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?

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

1 Answer 1


$$\exp\left(\sum\limits_{n=0}^{+\infty}a_n X^n\right)=\exp\left({\lim\limits_{n\to +\infty}\sum\limits_{k=0}^{n}a_k X^k}\right)=\lim\limits_{n \to +\infty}\exp \left(\sum\limits_{k=0}^{n}a_k X^k\right)=\lim\limits_{n \to \infty}\prod\limits_{k=0}^{n}\exp \left(a_k X^k\right)=\\=\prod\limits_{n=0}^{+\infty}\exp \left(a_n X^n\right) $$ We changed $\lim$ and $\exp$ because of the fact that $\exp x$ is continuous

  • $\begingroup$ Okay great thanks. Do you have a proof or reference that exp is continuous in the formal power series ring? $\endgroup$ Aug 27, 2014 at 11:28
  • $\begingroup$ 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 $\endgroup$
    – cool
    Aug 27, 2014 at 11:41
  • $\begingroup$ You'd better have $a_0=0$ for this to make sense, even with formal power series. $\endgroup$ Aug 27, 2014 at 11:47
  • $\begingroup$ @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! $\endgroup$ Aug 27, 2014 at 11:50
  • $\begingroup$ 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 $a_0=1$. $\endgroup$ Aug 27, 2014 at 11:52

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