FS5

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Find the Fourier series for f(x)={\begin{cases}-1&-3\leq x<0\\1&0<x\leq 3\end{cases}} on [-3,3]\,

a_{k}={\frac  {1}{3}}\int _{{-3}}^{3}f(x)\cos({\frac  {k\pi x}{3}})\,dx={\frac  {1}{3}}\int _{{-3}}^{0}-\cos({\frac  {k\pi x}{3}})\,dx+{\frac  {1}{3}}\int _{0}^{3}\cos({\frac  {k\pi x}{3}})\,dx\,

={\frac  {1}{k\pi }}\left[\sin({\frac  {k\pi x}{3}})\right]_{{-3}}^{0}-{\frac  {1}{k\pi }}\left[\sin({\frac  {k\pi x}{3}})\right]_{0}^{3}=0\,

a_{0}={\frac  {1}{3}}\int _{{-3}}^{0}-\,dx+{\frac  {1}{3}}\int _{0}^{3}\,dx=-1+1=0\,

b_{k}={\frac  {1}{3}}\int _{{-3}}^{3}f(x)\sin({\frac  {k\pi x}{3}})\,dx={\frac  {1}{3}}\int _{{-3}}^{0}-\sin({\frac  {k\pi x}{3}})\,dx+{\frac  {1}{3}}\int _{0}^{3}\sin({\frac  {k\pi x}{3}})\,dx\,

={\frac  {1}{k\pi }}\left[\cos({\frac  {k\pi x}{3}})\right]_{{-3}}^{0}-{\frac  {1}{k\pi }}\left[\cos({\frac  {k\pi x}{3}})\right]_{0}^{3}\,

={\frac  {1}{k\pi }}-{\frac  {(-1)^{k}}{k\pi }}-{\frac  {(-1)^{k}}{k\pi }}+{\frac  {1}{k\pi }}={\frac  {2-2(-1)^{k}}{k\pi }}\,\,\,k=1,2,3,...\,

={\frac  {4}{(2k+1)\pi }},\,\,\,k=0,1,2,...\,

So the Fourier series is:

f(x)=\sum _{{k=0}}^{\infty }{\frac  {4}{(2k+1)\pi }}\sin \left({\frac  {(2k+1)\pi x}{3}}\right)\,


Fourier Series

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