Fourier series coefficient (half range)

jmher0403
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Homework Statement



f(x) = 1, 0<x<1

Extend f(x) t generate an even function P(x) and find Fourier coefficients

Homework Equations



an = 2/T ∫ P(x)cos(2nx/T) dx


The Attempt at a Solution



P(x) = 1, -1<x<1
0, -2<x<-1 , 1<x<2

even function so b0 = 0

Average of P(x) over T = 0.5

an = 2/n∏ sin (n∏x)

I got right upto here...

answer for the exercise says

an = 0 when n even
2/n∏ when n=1,5,9,13...
-2/n∏ when n = 3,7,11,15...

I am confused because isn't all mutiples of pi in a sine function all equal to 0?

Please help :(
 
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jmher0403 said:

Homework Statement



f(x) = 1, 0<x<1

Extend f(x) t generate an even function P(x) and find Fourier coefficients

Homework Equations



an = 2/T ∫ P(x)cos(2nx/T) dx

The Attempt at a Solution



P(x) = 1, -1<x<1
0, -2<x<-1 , 1<x<2

What does T represent and what is it in this problem? Is it the full period? If so, with your choice of P(x), apparently T= 4? Is that what you used in your formulas? If you are going to extend f(x) = 1 on (0,1) to an even function, why not just use f(x) = 1 on (-1,1) for your function? I don't think you have given us a complete statement of the problem.
 
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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