Question 6.8.4: Find the nth-order Fourier approximation to the function f (...

Compute

\frac{a_{0}}{2}=\frac{1}{2} \cdot \frac{1}{\pi} \int_{0}^{2 \pi} t d t=\frac{1}{2 \pi}\left[\left.\frac{1}{2} t^{2}\right|_{0} ^{2 \pi}\right]=\pi.

and for k > 0, using integration by parts,

a_{k}=\frac{1}{\pi} \int_{0}^{2 \pi} t \cos k t d t=\frac{1}{\pi}\left[\frac{1}{k^{2}} \cos k t+\frac{t}{k} \sin k t\right]_{0}^{2 \pi}=0.

b_{k}=\frac{1}{\pi} \int_{0}^{2 \pi} t \sin k t d t=\frac{1}{\pi}\left[\frac{1}{k^{2}} \sin k t-\frac{t}{k} \cos k t\right]_{0}^{2 \pi}=-\frac{2}{k}.

\pi-2 \sin t-\sin 2 t-\frac{2}{3} \sin 3 t-\cdots-\frac{2}{n} \sin n t.

Figure 3 shows the third- and fourth-order Fourier approximations of f .