68.1. HERMITE POLYNOMIALS 2315
Now the Hermite polynomials are the coefficients of the power series of this function ex-panded in powers of t. Thus the nth one of these is
Hn (x) = exp(
x2
2
)1n!
dn
dtn
(exp(−1
2(x− t)2
))|t=0 (68.1.1)
and
exp(
tx− t2
2
)=
∞
∑n=0
Hn (x) tn (68.1.2)
Note that H0 (x) = 1,
H1 (x) = exp(
x2
2
)ddt
(exp(−1
2(x− t)2
))|t=0
= −e−12 (t−x)2
e12 x2
(t− x) |t=0 = x
From 68.1.2, differentiating both sides formally with respect to x,
t exp(
tx− t2
2
)=
∞
∑n=1
H ′n (x) tn
and so∞
∑n=0
Hn (x) tn = exp(
tx− t2
2
)=
∞
∑n=1
H ′n (x) tn−1 =∞
∑n=0
H ′n+1 (x) tn
showing thatH ′n (x) = Hn−1 (x) ,n≥ 1, H0 (x) = 0,H1 (x) = x
which could have been obtained with more work from 68.1.1. Also, differentiating bothsides of 68.1.2 with respect to t,
−exp(
tx− t2
2
)(t− x) =
∞
∑n=0
nHn (x) tn−1
Thus
(x− t)∞
∑n=0
Hn (x) tn =∞
∑n=0
nHn (x) tn−1 =∞
∑n=0
(n+1)Hn+1 (x) tn
and so∞
∑n=0
xHn (x) tn−∞
∑n=0
Hn (x) tn+1 =∞
∑n=0
(n+1)Hn+1 (x) tn
and so∞
∑n=0
xHn (x) tn−∞
∑n=1
Hn−1 (x) tn =∞
∑n=0
(n+1)Hn+1 (x) tn
Thus for n≥ 1,xHn (x)−Hn−1 (x) = (n+1)Hn+1 (x)