Solve Fourier Coefficients: Find Fn If f[t+T/2]=f(t)

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Hi
I am dealing with problem that says f(t)=ƩFn.exp(inwt) . f(t)=f(t+T)

show that if f(t)=f[t+T/2) then Fn is zero for odd n?

Attempt:

I wrote formula for Fn=1/T∫f(t).exp(-inwt) and then just replace f(t) by f(t+T). but I do not get anything, I do not know how I should approach this problem.
Any help would be highly appreciated?
 
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chessmath said:
Hi
I am dealing with problem that says f(t)=ƩFn.exp(inwt) . f(t)=f(t+T)

show that if f(t)=f[t+T/2) then Fn is zero for odd n?

Attempt:

I wrote formula for Fn=1/T∫f(t).exp(-inwt) and then just replace f(t) by f(t+T).

Try using f(t) = f(t + T/2).
 
How?
First change f(t)=f(t+T/2) and then calculate Fourier coefficient?
 
Problem is solved.Thanks.
is there anyway to mark post as solved or answered.
 
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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