# taylors series, complex variables

• Mar 25th 2010, 01:10 PM
stumped765
taylors series, complex variables
expand the function f(z)=1/z^2 in a taylors series around z=1. What is the radius of convergence?

Attempted soln: Since 1/z^2 has a singularity at 0, isnt the radius of convergence just R=1-0 ?
• Mar 25th 2010, 04:04 PM
tonio
Quote:

Originally Posted by stumped765
expand the function f(z)=1/z^2 in a taylors series around z=1. What is the radius of convergence?

Attempted soln: Since 1/z^2 has a singularity at 0, isnt the radius of convergence just R=1-0 ?

Well, yes...at most, but it could be less, right? You'll have to check it with the Taylor series directly.

Tonio
• Mar 25th 2010, 05:55 PM
stumped765
i found the taylor expansion to be
SUM(n=0 to infinity) (-1)^n (n+1)*(z-1)^n

does that sound right?
• Mar 25th 2010, 06:07 PM
tonio
Quote:

Originally Posted by stumped765
i found the taylor expansion to be
SUM(n=0 to infinity) (-1)^n (n+1)*(z-1)^n

does that sound right?

Sounds, looks and is right...and now verify that the conv. radius indeed is 1. Good work.

Tonio
• Mar 29th 2010, 12:40 PM
stumped765
is there any way to find radius of convergence without using the ratio/comparison tests?
• Mar 29th 2010, 12:58 PM
lilaziz1
You could do this:

$
f(x) = \frac{1}{z^2} = \frac{1}{1 + (z^2 - 1)}$

$\ \ \ \mid R \mid \leq 1$

$\ \ \ R = (z^2 - 1)$

$\ \ \ \mid (z^2 - 1) \mid \ \ \leq 1
$