# Radius of convergence of the inverse of a power series

Let $a = \sum_k a_k X^k \in \mathbb C [\![ X ]\!]$ with $a_0 = 1$ and convergence radius $\rho_a > 0$. I want to show that the convergence radius of the inverse $b = \sum_k b_k X^k \in \mathbb C [\![ X ]\!]$ is also greater than 0. How can I do that?

-
are you looking at the inverse or at $1/a$ (the reciprocal)? – user20266 Jun 15 '12 at 17:19
Hi @Thomas: I know you can answer both questions! Why don't you undelete your answer? – Georges Elencwajg Jun 15 '12 at 18:10
It is my impression the OP does not know that holomorphic functions admit a power series representation but is looking for an answer using sequence manipulation. That I could, theoretically, also provide, but it is rather elaborate. Sure, I can undelete it. – user20266 Jun 15 '12 at 18:20
I'm looking for the inverse in the ring of formal power series, i.e. a $b = \sum_k b_k X^k$ such that $a b = 1$, where $1 = c = \sum_k c_k X^k$ means, that $c_0 = 1$ and $c_k = 0$ for $k \geq 1$. It's easy to find these $b_k \in \mathbb C$. But is it possible to argue directly for example with the formula from Hadamard for the convergence radius of power series? – aexl Jun 15 '12 at 23:56

## 1 Answer

$a$ defines a holomorphic function $f(z) =\sum a_k z^k$ in a neighbourhood of $0$. If you are interested in $g$ such that $f\circ g = id$ (the inverse of $f$), then you will need to make sure that $\frac{df}{dz}(0)= a_1\neq 0$. In that case, by the inverse function theorem it has, locally, a holomorphic inverse. Holomorphic implies positive radius of convergence.

A similar reasoning applies if you are interested in $g$ such that $f\cdot g=1$, the reciprocal. This is well defined near the origin if $a_0\neq 0$, so I guess you are aiming at this. As in the case of the inverse, the reciprocal is complex differentiable near the origin, hence holomorphic, hence admits a power series expansion with positve radius of convergence.

I do assume, however, that you are looking for a more basic reasoning not involving knowledge about holomorphic functions. It is possible (this was shown to me 30 years ago in my first lecture on analysis ;-) to recursively find expressions for the power series coefficients of $1/f= g$ by analyzing the equation $f\cdot g= 1$ and using uniqueness results about the coefficients of power series (the rhs being a trivial power series), which allow calculating the radius of convergence using the usual formula for it involving $|c_k|^{1/k}$. For now, I consider it a too tedious task to work out the details of this, my apologies.

-