The Hermite polynomials H_{n}(x) are defined through

H_{n}(x) = (−1)^{n}e^{x^{2}} · (\frac{d}{dx})^{n} (e^{−x^{2}})

Theorem:

H_{n}(x) satisfies the recurrence relation

H_{0}(x) = 1
H_{n+1}(x) = 2x · H_{n}(x) − 2n · H_{n−1}(x)

for n ≥ 0 and the differential equation Hn^{′′}(x) − 2x · Hn^{′}(x) + 2n · H_{n}(x) = 0 for all n ≥ 0.

Proof:

Let F := ∑_{n=0}^{∞} \frac{H_{n}(x)}{n!}z^{n} be the exponential generating function of the sequence of polynomials. Then, because the Taylor series development theorem holds in formal power series rings (see [1], section 2.16), we can simplify

F = e^{x^{2}} · ∑_{n=0}^{∞} \frac{1}{n!} (\frac{d}{dx})^{n} (e^{−x^{2}}) · (−z)^{n}
= e^{x^{2}} · e^{−(x−z)^{2}}
= e^{2xz−z^{2}}

It follows that \frac{d}{dz}F = (2x − 2z) · F. This is equivalent to the claimed recurrence.

Starting from this equation, we compute a linear relation for the partial derivatives of F. Write ∂_{x} = \frac{d}{dx} and Δ_{z} = z \frac{d}{dz}. One computes

F = 1 · F
∂_{x}F = 2z · F
∂_{x}^{2}F = 4z^{2} · F
Δ_{z}F = (2xz − 2z^{2}) · F
∂_{x}Δ_{z}F = (2z + 4xz^{2} − 4z^{3}) · F
Δ_{z}^{2}F = (2x · z + (4x^{2} − 4) · z^{2} − 8x · z^{3} + 4 · z^{4}) · F

Solve a homogeneous 5 × 6 system of linear equations over Q(x) to get

(−2x) · ∂_{x}F + ∂_{x}^{2}F + 2 · Δ_{z}F = 0

This is equivalent to the claimed equation Hn^{′′}(x)−2x·Hn^{′}(x)+2n·H_{n}(x) = 0.

References

1 Bruno Haible: D-finite power series in several variables. Diploma thesis, University of Karlsruhe, June 1989. Sections 2.15 and 2.22.