Can someone provide a proof for the fact that the radius of convergence of the power series of an analytic function is the distance to the nearest singularity? I've read the identity theorem, but I don't see how it implies that the two functions must be equal everywhere.
[Math] Proof that Radius of Convergence Extend to Nearest Singularity
analyticitycomplex-analysis
Related Solutions
More precisely, the radius of convergence is the radius of the largest open disk centred at the expansion point on which there is an analytic function that coincides with your function near the expansion point. If the series around $0$ has radius of convergence $R$, then the corresponding analytic function is analytic in the disk around $0$ of radius $R$, and therefore in the disk around $a$ of radius $R - |a|$.
However there is a loophole: who says that function is the same as the function you are expanding around $a$? You might have chosen a function that has a branch cut that comes between $0$ and $a$ (with branch point outside the big disk). If the branch cut had been chosen differently, you'd have a function analytic in the big disk. But the way you chose it, the function near $z=a$ is on a different branch, and this branch might have a singularity near $z=a$ that the other branch does not. For example, this can occur with a function of the form
$$ f(z) = \frac{1}{\sqrt{z-p} - q} $$ where $p$ is in the first quadrant and you choose the principal branch of the square root.
EDIT: For concreteness, let's take $$ f(z) = \frac{1}{\sqrt{z} + 1 - 10 i} $$ with the principal branch. It has a branch point at $0$. Note that the principal branch of $\sqrt{z}$ has real part $\ge 0$, so the denominator is never $0$. However, other branches may have a pole at $z = (1-10i)^2 = -99-20i$. The Taylor series around $z=-99-20i$ has radius $101$ (the distance to the branch point at $0$). But the Taylor series around $z=-99+20i$ has radius only $40$, because the analytic continuation in a disk around $-99+20i$ will run into a pole at $-99-20i$.
More precisely, the radius of convergence of the series about $z=a$ is the largest $r$ such that there is an analytic function in the open disk of radius $r$ about $a$ which has the given power series as its Taylor series around $z=a$. In particular, removable singularities don't count, and branch cuts don't count if they can be moved out of the way.
EDIT: To prove this, you can proceed as follows:
- If $f$ is analytic in the open disc of radius $r$ about $a$, use Cauchy's estimates to show that for any $r_1 < r$ the radius of convergence of the Taylor series of $f$ around $z=a$ is at least $r_1$. Thus the radius of convergence is at least $r$.
- Conversely, if the series $\sum_{k} c_k (z-a)^k$ has radius of convergence $\ge r$, then this series converges uniformly on compact subsets of the disk of radius $r$ about $a$, and the sum of the series is therefore an analytic function on that disk; moreover (using Cauchy's formula) the Taylor series of this function around $z=a$ is the given series.
Best Answer
http://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions
I wrote the initial draft of the article linked above in February 2004, and mostly it's still as I wrote it, although others have contributed.
Postscript:
Let $C$ be a positively oriented circle centered at $a$ that encloses a point $z$ that is closer to $a$ then is any place where $f$ blows up, and that does not enclose, nor pass through, any point where $f$ blows up.
\begin{align}f(z) &{}= {1 \over 2\pi i}\int_C {f(w) \over w-z}\,dw \tag1 \\[10pt] &{}= {1 \over 2\pi i}\int_C {f(w) \over (w-a)-(z-a)} \,dw \tag2 \\[10pt] &{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{1 \over 1-{z-a \over w-a}}f(w)\,dw\tag3 \\[10pt] &{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{\sum_{n=0}^\infty\left({z-a \over w-a}\right)^n} f(w)\,dw\tag4 \\[10pt] &{}=\sum_{n=0}^\infty{1 \over 2\pi i}\int_C {(z-a)^n \over (w-a)^{n+1}} f(w)\,dw\tag5 \\[10pt] & = \sum_{n=0}^\infty (z-a)^n \underbrace{{1 \over 2\pi i}\int_C {f(w) \over (w-a)^{n+1}} \,dw}_{\text{No $z$ appears here!}}.\tag6 \end{align}
Step $(1)$ above is Cauchy's formula.
Step $(4)$ is summing a geometric series.
Step $(6)$ can be done because "$(z-a)^n$" has no $w$ in it; thus does not change as $w$ goes around the circle $C$.
The fact that no "$z$" appears in the expression in $(6)$ where that is noted, means that the last expression is a power series in $z-a$.