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Find the Fourier series for f(x)=x\, on [0,1]\,.

A general formula for the Fourier series of a function on an interval [c,c+T]\, is:

f(x)={\frac  {a_{0}}{2}}+\sum _{{n=1}}^{\infty }a_{n}\cos \left({\frac  {2n\pi x}{T}}\right)+b_{n}\sin \left({\frac  {2n\pi x}{T}}\right)\,

a_{n}={\frac  {2}{T}}\int _{c}^{{c+T}}f(x)\cos \left({\frac  {2n\pi x}{T}}\right)\,dx\,

b_{n}={\frac  {2}{T}}\int _{c}^{{c+T}}f(x)\sin \left({\frac  {2n\pi x}{T}}\right)\,dx\,

In the current problem, c=0\, and T=1\,.

f(x)={\frac  {a_{0}}{2}}+\sum _{{n=1}}^{\infty }a_{n}\cos 2n\pi x+b_{n}\sin 2n\pi x\,

The function f(x)=x\, is odd, so the cosine coefficients will all equal zero. Nevertheless, a_{0}\, should still be calculated separately.

a_{0}=2\int _{0}^{1}x\,dx=1\,

b_{n}=2\int _{0}^{1}x\sin(2n\pi x)\,dx\,

=2\left(\left[{\frac  {-x\cos 2n\pi x}{2n\pi }}\right]_{0}^{1}+\int _{0}^{1}{\frac  {\cos 2n\pi x}{2n\pi }}\,dx\right)\,

={\frac  {-1}{n\pi }}+\left[{\frac  {\sin 2n\pi x}{(2n\pi )^{2}}}\right]_{0}^{1}={\frac  {-1}{n\pi }}\,

So the Fourier series for f(x)\, is

f(x)={\frac  {1}{2}}-\sum _{{n=1}}^{\infty }{\frac  {\sin 2n\pi x}{n\pi }}\,

Fourier Series

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