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--[[User:Dgoodin|Dgoodin]] 10:46, 8 November 2009 (UTC) | --[[User:Dgoodin|Dgoodin]] 10:46, 8 November 2009 (UTC) | ||
| − | for 10.2 compare <math>\sum_{n=0}^\infty n^nz^{n}</math> to <math>\sum_{n=0}^\infty c^nz^{n}</math> where <math>c</math> is a constant to establish the inequality in the RoC's. Then let <math>c \rightarrow \infty</math> to squeeze the RoC of the first | + | for 10.2 compare <math>\sum_{n=0}^\infty n^nz^{n}</math> to <math>\sum_{n=0}^\infty c^nz^{n}</math> where <math>c</math> is a constant to establish the inequality in the RoC's. Then let <math>c \rightarrow \infty</math> to squeeze the RoC of the first power series to zero.--[[User:Rgilhamw|Rgilhamw]] 14:21, 8 November 2009 (UTC) |
Revision as of 09:22, 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)
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)
