Converting complex power series into a function

by beefcake24
Tags: complex, complex analysis, integration, power series, taylor series
beefcake24 is offline
Sep23-11, 04:37 AM
P: 16
Hey guys, sorry for sending out so many questions so fast. I just discovered this site, and it looks great. Plus, I have my first complex analysis midterm tomorrow, so I'm pretty stressed (you'd think after 4 years of math/econ/computer science you'd get used to it but there's nothing like the first hard midterm of the semester to rattle your nerves :-) ). Anyway, I only have one last question, I promise.

Find the region of convergence of (sum of n from 1 to infinity) (n^2)*z^n, where z is a complex number. Then find explicitly the function to which this power series is convergent in the region of convergence.

The region of convergence was easy to do, I got |z| < 1 (i.e. the open unit disk). However, I'm completely stumped on the second part on how to represent the explicit function that this power series converges to. I think it either has to do something with the sum of the integrals or with the taylor expansion of that function to match this power series, but I have no idea where to go from there.

If you know how to do this or at least can point me in the general direction, it would be greatly appreciated, and I will try to reciprocate and answer any questions I see that I know how to do.

Thanks guys, and by the way, these forums are great, I love it when intelligent people get together and share knowledge, and I'm so happy I found this site. Definitely beats yahoo answers haha.
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CompuChip is offline
Sep23-11, 07:53 AM
Sci Advisor
HW Helper
P: 4,301
I'd probably start with something like
[tex]\sum_{n = 0}^\infty z^n = \frac{1}{1 - z}[/tex]
for |z| < 1, and look at its derivatives to get the factors of n in front.

[tex]\frac{d^2}{dx^2} \sum_{n = 0}^\infty z^n = \sum_{n = 0}^\infty n^2 z^{n - 2} + \frac{1}{z} \sum_{n = 0}^\infty n z^{n - 1}[/tex]
and then do a bit of shifting and some more magic with the first derivative.

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