- #1

Gib Z

Homework Helper

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I was just messing around on another problem I was trying to solve for someone in the homework forums when I stumbled on this:

If we denote the n-th derivative the same way we do for exponents, eg the second derivative of f will be denoted by f^2, and the original function as its zero-th derivative, and define a function f, that is the product of two functions u and v, ie [tex]f(x) = u(x)\cdot v(x)[/tex].

Then the following holds:

[tex]f^n (x) = \sum_{k=0}^n \frac{n!}{k! (n-k)!} u^{n-k} v^k[/tex]

I'm guessing if we chose to use induction, that theorem would be not too difficult to prove since it would be analogous to the proof of the binomial theorem. Also in the deeps on my mind it looks somewhat similar to a repeated differentiation formula, but that may have been more general (any function f(x), not necessarily product of u and v) or even something else.

Could someone either tell me that repeated differentiation formula, post a nice proof, or just comment, because commentaries are welcome =]

EDIT: I was thinking of making my reply to you the 2000th post, but I thought best leave it till tomorrow (its getting quite late here) to use it on a post that my enlighten someone else who needs help =] Thank you very much for the congrats arildno, and everyone else on these forums that makes it the wonderful, pleasant and enlightening place it is! Heres to 2000 more =D

PS. For frequent posters, if you are able to, send it some Contributions to Greg, it feels nice giving a little bit back !

If we denote the n-th derivative the same way we do for exponents, eg the second derivative of f will be denoted by f^2, and the original function as its zero-th derivative, and define a function f, that is the product of two functions u and v, ie [tex]f(x) = u(x)\cdot v(x)[/tex].

Then the following holds:

[tex]f^n (x) = \sum_{k=0}^n \frac{n!}{k! (n-k)!} u^{n-k} v^k[/tex]

I'm guessing if we chose to use induction, that theorem would be not too difficult to prove since it would be analogous to the proof of the binomial theorem. Also in the deeps on my mind it looks somewhat similar to a repeated differentiation formula, but that may have been more general (any function f(x), not necessarily product of u and v) or even something else.

Could someone either tell me that repeated differentiation formula, post a nice proof, or just comment, because commentaries are welcome =]

EDIT: I was thinking of making my reply to you the 2000th post, but I thought best leave it till tomorrow (its getting quite late here) to use it on a post that my enlighten someone else who needs help =] Thank you very much for the congrats arildno, and everyone else on these forums that makes it the wonderful, pleasant and enlightening place it is! Heres to 2000 more =D

PS. For frequent posters, if you are able to, send it some Contributions to Greg, it feels nice giving a little bit back !

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