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Let $f(re^{i \theta} ) = \sum_{n=0}^{\infty} a_n r^n e^{i \theta n} $ where this power series has radius of convergence $R > r > 0$. I am trying to show that $\frac{1}{2\pi} \int_0^{2 \pi} |f(r e^{ i \theta } ) |^2 d \theta = \sum_{n=1}^{\infty} |a_n|^2 r^{2n} $

Attempt:

I am trying to expand $f \overline{f} = |f|^2 $ and then integrate term by term. I know

$$ f \overline{f} = \left( \sum_{n=0}^{\infty} a_n r^n e^{i \theta n} \right) \left( \sum_{n=0}^{\infty} \overline{a_n} r^n e^{-i \theta n } \right) $$

But it seems cumbersome to expand this product. IS there a trick to expand this nicely?

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You can indeed expand the product and integrate term by term. You have:

$$ f(re^{i\theta}) \overline{f(re^{i\theta})} = \sum_{n_1, n_2} a_{n_1} \overline{a_{n_2}} r^{n_1 + n_2} e^{i \theta (n_1 - n_2)}. $$

Note that

$$ \frac{1}{2\pi} \int_{0}^{2\pi} e^{i \theta (n_1 - n_2)} d\theta = \begin{cases} \left[ \frac{e^{i \theta (n_1 - n_2)}}{i (n_1 - n_2)} \right]_{\theta = 0}^{\theta = 2\pi} & n_1 \neq n_2 \\ 1 & n_1 = n_2 \end{cases} $$

so that the integral of the non-diagonal terms vanish and so you have

$$ \frac{1}{2\pi} \int_0^{2\pi} |f(re^{i\theta})| d\theta = \sum_{n_1,n_2}^{\infty} a_{n_1} \overline{a_{n_2}} r^{n_1+n_2} \frac{1}{2\pi} \int_0^{2\pi} e^{i \theta (n_1 - n_2)} d\theta = \sum_{n_1,n_2}^{\infty} a_{n_1} \overline{a_{n_2}} r^{n_1+n_2} \delta_{n_1 n_2} = \sum_{n=0}^{\infty} |a_n|^2 r^{2n}. $$

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  • $\begingroup$ $\delta_{n_1n_2} = 1 $ if $n_1=n_2$ and $0$ otherwise ? $\endgroup$ – user139708 Nov 10 '15 at 9:39
  • $\begingroup$ Yes, I've added the reason for that. $\endgroup$ – levap Nov 10 '15 at 9:42

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