Find the Fourier Series of the following function.

Question

I have been given the following question.

"The function $f(x)$ is odd, has a period $2\pi$ and satisfies:

$$f(x)=\begin{cases} 1 & 0\lt x \lt \pi \\ -1 & \pi \lt x \lt 2\pi \end{cases}$$

Find its fourier series.

As it is an odd function I have put that $a_0=0$ and $a_n=0$.

I have worked out that $b_n=\frac{2}{n\pi}(-1)^{n+1}$

Thus the solution $f(x)$ would be $ \sum_{n=1}^{\infty}b_n \sin(nx)$

Is this correct? I cannot find the solution.

Thanks.

Answer 1

$b_n=\frac 1\pi \int_0^{2\pi}f(x)\sin nx dx$

$b_n=\frac 1\pi \int_0^{\pi}\sin nx dx-\frac 1\pi \int_0^{\pi}\sin nx dx$

$b_n=\frac 1\pi \left|-\frac{\cos nx}{n}\right|^{\pi}_{0}-\frac 1\pi \left|-\frac{\cos nx}{n}\right|^{2\pi}_{\pi}$

$b_n=\frac {-1}{n\pi} \left|\frac{\cos nx}{n}\right|^{\pi}_{0}+\frac {1}{n\pi} \left|\frac{\cos nx}{n}\right|^{2\pi}_{\pi}$

$b_n=\frac{-1}{n\pi}((-1)^ n-1)+\frac{1}{n\pi}(1-(-1)^n)$

$b_n=\frac{1}{n\pi}(1-(-1)^ n)+\frac{1}{n\pi}(1-(-1)^n)$