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

• September 15th 2010, 09:28 PM
simkate96
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 Pascal’s Idenity for 1<(or=)k<(or=)m
• September 15th 2010, 10:54 PM
Prove It
What are $f(x)$ and $g(x)$?

Also this is near impossible to read.

Are you asking to show this?

$\displaystyle{h^n(x) = \sum_{k = 0}^n\frac{n!}{k!(n-k)!}f^{n-k}g^k(x)}$
• September 16th 2010, 05:34 AM
tonio
Quote:

Originally Posted by simkate96
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 Pascal’s Idenity for 1<(or=)k<(or=)m

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

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