What Are the Fourier Coefficients for the Half-Range Cosine Series of f(x)=x?

AI Thread Summary
The discussion focuses on finding the Fourier coefficients for the function f(x) = x defined on the interval 0 < x < L. The initial calculations yielded a_0 = L and a_n = 4L/n^2 pi^2, but the correct values are a_0 = L/2 and a_n (for odd n) = -4L/n^2 pi^2, with a_n being zero for even n. The correction is supported by the observation that shifting the function creates an odd function, which only requires sine terms in its Fourier series representation. Participants emphasize the need for detailed calculations to identify mistakes in the original attempt.
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Homework Statement



The function f(x) is defined on the interval 0<x<L by f(x)=x. It can be represented by the Fourier cosine series

f(x) = a_0 + sum a_n cos(n*pi*x / L)

Find its Fourier coefficients a_0 and a_n.


Homework Equations



Multiply both sides by cos(n*pi*x / L) and integrate from L to 0. Then integration by parts.


The Attempt at a Solution



I got a_0 = L and a_n = 4L/ n^2 pi^2

The answers should be : a_0 = L/4 and a_n = -2L/n^2 pi^2 for n = odd and a_n=0 for n = even.

thanks!
 
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The give answer is not quite correct, but neither is yours.
a_0 should be L/2 and a_n(odd) = -4L/(n^2 pi^2).
The value for a_0 can be seen to be correct by shifting the function by L/2 to the left. Then it is an odd function up to the shift upwards by L/2. So it can be represented by only sines, which explains that even values of the cosine are 0.

If you want us to find your mistake you have to post your calculation.
 
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