\documentclass[a4paper,12pt]{article}
\newcommand\ds{\displaystyle}
\begin{document}

\parindent=0pt

QUESTION

\begin{description}

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

$f(t)=\left\{\begin{array}{cl}1&-\pi\leq t<0,\\0&0\leq
t<\pi,\end{array}\right.$ and $f(t+2\pi)=f(t)$ for all $t$.

\end{description}


\bigskip

ANSWER

\begin{description}

\item[(b)]
$ $

\setlength{\unitlength}{0.5in}

\begin{picture}(6,3)

\put(0,1){\line(1,0){6}}

\put(3,0){\vector(0,1){3}}

\put(2,2){\line(1,0){1}}

\put(4,2){\line(1,0){1}}

\put(1,1.02){\line(1,0){1}}

\put(3,1.02){\line(1,0){1}}

\put(2.8,2.1){1}

\put(.8,0.7){$-2\pi$}

\put(1.8,0.7){$-\pi$}

\put(3.9,0.7){$\pi$}

\put(4.9,0.7){$2\pi$}

\put(5.5,2,5){period $=2\pi=T$}

\end{picture}

Fourier series
$\ds\frac{1}{2}a_0+\sum_{n=1}^\infty(a_n\cos(nt)+b_n\sin(nt))$.

\begin{eqnarray*}
a_0=\frac{2}{2\pi}\int_{-\pi}^\pi
f(t)\,dt&=&\frac{1}{\pi}\int_{-\pi}^01\,dt+\frac{1}{\pi}\int_0^\pi0\,dt\\
&=&\frac{1}{\pi}[t]_{-\pi}^0=\frac{1}{\pi}(0-(-\pi))=1
\end{eqnarray*}

\begin{eqnarray*}
a_n=\frac{2}{2\pi}\int_{-\pi}^\pi
f(t)\cos(nt)\,dt&=&\frac{1}{\pi}\int_{-\pi}^0\cos(nt)\,dt+\frac{1}{\pi}\int_0^\pi0\,dt\\
&=&\frac{1}{\pi}\left[\frac{\sin(nt)}{n}\right]_{-\pi}^0\\
&=&\frac{1}{\pi n}(\sin(0)-\sin(-n\pi))=0
\end{eqnarray*}

\begin{eqnarray*}
b_n=\frac{2}{2\pi}\int_{-\pi}^\pi
f(t)\sin(nt)\,dt&=&\frac{1}{\pi}\int_{-\pi}^0\sin(nt)\,dt+\frac{1}{\pi}\int_0^\pi0\,dt\\
&=&\frac{1}{\pi}\left[-\frac{\cos(nt)}{n}\right]_{-\pi}^0\\
&=&-\frac{1}{\pi n}[\cos(0)-\cos(-n\pi)]\\ &=&-\frac{1}{\pi
n}[1-\cos(n\pi)]\\ &=&-\frac{1}{\pi n}[1-(-1)^n]\\
&=&\left\{\begin{array}{cl}0&n\textrm{ even}\\-\frac{2}{\pi n}&n
\textrm{ odd}\end{array}\right.
\end{eqnarray*}

Therefore the Fourier series is
$$\frac{1}{2}-\frac{2}{\pi}\sum_{n\ {\rm
odd}}\frac{1}{n}\sin(nt)=\frac{1}{2}-\frac{2}{\pi}
\sum_{m=0}^\infty\frac{1}{(2m+1)}\sin((2m+1)t)$$


\end{description}




\end{document}
