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I would like to get the closed form for this generating function, assuming the $k$ is given up front / held constant: $\sum_{n=0}^{\infty} n^k x^n$

However I don't know if this is too advanced for me or if I am in over my head because I can't seem to get anywhere with this. One thing I did find, playing around with derivatives, is that if:

$F_0(x) = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$

$F_1(x) = xF'_0(x) = \sum_{n=0}^{\infty} n x^{n}$

$F_2(x) = xF'_1(x) = \sum_{n=0}^{\infty} n^2 x^{n}$

$F_3(x) = xF'_2(x) = \sum_{n=0}^{\infty} n^3 x^{n}$

$F_k(x) = xF'_{k-1}(x) = \sum_{n=0}^{\infty} n^k x^{n}$ and this is what I want to compute the closed form for

Does a closed form exist for this or do I have to compute this in $O(k)$ time by going through each level, computing the derivative, multiplying by $x$, and so on?

KaliMa
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1 Answers1

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A closed form does exist. One has $$ \sum_{n=0}^\infty n^px^n=\frac1{1-x}\:\omega_p\left(\frac{x}{1-x} \right), \quad |x|<1, $$ with the polynomials $$ \omega_n\left(x \right)=\sum_{k=0}^n \begin{Bmatrix}n\\k\end{Bmatrix}k!x^k $$ where $\displaystyle \begin{Bmatrix}n\\k\end{Bmatrix}$ are the Stirling numbers of the second kind.

A proof can be found in Boyadzhiev's paper. A related formula: eulerian polynomials.

Olivier Oloa
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  • I am guessing this is not something I can realistically expect to derive on my own? I am not understanding much of this paper. Is there no easier derivation? – KaliMa Jul 27 '16 at 21:24
  • @KaliMa I'm afraid it won't be much simpler. – Olivier Oloa Jul 27 '16 at 21:32
  • So is my derivative relationship more complex than this / is it a wasted observation? I don't see that in the paper anywhere – KaliMa Jul 27 '16 at 21:35
  • @KaliMa: You might find this step-by-step answer helpful. – Markus Scheuer Jul 28 '16 at 06:35
  • @MarkusScheuer That answer is great, but my difficulty lies in finding some way to handle $(zD_z)^k$ (such as in the example in my post where I have to go through each level). For example deriving some closed form or summation for applying $(zD_z)^k$ to $\frac{1}{1-z}$. Let's say $k=30$. Would you manually derivate $\frac{1}{1-z}$ that many times or is there a better way? – KaliMa Jul 28 '16 at 07:49
  • @KaliMa Observe that, by using the formula I gave in my answer above, we just get $$ \sum_{n=0}^\infty n^{30}x^n=\sum_{k=0}^{30} \begin{Bmatrix}30\k\end{Bmatrix}\frac{k!:x^k}{(1-x)^{k+1}}. $$ – Olivier Oloa Jul 28 '16 at 07:57
  • @OlivierOloa I don't understand why the formula works, though, or how it can be derived. I don't understand that paper (it's too dense for me and I have no idea where I need to be looking) – KaliMa Jul 28 '16 at 08:01
  • @KaliMa I understand what you mean. Sometimes one has to accept to take time to really understand what's going on in a mathematical subject. I think time might be very often our friend. – Olivier Oloa Jul 28 '16 at 08:04
  • Is this Stirling equation related to Faulhaber's formula in any way? – KaliMa Jul 28 '16 at 08:07
  • The answer is yes, by a Cauchy product. One may use that $\sum_{n\ge 0} a_n x^n \cdot \sum_{n\ge 0} x^n= \sum_{n\ge 0} (\sum_{0\le k \le n} a_k) x^n$. Applied to $a_n=n^p$. – Olivier Oloa Jul 28 '16 at 08:13
  • @KaliMa: I think that Olivier Oloas answer provides a nice manageable solution. Note that the operator $zD_z$ and the Stirling numbers are intimately connected. – Markus Scheuer Jul 28 '16 at 09:07
  • I am sure it is manageable to someone who is already familiar with much of this, but to me it looks like it has a ton of superfluous information that is really distracting for me; I need something a little more focused on this specific question – KaliMa Jul 28 '16 at 09:09