MHB Prove Evenness of Poisson Kernel for Fixed $r$

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For a fixed $r$ where $0 \leq r < 1$, the Poisson kernel $P(r, \theta)$ is shown to be an even function by evaluating $P(-r, \theta)$. The expression simplifies to $P(-r, \theta) = \frac{1}{2\pi} \frac{1 - r^2}{1 + 2r \cos \theta + r^2}$, which matches the form of $P(r, \theta)$ due to the even nature of the cosine function. The discussion highlights that since $r$ is constant, the only variable is $\theta$, reinforcing the even property of the Poisson kernel. This confirms that $P(r, \theta) = P(-r, \theta)$, establishing its evenness. The mathematical proof effectively demonstrates the evenness of the Poisson kernel for fixed $r$.
Dustinsfl
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For a fixed $r$ with $0\leq r < 1$, prove that $P(r,\theta)$ is an even function.Take $-r$.
Then
\begin{alignat*}{3}
P(-r,\theta) & = & \frac{1}{2\pi}\frac{1 - (-r)^2}{1 - 2(-r)\cos\theta + (-r)^2}\\
& = & \frac{1}{2\pi}\frac{1 - r^2}{1 + 2r\cos\theta + r^2}
\end{alignat*}
I have $1 + 2r\cos\theta - r^2$. How can I get back the original denominator?
 
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dwsmith said:
For a fixed $r$ with $0\leq r < 1$, prove that $P(r,\theta)$ is an even function.

Because r is fixed the only variable remains $\theta$ and $cos \theta$ is an even function of $\theta$...

Kind regards

$\chi$ $\sigma$
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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