Page 320 - Elements of Distribution Theory

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

306 Orthogonal Polynomials

For each k = 1,..., n, λ k is given by
∞
p n (x)

λ k = dF(x).
−∞ (x − x k )p (x k )

n
Proof. Fix n and consider a polynomial f of degree less than or equal to 2n − 1. Note

that, by Theorem 10.4, the zeros of p n are simple and, hence, |p (x j )| > 0, j = 1,..., n.
n
Define a function h by
n

p n (x)
h(x) = f (x j ) .
(x − x j )p (x j )

j=1 n
Note that
p n (x) = α(x − x 1 ) ··· (x − x n )
for some constant α = 0; this may be seen by noting that the function on the right is a
polynomial of degree n with the same zeros as p n . Hence, for each j = 1, 2,..., n,
p n (x)
(x − x j )
is a polynomial of degree n − 1so that h is also a polynomial of degree n − 1. It follows
that h − f is a polynomial of degree less than or equal to 2n − 1. Note that, for each
j = 1, 2,..., n,
p n (x k )

lim h(x) = f (x j ) + f (x k ) = f (x k );

x→x k (x k − x j )p (x j )
j =k n
hence, h − f has zeros at x 1 ,..., x n .It follows that
h(x) − f (x) = (x − x 1 ) ··· (x − x n )q(x) ≡ p n (x)r(x) (10.6)
where q and r are polynomials each of degree at most n − 1.
Writing

f (x) = h(x) − p n (x)r(x),
and using the fact that r is a polynomial of degree at most n − 1,
∞ ∞ ∞ ∞

f (x) dF(x) = h(x) dF(x) − p n (x)r(x) dF(x) = h(x) dF(x).
−∞ −∞ −∞ −∞
The result now follows from the fact that
n n
∞ ∞

p n (x)
h(x) dF(x) = f (x j ) dF(x) = λ j f (x j )

n
−∞ j=1 −∞ (x − x j )p (x j ) j=1
where λ 1 ,...,λ n are given in the statement of the theorem.
Example 10.5 (Legendre polynomials). Consider the Legendre polynomial
3 2 1
P 2 (x) = x − ,
2 2