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anonymous
 3 years ago
Prove Leibniz Formular:product rule
anonymous
 3 years ago
Prove Leibniz Formular:product rule

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anonymous
 3 years ago
Best 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\[f^{(n)}=\frac{ d^{(n)} f}{ dx^{n} }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am stuck on \[\left(\begin{matrix}m \\ k1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=?\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://physicspages.com/2011/03/22/generalizedproductruleleibnizsformula/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\left(\begin{matrix}m \\ k1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=\frac{ m! }{ (k1)!(mk+1)! }+\frac{ m! }{ k!(mk!) }\]\[=\frac{ m! }{ (k1)!(mk)! }\left[ \frac{ 1 }{ mk+1 }+\frac{ 1 }{ k } \right]\]\[=\frac{ m! }{ (k1)!(mk)! }\left[ \frac{ k+mk+1 }{ k(mk+1) } \right]\]\[=\frac{ m! }{ (k1)!(mk)! }.\frac{ m+1 }{ k(mk+1) }\]\[=\frac{ (m+1)! }{ k!(mk+1)! }=\left(\begin{matrix}m+1 \\ k\end{matrix}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Jonask hope this helped

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks guys ,appreciated
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