pon prov $1 &gt;/dev/null 2&gt;&amp;1 &amp;   Problem:  The udev event fires, and newPPP.sh is running, but newPPP.sh process is killed after ~4-5s. ppp does not have time to connect (its timeout is 10s for dial up).  How can I run a long time process, that will not be killed?  I tried using nohup, but it didn't work either.  System: Arch Linux  Update  I found a solution , thanks to .  I use at now to run my job detached from udev process.  But the one question remains unanswered: Why do nohup and &amp; not work? Proving a limit of a trigonometric function: $\lim_{x \to 2/\pi}\lfloor \sin \frac{1}{x} \rfloor=0$ De Rham Cohomology of the tangent bundle of a manifold I would like to compute the de Rham cohomology of the tangent bundle $TM$ of a manifold $M$. It seems to me that we can just homotopy each fibre to a point, and that this would give a homotopy equivalence between $TM$ and $M$. Since de Rham cohomology is a homotopy invariant, this would imply that $H^*_{\text{dR}}(TM) \cong H^*_{\text{dR}}(M)$.     Is it always true that the tangent bundle of a manifold is homotopic to the manifold itself by the homotopy of each fibre with a point? Is trace of regular representation in Lie group a delta function? My major is physics. I need to use some tools in group theory, but I am really  confused by the trace in compact infinite groups. The following is my question:  In discrete group theory, the irreducible representation of identity group element $e$ is always an identity matrix. So the trace of $e$ under  regular representation is the order of the group : $\chi(e) = |G|$. I hope to get  similar result for Lie group. For example, SO(2) has infinite number of irreducible representations, $D^{(m)}$, where $m=0,\pm 1, \pm 2,\cdots $. All of them are 1-dimensional $D^{(m)} = e^{im\phi}$. Here $\phi$ is the rotation angle. For element $R(\phi)$ in SO(2), the trace in $m$th irreducible representation should be  $\chi(R(\phi))=e^{im\phi}$, if we sum all these traces up, we get the trace in regular representation. So what is the trace of $e$ in the regular representation ? It seems to be infinite. g++ undefined reference to typeinfo When is "seems to be" used instead of "seems"? The baby seems happy/comfortable. The baby seems to be happy/comfortable.   Are these both correct? If both are correct, what are the differences between them? I computed the homology groups of the torus, can someone tell me if this is correct? The calculation, not the result that is. Thanks!  The cells of $T^2$ are $e^0, e^1_a, e^1_b, e^2$  The chain groups are  $$ C_0(T^2) = \{ k e^0 |&#160;k \in \mathbb{Z} \} = \mathbb{Z}$$  $$ C_1(T^2) = \{&#160;k_1 e^1_a + k_2 e^1_b |&#160;k_1 , k_2 \in \mathbb{Z} \} = \mathbb{Z} \oplus \mathbb{Z}$$  $$ C_2(T^2) = \mathbb{Z}$$  $$ C_k(T^2) = \{ 0 \} , k &gt; 2$$  Now the homology groups:  $$ H_0(T^2) = \ker \partial_0 / im \partial_1 = \mathbb{Z} / 0 = \mathbb{Z}$$  where $im \partial_1 = 0$ because there is no chain in $C_1(T^2)$ whose boundary is a zero-chain in $C_0(T^2)$. (Is this reasoning correct?)  $$ H_1(T^2) = \ker \partial_1 / im \partial_2 = \mathbb{Z} \oplus \mathbb{Z}$$  where $\ker \partial_1 =  \mathbb{Z} \oplus \mathbb{Z} $, i.e. again everything maps to zero because there is no element in $C_1$ whose boundary maps to an element in $C_0$.  $im \partial_2 = 0$ again because there is no element in $C_2$ whose boundary is an element of $C_1$.  I don't want to use Hurewicz to do $H_1$.  $$ H_2(T^2) = \mathbb{Z}$$ using similar arguments as above.  Thanks for your help.  