Question 10.45: Prove that Pn(z) = F (-n, n + 1; 1; 1 - z/2), n = 0, 1, 2, 3...
Prove that P_{n}(z)=F\left(-n, n+1 ; 1 ; \frac{1-z}{2}\right), n=0,1,2,3, \ldots
The Legendre polynomials P_{n}(z) are of degree n and have the value 1 for z=1. Similarly, from (10.29), page 328 ,
F(a, b ; c ; z)=1+\frac{a \cdot b}{1 \cdot c} z+\frac{a(a+1) b(b+1)}{1 \cdot 2 \cdot c(c+1)} z^2+\cdots (10.29)
it is seen that
F\left(-n, n+1 ; 1 ; \frac{1-z}{2}\right)=1-\frac{n(n+1)}{2}(1-z)+\frac{n(n-1)(n+1)(n+2)}{16}(1-z)^{2}+\cdots
is a polynomial of degree n having the value 1 for z=1.
The required result follows if we show that P_{n} and F satisfy the same differential equation. To do this, let (1-z) / 2=u, i.e., z=1-2 u, in Legendre’s equation (10.25), page 327,
(1-z^{2})Y^{\prime\prime}-2z Y^{\prime}+n(n+1)Y=0 (10.25)
to obtain
u(1-u) \frac{d^{2} Y}{d u^{2}}+(1-2 u) \frac{d Y}{d u}+n(n+1) Y=0
But this is the hypergeometric equation (10.30), page 328,
z(1-z)Y^{\prime\prime}+\{c-(a+b+1)z\}Y^{\prime}-a b Y=0 (10.30)
with a=-n, b=n+1, c=1, and u=(1-z) / 2. Hence the result is proved.