| Line 7: | Line 7: | ||
<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> | ||
| + | == Problem 1 == | ||
| + | |||
| + | == Problem 2 == | ||
| + | |||
| + | == Problem 3 == | ||
| + | |||
| + | Difference quotient should include a special case when <math>f(z)=f(z_0)</math>. | ||
| + | |||
| + | == Problem 4 == | ||
| + | |||
| + | == Problem 5 == | ||
| + | |||
| + | Use Problem 4. | ||
== Problem 6 == | == Problem 6 == | ||
Revision as of 21:02, 14 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
Just throwing some stuff here for test purpose:
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.
