I am looking at a couple different page on the definition of Radius of Convergence, specifically for Taylor series. I first learned it as follows: For a power series
$$\sum_{k=0}^\infty a_k (z-z_0)^k$$
the radius of convergence is a unique real number $R\in\mathbb R \cup \{0,\infty\}$ where the sum converges when $|z-z_0|<R$ and diverges when $|z-z_0|>R$.
However, I was told a different definition/convention specific to Taylor series: For a function $f:U\to\mathbb C$ and any $z_0\in U$ we say the radius of convergence for the Taylor series centered at $z_0$ is the largest $R$ for which the Taylor series converges to $f$ on $D(z_0;R)$. So it not only needs to converge, it has to converge to $f$.
This two definitions are clearly different, for example, consider the function $g:\mathbb C\setminus\{1\}$ where $g(z)=0$. The Taylor series of $g$ centered at $z_0=0$ is $0$. Using the first definition we know that the radius of convergence of this Taylor series is $\infty$. Using the second definition we see that the radius of convergence is actually $1$, since it does not converge to $g$ at $z=1$.
This is confusing already but upon a bit of searching, it seems like both of these definition contradicts with this fact in the wikipedia article: https://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions which says: "the radius of convergence is always the distance from the center $a$ to the nearest non-removable singularity;"
Using the same $g$ as above, this fact produces contradictory result with the second definition. Consider $h:\mathbb C\setminus \mathbb R^-\to \mathbb C$ where $h(z)=0$, $h$ has no removable singularity since none of them are isolated. Now the Taylor series of $h$ at $z_0=1$ is $0$ and has radius of convergence $\infty$. However, the distance between $z_0$ and the nearest non-removable singularity is $1$, since $\mathbb R^-$ is a set of non-removable singularities. So $h$ shows that the fact is contradictory with the first definition.
Given those, I have the following questions:
- Is anything I have stated incorrect?
- Is it a convention to define radius of convergence differently for Taylor series? (instead of the series converging, it has to converge to the function where the series is defined on)
- Is the fact listed in wikipedia correct? Is it yet another convention for "radius of convergence"?
Best Answer
Radius of convergence is a property of a power series, not of a function. Your first definition is correct, your second is not. The Wikipedia statement is misleading. What is true is that if the radius of convergence is $R$ (with $0 < R < \infty$), the Taylor series converges on the open disk of radius $R$ centered at $a$ to a function $f$ analytic on that disk, but it is impossible to extend $f$ to be analytic on a disk centered at $a$ with any larger radius; this implies that there is some point $p$ at distance $R$ from $a$ such that there is no way to extend $f$ to be analytic in a neighbourhood of $p$. In that sense, $R$ is the distance to the closest non-removable singularity. However, if you start with a function $g$ that has the same Taylor series, there is no reason to expect $g$ to agree with $f$ on that disk.
[EDIT] The exception is if $g$ is assumed to be analytic everywhere except for isolated singularities. Then the radius of convergence is, in fact, the distance from $a$ to the closest non-removable singularity of $g$. On the other hand if, say, $g$ has a branch cut, there is no reason that branch cut can't wander inside the disk of radius $R$ around $a$.