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<math>f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz.</math> | <math>f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz.</math> | ||
| − | Stirling's Formula | + | Here is Stirling's Formula: |
| − | <math>\lim_{n\to\infty}\frac{n!}{ | + | <math>\lim_{n\to\infty}\frac{n!}{\sqrt{n}\left(\frac{n}{e}\right)^n} =\sqrt{2\pi}.</math>--[[User:Bell|Steve Bell]] 10:28, 19 January 2011 (UTC) |
== Problem 1 == | == Problem 1 == | ||
Latest revision as of 10:25, 19 January 2011
Contents
Homework 1 collaboration area
Feel free to toss around ideas here.--Steve Bell
Here is my favorite formula:
$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz. $
Here is Stirling's Formula:
$ \lim_{n\to\infty}\frac{n!}{\sqrt{n}\left(\frac{n}{e}\right)^n} =\sqrt{2\pi}. $--Steve Bell 10:28, 19 January 2011 (UTC)
Problem 1
Problem 2
Problem 3
Difference quotient should include a special case when $ f(z)=f(z_0) $.
Problem 4
Problem 5
Use Problem 4.
Problem 6
Hint: Notice that if
$ |a_n r^n|< M, $
then
$ |a_n z^n|=|a_n r^n|\left(\frac{|z|}{r}\right)^n< M\left(\frac{|z|}{r}\right)^n, $
and you can compare the series to a convergent geometric series if
$ |z|<r. $
About the trick in the Problem 6, one direction is easy;
The other direction can be proved using a trick by considering $ r-\epsilon $ where $ \epsilon>0 $ is some arbitrarily small quantity. This yields a convergent geometric series, which serves as an upper-bound of the original absolute series. Finally, let $ \epsilon $ go to zero. Result from Problem 5 is involved.
