Proof of Positive Values for P(r,θ)

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SUMMARY

The discussion focuses on proving that the function \( P(r,\theta) \) is positive for all values of \( r \) and \( \theta \) within the specified ranges, specifically \( 0 \leq r < 1 \) and \( -\pi \leq \theta \leq \pi \). The formula provided is \( P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2} \). The analysis indicates that the denominator achieves its minimum at \( \theta = 0 \), where it simplifies to \( (1-r)^2 \), confirming that both the numerator and denominator are positive under the given conditions.

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Dustinsfl
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$$
P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}
$$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?
 
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dwsmith said:
$$
P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}
$$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 $\chi$ $\sigma$
 

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