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Consider the origin, z=0. The problem notes that the function is equal to 1 at z=0, so this is fine. Then we simply have to worry about the cosine, since the sine doesn't blow up anywhere. I ended up eventually just getting that the closest singularity to the origin was at <math>\frac{\pi}{2}</math>. Thoughts? --[[User:Adbohn|Adbohn]] 23:57, 11 December 2009 (UTC) | Consider the origin, z=0. The problem notes that the function is equal to 1 at z=0, so this is fine. Then we simply have to worry about the cosine, since the sine doesn't blow up anywhere. I ended up eventually just getting that the closest singularity to the origin was at <math>\frac{\pi}{2}</math>. Thoughts? --[[User:Adbohn|Adbohn]] 23:57, 11 December 2009 (UTC) | ||
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| + | Yes I believe what you stated works. Another way to think about it is that the tan function goes to infinity as theta goes to <math>\frac{\pi}{2}</math>. Since a series converges absolutely and uniformly with in it's RoC and it converges for every value on the real line up to <math>\frac{\pi}{2}</math> then it must converge at all points in that open disc away from the boundary.--[[User:Rgilhamw|Rgilhamw]] 17:14, 12 December 2009 (UTC) | ||
Revision as of 12:14, 12 December 2009
Discussion area to prepare for the Final Exam
On problem 2. I am breaking the curve $ \gamma $ up into two piece wise curves $ \gamma_1 $ and $ \gamma_2 $ that meet when the curve $ \gamma $ crosses the negative real axis at the point $ z_0 $. I then am taking the principle branch of log as an analytic function to evaluate the two curves with the Log $ {z_0} $ values dropping out. My worry is that since $ z_0 $ sits on the branch cut that the function won't be analytic for one of the endpoints of the curves. Am I getting myself into trouble with this?--Rgilhamw 21:13, 8 December 2009 (UTC)
for anyone who had the same question, Prof. Bell covered this in class today. Having the point where the curves break on the branch cut will not work, so it needs to be chopped up into more piece-wise curves.--Rgilhamw 18:29, 9 December 2009 (UTC)
I was wondering what anyone else did for problem 9, or even how they started it. I'm not sure what is meant by f is a rational function... A little help for a jump start would be nice.--Achurley 17:53, 11 December 2009 (UTC)
$ f $ is a rational function means that $ f $ can be expressed as the division of two polynomials. For problem 9, express $ sin(\theta) $ in terms of $ exp(i\theta) $ and use the substitution $ exp(i\theta)=z $. This expresses the integral in the complex plane along the unit circle in the counterclockwise direction.--Phebda 22:31, 11 December 2009 (UTC)
I had an idea for number 6 for finding the radius of convergence, and I wanted to see if anyone else agrees.
Using the trick from exam two, I want to draw the biggest circle with the center at z = 0, since that is the center of the power series, such that there are no singularities enclosed within the circle.
$ \frac{tan(z)}{z} = \frac{sin(z)}{z*cos(z)} $.
Consider the origin, z=0. The problem notes that the function is equal to 1 at z=0, so this is fine. Then we simply have to worry about the cosine, since the sine doesn't blow up anywhere. I ended up eventually just getting that the closest singularity to the origin was at $ \frac{\pi}{2} $. Thoughts? --Adbohn 23:57, 11 December 2009 (UTC)
Yes I believe what you stated works. Another way to think about it is that the tan function goes to infinity as theta goes to $ \frac{\pi}{2} $. Since a series converges absolutely and uniformly with in it's RoC and it converges for every value on the real line up to $ \frac{\pi}{2} $ then it must converge at all points in that open disc away from the boundary.--Rgilhamw 17:14, 12 December 2009 (UTC)
