Question

Find the Fourier expansion for: f:R->R periodic T=2pi , f(x)=3x, x is between [-pi,pi]

Answer 1

Why was this question closed?

Answer 2

no one solve this and i have another question

Answer 3

Anyway you can just go back to the definition of fourier coefficients:

\[a_n = \int_{-\pi}^{\pi} 3x*sin(nx)dx\]

\[b_n = \int_{-\pi}^{\pi} 3x*cos(nx)dx\]

then evaluate the integrals. Use the substitution t = nx and integration by parts with u being the linear term and v' the trigonometeric part.

Answer 4

yeah, and it's odd so you ignore the cosine term

Answer 5

Right, that saves a lot of work :)

Answer 6

so just the first integral is neded?

Answer 7

Yes, \(b_n\) = 0 for every n.

Answer 8

thank you so much guys