Given a function $f(x)=\sum_{n=1}^{\infty}\frac{c_n}{x^n n!}$, where $c_n$ are constants, we want to find the formal series expansion of the function $g(x)=\exp(f(x))$ in terms of $x$.
I want to verify this derivation (found somewhere):
Write $f(x)=\sum_{n=1}^{\infty}\frac{k_n }{x^{2n} n!}x^n$ where $k_n=\frac{c_n }{x^{2n}}$ then use the general chain rule formula [http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula] to get \begin{equation} g(x)=1+\sum_{s=0}^{\infty} \frac{x^s}{s!}\bigg[\frac{d^s}{dx^s}\exp\{f(x)\}\bigg]_{x=0} \end{equation} \begin{equation} =1+\sum_{s=0}^{\infty} x^s\sum_{\{v_m\}}\frac{1}{v_m!}\bigg(\frac{k_n}{m!}\bigg)^{v_m} \end{equation}
I have two questions regarding the derivation of this result:
1) The first regards derivative $\frac{d^s}{dx^s}\exp\{f(x)\}=\sum_{\{v_m\}}\frac{s!}{v_m!}\bigg(\frac{k_n}{m!}\bigg)^{v_m}$ which it looks like they are treating $k_n$ as a term not depending on $x$. Does that make sense?
2) Is the evaluation of derivative $\frac{d^s}{dx^s}\exp\{f(x)\}$ at $x=0$ allowed? Doesn't that create division by zero?
The result I am looking at is at pg.7 of this article [http://arxiv.org/abs/astro-ph/9711239]. Thank you.