Page 543 - Advanced_Engineering_Mathematics o'neil

15.2 Legendre Polynomials 523

By comparing coefficients in these two Maclaurin series for 1/(1 + t), we conclude that
P n (−1) = (−1) .
n

15.2.2 A Recurrence Relation for Legendre Polynomials
There is a recurrence relation for Legendre polynomials that gives P n+1 (x) in terms of P n (x) and
P n−1 (x).

THEOREM 15.4 A Recurrence Relation

If n is a positive integer, then
(n + 1)P n+1 (x) − (2n + 1)xP n (x) + nP n−1 (x) = 0. (15.6)

Proof Begin with
∂L x − t
= .
∂t (1 − 2xt + t )
2 3/2
2
Multiply this equation by 1 − 2xt + t to obtain
∂L
2
(1 − 2xt + t ) (x,t) − (x − t)L(x,t) = 0.
∂t
∞ n

Substitute L(x,t) = n=0 P n (x)t into this equation to obtain
∞ ∞
n−1 n
2
(1 − 2xt + t ) nP n (x)t − (x − t) P n (x)t = 0.
n=1 n=0
Carry out the indicated multiplications to write
∞ ∞ ∞

n
nP n t n−1 − 2nx P n (x)t + nP n (x)t n+1 −
n=1 n=1 n=1
∞ ∞

n n+1
xP n (x)t + P n (x)t = 0.
n=0 n=0
n
Rearrange the series where necessary to have t as the power of t in each summation:
∞ ∞ ∞

n n n
(n + 1)P n+1 (x)t − 2nx P n (x)t + (n − 1)P n−1 (x)t
n=0 n=1 n=2
∞ ∞

n n
− xP n (x)t + P n−1 (x)t = 0.
n=0 n=1
Combine these summations from n = 2 on, writing the terms for n = 0 and n = 1 separately, to
obtain
P 1 (x) + 2P 2 (x)t − 2xP 1 (x)t − xP 0 (x) − xP 1 (x)t + P 0 (x)t
∞

n
+ [(n + 1)P n+1 (x) − 2nx P n (x) + (n − 1)P n−1 (x) − xP n (x) + P n−1 (x)]t = 0.
n=2
For this power series in t to be zero for all t in some open interval about 0, the coefficient of
each power of t must equal 0. Therefore
P 1 (x) − xP 0 (x) = 0
2P 2 (x) − 2xP 1 (x) − xP 1 (x) + P 0 (x) = 0

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October 14, 2010 15:20 THM/NEIL Page-523 27410_15_ch15_p505-562