Rewrite formula using exponential generating functions

Question

In the equation below I want to extract $b_k$.

$$\frac{a_n}{n!} = \sum_{k=0}^{n}\frac{b_k}{k!(n-k)!}$$

For all other exercises in the book I had to use $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$ or its generating function, but I just don't see how I can use that now, to rewrite the function. I may be looking at the problem from a completely wrong direction.

Does anyone have a hint or a push in the right direction?

Thanks in advance.

Answer 1

Suppose that: $$ f(z) = \sum_{n\geq 0}\frac{a_n}{n!}z^n,\qquad g(z)=\sum_{n\geq 0}\frac{b_n}{n!}z^n. $$ Then: $$ [z^m]g(z)e^z = \sum_{n=0}^{m}\frac{b_n}{n!}\cdot\frac{1}{(m-n)!}=\frac{a_m}{m!}=[z^m]f(z) $$ so $f(z)=e^z g(z)$, from which $g(z)=e^{-z} g(z)$. Since: $$ e^{-z}=\sum_{n\geq 0}\frac{(-1)^n}{n!}z^n,$$ it follows that: $$ \frac{b_n}{n!}=[z^n]g(z)=[z^n]e^{-z}f(z) = \sum_{m=0}^{n}\frac{a_m}{m!}\cdot\frac{(-1)^{m-n}}{(m-n)!}, $$ or:

$$ b_n = \sum_{m=0}^{n}\binom{n}{m}a_m (-1)^{m-n}.$$

Extraction complete.