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Is there a closed form to the sum $$ \sum_{n=0}^{\infty} \frac{x^n}{n!}n^p, $$ where $p$ is a positive integer? I know that for $p = 1$ the series sums to $xe^x$, but not sure if there's a simple expression for $p > 1$. If there isn't a closed form, is this series related to any special functions? Thanks!

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  • $\begingroup$ Just to add that these are Touchard polynomials. And this problem is very natural considering Moments of Poisson random variables. $\endgroup$
    – Phicar
    Oct 31, 2020 at 19:44

3 Answers 3

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Hint: here are the first few values for $p=0,\ldots,5$: $$ e^x\cdot\left\{1,x,x^2+x,x^3+3 x^2+x,x^4+6 x^3+7 x^2+x,x^5+10 x^4+25 x^3+15 x^2+x\right\} $$You may recognize the coefficients of these polynomials; they are the Stirling numbers of the second kind. The recurrence relation can be proved by induction using the power rule. So in summary, $$ \sum_{n=0}^{\infty}\frac{x^n}{n!}n^p = e^x \cdot \sum_{k=1}^p \left\{ \begin{array}{cc} p \\ k \end{array} \right\} x^k $$

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BTW, the proof of this identity goes like this:

$$\sum_{n\ge 0} n^p \frac{x^n}{n!} = p! [z^p] \sum_{n\ge 0} \frac{x^n}{n!} \exp(nz) = p! [z^p] \exp(x\exp(z)) \\ = \exp(x) p! [z^p] \exp(x(\exp(z)-1)).$$

Observe that $\exp(z)-1 = z + \cdots$ so only the initial segment of the outer exponential contributes:

$$\exp(x) p! [z^p] \sum_{k=0}^p \frac{x^k (\exp(z)-1)^k}{k!}.$$

Thus when $p=0$ only $k=0$ contributes and we get $\exp(x)$ as expected. Otherwise with $p\ge 1$ we find

$$\exp(x) \sum_{k=1}^p x^k {p\brace k}$$

as claimed. Here we have used the fact that set partitions have combinatorial class

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\mathcal{U}\times\textsc{SET}_{\ge 1}(\mathcal{Z}))$$

which gives the EGF

$$\exp(u(\exp(z)-1))$$

so that

$$\sum_{n\ge k} {n\brace k} \frac{z^n}{n!} = [u^k] \exp(u(\exp(z)-1)) = \frac{(\exp(z)-1)^k}{k!}.$$

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Note that in general $$\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n} f(x) = \sum_{k=1}^n {n \brace k} \, x^k f^{(k)}(x)$$ which can be derived by noting the general form $$\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n} f(x) = \sum_{k=1}^n a_k(n) \, x^k f^{(k)}(x)$$ and then proceed inductively i.e. $$\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n+1} f(x) = xf^{(1)}(x) + \sum_{k=2}^n \left\{k \, a_k(n) + a_{k-1}(n)\right\} x^k f^{(k)}(x) + x^{n+1} f^{(n+1)}(x) \\ = \sum_{k=1}^{n+1} \left\{ k \, a_k(n) + a_{k-1}(n) \right\} x^k f^{(k)}(x) \stackrel{!}{=} \sum_{k=1}^{n+1} a_k(n+1) \, x^k f^{(k)}(x) $$ where $a_1(n)=1$, $a_n(n)=1$ and $a_0(n)=a_{n+1}(n)=0$ with $n\in \mathbb{N}$.

Finally check that the Stirling Numbers of Second kind $${n\brace k}=\frac{1}{k!} \sum_{i=0}^k (-1)^{k-i} \binom{k}{i} i^n$$ solve the recurrence $$a_k(n+1)=k \, a_k(n) + a_{k-1}(n) \, .$$

Your special case is $f(x)=e^x$.

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