Difference between revisions of "HWK8MA425Fall2009" - Rhea

Revision as of 10:48, 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 * (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)

for 10.2 compare $ \sum_{n=0}^\infty n^nz^{n} $ to $ \sum_{n=0}^\infty c^nz^{n} $ where $ c $ is a constant to establish the inequality in the RoC's. Then let $ c \rightarrow \infty $ to squeeze the RoC of the first power series to zero.--Rgilhamw 14:21, 8 November 2009 (UTC)

did any one get this as the answer for 8.1?

$ \sum_{n=0}^\infty (1-z)(z-z0)^{2k} $

--Kfernan 15:41, 8 November 2009 (UTC)

Also for problem 10.2 instead of using the comparison test you can just use the Hadamard Theorem since your $ An=n^n $ Then $ \frac{1}{lim sup(An^{1/n})} = \frac{1}{n} $ which equals zero as n tends to infinity --Kfernan 15:47, 8 November 2009 (UTC)