Does $x\left(\frac{d}{dx}\left(\cdots x \left(\frac{d}{dx} \left( \frac{x}{1-x}\right)\right)\cdots\right)\right)$ have a closed form expression?

Question

Does $\displaystyle \underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}}$ have a closed form?

I am curious about this because of this question. Testing this out for a few values of $m$ I get:

$m = 1: \dfrac{x}{(x-1)^2}$

$m = 2: \dfrac{-x^2-x}{(x-1)^3}$

$m = 3: \dfrac{x^3+4x^2+x}{(x-1)^4}$

$m = 4: \dfrac{-x^4-11x^3-11x^2-x}{(x-1)^5}$

$m = 5: \dfrac{x^5+26x^4+66x^3+26x^2+x}{(x-1)^6}$

$\vdots$

It is easy to see that the closed form will have a $(-1)^{n+1}$ in it and seeing how some of the terms look symmetric, it will probably have something like $(x+1)^n$, but I don't see a way to find the closed form.

Edit: I want to prove that $$\operatorname{Li}_{-a}(z)= \displaystyle \sum_{k = 1}^{\infty} k^m x^k=\frac 1{(1-z)^{m+1}}\sum_{k=0}^{m-1} E(m,k)\, z^{m-k}$$ with $E(a,k)$ the Eulerian numbers.

The first few terms seem to correspond to http://oeis.org/wiki/Eulerian_numbers,_triangle_of — Kenneth Goodenough, Apr 28 '16 at 22:47
@MattSamuel Are you hinting at my question for the polylogarithm? — Puzzled417, Apr 28 '16 at 23:41
So is that a new question or does it have something to do with the first part? — fred, Apr 29 '16 at 00:03
It is new but it has some relation to the first part. In particular, you will need to invoke that they are equal since it is a closed form for it. — Puzzled417, Apr 29 '16 at 00:04
Using mathematica for $m=2$, I get $\sum {k=1}^{\infty } k^m x^k = \frac{x^2+x}{(1-x)^3}$, but $\frac{1}{(1-x)^{m+1}} \sum{k=0}^{m-1} E(m,k) x^{m-k} = 0$. So don't think that what you are trying to prove is true. (You also change from using $x$ to $z$). — fred, Apr 29 '16 at 01:06

JJacquelin · Answer 1 · 2016-04-29T07:36:25.677

Note : $$\operatorname{Li}_{-m}(z)= \displaystyle \sum_{k = 1}^{\infty} k^m z^k=\frac 1{(1-z)^{m+1}}\sum_{k=0}^{m-1} E(m,k)\, z^{m-k}$$ with $E(m,k)$ the Eulerian numbers.

This relationship is wellknown. For example, see eqs.(4) and (5) in : http://mathworld.wolfram.com/EulerianNumber.html

With the correspondance of symbols : $z=r$ , $m=n$ , $k=i$ , $E(m,k)=\Big\langle \begin{array}{c} n \\ i \end{array} \Big\rangle$

Answer of the question about closed form :

The change of variable $x=e^t$ simplifies a lot the equation :

$$\displaystyle \underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}\; ,\; m>0} = \frac{d^m}{dt^m}\left(\frac{1}{1-e^t} \right)$$

I don't think that a simpler expression exists for the $m$-th derivative (other than the polylogarithm in this particular case, of course).

How do we prove that $$\dfrac{d^m}{dt^m}\left(\dfrac{1}{1-e^t} \right) = \dfrac{1}{(1-e^t)^{m+1}}\sum_{k=0}^{m-1} A(m,k) (e^t)^{m-k}?$$ — Puzzled417, Apr 29 '16 at 14:50
$\frac{1}{1-e^t}=1+\sum_{k=1}^{\infty}e^{kt}\qquad ;\qquad \frac{d^m}{dt^m}\left(\frac{1}{1-e^t}\right)=\sum_{k=1}^{\infty}k^me^{kt}=\text{Li}_{-m}(e^t)\quad$Then use the Eqs.(4) and (5) from http://mathworld.wolfram.com/EulerianNumber.html — JJacquelin, Apr 29 '16 at 15:33

Does $x\left(\frac{d}{dx}\left(\cdots x \left(\frac{d}{dx} \left( \frac{x}{1-x}\right)\right)\cdots\right)\right)$ have a closed form expression?

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