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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|Rgilhamw]] 16:35, 14 November 2009 (UTC) | 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|Rgilhamw]] 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, the <math>F=U_x - iU_y</math> is analytic. If <math>U\equiv0</math> on <math>D_r (z_o)</math>, a 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|Achurley]] 20:23, 15 November 2009 (UTC) | ||
Revision as of 16:23, 15 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?--Rgilhamw 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, the $ F=U_x - iU_y $ is analytic. If $ U\equiv0 $ on $ D_r (z_o) $, a 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.--Achurley 20:23, 15 November 2009 (UTC)
