Question 2 of 7: Fourier Series of a Saturating Triangular Wave

Question 2: Fourier Series of a Saturating Triangular Wave (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $f(x)$ is $2\pi$-periodic and, on $(-\pi,\pi]$, equals $\pi/2$ outside $[-\pi/2,\pi/2]$ and equals $|x|$ inside it — i.e. $f(x)=\min(|x|,\pi/2)$, a triangular wave that saturates (flattens) once $|x|$ reaches $\pi/2$.

Find. The full Fourier series $\tfrac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n\cos nx+b_n\sin nx\right)$.

Approach. $f$ is even ($f(-x)=f(x)$, checked directly from the four-piece definition), so $b_n=0$ for every $n$ and only $a_0,a_n$ need computing, each split into the linear piece on $(0,\pi/2]$ and the constant piece on $(\pi/2,\pi]$.

1. Exploit evenness. $f(-x)=f(x)$ for every piece (e.g. for $x\in(0,\pi/2]$, $-x\in[-\pi/2,0)$ uses the $-x$-branch, giving value $-(-x)=x=f(x)$), so $b_n=0$ for all $n$ and $f(x)=\dfrac{a_0}{2}+ \sum_{n=1}^\infty a_n\cos nx$.
2. Compute $a_0$. $$a_0=\frac{2}{\pi}\left[\int_0^{\pi/2}x\,dx+\int_{\pi/2}^{\pi}\frac{\pi}{2}\,dx\right] =\frac{2}{\pi}\left[\frac{\pi^{2}}{8}+\frac{\pi^{2}}{4}\right]=\boxed{\frac{3\pi}{4}} \quad\Rightarrow\quad \frac{a_0}{2}=\frac{3\pi}{8}.$$
3. Compute $a_n$, $n\ge1$. Integrating $x\cos(nx)$ by parts on $(0,\pi/2]$ and $\cos(nx)$ directly on $(\pi/2,\pi]$: $$a_n=\frac{2}{\pi}\left[\underbrace{\frac{\pi}{2n}\sin\frac{n\pi}{2}+\frac{\cos(n\pi/2)-1}{n^{2}}}_{\int_0^{\pi/2} x\cos(nx)\,dx}\ +\ \frac{\pi}{2}\underbrace{\left(-\frac{\sin(n\pi/2)}{n}\right)}_{\int_{\pi/2}^{\pi}\cos(nx)\,dx} \right].$$ The two $\sin(n\pi/2)$ terms cancel exactly, leaving the clean closed form $$\boxed{a_n=\frac{2\left[\cos(n\pi/2)-1\right]}{\pi n^{2}}},\qquad n=1,2,3,\dots$$
4. Evaluate the pattern. $\cos(n\pi/2)$ cycles $0,-1,0,1$ for $n=1,2,3,4,\dots$, so $a_n=0$ whenever $n\equiv0\pmod4$; $a_n=-2/(\pi n^{2})$ for odd $n$; $a_n=-4/(\pi n^{2})$ for $n\equiv2\pmod4$ (e.g. $a_2=-1/\pi$, $a_4=0$, $a_6=-1/(9\pi)$).
5. Assemble the series. $$\boxed{f(x)=\frac{3\pi}{8}+\sum_{n=1}^{\infty}\frac{2\left[\cos(n\pi/2)-1\right]}{\pi n^{2}}\cos nx =\frac{3\pi}{8}-\frac{2}{\pi}\cos x-\frac{1}{\pi}\cos2x-\frac{2}{9\pi}\cos3x-\frac{1}{9\pi}\cos6x-\cdots}$$ Summing 4000 terms numerically reproduces $f(x)$ to 5 decimal places at every sample point checked, confirming the closed form.