The formula for $g\frac{d}{dg}g\frac{d}{dg}...g\frac{d}{dg}f(g)$

Question

Yesterday, I asked if there is a formula given by a finite sum for the expression. Having experimented with Wolfram Alpha I found that it can be represented as

$$ \sum_{j=0}^{n}a_{j}g^{j}\frac{d^{j}}{dg^{j}}f(g),$$ where $a^{j}$ is given by the coefficient attached to $z^{n-1}$ in the sum expansion of the generating function $$\frac{z^{j}}{(1-jz)(1-(j-1)z)...(1-z)},$$

assuming I did not make any mistakes.

How does one go about proving it?

In the title there are $n$ iterations of $g\frac{d}{dg}$.

Answer 1

I have come up with an answer.

$$ a_{j} = {n \brace j}. $$

The equation holds for $n = 1$. I shall now prove that if it holds for $n$, it must hold for $n+1$.

Notice that Stirling numbers of the second kind have the property $$ {n + 1\brace j} = j{n \brace j} + {n \brace j - 1} $$ which I used for the proof.

$$ g\frac{d}{dg}\sum_{j = 1}^{n}{n \brace j}g^{j}\frac{d^{j}}{dg^{j}}f(g) = \sum_{j = 1}^{n}{n \brace j}jg^{j}\frac{d^{j}}{dg^{j}}f(g) + \sum_{j = 1}^{n}{n \brace j}g^{j+1}\frac{d^{j+1}}{dg^{j+1}}f(g) = \sum_{j = 1}^{n}{n \brace j}jg^{j}\frac{d^{j}}{dg^{j}}f(g) + \sum_{j = 1}^{n-1}{n \brace j}g^{j+1}\frac{d^{j+1}}{dg^{j+1}}f(g) + g^{n+1}\frac{d^{n+1}}{dg^{n+1}}f(g) = \sum_{j = 1}^{n}{n \brace j}jg^{j}\frac{d^{j}}{dg^{j}}f(g) + \sum_{j = 1}^{n}{n \brace j-1}g^{j}\frac{d^{j}}{dg^{j}}f(g) + g^{n+1}\frac{d^{n+1}}{dg^{n+1}}f(g) = g^{n+1}\frac{d^{n+1}}{dg^{n+1}}f(g) + \sum_{j = 1}^{n}{n + 1 \brace j}g^{j}\frac{d^{j}}{dg^{j}}f(g) = \sum_{j = 1}^{n + 1}{n + 1 \brace j}g^{j}\frac{d^{j}}{dg^{j}}f(g). $$ This is the end of the proof.

Answer 2

I do not prove the formula, but I relate your expression with known quantitites. A recursion yields $$\Big(x\frac{\mathrm{d}}{\mathrm{d}x}\Big)^nf(x) = \sum_{k=0}^n a_{n,k}x^kf^{(k)}(x),$$ where $a_{0,0} = 1$, $a_{0,k}=0$ for all integer $k \ne 0$ and $a_{n,k+1}=ka_{n,k}+a_{n,k-1}$ for all $n \ge 0$ and $k \in \mathbb{Z}$. These coefficients are Stirling numbers of second kind https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind. This reference gives the formula (with $k \ge 0$ fixed). $$\sum_{n \ge k} a_{n,k}x^{n-k}= \prod_{r=1}^k \frac{1}{1-rx}.$$ which can be written $$\sum_{n \ge k} a_{n,k}x^n= \frac{x^k}{\prod_{r=1}^k (1-rx)}.$$ This is almost your formula.

Remark: Stirling numbers are involved when one changes of basis in $\mathbb{R}[X]$. Applying the derivation formula to $f : x \mapsto x^d$, with $d \in \mathbb{N}$, yields $$d^nx^d = \sum_{k=0}^n a_{n,k}x^kd(d-1)\cdots(d-k+1)x^{d-k}.$$ Since it holds for all real number $x$, $$d^n = \sum_{k=0}^n a_{n,k}d(d-1)\cdots(d-k+1).$$ Both sides are polynomial functions of $d$. Two polynomials who coincide on $\mathbb{N}$ are equal hence $$X^n = \sum_{k=0}^n a_{n,k}X(X-1)\cdots(X-k+1).$$