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?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|||||||||||
|
|
$a$ 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. |
||||
|
|
