Question #91733
Obtain the Fourier cosine series for the following function:


F(x)= 1 for (0 less than equal to x< 1)

0 for (1 less than equal to x < 4 )
1
Expert's answer
2019-07-19T11:12:23-0400

Formula for Fourier cosine series is

F(x)a0+n=1ancos(nπLx)F(x) \approx a_{0} +\sum_{n=1}^{ \infty}a_{n}cos(\frac{n*\pi}{L}x)

where

a0=12LLLf(x)dxa_{0} = \frac{1}{2L} \int_{-L}^{L}f(x) dx

and

an=1LLLf(x)cos(nπxL)dxa_{n} = \frac{1}{L} \int_{-L}^{L}f(x)cos(\frac{n\pi x}{L}) dx


if function is even, then

LLf(x)dx=20Lf(x)dx\int_{-L}^{L}f(x) dx = 2\int_{0}^{L}f(x) dx

a0=14011dx=14a_{0} = \frac{1}{4} \int_{0}^{1}1 dx = \frac{1}{4}


an=1201cos(nπx4)dx=2sin(πn4)πna_{n} = \frac{1}{2}\int_{0}^{1}cos(\frac{n \pi x}{4}) dx=\frac{2sin(\frac{\pi n}{4})}{\pi n}

so Fourier cosine series is


F(x)=14+n=12sin(πn4)πnF(x) = \frac{1}{4} +\sum_{n=1}^{ \infty}\frac{2sin(\frac{\pi n}{4})}{\pi n}



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