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[SOLVED] Poisson's kernel

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
P(r,\theta) = \frac{1}{2\pi}\sum_{n = -\infty}^{\infty}r^{|n|}e^{in\theta} \overbrace{=}^{\mbox{?}} \frac{1}{\pi}\left[\frac{1}{2} + \sum_{n=1}^{\infty}r^n\cos n\theta\right]
$$

Is this true?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
$$
P(r,\theta) = \frac{1}{2\pi}\sum_{n = -\infty}^{\infty}r^{|n|}e^{in\theta} \overbrace{=}^{\mbox{?}} \frac{1}{\pi}\left[\frac{1}{2} + \sum_{n=1}^{\infty}r^n\cos n\theta\right]
$$

Is this true?
Yes it is. In the sum $\frac{1}{2\pi}\sum_{n = -\infty}^{\infty}r^{|n|}e^{in\theta}$, the "middle" term, with index $n=0$, gives $\frac1{2\pi}$. For the remaining terms, pair off the terms with indices $n$ and $-n$ and use the fact that $\frac12\bigl(e^{in\theta} + e^{-in\theta}\bigr) = \cos n\theta.$
 

chisigma

Well-known member
Feb 13, 2012
1,704
If that is of some interest, if $|r|<1$ then...


$\displaystyle \sum_{n=1}^{\infty} r^{n}\ \cos n \theta = \frac{1}{2}\ \sum_{n=1}^{\infty} r^{n}\ e^{i n \theta} + \frac{1}{2}\ \sum_{n=1}^{\infty} r^{n}\ e^{- i n \theta}=$

$\displaystyle = \frac{r}{2}\ (\frac{e^{i \theta}}{1-r e^{i \theta}} + \frac{e^{-i \theta}}{1-r e^{-i \theta}}) = \frac{r\ (\cos \theta-r)}{1+r^{2}-r \cos \theta}$ (1)

... so that...

$\displaystyle P(r, \theta)= \frac{1}{\pi}\ \{\frac{1}{2} + \frac{r\ (\cos \theta-r)}{1+r^{2}-r \cos \theta}\}$ (2)

Kind regards

$\chi$ $\sigma$