Triangle Wave Fourier Series

Relevant wiki: Fourier Series

The period of the given function is 2. The following plot should help elucidate the name:

One can see that the function is odd; thus all cosine terms vanish. One is left computing the sine coefficients:

$b_k = \frac{2}{T} \int_0^T f(x) \sin (\frac{2\pi k x}{T}) dx.$

It helps to break up the above function into two functions and two integrals, plugging in $T=2$ :

$b_k = \int_{-.5}^{.5} 2x \sin (\pi k x) dx + \int_{.5}^{1.5} (2-2x)\sin (\pi k x) dx .$

Evaluating the first and second using integration by parts:

$\begin{aligned} \int_{-.5}^{.5} 2x \sin (\pi k x) dx &= \bigl.-\frac{2x}{\pi k} \cos (\pi k x) \bigr|_{-1/2}^{1/2} + \frac{2}{\pi k} \int_{-.5}^{.5} \cos (\pi k x) dx \\ &=\bigl.-\frac{2x}{\pi k} \cos (\pi k x) \big|_{-1/2}^{1/2} + \bigl.\frac{2}{\pi^2 k^2} \sin (\pi k x)\bigr|_{-1/2}^{1/2} \end{aligned}$

$\begin{aligned} \int_{.5}^{1.5} (2-2x)\sin (\pi k x) dx &= \bigl.\frac{-(2-2x)}{\pi k} \cos (\pi k x) \bigr|_{1/2}^{3/2} - \frac{2}{\pi k} \int_{.5}^{1.5} \cos (\pi k x) dx \\ &=\bigl.\frac{-(2-2x)}{\pi k} \cos (\pi k x) \bigr|_{1/2}^{3/2} - \frac{2}{\pi^2 k^2} \bigl. \sin (\pi k x) \bigr|_{1/2}^{3/2} \end{aligned}$

Combining all expressions and checking when the integer multiples of $\pi/2$ give $-1$ , $1$ , or $0$ gives the result (only the second term from each integral ends up contributing):

$b_k = \frac{8}{k^2\pi ^2} (-1)^{(k-1)/2}, \qquad k \text{ odd},$

and $b_k = 0$ when $k$ even, thus yielding the Fourier series as stated.