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Given the power series: $f(z) = \sum_{n=0}^{\infty}a_{n} z^{n}$, with radius of convergence R, prove that for r < R the following holds:

$$\frac{1}{2\pi } \int_{0}^{2\pi} | f(r e^{i\theta }) |^{2} d \theta = \sum_{n=0}^{\infty }|a_{n}|^{2} r^{2n}$$

Federica Maggioni
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GAJO
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  • Your question is phrased as an isolated problem, without any further information or context. This does not match MSE quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. Making these improvements will attract more appropriate answers and make the question more valuable for future MSE visitors. – Antonio Vargas Apr 26 '13 at 18:57
  • @AntonioVargas, is this comment an outcome of the recent discussion on meta? Is this nw the official policy for questions showing little effort? – Richard Nash Apr 26 '13 at 19:08
  • @RichardNash I am not an official, and I don't believe it's official policy, but it is indeed my policy. – Antonio Vargas Apr 26 '13 at 19:12
  • @RichardNash, what Antonio wrote has been the official unwritten policy of the site, or more accurately: of most long terms users and officials. Please do note Antonio wrote "it may attract downvotes or be closed", not that if will be for sure. – DonAntonio Apr 26 '13 at 19:15
  • @DonAntonio I think what Antonio wrote closely resembles an answer on meta. :) – Richard Nash Apr 26 '13 at 19:17
  • @RichardNash, right, I copied the comment template from there. It's much more eloquent than I could hope to be :) – Antonio Vargas Apr 26 '13 at 19:17
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    Most probably, @RichardNash . In fact, I find that a very compelling, decent and useful piece of advice for new posters here. – DonAntonio Apr 26 '13 at 19:18
  • Here is same problem. Do not forget to up vote this answer if you benefit of it or any other answer. – Mhenni Benghorbal Apr 26 '13 at 19:36

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