Complex Fourier Series for f(t)=2sin(πt) with Periodicity

twoscoops
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Homework Statement


Find the complex Fourier series of the periodic function

f(t)=2sin(πt) 0 < t < 1 and f(t+1) = f(t) for all t. (π is pi)

Homework Equations


http://upload.wikimedia.org/math/9/d/7/9d7f73fbcba87cbff485e66646aa541d.png
http://upload.wikimedia.org/math/5/2/8/52890b286b5e8481ee9d4d56f45081ac.png
http://upload.wikimedia.org/math/b/0/6/b06b197a31293ddd3a9b3812f419259d.png

The Attempt at a Solution


ive tried many times using the forumla's and trying to rearrange to get exponentials but i always end up with something that looks really wrong, can someone point out any simplifications that can be made and do i need to integrate by parts (even though i can't see how that can be done). Any help will be greatly appriciated. Thanks
 
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I haven't worked with Fourier analysis or complex variables, but I would say to work with the first equation, the c_n. You can use integration by parts by treating i and n as constants and plugging in f(t) into the spot for f(x). That's where I would start, since after you find c_n the rest is cake.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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