91Ó°ÊÓ

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,

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

$\frac{1}{\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}\frac{dP}{d\mathrm{θ}}\right)=-I\left(I+1\right)P$

Here, $I=3$

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

Multiply by $\sin \mathrm{θ}$on both the sides,

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

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

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,

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

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