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:

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

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

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

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

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


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.

  • 1
    $\begingroup$ The first few terms seem to correspond to oeis.org/wiki/Eulerian_numbers,_triangle_of $\endgroup$ Apr 28, 2016 at 22:47
  • $\begingroup$ @MattSamuel Are you hinting at my question for the polylogarithm? $\endgroup$
    – Puzzled417
    Apr 28, 2016 at 23:41
  • $\begingroup$ What is $Li_{-a}(z)$, then? $\endgroup$
    – fred
    Apr 29, 2016 at 0:01
  • $\begingroup$ @fred That is the polylogarithm function. $\endgroup$
    – Puzzled417
    Apr 29, 2016 at 0:02
  • $\begingroup$ So is that a new question or does it have something to do with the first part? $\endgroup$
    – fred
    Apr 29, 2016 at 0:03

1 Answer 1


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).

  • $\begingroup$ How do we prove that $$\dfrac{d^m}{dt^m}\left(\dfrac{1}{1-e^t} \right) = \dfrac{1}{(1-e^t)^{m+1}}\sum_{k=0}^{m-1} A(m,k) (e^t)^{m-k}?$$ $\endgroup$
    – Puzzled417
    Apr 29, 2016 at 14:50
  • $\begingroup$ $\frac{1}{1-e^t}=1+\sum_{k=1}^{\infty}e^{kt}\qquad ;\qquad \frac{d^m}{dt^m}\left(\frac{1}{1-e^t}\right)=\sum_{k=1}^{\infty}k^me^{kt}=\text{Li}_{-m}(e^t)\quad$Then use the Eqs.(4) and (5) from mathworld.wolfram.com/EulerianNumber.html $\endgroup$
    – JJacquelin
    Apr 29, 2016 at 15:33

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