Fourier Sine Coefficients

Fourier Series

Any periodic function can be expressed as a sum of sines and cosines — the Fourier series:

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[a_n \cos(nx) + b_n \sin(nx)\right]$

for functions on $[-\pi, \pi]$. The coefficients are:

$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos(nx)\,dx \qquad b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)\,dx$

Sine Coefficients of $f(x) = x$

Since $f(x) = x$ is an odd function, all cosine coefficients $a_n = 0$. The sine coefficients are:

$b_n = \frac{2(-1)^{n+1}}{n}$

So: $x = 2\sin(x) - \sin(2x) + \frac{2}{3}\sin(3x) - \cdots \quad \text{on } (-\pi, \pi)$

Numerical Computation

Approximate the integral using the left Riemann sum with $N$ steps on $[-\pi, \pi]$:

$b_n \approx \frac{1}{\pi} \sum_{i=0}^{N-1} f(x_i) \sin(n x_i) \cdot h \quad \text{where } h = \frac{2\pi}{N}$

Your Task

Implement double fourier_bn(f, n, steps) that numerically computes the $n$-th Fourier sine coefficient of $f$ on $[-\pi, \pi]$.