FS8

Find the Fourier series for $f(x) = x\,$ on $[0,1]\,$ .

A general formula for the Fourier series of a function on an interval $[c,c+T]\,$ is:

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

$a_n = \frac{2}{T}\int_c^{c+T} f(x) \cos\left(\frac{2n\pi x}{T}\right)\,dx\,$

$b_n = \frac{2}{T}\int_c^{c+T} f(x) \sin\left(\frac{2n\pi x}{T}\right)\,dx\,$

In the current problem, $c = 0\,$ and $T=1\,$ .

$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos 2n\pi x + b_n \sin 2n\pi x\,$

The function $f(x)=x\,$ is odd, so the cosine coefficients will all equal zero. Nevertheless, $a_0\,$ should still be calculated separately.

$a_0 = 2 \int_0^1 x\,dx = 1\,$

$b_n = 2\int_0^1 x\sin(2n\pi x)\,dx\,$

$= 2\left(\left[\frac{-x \cos 2n\pi x}{2n\pi}\right]_0^1 + \int_0^1 \frac{\cos 2n\pi x}{2n\pi}\,dx\right)\,$

$= \frac{-1}{n\pi} + \left[\frac{\sin 2n\pi x}{(2n\pi)^2}\right]_0^1 = \frac{-1}{n\pi}\,$

So the Fourier series for $f(x)\,$ is

$f(x) = \frac{1}{2} - \sum_{n=1}^\infty \frac{\sin 2n\pi x}{n\pi}\,$