Page 546 - Advanced_Engineering_Mathematics o'neil

526 CHAPTER 15 Special Functions and Eigenfunction Expansions

Then, if n > m,wehave

1 m 1

q(x)P n (x)dx = c k P k (x)P n (x)dx = 0
−1 −1
k=0
because each k < n.

1 2
Using this result, we can derive a simple expression for P (x)dx, the denominator in the
−1 n
expression for the Fourier-Legendre coefficients of any function.

THEOREM 15.6

For n = 0,1,2,···,

1 2
2 .
P (x)dx =
n
−1 2n + 1
Proof Let A n denote the coefficient of x in P n (x) and let
n
1
2
p n = P (x)dx.
n
−1
n
The highest power term in P n (x) is A n x , while the highest power term in P n−1 (x) is A n−1 x n−1 .
n
Therefore all terms involving x cancel in the polynomial
A n
q(x) = P n (x) − xP n−1
A n−1
so q(x) has degree at most n − 1. Write

A n
P n (x) = q(x) + xP n−1 (x).
A n−1
1
Because q(x)P n (x)dx = 0 by Theorem 15.5, then
−1
1
p n = P n (x)P n (x)dx
−1
1

A n
= P n (x) q(x) + xP n−1 (x) dx
−1 A n−1
1
A n
= xP n (x)P n−1 (x)dx.
A n−1 −1
Now use the recurrence relation to write
n + 1 n
xP n (x) = P n+1 (x) + P n−1 (x).
2n + 1 2n + 1
Then
n + 1 n
xP n (x)P n−1 (x) = P n+1 (x)P n−1 (x) + P 2 (x).
2n + 1 2n + 1 n−1

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October 14, 2010 15:20 THM/NEIL Page-526 27410_15_ch15_p505-562