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III) Integrate along the <math>\frac{\pi}{8}</math> line. Since the total integral along the curve equals zero, this integral must be the negative of the integral found in (I). To do this integral, let <math>z=texp(i\frac{\pi}{8})</math>. The real part is what we are looking for. Hint: <math>cos(\frac{\pi}{8})=\frac{1}{2}\sqrt{2+\sqrt{2}}</math>, and <math>sin(\frac{\pi}{8})=\frac{1}{2}\sqrt{2-\sqrt{2}}</math>. | III) Integrate along the <math>\frac{\pi}{8}</math> line. Since the total integral along the curve equals zero, this integral must be the negative of the integral found in (I). To do this integral, let <math>z=texp(i\frac{\pi}{8})</math>. The real part is what we are looking for. Hint: <math>cos(\frac{\pi}{8})=\frac{1}{2}\sqrt{2+\sqrt{2}}</math>, and <math>sin(\frac{\pi}{8})=\frac{1}{2}\sqrt{2-\sqrt{2}}</math>. | ||
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Revision as of 06:56, 20 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 cant evaluate the integral around the curved portion for problem VI.12.2 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. Also on your lecture on the 14th when you started problem VI.12.2 the integral over the lower curve, you evaluated it as $ sqrt(pi)/2 $ shouldn't it be $ sqrt(pi)/2sqrt(2) $ ?
--Kfernan 10:05, 20 October 2009 (UTC)
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 equals zero.
III) Integrate along the $ \frac{\pi}{8} $ line. Since the total integral along the curve equals zero, this integral must be the negative of the integral found in (I). To do this integral, let $ z=texp(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
