<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand\ds{\displaystyle}
\begin{document}

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QUESTION

\begin{description}

\item[(b)]
Derive the Fourier series for the periodic function $f(t)$ which
is defined by

$f(t)=t$ for $-\pi&lt;t\leq\pi$,\hspace{1cm} and $f(t+2\pi)=f(t)$ for
all $t$.

\end{description}

\bigskip

ANSWER

\begin{description}

\item[(b)]
Period=$2\pi$. The function is odd (from the graph) and this can
be used to simplify the calculation of coefficients.

\begin{center}
\epsfig{file=17x-2000-5.eps, width=40mm}
\end{center}


$a_0=\frac{1}{\pi}\int_{-\pi}^\pi f(t)\,dt=0,$ since
$\ds\int_{-a}^af(t)\,dt=0$ when $f(t)$ is odd.

$a_n=\frac{1}{\pi}\int_{-\pi}^\pi f(t)\cos(nt)\,dt=0,$ (integrand
is again a odd function).

\begin{eqnarray*}
b_n&amp;=&amp;\frac{1}{\pi}\int_{-\pi}^\pi f(t)\sin(nt)\,dt\\
&amp;=&amp;\frac{2}{\pi}\int_0^\pi f(t)\sin(nt)\,dt\ \ {\rm (even\
integrand\ as\ product\ of\ odd\ functions)}\\
&amp;=&amp;\frac{2}{\pi}\int_0^\pi t\sin(nt)\,dt\\
&amp;=&amp;\frac{2}{\pi}\left\{\left[t\left(-\frac{\cos(nt)}{n}\right)\right]_0^\pi
-\int_0^\pi-\frac{\cos(nt)}{n}.1\,dt\right\}\\
&amp;=&amp;\frac{2}{\pi}\left\{-\frac{\pi}{n}\cos(n\pi)+\frac{1}{n}
\left[\frac{\sin(nt)}{n}\right]_0^\pi\right\}\\
&amp;=&amp;-\frac{2}{n}(-1)^n+\frac{2}{\pi n^2}(0-0)=\frac{2}{n}(-1)^{n+1}
\end{eqnarray*}

Therefore $\ds
f(t)\sim\sum_{n=1}^\infty\frac{2(-1)^{n+1}}{n}\sin(nt)$

\end{description}




\end{document}
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