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)
