| Line 6: | Line 6: | ||
<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> | ||
| + | |||
| + | Just throwing some stuff here -- 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. | ||
[[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 20:52, 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. $
Just throwing some stuff here -- 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.
