## ghdoru11 one year ago Find the Fourier expansion for: f:R->R periodic T=2pi , f(x)=3x, x is between [-pi,pi]

1. beginnersmind

2. ghdoru11

no one solve this and i have another question

3. beginnersmind

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.

4. IrishBoy123

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

5. beginnersmind

Right, that saves a lot of work :)

6. ghdoru11

so just the first integral is neded?

7. beginnersmind

Yes, $$b_n$$ = 0 for every n.

8. ghdoru11

thank you so much guys