Page 321 - Elements of Distribution Theory

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052184472Xc10 CUNY148/Severini May 24, 2005 2:50

10.2 General Systems of Orthogonal Polynomials 307

√
which has zeros at ±1/ 3. Hence, we may write
3 √ √
P 2 (x) = (x − 1/ 3)(x + 1/ 3).
2
Then
1 1 P 2 (x) 1
λ 1 = √ √ dx =
2 −1 (x − 1/ 3)P (−1/ 3) 2

2
and
1 1 P 2 (x) 1
λ 2 = √ √ dx = .
2 −1 (x + 1/ 3)P (1/ 3) 2

2
Hence, any polynomial f of degree 3 or less can be calculated by
√ √
1
f (1/ 3) + f (−1/ 3)
f (x) dF(x) = .
−1 2
3
2
This may be verified directly by integrating 1, x, x , and x .
The method described in Theorem 10.5 can also be used as the basis for a method of
numerical integration; this will be discussed in Section 10.4.

Completeness and approximation
Let {p 0 , p 1 ,...} denote orthogonal polynomials with respect to F.We say that {p 0 , p 1 ,...}
is complete if the following condition holds: suppose f is a function such that

∞
2
f (x) dF(x) < ∞
−∞
and
∞

f (x)p n (x) dF(x) = 0, n = 0, 1,... ;
−∞
then f = 0 almost everywhere (F).

Example 10.6 (Completeness of the Legendre polynomials). Let F denote the distribution
function of the uniform distribution on (−1, 1) and let f denote a function such that
1
2
f (x) dF(x) < ∞
−1
and
1
f (x)P n (x) dF(x) = 0, n = 0, 1,...
−1
where P n denotes the nth Legendre polynomial.
Let q n be an abritrary polynomial of degree n. Then
1 1 1
2 2 2
( f (x) − q n (x)) dF(x) = f (x) dF(x) + q n (x) dF(x).
−1 −1 −1