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Math Help - Fourier Serie

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    Fourier Serie

    How do I determine the fourier series of interval function: f(x)= 8+2(x/9) , 0<x<4
    , -8+2(x/9) , -4<x<0
    and in addition we know: f(x+8)=f(x)

    Please help me.
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    Quote Originally Posted by DanielM99 View Post
    How do I determine the fourier series of interval function: f(x)= 8+2(x/9) , 0<x<4
    , -8+2(x/9) , -4<x<0
    and in addition we know: f(x+8)=f(x)

    Please help me.
    f(x)=\sum_{n=0}^{\infty}b_n\sin\left( \frac{n\pi}{p}x\right)

    where b_n=\frac{2}{p}\int_{0}^{p}f(x) \sin\left( \frac{n\pi}{2}x\right)dx

    Our integral is

    b_n=\frac{1}{2}\int_{0}^{4}\left( 8+\frac{2x}{9}\right) \sin\left( \frac{n\pi}{p}x\right)dx =

    =-\frac{4}{n\pi}\left(8+\frac{2x}{9} \right)\cos\left( \frac{n\pi}{2}x\right)+\frac{2}{9}\cdot \frac{4^2}{n^2\pi^2}\sin\left( \frac{n\pi}{2}x\right) \big]_{0}^{4}

    =-\frac{4}{n\pi}\left( \frac{80}{9}\right)\cos(2\pi n)+\frac{32}{n\pi}= -\frac{32}{9 \pi n}

    so finally we get

    f(x)=\sum_{n=0}^{\infty} = -\frac{32}{9 \pi n} \sin\left( n \frac{\pi }{4}x\right)
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