Finding Fourier Series of sin(a*pi*t): Results & Confirmation

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Homework Statement



Find the Fourier series for: sin(a*pi*t). Consider what happens when a -> 1/L

Homework Equations





The Attempt at a Solution



I keep getting zeros for a_o, a_n, and b_n.

I though that atleast b_n should give me something, can anyone also confirm this?
 
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What is L? Are you trying to find an expansion on (-L,L) using terms like \sin{(n\pi x/l)} And are you assuming in your calculations that a is an integer? If a is not an integer you should get lots of bn terms. Make sure you haven't assumed that terms like \sin{a\pi} are zero in your calculations. Hard to guess without seeing your work.
 
Sorry for wasting your time, but I didn't feel like typing up a page of work.
 
It might not be wasting either of our times. A common mistake students make when, for example, trying to find the Fourier series for sin(3x) on (-\pi,\pi) is to think the forumla for

b_n=\frac 1 \pi \int_{-\pi}^\pi \sin{(3x)}\sin{(nx)}\ dx

works when n = 3, which it doesn't. So they are puzzled why all the bn are zero. Your question made me think you might be making one or both that type of error or assuming a is an integer.
 
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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