| Line 22: | Line 22: | ||
--------------------------------------------------------------------------------- | --------------------------------------------------------------------------------- | ||
| + | |||
| + | |||
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? | ||
| Line 32: | Line 34: | ||
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) | 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) | ||
| + | |||
| + | did any one get this as the answer for 8.1? | ||
| + | |||
| + | <math>\sum_{n=0}^\infty (1-z)(z-z0)^(2k)</math> | ||
| + | |||
| + | --[[User:Kfernan|Kfernan]] 15:41, 8 November 2009 (UTC) | ||
Revision as of 10:41, 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)
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)
