MHB Proof of Positive Values for P(r,θ)

  • Thread starter Thread starter Dustinsfl
  • Start date Start date
  • Tags Tags
    Positive Proof
Dustinsfl
Messages
2,217
Reaction score
5
$$
P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}
$$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?
 
Physics news on Phys.org
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 $\chi$ $\sigma$
 
Back
Top