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10.3 Classical Orthogonal Polynomials 313

Theorem 10.7. For each n = 0, 1,..., the nth Hermite polynomial is given by (10.9).

Proof. According to Theorem 10.3, the polynomials given by (10.9) are orthogonal with
respect to the standard normal distribution provided that
1 d n
φ(x)
φ(x) dx n
is a polynomial of degree n and that
d n
j
lim x φ(x) = 0 (10.10)
dx n
|x|→∞
for all n and j. Once these are established, the result follows provided that the coefficient
n
of x in (10.9) is 1.
For each n = 0, 1,..., define a function q n by
d n
φ(x) = q n (x)φ(x).
dx n
Note that q 0 (x) = 1 and, hence, is a polynomial of degree 0. Assume that q m is a polynomial
of degree m;we will show that this implies that q m+1 is a polynomial of degree m + 1. It
will then follow by induction that q n is a polynomial of degree n for n = 0, 1,....
Note that
d m+1 d
φ(x) = q m+1 (x)φ(x) = q m (x)φ(x)
dx m+1 dx
so that

q m+1 (x)φ(x) = q (x)φ(x) − xq m (x)φ(x);
m
hence,

q m+1 (x) = q (x) − xq m (x).
m
Under the assumption that q m is a polynomial of degree m,it follows that q m+1 is a poly-
nomial of degree m + 1. Hence, for all n = 0, 1,..., q n is a polynomial of degree n.
Since
j
lim x φ(x) = 0
|x|→∞
for all j = 0, 1,..., (10.9) follows from the fact that
d n
φ(x) = q n (x)φ(x)
dx n
where q n is a polynomial of degree n.
m
Finally, note that, if the coefficient of x in q m (x)is1, then, since

q m+1 (x) = q (x) − xq m (x),
m
the coefficient of x m+1 in q m+1 (x)is −1. The result now follows from the facts that q 0 (x) = 1
n
and H n (x) = (−1) q n (x).
The following corollary gives another approach to constructing the Hermite polynomials;
the proof follows from Theorem 10.7 and is left as an exercise.