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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: $$\frac12+\sum_{n=1}^{\infty}r^n\cos n(\theta-\phi)=\frac{1-r^2}{2(1-2r\cos(\theta-\phi)+r^2)}.$$

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    $$\sum_{n=1}^{\infty}r^n=\frac{r}{1-r}$$. Could you check your summation formula?2011-11-22
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    @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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    Consider the sum $$\Re\left(\sum_{n=1}^{\infty}(r\exp\,iu)^n\right)$$2011-11-22
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    @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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    You 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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    @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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    Right. 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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    @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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    Feel free to write up your results as an answer. :)2011-11-22
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    Have 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

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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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    There should be a $-1$ somewhere within the $\Re(\cdot)$... ;)2011-11-22
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    Is that better? :)2011-11-22
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    Nono, 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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    @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