
#1
Nov311, 09:24 AM

P: 169

1. The problem statement, all variables and given/known data
Consider the 2[itex]\pi[/itex]periodic function f(t) = t t in [Pi;Pi] a) show that the real fouier series for f(t) is: [itex]f(t) ~ \sum\limits_{n=1}^{\infty}\frac{2}{n}(1)^{n+1}\sin nt[/itex] b) Use the answer to evaluate the following : [itex]\sum\limits_{n=1}^{\infty}\dfrac{(1)^{n+1}}{2n1}[/itex] Hint: Use Fouier's law with t = [itex]\pi[/itex]/2 2. Relevant equations Fouiers Law? I'm danish, and therefore i'm not really sure what it's called. 3. The attempt at a solution Part a i have done by finding the coefficients. Part b) I can't see where the problem in part b and the answer to a relates. I've tried with Maple 15 to calculate the value and i'm getting Pi/4, but i keep getting something different for the series from a) Please Help me, i would really like to understand this as i'm studying physics. 



#2
Nov311, 10:57 AM

P: 6

The answer should be pi/4 ...
from fourier series we got ... t = [itex]\sum\frac{2}{n}(1)^{n+1}sinnt[/itex] for t = [itex]\frac{\pi}{2}[/itex] , [itex]\frac{\pi}{2}[/itex] = 2  2/3 + 2/5  2/9 + ...... which followed , [itex]\frac{\pi}{2}[/itex] = 2 * [itex]\sum\frac{(1)^{n+1}}{2n  1}[/itex] hence, [itex]\sum\frac{(1)^{n+1}}{2n  1}[/itex] = [itex]\frac{\pi}{4}[/itex] 



#3
Nov311, 11:41 AM

P: 169

Hmm, isn't that a backwards proof? Can't i use something like the even terms of sin(n*Pi/2) = 0 and the odd ones alternates between 1 and 1
And how is t = ... isn't it f(t)? 


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