\begin{align*}
a_n \amp = \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{t}{2}\cos nt\,dt\\
\amp = \left.\left(\frac{t\sin nt }{2\pi n} + \frac{\cos nt}{2\pi n^2}\right)\right|_{-\pi}^{\pi}\\
\amp = 0 \quad \text{for} \quad n=1,2,\ldots
\end{align*}
\begin{align*}
b_n \amp = \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{t}{2}\sin nt\,dt\\
\amp = \left.\left(\frac{ -t\cos nt}{2\pi n}+\frac{\sin nt}{2\pi n^2}\right)\right|_{-\pi}^{\pi}\\
\amp =\frac{-\cos n\pi}{n}\\
\amp =\frac{(-1)^{n+1}}{n} \quad \text{for} \quad n=1,2,\ldots\text{.}
\end{align*}
The coefficient \(a_0\) is computed by the calculation
\begin{equation*}
a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{t}{2}\,dt = \frac{t^2}{4\pi}\bigg|_{-\pi}^{\pi}=0\text{.}
\end{equation*}
Substituting the coefficients
\(a_j\) and
\(b_j\) in Equation
(11.1.1) produces the required solution. The graphs of
\(U(t)\) and the first three partial sums
\(S_1(t) = \sin t, \, S_2(t) = \sin
t-\frac{1}{2}\sin 2t\text{,}\) and
\(S_3(t) = \sin t-\frac{1}{2}\sin 2t + \frac{1}{3}\sin 3t\) are shown in
Figure 11.1.7.