# 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?

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 oeis.org/wiki/Eulerian_numbers,_triangle_of Apr 28, 2016 at 22:47
• @MattSamuel Are you hinting at my question for the polylogarithm? Apr 28, 2016 at 23:41
• What is $Li_{-a}(z)$, then?
– fred
Apr 29, 2016 at 0:01
• @fred That is the polylogarithm function. Apr 29, 2016 at 0:02
• So is that a new question or does it have something to do with the first part?
– fred
Apr 29, 2016 at 0:03

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

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 $$\dfrac{d^m}{dt^m}\left(\dfrac{1}{1-e^t} \right) = \dfrac{1}{(1-e^t)^{m+1}}\sum_{k=0}^{m-1} A(m,k) (e^t)^{m-k}?$$ Apr 29, 2016 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 mathworld.wolfram.com/EulerianNumber.html Apr 29, 2016 at 15:33