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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0$$\sum_{n=1}^{\infty}r^n=\frac{r}{1-r}$$. Could you check your summation formula? – 2011-11-22
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0@J.M. When you posted that I noticed something was wrong with it, I wrote $cos (\theta-\phi)$ when I meant $cos n(\theta-\phi)$ – 2011-11-22
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0Consider the sum $$\Re\left(\sum_{n=1}^{\infty}(r\exp\,iu)^n\right)$$ – 2011-11-22
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0@J.M.: So $\Re\left(\sum_{n=1}^{\infty}(r\exp\,i(\theta-\phi)^n\right)=\sum_{n=1}^{\infty}r^n\cos n(\theta-\phi)$ but how to we evaluate the LHS? – 2011-11-22
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0You use the first summation formula I gave. It's still a geometric series... after which, it's all algebra and tears... – 2011-11-22
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0@J.M. I see, but how do we know $|r\exp\ i(\theta-\phi)|<1$, surely this only holds when $|r|<1$? (and I see what you mean about algebra and tears...) – 2011-11-22
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0Right. Your original cosine series doesn't make sense either if $|r| \geq 1$, and we know that the cosine is always between $-1$ and $1$... – 2011-11-22
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0@J.M.: You're right! that does make sense. Also just to let you know, I've done enough of the algebra.. with a little help from wolfram, and it all pans out rather nicely! Thanks so much! – 2011-11-22
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0Feel free to write up your results as an answer. :) – 2011-11-22
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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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0There should be a $-1$ somewhere within the $\Re(\cdot)$... ;) – 2011-11-22
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0Is that better? :) – 2011-11-22
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0Nono, see, the usual geometric series is $$\frac1{1-r}=\sum_{k=\color{red}{0}}^\infty r^k$$, but what you need to use there is $$\frac1{1-r}-1=\sum_{k=\color{red}{1}}^\infty r^k$$... yes, indices matter! – 2011-11-23
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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