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304 Orthogonal Polynomials
n
Suppose g is a polynomial of degree r. Then g(x) is a polynomial of degree nr and
d n n
g(x)
dx n
is a polynomial of degree n(r − 1); hence, g must be a polynomial of degree 2. Since all
polynomials are bounded on [−1, 1], in order to satisfy the second condition, it suffices that
lim g(x) = lim g(x) = 0.
x→1 x→−1
2
Writing g(x) = ax + bx + c,we need a + c = 0 and b = 0; that is, g is of the form
2
c(x − 1). It follows that orthogonal polynomials with respect to the uniform distribution
on (−1, 1) are given by
d n 2 n
(x − 1) .
dx n
Note that
d n 2 n n
(x − 1) = n!(2x) + Q(x)
dx n
2
where Q(x)isa sum in which each term contains a factor x − 1. Hence, the standardized
polynomials that equal 1 at x = 1 are given by
1 d n 2 n
(x − 1) .
n!2 dx n
n
Zeros of orthogonal polynomials and integration
Consider a function g : R → R.A zero of g is a number r, possibly complex, such that
g(r) = 0. If g is a polynomial of degree n, then g can have at most n zeros. A zero r is said
to have multiplicity α if
g(r) = g (r) =· · · = g (α−1) (r) = 0
(α)
and g (r) = 0. A zero is said to be simple if its multiplicity is 1.
Let g denote an nth degree polynomial and let r 1 ,...,r m denote the zeros of g such that
m
r j has multiplicity α j , j = 1,..., m. Then α j = n and g can be written
j=1
α 1
g(x) = a(x − r 1 ) ··· (x − r m ) α m
for some constant a.
The zeros of orthogonal polynomials have some useful properties.
Theorem 10.4. Let {p 0 , p 1 ,...} denote orthogonal polynomials with respect to F and let
[a, b], −∞ ≤ a < b ≤∞, denote the support of F. Then, for each n = 0, 1,..., p n has
n simple real zeros, each of which takes values in (a, b).
Proof. Fix n. Let k denote the number of zeros in (a, b)at which p n changes sign; hence,
0 ≤ k ≤ n.
Assume that k < n and and let x 1 < x 2 < ··· < x k denote the zeros in (a, b)at which
p n changes sign. Consider the polynomial
f (x) = (x − x 1 )(x − x 2 ) ··· (x − x k ).