How to Calculate center of mass for 20 dimensions particles I have a problem with calculation the center of mass of 20-dimension particles, some thing like this: A = [1 6 8 54 6 8 5 4 8 9 6 4 7 9 6 6 3 8 43 9] , Mass = 0.25 B = [2 6 3 4 6 8 4 4 8 5 6 4 2 2 6 6 3 8 1 1] , Mass = 0.6 C = [4 3 4 53 6 2 5 21 8 1 6 2 37 2 6 2 3 1 43 9] , Mass = 0.05 Could anyone guide me to find the solution for center of mass in each dimension (x1, x2, ... x20)? Thanks
Center of mass in n dimension, possible? Is it possible to calculate the center of mass for some 20-dimension ($x_i, i=1,\ldots,20$) particles? If it is possible, how can I do that?
Is there a way to start over?  No spoilers here, but after a certain point in the game I wish I had done something different, so I'd like to perhaps start over and make some different decisions based on my current game play experience. Duplicated Onedrive icon in explorer I've installed windows 10 on a new pc, with a secondary hard disk containing my onedrive folder, taken from my old pc. I've configured onedrive in order to use that folder, and it's working fine.  The problem is that when I open Explorer I can see two onedrive icons on the sidebar, both of them pointing at the same folder.  How can I remove the duplicated onedrive entry from the sidebar? e.force:navigateToURL: [Cannot read property 'setParams' of undefined] Maximum value of $\sin A+\sin B+\sin C$? What is the maximum value of $\sin A+\sin B+\sin C$ in a triangle $ABC$. My book says its $3\sqrt3/2$ but I have no idea how to prove it.   I can see that if $A=B=C=\frac\pi3$ then I get $\sin A+\sin B+\sin C=\frac{3\sqrt3}2$. And also  maximum is attained for $a=b=c$. But this does not give me any idea for the proof.  Can anyone help? Does photon have size measurement because of its particle nature Does photon have size measurement because of its particle nature like electron's 3.86*10^-13m etc.. What do I do when glitched? I was playing Grand Theft Auto Online on Xbox 360, and a player teleported to me.  I shot them, but they placed a campfire on me.  Now, I am constantly being burnt to death. How do I get out? Find all integer solutions to $x^2+4=y^3$. I'd like to find out why  \begin{align} \sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6 \end{align}  I tried to rewrite it into a geometric series  \begin{align} \sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2 \end{align}  But I don't know what to do with the $n^2$. Error : Warning: strpos() [function.strpos]: Offset not contained in string - can not find solution How do I test asynchronous Apex? This is a canonical question and answer developed by the community to help address common questions. If you've been directed here, or your question has been closed as a duplicate, please look through the resources here and use them to shape more specific questions. To browse all canonical questions and answers, including more unit test resources, navigate to the  tag.  I'm testing asynchronous code - like Batch Apex, Queueable Apex, @future methods, or Scheduled Apex - and I'm getting unexpected results. It seems like portions of my code aren't running, or I'm getting confusing errors about batch invocations. How do I effectively test this kind of code? Let the search find questions by ID Add desktop notifications for Chrome or Firefox Can I safely charge my laptop with a non-standard, third-party charger? I have a Toshiba Satellite laptop. My charger has stopped working.  I have access to a Lenovo charger. Can I use this charger on my laptop? Draw a vertical line over the entries of a column in an array I'd like to draw vertical/horizontal lines over the entries of a column/row in an array. I find the method to draw horizontal lines at . But when I used the same way to draw the vertical lines. It didn't look well. My codes are:  \usepackage{tikz} \usetikzlibrary{calc} \newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};} \newcommand{\DrawLine}[3][]{% \begin{tikzpicture}[overlay,remember picture]     \draw [#1] ($(#2)+(0,0.6ex)$) -- ($(#3)+(0,0.6ex)$); \end{tikzpicture}% }%  \begin{displaymath} \left[ \begin{array}{cccc} \tikzmark{11}{15} &amp; 0 &amp; \tikzmark{13}{0} &amp;0\tikzmark{14} \\ 0 &amp; 50 &amp; 20 &amp; 25 \\ 35 &amp; 5 &amp; 0 &amp; 10\\ \tikzmark{41}{0} &amp; 65 &amp; \tikzmark{43}{50} &amp; 65 \end{array} \right]. \DrawLine[black,thick]{11}{14} \DrawLine[black,thick]{11}{41} \DrawLine[black,thick]{13}{43} \end{displaymath}   The result looks like     But the result I want is     Does anyone know how to fix this? Thank you. Can I travel within the Schengen area if I currently don't have a valid visa or residency permit? 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How to derive $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$ indicates that the following relation is &quot;easily shown&quot;: $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$, however I'm having some trouble showing it.  I think I'm just messing up the multivariable Taylor expansion (of $F(\vec p)$).  Can one of you walk me through it or link me to site that will?  Thanks.  Edit: Here's what I get (without using $x=i\hbar \frac \partial {\partial p}$ which I haven't proven yet): $$F(\vec p) = F(\vec 0) + \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}p_j + \frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} p_j p_k + \dots$$ so $$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$= x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla \psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla \psi)_j (\nabla \psi)_k + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla x_i\psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla x_i\psi)_j (\nabla x_i\psi)_k + \dots]$$ where $$(\nabla x_i\psi)_j=\frac {\partial x_i}{\partial x_j}\psi + x_i\frac {\partial \psi}{\partial x_j} = \delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}$$ and $$(\nabla x_i\psi)_j(\nabla x_i\psi)_k=(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j})(\delta_{ik}\psi + x_i\frac {\partial \psi}{\partial x_k}) = \delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}$$ Thus: $$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$=  x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}\frac {\partial \psi}{\partial x_j} + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} \frac {\partial \psi}{\partial x_j} \frac {\partial \psi}{\partial x_k} + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}) + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}) + \dots]$$ From here it doesn't look like those higher order terms are all going to cancel out.