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To find the radius of convergence of <math>\sum_{n=0}^\infty (n!)z^{n!}</math>, you'll need to use the Ratio Test. | To find the radius of convergence of <math>\sum_{n=0}^\infty (n!)z^{n!}</math>, you'll need to use the Ratio Test. | ||
| − | <math>\frac{u_{n+1}}{u_n}=\frac{(n+1)!z^{(n+1)!}{n!z^{n!}=(n+1)z^{n\cdot n!</math>. | + | <math>\frac{u_{n+1}}{u_n}=\frac{(n+1)!z^{(n+1)!}}{n!z^{n!}}=(n+1)z^{n\cdot n!}</math>. |
| − | Ask yourself what that does as n goes to infinity in case |z|<1, =1, >1. | + | Ask yourself what that does as n goes to infinity in case |z|<1, =1, >1. You might recall that <math>nr^n\to 0</math> as <math>n\to\infty</math> if <math>|r|<1</math>. --[[User:Bell|Steve Bell]] |
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| + | On Problem 1 from the practice problems I was thinking Liouville's theorem, but that is for a bounded entire function. I seem to vaguely recall this being shown with an analytic function which is bounded over a closed region, but I can't seem to find it in my notes. Anyone have this?--[[User:Rgilhamw|Robert Gilham Westerman]] 16:35, 14 November 2009 (UTC) | ||
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| + | On 30 October we talked about this in class... Prof. Bell gave us a Theorem: If U is harmonic on domain <math>\Omega</math> and vanishes on an open subset of <math>\Omega</math> then <math>U\equiv0</math>. The proof was this: | ||
| + | If U is harmonic, then <math>F=U_x - iU_y</math> is analytic. If <math>U\equiv0</math> on <math>D_r(z_o)</math>, an open subset of <math>\Omega</math>, then <math>F\equiv0</math> too. By the Identity Theorem, <math>F\equiv0</math> on <math>\Omega</math>. So, <math>\nabla U\equiv0</math> on <math>\Omega</math>; hence U is a constant. <math>U(z_o) = 0</math>, so the constant is 0.--[[User:Achurley|Aaron Hurley]] 20:23, 15 November 2009 (UTC) | ||
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| + | Did anyone get <math>\frac{\sqrt{7}}{2}</math> as the RoC for number 10?--[[User:Rgilhamw|Rgilhamw]] 18:25, 17 November 2009 (UTC) | ||
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| + | I got the same for number 10. I'm having troubles with Chapter 7 problems 11.3, 13.1-2, 14.1, 16.1-2, any advice how to get started?--[[User:Pstechsc|Pstechsc]] 19:45, 17 November 2009 (UTC) | ||
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| + | I am struggling with 16.2 as well. My guess is to write f as a polynomial of degree N, but I am not sure where to go after that.--[[User:Phebda|Phebda]] 23:41, 17 November 2009 (UTC) | ||
Latest revision as of 19:41, 17 November 2009
Discussion area to prepare for Exam 2
To find the radius of convergence of $ \sum_{n=0}^\infty (n!)z^{n!} $, you'll need to use the Ratio Test.
$ \frac{u_{n+1}}{u_n}=\frac{(n+1)!z^{(n+1)!}}{n!z^{n!}}=(n+1)z^{n\cdot n!} $.
Ask yourself what that does as n goes to infinity in case |z|<1, =1, >1. You might recall that $ nr^n\to 0 $ as $ n\to\infty $ if $ |r|<1 $. --Steve Bell
On Problem 1 from the practice problems I was thinking Liouville's theorem, but that is for a bounded entire function. I seem to vaguely recall this being shown with an analytic function which is bounded over a closed region, but I can't seem to find it in my notes. Anyone have this?--Robert Gilham Westerman 16:35, 14 November 2009 (UTC)
On 30 October we talked about this in class... Prof. Bell gave us a Theorem: If U is harmonic on domain $ \Omega $ and vanishes on an open subset of $ \Omega $ then $ U\equiv0 $. The proof was this: If U is harmonic, then $ F=U_x - iU_y $ is analytic. If $ U\equiv0 $ on $ D_r(z_o) $, an open subset of $ \Omega $, then $ F\equiv0 $ too. By the Identity Theorem, $ F\equiv0 $ on $ \Omega $. So, $ \nabla U\equiv0 $ on $ \Omega $; hence U is a constant. $ U(z_o) = 0 $, so the constant is 0.--Aaron Hurley 20:23, 15 November 2009 (UTC)
Did anyone get $ \frac{\sqrt{7}}{2} $ as the RoC for number 10?--Rgilhamw 18:25, 17 November 2009 (UTC)
I got the same for number 10. I'm having troubles with Chapter 7 problems 11.3, 13.1-2, 14.1, 16.1-2, any advice how to get started?--Pstechsc 19:45, 17 November 2009 (UTC)
I am struggling with 16.2 as well. My guess is to write f as a polynomial of degree N, but I am not sure where to go after that.--Phebda 23:41, 17 November 2009 (UTC)
