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Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[(fg)^{(n)}=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)f^{k}g^{nk}\] where\[f^{(n)}=\frac{ d^{(n)} f}{ dx^{n} }\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
link can also help
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
I am stuck on \[\left(\begin{matrix}m \\ k1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=?\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
http://physicspages.com/2011/03/22/generalizedproductruleleibnizsformula/
 one year ago

Hares333 Group TitleBest ResponseYou've already chosen the best response.1
http://www.math.ucsd.edu/~wgarner/math20a/prodrule.htm
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.0
\[\left(\begin{matrix}m \\ k1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=\frac{ m! }{ (k1)!(mk+1)! }+\frac{ m! }{ k!(mk!) }\]\[=\frac{ m! }{ (k1)!(mk)! }\left[ \frac{ 1 }{ mk+1 }+\frac{ 1 }{ k } \right]\]\[=\frac{ m! }{ (k1)!(mk)! }\left[ \frac{ k+mk+1 }{ k(mk+1) } \right]\]\[=\frac{ m! }{ (k1)!(mk)! }.\frac{ m+1 }{ k(mk+1) }\]\[=\frac{ (m+1)! }{ k!(mk+1)! }=\left(\begin{matrix}m+1 \\ k\end{matrix}\right)\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.0
@Jonask hope this helped
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
thanks guys ,appreciated
 one year ago
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