Help with a certain derivative of a log of sum -- problem rooted in Poisson distributions that I think may have an elegant solution

Question

For positive integers $y$, I would like to evaluate $$\frac{d}{dx} \log \sum_{m=0}^\infty \frac{x^m m^y}{m!}.$$ That's the whole question. It's a small bit of a research problem.

Symbolic math programs are capable of this calculation for fixed $y$, but I cannot discern a pattern.

$$y=1: \quad \frac{1+x}{x}$$ $$y=2: \quad \frac{1+3x+x^2}{x+x^2}$$ $$y=3: \quad \frac{1+7x+6x^2+x^3}{x+3x^2+x^3}$$ $$y=4: \quad \frac{1+16x+25x^2+10x^3+x^4}{x+7x^2+6x^3+x^4}$$ $$y=5: \quad \frac{1 + 31 x + 90 x^2 + 65 x^3 + 15 x^4 + x^5}{x + 15 x^2 + 25 x^3 + 10 x^4 + x^5}$$ $$y=6: \quad \frac{1 + 63 x + 301 x^2 + 350 x^3 + 140 x^4 + 21 x^5 + x^6}{x + 31 x^2 + 90 x^3 + 65 x^4 + 15 x^5 + x^6}$$

Answer 1

As pointed out by @Claude Leibovici, it suffices to identify the polynomial

$$ \sum_{m=0}^{\infty} \frac{m^y}{m!}x^m. $$

This is related to the Stirling numbers of the second kind $\left\{{n \atop k}\right\}$ in the following way: It is well-known that

$$ z^n = \sum_{k=0}^{n} \left\{{n \atop k}\right\}(z)_k, $$

where $(z)_k = z(z-1)\dots(z-k+1)$ is the falling factorial. Replacing $z$ and $n$ by $m$ and $y$,

$$ m^y = \sum_{k=0}^{y} \left\{{y \atop k}\right\}\frac{m!}{(m-k)!}, $$

where we regard $\frac{1}{(m-k)!} = 0$ if $k > m$. Then

\begin{align*} \sum_{m=0}^{\infty} \frac{m^y}{m!}x^m &= \sum_{m=0}^{\infty} \left( \sum_{k=0}^{y} \left\{{y \atop k}\right\}\frac{m!}{(m-k)!} \right) \frac{x^m}{m!} \\ &= \sum_{k=0}^{y} \left\{{y \atop k}\right\} \sum_{m=k}^{\infty} \frac{x^m}{(m-k)!} \\ &= \sum_{k=0}^{y} \left\{{y \atop k}\right\} x^k e^x. \end{align*}

This kind of polynomials arises in various fields so often that it is given a special name. Indeed, the family of polynomials defined by

$$ T_n(x) := \sum_{k=0}^{n} \left\{ {n \atop k} \right\} x^k $$

is called the Touchard polynomials. Then using this,

$$ \frac{\mathrm{d}}{\mathrm{d}x} \sum_{m=0}^{\infty} \frac{m^y}{m!}x^m = \sum_{m=0}^{\infty} \frac{m^{y+1}}{m!}x^{m-1} = \frac{1}{x}T_{y+1}(x)e^x $$

and therefore

$$ \frac{\mathrm{d}}{\mathrm{d}x} \log \left( \sum_{m=0}^{\infty} \frac{m^y}{m!}x^m \right) = \frac{T_{y+1}(x)}{x T_y(x)}. $$

Answer 2

For a fixed integer value of $y \geq 1$, we have $$\sum_{m=0}^\infty \frac{x^m m^y}{m!}=e^x x P_{y-1}(x)$$ $$ \log\left( \sum_{m=0}^\infty \frac{x^m m^y}{m!}\right)=x+\log(x)+\log\big(P_{y-1}(x)\big)$$ $$\frac d {dx}\log\left( \sum_{m=0}^\infty \frac{x^m m^y}{m!}\right)=1+\frac 1 x+\frac{P'_{y-1}(x) } {P_{y-1}(x) }$$ So, I think that the problem is "just" to identify what are the polynomials $P_{y-1}(x)$. Listing them $$\left( \begin{array}{cc} 1 & 1 \\ 2 & x+1 \\ 3 & x^2+3 x+1 \\ 4 & x^3+6 x^2+7 x+1 \\ 5 & x^4+10 x^3+25 x^2+15 x+1 \\ 6 & x^5+15 x^4+65 x^3+90 x^2+31 x+1 \\ 7 & x^6+21 x^5+140 x^4+350 x^3+301 x^2+63 x+1 \\ 8 & x^7+28 x^6+266 x^5+1050 x^4+1701 x^3+966 x^2+127 x+1 \\ 9 & x^8+36 x^7+462 x^6+2646 x^5+6951 x^4+7770 x^3+3025 x^2+255 x+1 \\ 10 & x^9+45 x^8+750 x^7+5880 x^6+22827 x^5+42525 x^4+34105 x^3+9330 x^2+511 x+1 \end{array} \right)$$

The coefficients seem to be the Stirling numbers of the second kind.

$$\mathcal{S}_{y+1}^{(y)}=\{1,3,6,10,15,21,28,36,45\}$$ $$\mathcal{S}_{y+2}^{(y)}=\{1,7,25,65,140,266,462,750\}$$ $$\mathcal{S}_{y+3}^{(y)}=\{1,15,90,350,1050,2646,5880\}$$ $$\mathcal{S}_{y+4}^{(y)}=\{1,31,301,1701,6951,22827\}$$ $$\mathcal{S}_{y+5}^{(y)}=\{1,63,966,7770,42525\}$$

I think that what are the polynomials is now clear.