Finding a Series Expansion for $(r^{-1} \partial_r)^n$ I wish to compute $(r^{-1} \partial_r)$ where $\partial_r = \frac{\partial}{\partial r}$ denotes a partial derivative in a variable $r$. I have constructed a table of the second through fifth powers.




$n$
$\left(r^{-1}\partial_{r}\right)^{n}$




$2$
$r^{-2}\partial_{r}^{2}-r^{-3}\partial_{r}$


$3$
$r^{-3}\partial_{r}^{3}-3r^{-4}\partial_{r}^{2}+3r^{-5}\partial_{r}$


$4$
$r^{-4}\partial_{r}^{4}-6r^{-5}\partial_{r}^{3}+15r^{-6}\partial_{r}^{2}-15r^{-7}\partial_{r}$


$5$
$r^{-5}\partial_{r}^{5}-10r^{-6}\partial_{r}^{4}+45r^{-7}\partial_{r}^{3}-105r^{-8}\partial_{r}^{2} + 105r^{-9}\partial_{r}$




Based on the data, the solution should assume the form
$$
(r^{-1} \partial_r)^n = \sum_{k=0}^{n-1} c_k r^{-(n+k)} \partial_r^{n-k}
$$
I do not know of a formula for the $c_k$. I was trying to look for a relationship with Stirling numbers or generating functions but did not have much luck.
 A: In this paper, you can find the formula
$$
(r^{\alpha  + 1} \partial _r )^n  = \sum\limits_{k = 0}^n {( - \alpha )^n r^{n\alpha  + k} C(n,k, - \alpha ^{ - 1} )\partial _r^k } 
$$
with several representations for the quantities $C(n,k,x)$, one of which is
$$
C(n,k,x) = \sum\limits_{r = k}^n {s(n,r)S(r,k)x^r } .
$$
Here $s$ and $S$ are the Stirling numbers of the first and second kind, respectively.
A: Plugging in your sequence of integers $1,-1,1,-3,3,1,-6,15,-15,\ldots$ into the Online Encyclopedia of Integer Sequences brings up sequence A001498, the coefficients of the Bessel polynomials $y_n$ written in increasing order.
From this description we can go to the Wikipedia page for the Bessel polynomials (linked above) to find the formula
$$
y_n(x) = \sum_{k=0}^n \frac{(n+k)!}{(n-k)! \ k!} \left ( \frac{x}{2} \right )^k
$$
from here it's not hard to guess that your formula should be
$$
\left ( r^{-1} \partial_r \right )^{n+1} = \sum_{k=0}^{n} (-1)^k \frac{(n+k)!}{(n-k)! \ k! \ 2^k} r^{-(n+k)} \partial_r^{n-k}
$$
NB, I haven't proven this formula is correct, but it shouldn't be hard to do that now that we know what it is.

I hope this helps ^_^
