# Convolution fourier series question

1. Dec 28, 2009

### nhrock3

P_r is defined as:
$$P_r(x)=\frac{1-r^2}{1-2r\cos x +r^2}$$
and
$$P_r(x)=\frac{1-r^2}{1-2r\cos x +r^2}=\sum_{n=-\infty}^{\infty}r{|n|}e^{inx}$$
and
$$f(x)=\sum_{-\infty}^{\infty}c_ne^{inx}$$
which is continues

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

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

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

??