Page 319 - Elements of Distribution Theory

P1: JZP
052184472Xc10 CUNY148/Severini May 24, 2005 2:50

10.2 General Systems of Orthogonal Polynomials 305

Since this is a polynomial of degree k,it follows from Theorem 10.1 that

b
f (x)p n (x) dF(x) = 0. (10.5)
a
Note that f (x) changes sign at each x j , j = 1,..., k,so that w(x) = f (x)p n (x)isalways
of the same sign. Without loss of generality, we assume that w(x) > 0 for all x. Hence,
b b
w(x) dF(x) = f (x)p n (x) dF(x) dx > 0.
a a
This contradicts (10.5) so that we must have k = n. Thus, p n has n simple zeros in (a, b);
however, p n has only n zeros so that all zeros of p n lie in (a, b) and are simple.

Example 10.4 (Legendre polynomials). The Legendre polynomial P 1 (x) = x has one
zero, at x = 0. The second Legendre polynomial,

3 2 1
P 2 (x) = x −
2 2
√
has zeros at x =±1/ 3. It may be shown that the third Legendre polynomial is given by
5 3 3
P 3 (x) = x − x,
2 2
√
which has zeros at x =± (.6) and x = 0.
Let p 0 , p 1 ,... denote orthogonal polynomials with respect to a distribution function F
and let x 1 denote the zero of p 1 . Let f (x) = ax + b, where a and b are constants. Then
f (x) = cp 1 (x) + d for some constants c and d; since p 1 (x 1 ) = 0we must have d = f (x 1 ).
It follows that

∞ ∞
f (x) dF(x) = c p 1 (x) dF(x) + f (x 1 ).
−∞ −∞
Since p 1 is orthogonal to all constant functions,
∞

p 1 (x) dF(x) = 0;
−∞
hence, for any linear function f ,
∞

f (x) dF(x) = f (x 1 ).
−∞
That is, the integral with respect to F of any linear function can be obtained by simply
evaluating that function at x 1 . The following result generalizes this method to an orthogonal
polynomial of arbitrary order.

Theorem 10.5. Let {p 0 , p 1 ,...} denote orthogonal polynomials with respect to F. For a
given value of n = 1, 2,..., let x 1 < x 2 <. . . < x n denote the zeros of p n . Then there exist
constants λ 1 ,λ 2 ,...,λ n such that, for any polynomial f of degree 2n − 1 or less,
n

∞

f (x) dF(x) = λ j f (x j ).
−∞ j=1