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Does anybody know how to attack problem 10.2? Also for problem 8.1, I am thinking the power series should just be <math>(z-zo)^{k} </math>Did anybody do it another way?
 
Does anybody know how to attack problem 10.2? Also for problem 8.1, I am thinking the power series should just be <math>(z-zo)^{k} </math>Did anybody do it another way?
 
--Adrian Delancy
 
--Adrian Delancy
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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 * (z-1)/(z-1)</math> , and used the fact that the geometric series (with coefficient 1, center 0) converges to <math>1/(z-1)</math>
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--[[User:Dgoodin|Dgoodin]] 10:46, 8 November 2009 (UTC)

Revision as of 05:46, 8 November 2009


Homework 8

HWK 8 problems

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 * (z-1)/(z-1) $ , and used the fact that the geometric series (with coefficient 1, center 0) converges to $ 1/(z-1) $

--Dgoodin 10:46, 8 November 2009 (UTC)

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