Generating function of $ \lim_{x\rightarrow 0} \frac{1}{n!} \frac{\partial^n}{\partial x^n} [(1+ax)^n f(x)] $

Question

Is there a closed form for the generating function (or exponential generating function) of the sequence $$s_n= \lim_{x\rightarrow 0}\frac{1}{n!} \frac{\partial^n}{\partial x^n}\left[(1+ax)^n f(x) \right]$$ where $f(x)$ is an analytic real function and $a$ is a real number?

Answer 1

Following Phira's suggestion, let's start with $f(x)=x^k$.

Note that $s_n=\delta_{nk}$ for $a=0$, which implies that the ordinary generating function is $f(x)$ for $a=0$. For $a\neq 0$, $$s_n = \lim_{x\rightarrow 0} \frac{1}{n!} \frac{d^n}{dx^n}\left[(1+ax)^n x^k \right] = \lim_{x\rightarrow 0} \frac{1}{n!} \sum_{i=0}^n \binom{n}{i} \left[\frac{d^{n-i}}{dx^{n-i}}(1+ax)^n\right] \frac{d^i}{dx^i} x^k = \lim_{x\rightarrow 0} \frac{1}{n!} \sum_{i=0}^n \binom{n}{i} \left[\frac{d^{n-i}}{dx^{n-i}}(1+ax)^n\right] i!\delta_{ik} = \lim_{x\rightarrow 0} \frac{1}{(n-k)! } \frac{d^{n-k}}{dx^{n-k}}\left[(1+ax)^n\right] = \binom{n}{k}a^{n-k}~~{\rm if}~~n\geq k$$ and $s_{n}=0$ for $n<k$.

The ordinary generating function is $$OG(z)=\sum_{n=0}^{\infty} s_n z^n = \sum_{n=k}^{\infty}\binom{n}{k} a^{n-k}z^n = z^k\sum_{n=0}^{\infty}\binom{n+k}{k} a^{n}z^n = \frac{z^k}{(1-az)^{k+1}}\,.$$

Now in the general case if $f(x)=\sum_k b_k x^k$, then $$ OG(z) = \sum_{k=0}^{\infty}\frac{b_k z^k}{(1-az)^{k+1}} = \frac{1}{1-az}f\left(\frac{z}{1-az}\right) $$

The exponential generating for $f(x)=x^k$ is $$EG(z) = \sum_{n=0}^{\infty} \frac{s_n}{n!} z^n = \sum_{n=k}^{\infty} \frac{a^{n-k} z^n}{k!(n-k)!} = \sum_{n=0}^{\infty} \frac{a^{n} z^{n+k}}{k! n!} = \frac{z^k}{k!}e^{az}\,.$$ and so for $f(x)=\sum_k b_k x^k$ $$EG(z) = e^{az}\sum_{k=0}^{\infty} \frac{b_k}{k!}z^k = e^{az} g(z) $$ where in the second line we have introduced $g(z)$, which we may identify as the exponential generating function of the sequence $b_k$. There is a simple result for the generalized hypergeometric functions $$ f(x) = {}_pF_q(a_1,\dots,a_p;b_1,\dots,b_q;x) \quad\longleftrightarrow\quad g(z)={}_pF_{q+1}(a_1,\dots,a_p;b_1,\dots,b_q,1;z) $$