68.1. HERMITE POLYNOMIALS 2317
,this by Leibniz formula. Thus this cancels with the first term to give
∂
∂xHn (x,λ ) =
(−λ )n nn!
(− 1
λ
)e
12λ
x2 ∂ n−1
∂xn−1
(e−
12
x2λ
)=
(−λ )n−1
(n−1)!e
12λ
x2 ∂ n−1
∂xn−1
(e−
12
x2λ
)≡ Hn−1 (x,λ )
In case of n = 1, this appears to also work. ∂
∂x H1 (x,λ ) = 1 = H0 (x,λ ) from the abovecomputations. This shows that
∂
∂xHn (x,λ ) = Hn−1 (x,λ )
Next, is the claim that
(n+1)Hn+1 (x,λ ) = xHn (x,λ )−λHn−1 (x,λ )
If n = 1, this says that
2H2 (x,λ ) = xH1 (x,λ )−λH0 (x,λ )
= x2−λ
and so the formula does indeed give the correct description of H2 (x,λ ) when n = 1. Thusassume n > 1 in what follows. The left side equals
(−λ )n+1
n!e
12λ
x2 ∂ n+1
∂xn+1
(e−
12λ
x2)
This equals(−λ )n+1
n!e
12λ
x2 ∂ n
∂xn
(− x
λe−
12
x2λ
)Now by Liebniz formula,
=(−λ )n+1
n!e
12λ
x2[− x
λ
∂ n
∂xn e−12
x2λ +n
(−1λ
)∂ n−1
∂xn−1
(e−
12
x2λ
)]=
(−λ )n+1
n!e
12λ
x2(− x
λ
∂ n
∂xn e−12
x2λ
)+
(−λ )n+1
n!e
12λ
x2n(−1λ
)∂ n−1
∂xn−1
(e−
12
x2λ
)= x
(−λ )n
n!e
12λ
x2 ∂ n
∂xn e−12
x2λ +
(−λ )n
(n−1)!e
12λ
x2 ∂ n−1
∂xn−1
(e−
12
x2λ
)= xHn (x,λ )−λHn−1 (x,λ )
which shows the formula is valid for all n≥ 1.Next is the claim that
Hn (−x,λ ) = (−1)n Hn (x,λ )