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QUESTION

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

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

\end{description}

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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.

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$a_0=\frac{1}{\pi}\int_{-\pi}^\pi f(t)\,dt=0,$ 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&=&\frac{1}{\pi}\int_{-\pi}^\pi f(t)\sin(nt)\,dt\\
&=&\frac{2}{\pi}\int_0^\pi f(t)\sin(nt)\,dt\ \ {\rm (even\
integrand\ as\ product\ of\ odd\ functions)}\\
&=&\frac{2}{\pi}\int_0^\pi t\sin(nt)\,dt\\
&=&\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\}\\
&=&\frac{2}{\pi}\left\{-\frac{\pi}{n}\cos(n\pi)+\frac{1}{n}
\left[\frac{\sin(nt)}{n}\right]_0^\pi\right\}\\
&=&-\frac{2}{n}(-1)^n+\frac{2}{\pi n^2}(0-0)=\frac{2}{n}(-1)^{n+1}
\end{eqnarray*}

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

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