Differentiating the Rodrigues identity gives

H_{0} = 1

and

H_{n + 1} = 2 t H_{n} - H_{n}^{'}

. Induct with the proposed finite sum. The coefficient at

t^{n + 1 - 2 j}

in this recurrence is the contribution from

2 t H_{n}

minus the derivative contribution with previous index

j - 1

\frac{n ! ( - 1 ) ^{j} 2 ^{n + 1 - 2 j}}{j ! ( n - 2 j )!} - \frac{n ! ( - 1 ) ^{j - 1} 2 ^{n - 2 j + 2}}{( j - 1 )! ( n - 2 j + 1 )!} .

Using a common denominator, this equals

(n + 1)! (- 1)^{j} 2^{n + 1 - 2 j} / [j! (n + 1 - 2 j)!]

. At an endpoint omit terms whose factorial indices would be negative; the same formula then covers the sole remaining term. Thus

H_{n} (t) = n! j = 0 \sum ⌊ n /2 ⌋ \frac{( - 1 ) ^{j} 2 ^{n - 2 j}}{j ! ( n - 2 j )!} t^{n - 2 j} .

The initial value and recurrence identify these polynomials with the given Rodrigues definition.

b

Solution

Substitute the finite formula and write

n = m + 2 j

. Under the allowed rearrangements,

\sum_{n \geq 0} \frac{H _{n} ( t ) z ^{n}}{n !} = \sum_{m, j \geq 0} \frac{( 2 t z ) ^{m}}{m !} \frac{( - z ^{2} ) ^{j}}{j !} = e^{2 t z} e^{- z^{2}} .

Hence $G (t, z) = e^{2 t z - z^{2}}$ , with the normalization of the physicists' Hermite polynomials used in the source.

Complete the square to obtain, initially for real

y, z

\int_{R} e^{- t^{2}} G (t, y) G (t, z) d t = e^{- y^{2} - z^{2}} \int_{R} e^{- t^{2} + 2 t (y + z)} d t = π e^{2 yz} .

y^{m} z^{n}

on the two sides. The left coefficient is

(m! n!)^{- 1} \int e^{- t^{2}} H_{m} (t) H_{n} (t) d t

, and the right is zero unless

m = n

, in which case it is

π 2^{n} / n!

. The polynomials are real on the real line, so these are the complex

L^{2}

ϕ_{n}

. Thus

⟨ ϕ_{m}, ϕ_{n} ⟩ = δ_{mn} π 2^{n} n!, ∥ ϕ_{n} ∥_{2} = π^{1/4} 2^{n} n! .

Polynomial times Gaussian decay also directly shows each

ϕ_{n}

lies in

L^{2}

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