(New page: Category:MA5530Spring2014Bell Category:MA530 Category:math Category:homework == Homework 2 collaboration area == Here it is again: <math>f(a)=\frac{1}{2\pi i}\int_\gam...) |
(Started Discussion of Problem 1) |
||
| Line 15: | Line 15: | ||
---- | ---- | ||
| + | '''Exercise 1''' | ||
| + | Suppose that <math>\varphi(z)</math> is a continuous function on the trace of a path <math>\gamma</math>. Prove that the function | ||
| + | |||
| + | <math>f(z)=\int_{\gamma}\frac{\varphi(\zeta)}{\zeta-z}d\zeta</math> | ||
| + | |||
| + | is analytic on <math>\mathbb{C}-\text{tr }\gamma</math>. | ||
| + | |||
| + | |||
| + | '''Discussion''' | ||
| + | |||
| + | -A function is said to be analytic on an open set <math>\Omega</math> if it is <math>\mathbb{C}</math>-differentiable at every point in <math>\Omega</math>. Since the definition of analytic involves an open set, to complete Exercise 1 we need to have an open set somewhere. Since <math>\text{tr }\gamma</math> is the continuous image of a compact set, it is compact. Since <math>\mathbb{C}</math> is a Hausdorff space, <math>\text{tr }\gamma</math> is closed. Alternately one can note that <math>\mathbb{C}</math> is homeomorphic to <math>\mathbb{R}^{2}</math> where we know that a compact set is closed and bounded. Since the complement of a closed set is open, <math>\mathbb{C}-\text{tr }\gamma</math> is open. The long and short of this is: <math>\mathbb{C}-\text{tr }\gamma</math> is an open set so one just has to show that <math>\varphi</math> is <math>\mathbb{C}</math>-differentiable at every point in this set in order to complete the exercise. | ||
| + | ---- | ||
[[2014_Spring_MA_530_Bell|Back to MA530, Spring 2014]] | [[2014_Spring_MA_530_Bell|Back to MA530, Spring 2014]] | ||
Revision as of 19:09, 28 January 2014
Homework 2 collaboration area
Here it is again:
$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz. $
This is the place.
Exercise 1
Suppose that $ \varphi(z) $ is a continuous function on the trace of a path $ \gamma $. Prove that the function
$ f(z)=\int_{\gamma}\frac{\varphi(\zeta)}{\zeta-z}d\zeta $
is analytic on $ \mathbb{C}-\text{tr }\gamma $.
Discussion
-A function is said to be analytic on an open set $ \Omega $ if it is $ \mathbb{C} $-differentiable at every point in $ \Omega $. Since the definition of analytic involves an open set, to complete Exercise 1 we need to have an open set somewhere. Since $ \text{tr }\gamma $ is the continuous image of a compact set, it is compact. Since $ \mathbb{C} $ is a Hausdorff space, $ \text{tr }\gamma $ is closed. Alternately one can note that $ \mathbb{C} $ is homeomorphic to $ \mathbb{R}^{2} $ where we know that a compact set is closed and bounded. Since the complement of a closed set is open, $ \mathbb{C}-\text{tr }\gamma $ is open. The long and short of this is: $ \mathbb{C}-\text{tr }\gamma $ is an open set so one just has to show that $ \varphi $ is $ \mathbb{C} $-differentiable at every point in this set in order to complete the exercise.
