Fourier Expansion for a One-Period Function with Period 2pi

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SUMMARY

The discussion focuses on finding the Fourier expansion for a one-period function with a period of 2π. Participants confirm that the integral for the coefficients can be split into two parts due to the even nature of the function, resulting in ak = 0. The proposed method for calculating bk involves integrating from 0 to π and from π to 2π, specifically using the expression 1/π [∫(0 to π) (x cos(kx) dx) + ∫(π to 2π) ((-x + 2π) cos(kx) dx)]. Additionally, it is emphasized that a0 is not equal to 0.

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Homework Statement



Given this one period graph of a function with period 2pi,

YpNoh.png


find a Fourier expansion.

Homework Equations


The Attempt at a Solution



So I was wondering if i can split up the integrals into two parts to find the coefficients. It is an even function so ak = 0.

for bk can I split the integral as follows

1/pi[int 0 to pi (x cos kx dx) + int pi to 2pi ((-x+2pi) cos kx dx]

?
 
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Yup, that's what you're supposed to do. Also, remember a0 isn't equal to 0.
 

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