91Ó°ÊÓ

The coefficients \({a_0}\) can be calculated as follows:

\(\begin{array}{c}\frac{{{a_0}}}{2} = \frac{1}{{2\pi }}\int_0^{2\pi } {f\left( t \right){\rm{d}}t} \\ = \frac{1}{{2\pi }}\left[ {\pi - \left( {2\pi - \pi } \right)} \right]\\ = 0\end{array}\)

The coefficients \({a_k}\) can be calculated as follows:

\(\begin{array}{c}{a_k} = \frac{1}{\pi }\int_0^{2\pi } {f\left( t \right)\cos ktdt} \\ = \frac{1}{\pi }\int_0^\pi {\cos kt{\rm{d}}t} - \frac{1}{\pi }\int_\pi ^{2\pi } {\cos kt{\rm{d}}t} \\ = \frac{1}{\pi }\left\{ {\left[ {\frac{{\sin kt}}{k}} \right]_0^\pi - \left[ {\frac{{\sin kt}}{k}} \right]_\pi ^{2\pi }} \right\}\\ = 0\end{array}\)

The coefficients \({b_k}\) can be calculated as follows:

\(\begin{array}{c}{a_k} = \frac{1}{\pi }\int_0^{2\pi } {f\left( t \right)\sin ktdt} \\ = \frac{1}{\pi }\int_0^\pi {\sin kt{\rm{d}}t} - \frac{1}{\pi }\int_\pi ^{2\pi } {\sin kt{\rm{d}}t} \\ = \frac{1}{\pi }\left\{ {\left[ { - \frac{{\cos kt}}{k}} \right]_0^\pi + \left[ {\frac{{\cos kt}}{k}} \right]_\pi ^{2\pi }} \right\}\\ = \frac{2}{{\pi k}} - \frac{2}{{\pi k}}\cos k\pi \\ = \left\{ {\begin{array}{*{20}{c}}{\frac{4}{{k\pi }}}&{{\rm{if}}\;{\rm{k}}\,{\rm{is odd}}}\\0&{{\rm{if}}\;{\rm{k}}\,{\rm{is even}}}\end{array}} \right.\end{array}\)

The Fourier function can be written as follows:

\({f_4}\left( t \right) = \frac{4}{\pi }\sin t + \frac{4}{{3\pi }}\sin 3t\)

And,

\({f_5}\left( t \right) = \frac{4}{\pi }\sin t + \frac{4}{{3\pi }}\sin 3t + \frac{4}{{5\pi }}\sin 5t\)

The figure below represents the graph of the curve \({f_4}\) in the interval \(\left[ {0,2\pi } \right]\).

The figure below represents the graph of the curve \({f_5}\) in the interval \(\left[ {0,2\pi } \right]\).

The figure below represents the graph of the curve \({f_5}\) in the interval \(\left[ { - 2\pi ,2\pi } \right]\).