Recent content by MATHMAN89

I got sin(theta)/(2-sin(theta)) as my answer. Sound right?

Aha. And then I have lim n →∞ of I am ( sum (e^iθ/2)^k , k = 1 to n) So this is a geometric series and would be equal to lim n →∞ of I am ( e^iθ/2 * (1-((e^iθ)/2)^n)/(1-(e^iθ)/2)

I only have one problem left I'm trying to do now, and it's not one that I posted because I thought it would be the easiest. Can you see https://www.physicsforums.com/showthread.php?p=3567888&posted=1#post3567888 ? I appreciate your help, Dick. You're the best!

Just checked on WOLFRAM. Correct! VICTORY!

I think I got it. I ended up with (1-i√3)/6 Sound right?? Should be if I did all the calculations right.

So I am left with z/((z-ei*pi)(z-e5i*pi/3) How can I evaluate this when I plug in? Just use Euler?

I see what you mean. If I do ((e^i*theta)/2)^k I would get (cos(kθ)+isin(kθ))/2^k I just need to get the numerator equal to sin(kθ) How do I do that?? I need to get rid of the cos (kθ) and the i in front of sin Could I do(e^i*theta - e^-i*theta / 2i)^k = sin(k*theta) ?

AHA! got it! Thanks

Hi Dick, I know how to do this problem in the real numbers, I use L' Hopital's rule twice and eventually get that the limit is e^-1/6. Do I need to do the problem differently, though, since z = x + iy ? If so, do I start with assuming x is 0 and then taking the limit as y approaches zero...

Thanks for your response. The problem can be rewritten as sum (ak*(1/2)^k) where ak=sin(kθ) Then I can use limsup ( n√abs(ak) = 1 Therefore my radius of convergence is 1. But now how do I use that to find the limit?

I did this but I still get something like e^ infinity*0 so do I use L'hopital's rule here?

Homework Statement The problem is as follows: lim(n→∞)(sum(sin(kθ)/2k,k=1,n)Homework Equations The Attempt at a Solution I feel like this should converge since sin oscillates between -1 and 1 and 2k keeps getting larger and larger. Do I have to use power series somehow? Any ideas, anyone?

Still not sure how that gets me at my answer.

sinz/z = 1 -z^2/2 +z^4/4! ... = cos z right?

Aha, OK. I've got the first one now. Thank you so much. I don't see how I get log(1+z) for the second one, though.