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522 CHAPTER 15 Special Functions and Eigenfunction Expansions

This result is useful in deriving properties of Legendre polynomials. We will not give a
complete proof, but we will derive some terms in the power series expansion of L(x,t) about
t = 0, suggesting why the theorem is true. Begin with the Maclaurin series for (1 − w) −1/2 :
1 1 3 15 105 945
4
2
5
3
√ = 1 + w + w + w + w + w + ···
1 − w 2 8 48 384 3840
2
for −1 <w < 1. Put w = 2xt − t to obtain
1 1 3
2 2
2
√ =1 + (2xt − t ) + (2xt − t )
1 − 2xt + t 2 2 8
15 105 945
2 5
2 4
2 3
+ (2xt − t ) + (2xt − t ) + (2xt − t ) + ··· .
48 384 3840
2
Now expand each of these powers of 2xt − t and collect the coefficient of each power of t to
obtain
1 1 3 3 5
2
3
4
3 3
√ =1 + xt − t − xt + t + x t
1 − 2xt + t 2 2 2 8 2
15 15 5 35 105
5
3 5
6
2 6
2 4
− x t + xt − t + x t + x t
4 8 16 4 16
35 35 63 315 315
5 5
4 6
8
3 7
7
− xt + t + x t − x t + x t
16 128 8 16 16
315 315 69
2 8
9
10
− x t + xt − t + ···
32 128 256

1 3 2 2 3 5 3 3
=1 + xt + − + x t + − x + x t
2 2 2 2
3 15 2 35 4 4 15 35 3 63 5 5

+ − x + x t + x − x + x t + ···
8 4 8 8 4 8
2
3
4
5
= P 0 (x) + P 1 (x)t + P 2 (x)t + P 3 (x)t + P 4 (x)t + P 5 (x)t + ··· .
We know that P n (1) = 1 because a 0 and a 1 in the derivation of these polynomial solutions
of Legendre’s equation were chosen for this purpose. Using the generating function, we can also
evaluate P n (−1).
COROLLARY 15.1
For every nonnegative integer n,
n
P n (−1) = (−1) .
Proof Evaluate
1 1
L(−1,t) = √ = √
1 + 2t + t 2 (1 + t) 2
∞
1
= = P n (−1)t n
1 + t
n=0
for −1 < t < 1. But, for these values of t, we have the geometric series
∞
1 n n
= (−1) t .
1 + t
n=0
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October 14, 2010 15:20 THM/NEIL Page-522 27410_15_ch15_p505-562