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[SOLVED] Poisson kernel

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}
$$


Prove that $P(r,\theta) > 0$ for all $r$ and $\theta$ where $0\leq r < 1$ and $-\pi\leq\theta\leq\pi$.

How can I start this?
 

chisigma

Well-known member
Feb 13, 2012
1,704
$$
P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}
$$


Prove that $P(r,\theta) > 0$ for all $r$ and $\theta$ where $0\leq r < 1$ and $-\pi\leq\theta\leq\pi$.

How can I start this?
The denominator has a minimum for $\theta=0$ where $\cos \theta=1$ and here the denoninator is $(1-r)^{2}$, so that if $0 \le r < 1$ numerator and denominator are both > 0...


Kind regards


$\chi$ $\sigma$