Page 547 - Advanced_Engineering_Mathematics o'neil

15.2 Legendre Polynomials 527

Then
1

A n
p n = xP n (x)P n−1 (x)dx
A n−1 −1
1 1
A n n + 1 n 2
= P n+1 (x)P n−1 (x)dx + P n−1 (x)dx
A n−1 2n + 1 −1 2n + 1 −1
n
A n
= p n−1 ,
A n−1 2n + 1
1
since P n+1 (x)P n−1 (x)dx = 0. But we know A n from Corollary 15.2. Substitute this into the
−1
last expression to obtain
1 · 3 · 5···(2n − 3)(2n − 1) (n − 1)! n
p n = p n−1
n! 1 · 3 · 5···(2n − 3) 2n + 1
2n − 1
= p n−1 .
2n + 1
Now work forward:
1 1 1 1 1 2
2
p 1 = p 0 = P 0 (x) dx = dx =
3 3 −1 3 −1 3
3 3 2 2
p 2 = p 1 = =
5 5 3 5
5 2
p 3 = p 2 =
7 7
7 2
p 4 = p 3 =
9 9
and so on. One can complete the proof by induction and obtain
2
p n = .
2n + 1
With this result, we can write the Fourier-Legendre coefficient of f (x) as
2n + 1 1
c n = f (x)P n (x)dx.
2 −1
Now here is an example of a Fourier-Legendre expansion.

EXAMPLE 15.5
We will write the Fourier-Legendre expansion of f (x) = cos(πx/2). Because cos(πx/2) is
continuous with a continuous derivative,
∞

cos(πx/2) = c n P n (x)
n=0
for −1 < x < 1, where
2n + 1 1
c n = cos(πx/2)P n (x)dx.
2 −1
There is no simple expression for c n for arbitrary n, so we will approximate this eigenfunction
expansion with the sum of the first six terms. We must therefore compute c 0 ,··· ,c 5 . Since we
know P 0 (x) through P 5 (x), these integrations can be carried out explicitly.

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October 14, 2010 15:20 THM/NEIL Page-527 27410_15_ch15_p505-562