I have an infinite summation I'm working with, but there's no 'nice' closed form for the coefficients of this sum without introducing some functionals.
But, I'm not sure if the way I've done so is necessarily correct.
The coefficients in the sum look as follows: $\frac{1}{m!} \ln^m(D_w + 1)[e^{e^w}] \rvert_{w = \ln(D_z)} [f(z)]\rvert_{z=a}$.
where $D_w, D_z$ are the derivative operators denoting $\frac{d}{dw}$, $\frac{d}{dz}$ respectively.
This makes me uneasy because I am applying the nth power of the log of the derivative operator to the function $e^{e^w}$, only to then substitute the value of $w$ for the log of another derivative operator $D_z$, and then applying that to the function $f(z)$.
Really what I'm trying to express is the following:
$\frac{\ln^m(D_w + 1)}{m!}[e^{e^w}] \rvert_{w = \ln(D_z)} [f(z)]\rvert_{z=a} = \sum_{n=m}^{\infty} s(n, m) \frac{D^n_w}{n!} [e^{e^w}] \rvert_{w = \ln(D_z)} [f(z)]\rvert_{z=a} $
$= \sum_{n=m}^{\infty} \frac{s(n, m)}{n!} e^{e^{w}} T_n(e^w) \rvert_{w = \ln(D_z)} [f(z)]\rvert_{z=a} = \sum_{n=m}^{\infty} \frac{s(n, m)}{n!} e^{D_z} T_n(D_z) [f(z)]\rvert_{z=a} $
$= \sum_{n=m}^{\infty} \frac{s(n, m)}{n!} e^{D_z} T_n(D_z) \rvert_{w = \ln(D_z)} [f(z)]\rvert_{z=a}$
$ = \sum_{n=m}^{\infty} \frac{s(n, m)}{n!} e^{D_z} \sum_{k=0}^{n} \left\{ {n \atop k} \right\} D_z^k [f(z)]\rvert_{z=a} =\sum_{n=m}^{\infty} \frac{s(n, m)}{n!} \sum_{k=0}^{n} \left\{ {n \atop k} \right\} f^{(k)}(a+1) $.
Here $s(n,m), \left\{ {n \atop k} \right\}, T_n(x), $ is the stirling number of the first kind, second kind, and touchard polynomial respectively.
I unfortunately wasn't able to find anything tractable about this sort of notation online. And so hope it's fine to post this question here. But if there are any resources someone could point me to, that would be great.
