Prove that the substitution of formal power series $F(G(x))=\sum_{k\geq0}f_k \frac{G(x)^k}{n!}$ converges for every $F$ if and only if $G(0)=0$
Asked
Active
Viewed 480 times
0
-
1See this answer for the appropriate notion of convergence. – Bill Dubuque May 02 '16 at 20:49
1 Answers
2
Let's call the degree (the $x$-adic valuation really) of a formal power series the lowest exponent appearing in a non-zero term.
Assume first that $G(0)=0$. That means that $\deg G(x)\ge1$. Therefore $\deg G(x)^k\ge k$ for all $k$. So if we look at terms of degree $\ell$ for some fixed natural number $\ell$, only the terms of $f_k G(x)^k/k!$ with $k\le \ell$ will contribute. But these are only finitely many formal power series, so there is no problem. Thus the sum converges.
On the other hand, if $G(0)\neq0$, then each and every one of the series $G(x)^k/k!$ has a non-zero constant term. So if $F(x)$ has infinitely many terms, then there are infinitely many non-zero constant terms in the sum, and thus the sum won't converge in the ring of formal power series.
Jyrki Lahtonen
- 133,153
-
1Wanted to nitpick about $\deg 0$, but as introduced here, its $x$-adic analogue covers comparisons with that too. Commenting this as a reminder for me actually. – ccorn Nov 17 '13 at 22:26
-
1