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Convolution fourier series question

  1. Dec 28, 2009 #1
    P_r is defined as:
    [tex]P_r(x)=\frac{1-r^2}{1-2r\cos x +r^2}[/tex]
    and
    [tex]P_r(x)=\frac{1-r^2}{1-2r\cos x +r^2}=\sum_{n=-\infty}^{\infty}r{|n|}e^{inx}[/tex]
    and
    [tex]f(x)=\sum_{-\infty}^{\infty}c_ne^{inx}[/tex]
    which is continues

    i need to prove that:
    [tex]f_r(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}p_r(t)dt=\sum_{n=1}^{\infty}c_nr^{|n|}e^{inx}[/tex]

    the solution says to use the convolution property
    [tex]c_n(f)=c_n[/tex]
    [tex]c_n(P_r)=r^{|n|}[/tex]
    [tex]c_n(f_r)=c_n r^{|n|}[/tex]

    but i cant see how the multiplication of those coefficient gives me the
    expression i needed to prove

    ?

    i only got the right side not the left integral

    ??
     
  2. jcsd
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