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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_0$$e^{-x}$$\mathit{E(xt)}$$\mathit{dx}$.

Hint: Use the equality $\mathit{n}$! = $\int^\infty_0$$e^{-x}$$\mathit{x}^n$$\mathit{dx}$

Thanks in advance!

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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.

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