explicit formula of $(x \frac{\partial}{\partial x})^n f(x)$

Question

Is there any explicit expression of

$$ (x \frac{\partial}{\partial x})^n f(x) $$

as function of $x$ and $\frac{\partial^{k}f}{\partial x^k} $$, $$ 1\leq k \leq n$.

Any idea

Thanks

Answer 1

I focused on it.

Let's suppose that it takes the following polynomial formula

$$ (x\frac{\partial}{\partial x})^n = \sum_{k=0}^{n} a_{k}^{n} x^k \frac{\partial^k}{\partial x^k} $$

when making calculation for the first values of n, $n=1, n=2, n=3 ...$ we notice that

$a_0^{n} = 0$

$a_1^{n} = 1$

and $\ a_n^n = 1$

For the other coefficients $2 \leq k \leq n-1$ we get

$a_2^n = \frac{1}{2} (2^n - 2)$

$a_3^n = \frac{1}{6} (3^n - 3 \ 2^n + 3)$

$a_4^n = \frac{1}{24} (4^n - 4 \ 3^n + 6 \ 2^n - 4)$

$a_5^n = \frac{1}{120} (5^n - 5 \ 4^n + 10 \ 3^n - 10 \ 2^n + 5)$

This is likely to be the following general formula $$ a_k^n = \frac{(-1)^k}{k!} \sum_{p=1}^{k}C_{k}^{p} (-1)^p p^n$$

Using the recurrence relation for $a_k^n$ coefficients $a_k^{n+1} = k a_k^n + a_{k-1}^n$, we can check the general formula.

Then we get,

$$ (x\frac{\partial}{\partial x})^n f = \sum_{k=1}^{n-1} \frac{(-1)^k}{k!} \sum_{p=1}^{k}C_{k}^{p} (-1)^p p^n x^k \frac{\partial^k f}{\partial x^k} + x^{n} \frac{\partial^n f}{\partial x^n} $$