Complex Analysis - Radius of convergence of a Taylor series

[Pyroadept]

Homework Statement

Find the radius of convergence of the Taylor series at 0 of this function

f(z) = [itex]\frac{e^{z}}{2cosz-1}[/itex]

Homework Equations

The Attempt at a Solution

Hi everyone,

Here's what I've done so far:

First, I tried to re-write it as a Laurent series to find where the closest singularity to 0 is.

e^z = Ʃ(x^n)/n!

cos z = Ʃ (-1)^n . x^2n / (2n)!

However, I'm a little unsure how to combine these into a single laurent series. Is this even going about the problem in the correct way?

Thanks

 

[Pyroadept]

Actually, I have had a wave of inspiration since - is this correct?

The singularities occur for 2.cos(z)-1 = 0 i.e. cos(z) = 1/2

This happens for z = pi/3 (+ 2k.pi, but this z is the smallest one)

So then

the distance from z=1 to z=pi/3 is:

√(1 - pi/3)^2) = 2pi/3

which is then the radius of convergence?

 

[jackmell]

If you're given the function, then the radius of convergence is the distance to the nearest singular point. That's different if you're only given the series and you don't know what function the series represents. In that second case, you have to compute the radius of convergence. So you got the function, from zero, how far is the nearest singular point?