User:Egm6322.s12.team2.Xia/RP

=R 9.4 -Verify the second homogeneous solutions of Legendre equation =

Given
Given the following equations $$\displaystyle Q_{n}(x)=P_{n}(x)\tanh^{-1}x-2\sum_{j=1,3\cdots}^{J}\frac{2n-2j+1}{(2n-j+1)j}P_{n-j}(x)$$ (9.4.1) $$\displaystyle J\,:=1+2\left[\frac{n-1}{2}\right]=1+2\mathrm{Int}(\frac{n-1}{2})$$ (9.4.2)

Find
Verify the expressions for $$\displaystyle Q_{0},\, Q_{1},\, Q_{2}$$ as shown on the following equations $$\displaystyle Q_{0}(x)=\frac{1}{2}\log(\frac{1+x}{1-x})=\tanh^{-1}x$$ (9.4.3) $$\displaystyle Q_{1}(x)=\frac{x}{2}\log(\frac{1+x}{1-x})-1=x\tanh^{-1}x-1$$ (9.4.4) $$\displaystyle Q_{2}(x)=\frac{1}{2}(3x^{2}-1)\log(\frac{1+x}{1-x})-\frac{3}{2}x= ?$$ (9.4.5)

Solution
The Legendre polynomial is as follows: $$\displaystyle {P}_{n}(x)=\sum^{m}_{r=0}(-1)^r \frac{(2n-2r)!x^{n-2r}}{2^{n}r!}(n-r)!(n-2r)! $$ (9.4.6) where $$\displaystyle m = \frac{n}{2}$$. Thus $$\displaystyle {P}_{-1}(x) = 0$$

$$\displaystyle {P}_{0}(x) = 1 $$

$$\displaystyle {P}_{1}(x) = x $$

$$\displaystyle {P}_{2}(x) = \frac{1}{2}(3x^2-1) $$

$$\displaystyle $$ (9.4.6) Thus from 9.4.1 we have $$\displaystyle Q_{0}(x)=\underbrace{P_{0}(x)}_{=1}\tanh^{-1}x-2\left(\frac{-2+1}{-1+1}\right)^{\dagger}\underbrace{P_{0-1}(x)}_{=0}=\tanh^{-1}x $$ $$\displaystyle Q_{1}(x)=\underbrace{P_{1}(x)}_{=x}\tanh^{-1}x-2\frac{2-2+1}{2-1+1}\underbrace{P_{1-1}(x)}_{=1}=x\tanh^{-1}x-1$$ $$\displaystyle Q_{2}(x)=\underbrace{P_{2}(x)}_{(3x^{2}-1)/2}\tanh^{-1}x-2\sum_{j=1,3}^{3}\frac{4-2j+1}{(4-j+1)j}P_{2-j}(x)$$ $$\displaystyle =\frac{1}{2}(3x^{2}-1)\tanh^{-1}x-2\left(\frac{3}{4}\underbrace{P_{1}(x)}_{=x}+\frac{4-6+1}{(4-3+1)\cdot3}\underbrace{P_{-1}(x)}_{=0}\right)$$ $$\displaystyle =\frac{1}{2}(3x^{2}-1)\tanh^{-1}x-\frac{3}{2}x$$ $$\displaystyle \dagger $$ Although $$\displaystyle P_{-1}(0)$$, the fact that 1-1=0  makes this term meaningless. We'd better define $$\displaystyle \sum_{j=1,3\cdots}^{J}\frac{2n-2j+1}{(2n-j+1)j}=0 $$, if $$\displaystyle n=0 $$