Edit  I posted the computations as an answer below. I got two up votes but I don't know by who so I'm not yet sure I can trust them... Proof on showing if F(x,y,z)=0 then product of partial derivatives (evaluated at an assigned coordinate) is -1 The task is as follows:     Given: $$F(x,y,z) = 0$$    Goal:  Show    $\frac{\partial  z}{\partial y}|_x \frac{\partial  y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1$   Here is my work so far:  (1) Differentiate with respect to y, I get:  $0 + F_2 + F_3  \frac{\partial z}{\partial y} = 0$  So $ F_3  \frac{\partial z}{\partial y} = - F_2$  (2) Differentiate with respect to x, I get:  $F_1 + F_2 \frac{\partial y}{\partial x}  + 0 = 0$  So $ F_2  \frac{\partial y}{\partial x} = - F_1$  (3) Differentiate with respect to z, I get:  $F_1  \frac{\partial x}{\partial z}    + 0  + F_3 = 0$  So $ F_1  \frac{\partial x}{\partial z} = - F_3$  (4) After some manipulations with the $F_i$, I get to the conclusion that $\frac{\partial z}{\partial y}* \frac{\partial y}{\partial x} * \frac{\partial x}{\partial z}  = -1$, so when evaluated with x, z, y respectively, conclusion is still true  My questions are:  (1) Is my proof correct?  (2) For example, when I differentiate with respect to y, I "let" $F_1$ be 0 and find partials for other coordinates.    I had a hard time trying to explain to my friend on the reason(s) why I can do such "let be 0" thing.  Although I think if I can't do that, then there is no way that I can reach the conclusion, but I somehow feel confused about the fact too.  Since my book is doing it that way, my understanding is that I can do such "let be 0" thing based on the independece of x with respect to y, when I differentiate with respect to y.  But is my thought ok?  Would someone please help me on this question? Thank you very much ^_^ Is Dragonstone unoccupied after Stannis Baratheon's defeat? Is anything known about Dragonstone after:      Stannis Baratheon is killed in his bid for the Iron Throne?   As the seat of the naval defense of King's Landing and the birthplace of Daenerys Targaryen, she has a vested interest in taking it.   At the conclusion of Season 6 Episode 10, is Dragonstone unoccupied? The cardinality of the set of all finite subsets of an infinite set Ierator and for-each loop both we use for traverse no of items, I have used for-each many times but I am not clear about iterator, means when we should use ierator. I am studying salesforce workbook example..  global class CustomIterable implements iterator&lt;Account&gt;{    public List&lt;Account&gt; accs {get; set;}   public Integer i {get; set;}    public CustomIterable(){      accs =[SELECT Name fROM Account limit 5];      i = 0;              }   global boolean hasNext(){          if(i &gt;= accs.size()) {          return false;         } else {          return true;         }    }    global Account next(){          if(i == 8){return null;}          i++;          return accs[i-1];     } }   //------------  @isTest(seealldata=true) global class IteratorClass{     public static testmethod void main(){          IteratorClass obj = new IteratorClass();         Iterator&lt;Account&gt; acs =  obj.Iterator();         system.debug(acs);      }     global Iterator&lt;Account&gt; Iterator(){             return new CustomIterable();          }  } RSA encryption with private key and decryption with a public key How were the solar masses and distance of the GW150914 merger event calculated from the signal? I was trying to prove following inequality:  $$|\sin n\theta| \leq n\sin \theta \ \text{for all n=1,2,3... and } \  0&lt;\theta&lt;π $$  I succeeded in proving this via induction but I didn't get "feel" over the proof. Are there other proof for this inequality? Is it possible to install Office 365 (2013) on Ubuntu 15.04 Vivid Vervet? I would like to install Office 2015/365 on Ubuntu 15.04. I tried to install Office using Wine, but that did not work. Please note that I do not want virtualisation unless it has a -like behaviour, where application windows are treated like regular Ubuntu application windows by the system. Group of even order contains an element of order 2 I am working on the following problem from group theory:     If $G$ is a group of order $2n$, show that the number of elements of $G$ of order $2$ is odd.   That is, for some integer $k$, there are $2k+1$ elements $a$ such that $a \in G,\;\; a*a = e$, where $e$ is the identity element of $G$.