I define a function on an open set of the complex plane to be analytic if about any point $z_0$ in that set it can be expanded as a power series in $(z - z_0)$ that converges in some neighbourhood of $z_0$. On the face of it, this definition doesn't tell us anything about the convergence of a given Taylor series throughout the open set in question.

For instance, if $f(z) = 1/(z - 1)$ and our open set is $\{ z: |z - 1/3| < 1/2\}$, then a Taylor expansion of $f(z)$ about $z = 3/4$ doesn't converge everywhere in our open set. It does converge in a neighbourhood of $z = 3/4$, of course, but not everywhere in the open set.

However, if our open set in question is a disc, then I believe that: the Taylor expansion of a function which is analytic in that open set about its centre converges everywhere in that open set. My question is: why is this?


1 Answer 1


The relevant result is the Cauchy Hadamard theorem.

Suppose you have a power series $f(z) = \sum_n f_n z^n$ (take $z_0 = 0$ for simplicity).

Define the radius of convergence by ${1 \over R} = \limsup_n \sqrt[n]{|f_n|}$. Then the series converges absolutely for $|z|<R$ and diverges for $|z|>R$.

Loosely this is because we are, roughly, looking at the series $\sum_{n \ge n_0} ({z \over R})^n$ for large $n_0$, and this converges absolutely for $|z|<R$ and diverges for $|z|>R$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .