N-th derivative of a product in binomial expansion?

Question

I believe that the following is true:

$$\frac{d^n}{dx^n}f(x)g(x)=\sum_{i=0}^{\infty}\frac{n!}{i!(n-i)!}f^{(n-m)}(x)g^{(m)}(x)$$

The rational part of the summation is binomial expansion constants and $f^{(n)}(x)=\frac{d^n}{dx^n}f(x)$

I have tested it for some values of $n$ where $f$ and $g$ are either polynomials or exponential functions and it appears to hold true.

The question is whether or not the above is true with a proof.

For those who concern, $n$ may or may not be a positive integer or even an integer at all because I wish to use this in Fractional Calculus allowing $n\in\mathbb{C}$.

@Bib-lost Well, its harder to write it out as $n(n-1)(n-2)(n-3)\dots(n-n)\dots$, but they are equivalent. — Simply Beautiful Art, Feb 26 '16 at 21:08
So how do you definie $n!$ if $n < 0$? What is (-2)! for example? — Bib-lost, Feb 26 '16 at 23:19
@Bib-lost Well, you'd never have to worry about that because $n-n=0$, which occurs if $n$ is a positive integer. Just know that $|(-2)!|=\infty$ so that the denominator is infinity and the whole term becomes $0$. — Simply Beautiful Art, Feb 27 '16 at 00:40
In any case, if the terms in the series all become zero after the $n$th term, it would be cleaner (in my opinion) to just write $\sum_{i=1}^n$, as suggested by the answers bellow. Now you're making it look like a convergent series, instead of just a finite sum. — Bib-lost, Feb 27 '16 at 08:46
@Bib-lost Well, it obviously depends on what $n$ actually is, which most would assume to be an integer. For my uses however, I am going to use it for non integer values for something related to Fractional Calculus, thus it won't be a finite summation for me. — Simply Beautiful Art, Feb 27 '16 at 12:59
Hmm, that does make your question considerably more difficult. Perhaps this is something you should mention in the question itself, as both answers (and I too) supposed $n$ to be a positive integer in your question. — Bib-lost, Feb 27 '16 at 14:06
just curious: how would you define the factorial of a non-integer? — user190080, Feb 27 '16 at 14:19

peter.petrov · Accepted Answer · 2016-02-26T20:43:47.790

1

Yes, this is a famous formula about the Nth-derivative of a product,
it's from Leibniz (so says my textbook at least).
It can be proved by induction without any obstacles.

It's just that the sum is finite.

N-th derivative of a product in binomial expansion?

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