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 ?

Printable View

- Mar 25th 2010, 01:10 PMstumped765taylors 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 PMtonio
- Mar 25th 2010, 05:55 PMstumped765
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 PMtonio
- Mar 29th 2010, 12:40 PMstumped765
is there any way to find radius of convergence without using the ratio/comparison tests?

- Mar 29th 2010, 12:58 PMlilaziz1
You could do this:

$\displaystyle

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

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

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

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

$