Chapter 1 - Further Problems for Chapter 1 - Page 20: 1.11

$a) $ $$ \rho(\theta) = \left\{ \begin{array}{ll} \frac{1}{\pi} & 0 \le \theta \le\pi \\ 0 & \text{otherwise} \\ \end{array} \right.$$ $b) $ $$\langle \theta \rangle = \frac{\pi}{2}$$ $$\langle \theta^2 \rangle = \frac{\pi^2}{3}$$ $$\sigma = \frac{\pi}{2\sqrt{3}}$$ $c) $ $$\langle \sin(\theta) \rangle = \frac{2}{\pi} $$ $$\langle \cos(\theta) \rangle = 0 $$ $$\langle \cos^2(\theta) \rangle = \frac{1}{2}$$

$a) $ $\because \text{It is equally likely that the pin can come to rest at any}$ $\text{angle between 0 and }\pi.$ $\therefore \text{The probability density }\rho(\theta) \text{ is independent of }\theta$ $$\int_{-\infty}^{\infty}\rho(\theta)d\theta = 1$$ $$\rho(\theta) = 0 \text{ for }\theta<0 \text{ and }\theta>\pi$$ $$\therefore \int_{0}^{\pi}\rho(\theta)d\theta = 1$$ $$\therefore \rho(\theta)\int_{0}^{\pi}d\theta = 1$$ $$\rho(\theta) = \frac{1}{\pi} \text{ for } 0\le\theta\le\pi$$ $$ \therefore \rho(\theta) = \left\{ \begin{array}{ll} \frac{1}{\pi} & 0 \le \theta \le\pi \\ 0 & \text{otherwise} \\ \end{array} \right.$$ $\text{Graph is enclosed in the image below}$ $b)$ $$\langle \theta \rangle = \int_0^{\pi} \theta \rho(\theta) d\theta$$ $$\langle \theta \rangle = \int_0^{\pi} \theta \frac{1}{\pi} d\theta$$ $$\langle \theta \rangle = \frac{1}{\pi}\left[ \frac{\theta^2}{2}\right]_0^{\pi}$$ $$\langle \theta \rangle = \frac{\pi}{2}$$ $$\langle \theta^2 \rangle = \int_0^{\pi} \theta^2 \rho(\theta) d\theta$$ $$\langle \theta^2 \rangle = \int_0^{\pi} \theta^2 \frac{1}{\pi} d\theta$$ $$\langle \theta^2 \rangle = \frac{1}{\pi}\left[ \frac{\theta^3}{3}\right]_0^{\pi}$$ $$\langle \theta^2 \rangle = \frac{\pi^2}{3}$$ $$\sigma^2 = \langle\theta^2\rangle - {\langle \theta \rangle}^2$$ $$\sigma^2 = \frac{\pi^2}{3} - {(\frac{\pi}{2})}^2$$ $$\sigma^2 = \frac{\pi^2}{3} - \frac{\pi^2}{4}$$ $$\sigma^2 = \frac{\pi^2}{12}$$ $$\sigma = \frac{\pi}{2\sqrt{3}}$$ $c)$ $$\langle \sin(\theta) \rangle = \int_0^{\pi} \sin(\theta)\rho(\theta)d\theta = \frac{1}{\pi}\left[-\cos \theta \right]_0^{\pi} = \frac{2}{\pi}$$ $$\langle \cos(\theta) \rangle = \int_0^{\pi} \cos(\theta)\rho(\theta)d\theta = \frac{1}{\pi}\left[\sin \theta \right]_0^{\pi} = 0$$ $$\langle \cos^2(\theta) \rangle = \int_0^{\pi} \cos^2(\theta)\rho(\theta)d\theta = \frac{1}{\pi}\int_0^{\pi} \left[\frac{1-\cos2\theta}{2}\right] d\theta = \frac{1}{\pi}\left[\sin \theta \right]_0^{\pi} = 0$$ $$\langle \cos^2(\theta) \rangle = \frac{1}{2}$$