Binomial Theorem of Differentiation?

Question

I noticed that $$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$ and it's had me scratching for a little bit. It's easy to see how the cross terms add up but can anyone draw a direct comparison between the behavior being exhibited here and the binomial expansion? It seems to work for the case equivalent to the multinomial theorem as well ($\frac{d^{m}}{dx^{m}} \prod_{k=1}^{n} f_k(x)$). Can someone produce a proof for this more general case (preferably by induction)? Any and all insights are welcome!

Answer 1

Taylor series expansion of $f(x+h)$ $$ f(x+h)=f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+.... $$

Taylor series expansion of $g(x+h)$ $$ g(x+h)=g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+.... $$

Taylor series expansion of $f(x+h)g(x+h)$ $$ f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+.... $$

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$$ f(x+h)g(x+h)=f(x)g(x)+h\frac{d}{dx} \left(f(x)g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x)g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x)g(x) \right)+....=(f(x)+h\frac{d}{dx} \left(f(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( f(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(x) \right)+....)(g(x)+h\frac{d}{dx} \left(g(x) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left( g(x) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( g(x) \right)+....) $$

If you order $h^n$, it will give you binom expansion

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

Because it has exactly the same coefficient of

$$ =(1+hx +\frac{h^2 x^2 }{2!}+\frac{h^3x^3 }{3!}+....)(1+hy +\frac{h^2 y^2 }{2!}+\frac{h^3y^3 }{3!}+....)=e^{hx}e^{hy}=e^{h(x+y)} $$

$$ e^{h(x+y)}=1+h(x+y) +\frac{h^2 (x+y)^2 }{2!}+\frac{h^3(x+y)^3 }{3!}+.... $$

If you order $h^n$ of $e^{hx}e^{hy}$ and equal to $h^n$ of $e^{h(x+y)}$ , it will give you binom expansion proof.

$$\frac{(x+y)^n }{n!}=\sum_{i=0}^n \frac{x^i y^{n-i}}{i! (n-i)!} $$

$$(x+y)^n=\sum_{i=0}^n \frac{n! x^i y^{n-i}}{i! (n-i)!} $$

$$(x+y)^n=\sum_{i=0}^n {n \choose i} x^i y^{n-i} $$

Thus it is also true $$\frac{d^{n}}{dx^{n}} f(x)g(x)=\sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x)$$

Answer 2

This is something you typically do by induction (I have fixed a misprint in your notation), starng with $$ \frac{d}{dx} \left( f(x) g(x) \right) = \frac{d}{dx}\left( f(x) \right)g(x) + f(x) \frac{d}{dx} \left( g(x) \right). $$ Here is as it goes: \begin{align}\frac{d^{n+1}}{dx^{n+1}} f(x)g(x)&= \frac{d}{dx} \sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i)}(x) \\&= \sum_{i=0}^n {n \choose i} f^{(i+1)}(x)g^{(n-i)}(x) + \sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n-i+1)}(x) \\&= \sum_{i=1}^{n+1} {n \choose i-1} f^{(i)}(x)g^{()n+1)-i)}(x) + \sum_{i=0}^n {n \choose i} f^{(i)}(x)g^{(n+1)-i)}(x)\\&= \sum_{i=0}^{n+1} \left( {n \choose i-1} + {n \choose i}\right) f^{(i)}(x)g^{(n+1)-i)}(x)\\&=\sum_{i=0}^{n+1} \left( {n+1 \choose 1}\right) f^{(i)}(x)g^{(n+1)-i)}(x).\end{align}