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Thread: h^(n)= ∑ (k=0 to n) n!/k!(n-k)!f^(n-k) * g^(k)(x) : Prove that the following is true

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    h^(n)= ∑ (k=0 to n) n!/k!(n-k)!f^(n-k) * g^(k)(x) : Prove that the following is true

    Prove that the following is true for all n ∈ N(natural numbers)
    h^(n)= ∑ (k=0 to n) n!/k!(n-k)!f^(n-k) * g^(k)(x)


    Note: We can assume Pascalís Idenity for 1<(or=)k<(or=)m
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    What are $\displaystyle f(x)$ and $\displaystyle g(x)$?

    Also this is near impossible to read.

    Are you asking to show this?

    $\displaystyle \displaystyle{h^n(x) = \sum_{k = 0}^n\frac{n!}{k!(n-k)!}f^{n-k}g^k(x)}$
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    Quote Originally Posted by simkate96 View Post
    Prove that the following is true for all n ∈ N(natural numbers)
    h^(n)= ∑ (k=0 to n) n!/k!(n-k)!f^(n-k) * g^(k)(x)


    Note: We can assume Pascalís Idenity for 1<(or=)k<(or=)m


    This is a very sloppy question: what are $\displaystyle h,\,f,\,g?$.

    I think you're asking to prove Leibnitz equality for the n-th derivative of the product of two derivable functions, and then induction is your friend.

    Tonio
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