# Help with an exponential/generating functions problem.

Is there anyone that would be willing to help me solve this problem? While I am comfortable with using ordinary generating functions, I do not have a lot of experience or intuition when it comes to exponential generating functions. Any tips or a push in the right direction would be much appreciated!

Let $$\mathit{A(t)}$$ be an ordinary generating function and let $$\mathit{E(t)}$$ be an exponential generating function for a sequence {$$\mathit{a_n}$$}.

Show that $$\mathit{A(t)}$$ = $$\int^\infty_0e^{-x}\mathit{E(xt)}\mathit{dx}$$.

Hint: Use the equality $$\mathit{n}$$! = $$\int^\infty_0e^{-x}\mathit{x}^n\mathit{dx}$$

Thanks in advance!

## 1 Answer

Let $$A(t) = \sum_{n\ge 0} a_n t^n$$ be the o.g.f. and $$E(t) = \sum_{n\ge 0}a_n \frac{t^n}{n!}$$ be the e.g.f.

Then we have \begin{align} \int_0^\infty e^{-x}E(xt) dx &= \int_0^\infty e^{-x} \sum_{n\ge 0}\frac{a_nt^nx^n}{n!} dx\\ &= \sum_{n\ge 0}\int_0^\infty e^{-x}\frac{a_nt^nx^n}{n!} dx\\ &= \sum_{n\ge 0}a_nt^n\frac{1}{n!}\int_0^\infty e^{-x} x^n dx\\ &= \sum_{n\ge 0}a_nt^n\frac{1}{n!} n!\\ &= \sum_{n\ge 0}a_nt^n\\ &= A(t) \end{align} Note we use the fact that $$\int (f_1(x) + f_2(x) + \cdots) dx = \int f_1(x) dx + \int f_2(x)dx + \cdots$$. Then since we are integrating with respect to $$x$$ we have $$\frac{a_nt^n}{n!}$$ as a constant we can pull out, and we use the hint.