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Let $f(z):=\frac{\sin z}{z(z-\pi)(z+\pi)}$. What is the radius of convergence of the series $$ \sum_{n=0}^{\infty} \frac{f^{(n)}(i)}{n!}(z-i)^n\quad\text{?} $$Explain.

My Attempt :

Note that the series in question is just the Taylor series of the function $f(z)$ about the point $i$. So the solution reduces to determining the distance between the closest non-removable isolated singularity and the point $0+i$. All of the isolated singularities are removable and we must conclude that the radius of convergence is $\infty$. I verify that the singularities are all removable below.

$$ \lim_{z\to 0} \frac{\sin z}{z(z-\pi)(z+\pi)} =-1/\pi^2 $$ $$ \lim_{z\to \pi} \frac{\sin z}{z(z-\pi)(z+\pi)} =-1/2\pi^2 $$ $$ \lim_{z\to -\pi} \frac{\sin z}{z(z-\pi)(z+\pi)} =-1/2\pi^2 $$

Bob
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    We say that the domain of convergence is $\mathbb{C}$, but that the radius of convergence is $+\infty$, or more commonly, that the series converges everywhere. – Travis Willse Mar 24 '15 at 04:45
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    See here. – science Mar 24 '15 at 05:19
  • Right. So is it fair to say that your link explains why, when I show that all my isolated singularities are removable, I also show that it is holomorphic everywhere? – Bob Mar 24 '15 at 21:08

3 Answers3

0

Note that

The radius of convergence is the distance to the nearest singularity for the function $f(z)$.

science
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$f(z)=\frac{\sin z}{z(z-\pi)(z+\pi)}=\sum_{n=0}^{\infty}a_n (z-i)^n$ ,where $a_n=\frac{f^n(i)}{n!}$ , provided $f(z)$ is analytic in the region containing $i\in \mathbb C$.

Now singularities of $f(z)$ are $0$ , $\pi$ , $-\pi$.

Now $f(z)$ is analytic in $C:|z-i|=R$ where $R>0$ such that $0,\pi,-\pi\not \in C$.

So, $R<1$. Because if $R\ge 1$ then the circle $C$ contains at least the point $0$.

i.e. , the radius of convergence is $1$.

Empty
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  • Why doesn't the fact that the singularities are removable play a part here? – Bob Mar 24 '15 at 16:40
  • Taylor series expansion valid for an analytic function on a given region..Not for a function having removable singularities.. – Empty Mar 24 '15 at 18:53
  • I guess I'm thinking of sin(z)/z which has an infinite radius of convergence around 0. You can either write the taylor series out to see this or you can take the limit. Now, is my problem different because I'm interested in the radius of convergence around "i"? – Bob Mar 24 '15 at 19:22
  • Also, try looking at the plot of my function [link]http://www.wolframalpha.com/input/?i=plot%20sin(z)%2F%5Bz(z-pi)(z%2Bpi)%5D There don't appear to be any poles – Bob Mar 24 '15 at 20:08
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After editing my attempt to change the RoC to $\infty$ from $\mathbb{C}$ I believe my attempt is correct.

Bob
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