(a) Use Exercise 28 to show that the power series in \(x\) for the general
solution of
$$
\left(1-x^{2}\right) y^{\prime \prime}-2 b x y^{\prime}+\alpha(\alpha+2 b-1)
y=0
$$
is \(y=a_{0} y_{1}+a_{1} y_{2},\) where
$$
y_{1}=\sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(2 j-\alpha)(2 j+\alpha+2
b-1)\right] \frac{x^{2 m}}{(2 m) !}
$$
and
$$
y_{2}=\sum_{m=0}^{\infty}\left[\prod_{j=0}^{m-1}(2 j+1-\alpha)(2 j+\alpha+2
b)\right] \frac{x^{2 m+1}}{(2 m+1) !}
$$
(b) Suppose \(2 b\) isn't a negative odd integer and \(k\) is a nonnegative
integer. Show that \(y_{1}\) is a polynomial of degree \(2 k\) such that
\(y_{1}(-x)=y_{1}(x)\) if \(\alpha=2 k\), while \(y_{2}\) is a polynomial of degree
\(2 k+1\) such that \(y_{2}(-x)=-y_{2}(-x)\) if \(\alpha=2 k+1\). Conclude that if
\(n\) is a nonnegative integer, then there's a polynomial \(P_{n}\) of degree \(n\)
such that \(P_{n}(-x)=(-1)^{n} P_{n}(x)\) and
$$
\left(1-x^{2}\right) P_{n}^{\prime \prime}-2 b x P_{n}^{\prime}+n(n+2 b-1)
P_{n}=0
$$
(c) Show that (A) implies that
$$
\left[\left(1-x^{2}\right)^{b} P_{n}^{\prime}\right]^{\prime}=-n(n+2
b-1)\left(1-x^{2}\right)^{b-1} P_{n}
$$
and use this to show that if \(m\) and \(n\) are nonnegative integers, then
$$
\begin{array}{l}
{\left[\left(1-x^{2}\right)^{b} P_{n}^{\prime}\right]^{\prime}
P_{m}-\left[\left(1-x^{2}\right)^{b} P_{m}^{\prime}\right]^{\prime} P_{n}=}
\\\
{[m(m+2 b-1)-n(n+2 b-1)]\left(1-x^{2}\right)^{b-1} P_{m} P_{n}}
\end{array}
$$
(d) Now suppose \(b>0\). Use (B) and integration by parts to show that if \(m
\neq n\), then
$$
\int_{-1}^{1}\left(1-x^{2}\right)^{b-1} P_{m}(x) P_{n}(x) d x=0
$$
(We say that \(P_{m}\) and \(P_{n}\) are orthogonal on (-1,1) with respect to the
weighting function \(\left.\left(1-x^{2}\right)^{b-1} .\right)\)
