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524 CHAPTER 15 Special Functions and Eigenfunction Expansions

and, for n = 2,3,···,
(n + 1)P n+1 (x) − 2nx P n (x) + (n − 1)P n−1 (x) − xP n (x) + P n−1 (x) = 0.

Thesegiveus
P 1 (x) = xP 0 (x) = x,
1 1
2
P 2 (x) = (3xP 1 (x) − P 0 (x)) = (3x − 1),
2 2
and, for n = 2,3,···,
(n + 1)P n+1 (x) − (2n + 1)xP n (x) + nP n−1 (x) = 0.
Since this equation is also valid for n = 1, the recurrence relation is proved.

When we consider eigenfunction expansions in terms of Legendre polynomials, we will
n
need to know the coefficient of x in P n (x). We can obtain this using the recurrence relation.

COROLLARY 15.2

n
For each positive integer n, the coefficient of x in P n (x) is
1 · 3···(2n − 1)
.
n!
This is the product of the odd integers from 1 through 2n − 1 inclusive, divided by the
product of the integers from 1 through n.

Proof Let A n be the coefficient of x in P n (x). In the recurrence relation (15.6), the highest
n
power of x that occurs is x n+1 , and this power occurs only in P n+1 (x) and in xP n (x). Therefore
the coefficient of x n+1 on the left side of the recurrence relation is
(n + 1)A n+1 − (2n + 1)A n .
This must equal zero, because the right side of the recurrence relation is 0, with no x n+1 term.
Thus
2n + 1
A n+1 = A n
n + 1
for n = 0,1,2,···. We know that A 0 = 1 because P 0 (x) = 1, so we can work back from this
recurrence relation for A n to obtain:
2n + 1 2n + 1 2(n − 1) + 1
A n+1 = A n = A n−1
n + 1 n + 1 (n − 1) + 1
2n + 1 2n − 1 2n + 1 2n − 1 2(n − 2) + 1
= A n−1 = A n−2
n + 1 n n + 1 n (n − 2) + 1
2n + 1 2n − 1 2n − 3 2n + 1 2n − 1 2n − 3 3
= A n−2 = ··· = ··· A 0
n + 1 n n − 1 n + 1 n n − 1 2
2n + 1 2n − 1 2n − 3 3
= ··· .
n + 1 n n − 1 2
Therefore
1 · 3 · 5···(2n − 1)(2n + 1)
A n+1 = ,
(n + 1)!
and this is the conclusion of the theorem, stated in terms of n + 1 instead of n.

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October 14, 2010 15:20 THM/NEIL Page-524 27410_15_ch15_p505-562