| Line 23: | Line 23: | ||
== Problem 6 == | == Problem 6 == | ||
| − | + | Hint: Notice that if | |
| + | |||
| + | <math>|a_n r^n|< M,</math> | ||
| + | |||
| + | then | ||
| + | |||
| + | <math>|a_n z^n|=|a_n r^n|\left(\frac{|z|}{r}\right)^n< | ||
| + | M\left(\frac{|z|}{r}\right)^n,</math> | ||
| + | |||
| + | and you can compare the series to a convergent geometric series | ||
| + | if | ||
| + | |||
| + | <math>|z|<r.</math> | ||
About the trick in the Problem 6, one direction is easy; | About the trick in the Problem 6, one direction is easy; | ||
Revision as of 08:39, 18 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. $
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.
