Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Jonask

  • 3 years ago

Prove Leibniz Formular:product rule

  • This Question is Closed
  1. Jonask
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[(fg)^{(n)}=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)f^{k}g^{n-k}\] where\[f^{(n)}=\frac{ d^{(n)} f}{ dx^{n} }\]

  2. Jonask
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    link can also help

  3. Jonask
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am stuck on \[\left(\begin{matrix}m \\ k-1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=?\]

  4. Jonask
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    http://physicspages.com/2011/03/22/generalized-product-rule-leibnizs-formula/

  5. Hares333
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    http://www.math.ucsd.edu/~wgarner/math20a/prodrule.htm

  6. sirm3d
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\left(\begin{matrix}m \\ k-1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=\frac{ m! }{ (k-1)!(m-k+1)! }+\frac{ m! }{ k!(m-k!) }\]\[=\frac{ m! }{ (k-1)!(m-k)! }\left[ \frac{ 1 }{ m-k+1 }+\frac{ 1 }{ k } \right]\]\[=\frac{ m! }{ (k-1)!(m-k)! }\left[ \frac{ k+m-k+1 }{ k(m-k+1) } \right]\]\[=\frac{ m! }{ (k-1)!(m-k)! }.\frac{ m+1 }{ k(m-k+1) }\]\[=\frac{ (m+1)! }{ k!(m-k+1)! }=\left(\begin{matrix}m+1 \\ k\end{matrix}\right)\]

  7. sirm3d
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Jonask hope this helped

  8. Jonask
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks guys ,appreciated

  9. Jonask
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    3 Medals

  10. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy