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--[[User:Kfernan|Kfernan]] 16:34, 20 October 2009 (UTC) | --[[User:Kfernan|Kfernan]] 16:34, 20 October 2009 (UTC) | ||
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| + | Yep, that looks right. Also, another interesting way to solve the problem is with the function <math>e^{iz^2}</math> instead of <math>e^{-z^2}</math>. Although the two quickly follow from each other.--[[User:Davis29|Matt Davis]] | ||
Revision as of 01:56, 21 October 2009
Homework 6
Professor Bell, could you post the notes from friday? Thanks. --Yu Suo 16:31, 18 October 2009 (UTC)
Yu, they are now on the MA 425 Home Page under
Lecture 10/16/2009 --Steve Bell
Professor Bell, You showed in class that we can't show that the integral around the curved portion for problem VI.12.2 goes to zero using the basic estimate because when $ \theta = \pi/4 $ it turns out to be one. Can we use the basic estimate method for VI.12.1 because now $ \theta= \pi/8 $, which should not cause a problem. --Kevin Fernandes 10:05, 20 October 2009 (UTC)
Kevin, yes, the basic estimate works just fine on the curved part for VI.12.1. --Steve Bell
Problem 12.1
I) Integrate along the real axis. Let $ z = t $.
II) Integrate along the curve from $ t=0 $ to $ t=\frac{\pi}{8} $ and show that this tends to zero.
III) Integrate along the $ \frac{\pi}{8} $ line. Since the total integral along the closed curve equals zero, this integral must be the negative of the integral found in (I). To do this integral, let $ z=t\exp(i\frac{\pi}{8}) $. The real part is what we are looking for. Hint: $ \cos(\frac{\pi}{8})=\frac{1}{2}\sqrt{2+\sqrt{2}} $, and $ \sin(\frac{\pi}{8})=\frac{1}{2}\sqrt{2-\sqrt{2}} $.
-Alex Krzywda
Note: In VI.12.1, it will be necessary to make a simple change of variables $ u=t/\sqrt{\sqrt{2}} $ to get the integral in the final answer. --Steve Bell
I got these answers for the second problem for both the integrals
$ \sqrt{\pi}/2\sqrt{2} $
Did any one get anything besides that?
--Kfernan 16:34, 20 October 2009 (UTC)
Yep, that looks right. Also, another interesting way to solve the problem is with the function $ e^{iz^2} $ instead of $ e^{-z^2} $. Although the two quickly follow from each other.--Matt Davis
