# Showing that $\exp(\sum_{n=1}^\infty a_nX^n)=\prod_{n=1}^\infty\exp(a_nX^n)$ for formal power series

I've just come across formal power series and am not very fluent with them yet.

I'd like to show that $\exp(\sum_{n=1}^\infty a_nX^n)=\prod_{n=1}^\infty\exp(a_nX^n)$.

Can anybody help?

• This identity holds modulo $X^N$ for all $N$. Aug 29, 2014 at 15:05
• @Simeon Oh of course! I feel so stupid. Thanks! Aug 29, 2014 at 16:38

$$\exp\left(\sum\limits_{n=0}^{+\infty}a_n X^n\right)=\exp\left({\lim\limits_{n\to +\infty}\sum\limits_{k=0}^{n}a_k X^k}\right)=\lim\limits_{n \to +\infty}\exp \left(\sum\limits_{k=0}^{n}a_k X^k\right)=\lim\limits_{n \to \infty}\prod\limits_{k=0}^{n}\exp \left(a_k X^k\right)=\\=\prod\limits_{n=0}^{+\infty}\exp \left(a_n X^n\right)$$ We changed $\lim$ and $\exp$ because of the fact that $\exp x$ is continuous
• As I know, $|\exp x -\sum\limits_{k=0}^{n}\frac{x^k}{k!}|<\varepsilon$, and usually one considers $\sum\limits_{n=0}^{+\infty}a_n x^n$ as a continuous function where the series converges, and the series for exp converges everywhere
• You'd better have $a_0=0$ for this to make sense, even with formal power series. Aug 27, 2014 at 11:47
• @Paul I'm considering $\exp$ as a function on the formal power series ring. So I'm considering $\sum_{n=0}^\infty a_nX^n$ as a variable, not a function itself. I think your answer above runs through fine so long as $\exp$ is continuous. I imagine it is! Aug 27, 2014 at 11:50
• Thanks @Ted, do you know if there's a way to make it work when $a_0 \neq 0$? In particular I'm interested in the case $a_0=1$. Aug 27, 2014 at 11:52