Thread: Radius of convergence of an analytic function

This is from a practice exam:

Let $\displaystyle f(z)=\sum_{k=0}^\infty a_kz^k$ have a radius of convergence 2, and let $\displaystyle g(z)=\sum_{k=0}^\infty b_k(z-3)^k$ have a radius of convergence 2. Let $\displaystyle I=\{z:f(z),g(z)\text{ both converge}\}$.

(a) If $\displaystyle f(z)=g(z)$ for $\displaystyle z\in I$, prove there is an analytic function $\displaystyle w$ for which $\displaystyle w(z)=f(z)$ when $\displaystyle |z|<2$ and $\displaystyle w(z)=g(z)$ when $\displaystyle |z-3|<2$.

(b) For the function $\displaystyle w$ of part (a), determine the radius of convergence $\displaystyle R$ for the power series $\displaystyle w(z)=\sum_{k=0}^\infty c_k\left(z-\frac{3}{2}\right)^k$.

(c) What relationship exists between the coefficient sequence $\displaystyle \{a_k\}$ and the coefficient sequence $\displaystyle \{b_k\}$?

Part (a) is fairly straightforward. Let $\displaystyle A=\{z:|z|<2\}$ and $\displaystyle B=\{z:|z-3|<2\}$. Then $\displaystyle A\cap B\subseteq I$, which means we can define $\displaystyle w$ by $\displaystyle w=f'\cup g'$, where $\displaystyle f'$ and $\displaystyle g'$ are the restrictions of $\displaystyle f$ and $\displaystyle g$ to $\displaystyle A$ and $\displaystyle B$, respectively. It follows immediately that $\displaystyle w$ is analytic on $\displaystyle A\cup B$.

However, I'm lost for parts (b) and (c). Any help would be much appreciated!

The series is convergent for the $\displaystyle z=-2+0.00000001$ and $\displaystyle z=5-0.0000000001$ for example, hence we can expect that the radius of convergence is greater than $\displaystyle \frac 72$. Now you have to show it's indeed $\displaystyle \frac 72$.

Hmm. I'm sorry to say I don't understand your reasoning here.

By Taylor's theorem $\displaystyle R\geq \sup\{r : D(3/2;r)\subseteq A\cup B\}=\frac{\sqrt{7}}{2}$. (Here I use the notation $\displaystyle D (z_0;r)=\{z:|z-z_0|<r\}$). However, I don't see how we can get $\displaystyle R\geq\frac{7}{2}$, nor $\displaystyle R\leq\frac{7}{2}$.

The series is convergent on set a set which a diameter $\displaystyle \geq 7$, because the series converges on points of the form $\displaystyle -2+\delta$ end $\displaystyle 5-\delta$ for $\displaystyle 1>\delta>0$

Oh, of course... I had forgotten every power series converges precisely on a disc and a subset of its boundary. Sorry about that.