| Line 14: | Line 14: | ||
About the trick in the Problem 6, one direction is easy using Problem 5; | About the trick in the Problem 6, one direction is easy using Problem 5; | ||
| − | The other direction can be proved using a trick by considering <math>r-\epsilon</math> where <math>\epsilon>0</math> is some arbitrarily small quantity. This yields a convergent geometric series, which serves as an upper-bound of the original absolute series. Finally, | + | The other direction can be proved using a trick by considering <math>r-\epsilon</math> where <math>\epsilon>0</math> is some arbitrarily small quantity. This yields a convergent geometric series, which serves as an upper-bound of the original absolute series. Finally, let <math>\epsilon</math> go to zero. |
[[2011 Spring MA 530 Bell|Back to the MA 530 Rhea start page]] | [[2011 Spring MA 530 Bell|Back to the MA 530 Rhea start page]] | ||
Revision as of 19:55, 14 January 2011
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 6
Just throwing some stuff here for test purpose:
About the trick in the Problem 6, one direction is easy using Problem 5;
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.
