Derive${\mathit{P}}_{\mathbf{3}}\left(x\right)$from the Rodrigues formula, and check that ${\mathit{P}}_{\mathbf{3}}\left(cos\theta \right)$satisfies the angular equation (3.60) for $\mathit{I}\mathbf{=}\mathbf{3}$. Check that ${\mathit{P}}_{\mathbf{3}}$and ${\mathit{P}}_{\mathbf{1}}$are orthogonal by explicit integration.

Write the formula of Rodrigues’s.

${\mathit{P}}_{\mathbf{I}}\left(x\right)\mathbf{=}\frac{\mathbf{1}}{{\mathbf{2}}^{\mathbf{I}}\mathbf{}\mathbf{I}\mathbf{!}}{\left(\frac{d}{dx}\right)}^{\mathbf{I}}{\left(x^2-1\right)}^{\mathbf{I}}$ …… (1)

Calculate the $P_3\left(x\right)$using equation (1).

Therefore,

$$\begin{gathered} P_3\left(x\right)=\frac{1}{2^{3}3!}{\left(\frac{d}{dx}\right)}^3{\left(x^2-1\right)}^3 \\ =\frac{1}{48}\frac{d^3}{d{x}^3}{\left(x^2-1\right)}^3 \\ =\frac{1}{8}\frac{d^2}{d{x}^2}x{\left(x^2-1\right)}^2 \end{gathered}$$

Solve as further,

$$\begin{gathered} P_3\left(x\right)=\frac{1}{8}\frac{d}{dx}\left[{\left(x^2-1\right)}^2+2x\left(x^2-1\right)2x\right] \\ =\frac{1}{8}\frac{d}{dx}\left[{\left(x^2-1\right)}^2+4x^2\left(x^2-1\right)\right] \\ =\frac{1}{8}\frac{d}{dx}\left[{\left(x^2-1\right)}^2+\left(4x^4-4x^2\right)\right] \\ =\frac{1}{8}\left(2\left(x^2-1\right)\left(2x\right)+4\left(4x^3\right)-8x\right) \end{gathered}$$

Solve as further,

$$\begin{gathered} P_3\left(x\right)=\frac{1}{8}\left(20x^3-4x-8x\right) \\ =\frac{5}{2}{x}^3-\frac{3}{2}x \end{gathered}$$

Therefore,

$P_3\left(\cos \theta \right)=\frac{5}{2}{\cos }^3\theta -\frac{3}{2}\cos \theta$

$\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{dP}{d\theta }\right)=-I\left(I+1\right)P$

Here, $I=3$

$$\begin{gathered} \frac{d{P}_3}{d\theta }=\frac{d}{d\theta }\left(\frac{5}{2}{\cos }^3\theta -\frac{3}{2}\cos \theta \right) \\ =\left[\left(\frac{5}{2}\right)3{\cos }^2\theta \left(-\sin \theta \right)-\frac{3}{2}\left(-\sin \theta \right)\right] \\ =\frac{-3}{2}\sin \theta \left[5{\cos }^2\theta -1\right] \end{gathered}$$

Multiply by $\sin \theta$on both the sides,

$$\begin{gathered} \sin \theta \frac{d{P}_3}{d\theta }=\frac{-3}{1}{\sin }^2\theta \left[5{\cos }^2\theta -1\right] \\ \frac{\partial }{\partial \theta }\left(\sin \theta \frac{d{P}_3}{d\theta }\right)=\frac{-3}{2}\frac{\partial }{\partial \theta }\left[{\sin }^2\theta \left[5{\cos }^2\theta -1\right]\right] \\ =\frac{-3}{2}\left[\left(5{\cos }^2\theta -1\right)2\sin \theta \cos \theta +{\sin }^2\theta \left[10\cos \theta \left(-\sin \theta \right)\right]\right] \\ =\frac{-3}{2}\left[\left(5{\cos }^2\theta -\cos \theta \right)2\sin \theta -{\sin }^3\theta 10\cos \theta \right] \\ \frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial P_3}{\partial \theta }\right)=\frac{-3}{2}\left[2\left(5{\cos }^3\theta -\cos \theta \right)-10{\sin }^2\theta \cos \theta \right] \\ =-3\times 4\left[\frac{5}{2}{\cos }^3\theta -\frac{3}{2}\cos \theta \right] \\ =-I\left(I+1\right)P_3 \end{gathered}$$

$$\begin{gathered} \frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial P_3}{\partial \theta }\right)=\frac{-3}{2}\left[2\left(5{\cos }^3\theta -\cos \theta \right)-10{\sin }^2\theta \cos \theta \right] \\ =-3\times 4\left[\frac{5}{2}{\cos }^3\theta -\frac{3}{2}\cos \theta \right] \\ =-I\left(I+1\right)P_3 \end{gathered}$$

Hence, it is proved.

Now,

$$\begin{gathered} P_1\left(x\right)=x \\ {P}_3\left(x\right)=\frac{5}{2}{x}^3-\frac{3}{2}x \end{gathered}$$

Using explicit integration the orthogonal condition can be derived as,

$$\begin{gathered} \underset{-1}{\overset{+1}{\int }}{P}_1\left(x\right)P_3\left(x\right)dx=0 \\ \underset{-1}{\overset{+1}{\int }}x\frac{1}{2}\left(5x^3-3x\right)dx \\ \frac{1}{2}\underset{-1}{\overset{+1}{\int }}\left(5x^4-3x^2\right)dx \\ \frac{1}{2}\left[5{\left(\frac{x^5}{5}\right)}_{-1}^{+1}-3{\left(\frac{x^3}{3}\right)}_{-1}^{+1}\right] \\ \frac{1}{2}\left[1+1-1\right]=0 \end{gathered}$$

Therefore, $P_1\left(x\right)$and $P_3\left(x\right)$are orthogonal.