Chapter 5.3, Problem 22P

To determine

To show: The two solution of the Legendre equation for $\left|x\right|<1$ are $y_1\left(x\right)=\left[\begin{array}{l}1-\frac{\alpha \left(\alpha +1\right)x^2}{2!}+\frac{\alpha \left(\alpha -2\right)\left(\alpha +1\right)\left(\alpha +3\right)x^4}{4!}+\dots +\\ {\displaystyle \sum _{m=3}^{\infty }\frac{{\left(-1\right)}^m\left(\alpha \right)\cdots \left(\alpha -2m+2\right)\left(\alpha +1\right)\cdots \left(\alpha +2m-1\right)x^{2m}}{\left(2m\right)!}}\end{array}\right]$ and $y_2\left(x\right)=\left[\begin{array}{l}x-\frac{\left(\alpha -1\right)\left(\alpha +2\right)x^3}{3!}+\frac{\left(\alpha -1\right)\left(\alpha -3\right)\left(\alpha +2\right)\left(\alpha +4\right)x^5}{5!}+\dots +\\ {\displaystyle \sum _{m=3}^{\infty }\frac{{\left(-1\right)}^m\left(\alpha -1\right)\cdots \left(\alpha -2m+1\right)\left(\alpha +2\right)\cdots \left(\alpha +2m\right)x^{2m+1}}{\left(2m+1\right)!}}\end{array}\right]$.