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| − | ^^^On 8.1, I think you just need to use what you already know about geometric series, as Taylor's Theorem isn't mentioned until a couple of sections after. I just wrote <math>z^k = z^k * ( | + | ^^^On 8.1, I think you just need to use what you already know about geometric series, as Taylor's Theorem isn't mentioned until a couple of sections after. I just wrote <math>z^k = z^k * (1-z)/(1-z)</math> , and used the fact that the geometric series (with coefficient 1, center 0) converges to <math>1/(1-z)</math> |
--[[User:Dgoodin|Dgoodin]] 10:46, 8 November 2009 (UTC) | --[[User:Dgoodin|Dgoodin]] 10:46, 8 November 2009 (UTC) | ||
Revision as of 06:47, 8 November 2009
Homework 8
NEWS FLASH: The due date for HWK 8 has been extended to Monday, Nov. 9
Hint for V.16.1: We know that
$ f(z)=\sum_{n=0}^\infty z^n=\frac{1}{1-z} $
if $ |z|<1 $. Notice that
$ f'(z)=\sum_{n=1}^\infty nz^{n-1} $,
and
$ f''(z)=\sum_{n=2}^\infty n(n-1)z^{n-2} $.
What are the power series for $ zf'(z) $ and $ z^2f''(z) $? How can you combine these to get the series in the question? --Steve Bell
Does anybody know how to attack problem 10.2? Also for problem 8.1, I am thinking the power series should just be $ (z-zo)^{k} $Did anybody do it another way? --Adrian Delancy
^^^On 8.1, I think you just need to use what you already know about geometric series, as Taylor's Theorem isn't mentioned until a couple of sections after. I just wrote $ z^k = z^k * (1-z)/(1-z) $ , and used the fact that the geometric series (with coefficient 1, center 0) converges to $ 1/(1-z) $
--Dgoodin 10:46, 8 November 2009 (UTC)
