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 ?
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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
i found the taylor expansion to be SUM(n=0 to infinity) (-1)^n (n+1)*(z-1)^n does that sound right?
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
is there any way to find radius of convergence without using the ratio/comparison tests?
You could do this:
Last edited by lilaziz1; March 29th 2010 at 05:21 PM.
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