# [SOLVED]Poisson's kernel

#### dwsmith

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

Is this true?

#### Opalg

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

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
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

$\chi$ $\sigma$