If $u(r,\theta)=\frac1{\pi}\int_0^{2\pi}\Bigg[ \frac12+\sum_{n=1}^{\infty}r^n\cos n(\theta-\phi) \Bigg]f(\phi)d\phi,$ can anyone help show me why this implies $u(r,\theta)=\frac{(1-r^2)}{2\pi}\int_0^{2\pi} \frac{f(\phi)}{1-2r\cos(\theta-\phi)+r^2}d\phi \,?$ I had previously derived the first equation, but am rather stuck on this step. I've been trying to show equivalently that: $\frac12+\sum_{n=1}^{\infty}r^n\cos n(\theta-\phi)=\frac{1-r^2}{2(1-2r\cos(\theta-\phi)+r^2)}.$
Obtaining Poisson's formula from a integral of summation
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real-analysis
fourier-series
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0Have done! So now if no one else answers it won't go unanswered. Thanks so much for your help, I would not have seen this on my own, I will remember this method. – 2011-11-22
1 Answers
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As we know $|r|<1$
$\frac12+\sum_{n=1}^{\infty}r^n\cos n(\theta-\phi)=\frac12+\Re\left(\sum_{n=1}^{\infty}(r\exp\,i(\theta-\phi))^n\right)=\frac12+\Re \Bigg( \frac{1}{1-r\exp\,i(\theta-\phi)}-1\Bigg) $ $=\frac{1-r^2}{2(1-2r\cos(\theta-\phi)+r^2)}$
By the geometric series summation and after a lot of algebra to find the real part. This result is equivalent to the desired result
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0@J.M.: Ah yes, correct now, that was rather stupid, but in my defense when I came back to correct it I had a few glasses of wine ;) – 2011-11-23