anonymous
  • anonymous
Prove Leibniz Formular:product rule
Calculus1
katieb
  • katieb
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anonymous
  • anonymous
\[(fg)^{(n)}=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)f^{k}g^{n-k}\] where\[f^{(n)}=\frac{ d^{(n)} f}{ dx^{n} }\]
anonymous
  • anonymous
link can also help
anonymous
  • anonymous
I am stuck on \[\left(\begin{matrix}m \\ k-1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=?\]

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anonymous
  • anonymous
http://physicspages.com/2011/03/22/generalized-product-rule-leibnizs-formula/
anonymous
  • anonymous
http://www.math.ucsd.edu/~wgarner/math20a/prodrule.htm
sirm3d
  • sirm3d
\[\left(\begin{matrix}m \\ k-1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=\frac{ m! }{ (k-1)!(m-k+1)! }+\frac{ m! }{ k!(m-k!) }\]\[=\frac{ m! }{ (k-1)!(m-k)! }\left[ \frac{ 1 }{ m-k+1 }+\frac{ 1 }{ k } \right]\]\[=\frac{ m! }{ (k-1)!(m-k)! }\left[ \frac{ k+m-k+1 }{ k(m-k+1) } \right]\]\[=\frac{ m! }{ (k-1)!(m-k)! }.\frac{ m+1 }{ k(m-k+1) }\]\[=\frac{ (m+1)! }{ k!(m-k+1)! }=\left(\begin{matrix}m+1 \\ k\end{matrix}\right)\]
sirm3d
  • sirm3d
@Jonask hope this helped
anonymous
  • anonymous
thanks guys ,appreciated
anonymous
  • anonymous
3 Medals

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