The Hermite polynomials {Hn(x)}(n = 0 to ∞) can be obtained from the generating function...

The Hermite polynomials {Hn(x)}(n = 0 to ∞) can be obtained from the generating function...

F(x, u) = exp (2xu − u^2) = Σ(n = 0 to ∞) u^n/n! Hn(x)

1.Prove the recurrence relation d/dx Hn(x) = 2nH(n−1(x)) for n = 0, 1, 2,
2.Show that the recurrence relation is satisfied for n=2 using using the Rodrigues’ formula

Hn(x) = (−1)^n exp(x^2) d^n/dx^n (exp(−x^2)

Since (d/dx) exp(2xu - u^2) = 2u * exp(2xu - u^2), we have
(d/dx)Σ(n = 0 to ∞) u^n/n! Hn(x) = 2u * Σ(n = 0 to ∞) u^n/n! Hn(x)
............................= Σ(n = 0 to ∞) 2u^(n+1)/n! Hn(x)
............................= Σ(n = 1 to ∞) 2u^n/(n-1)! H(n-1(x)), via re-indexing
............................= Σ(n = 1 to ∞) 2n u^n/n! H(n-1(x))
............................= Σ(n = 0 to ∞) 2n u^n/n! H(n-1(x)), via adding '0'

Equating like coefficients yields (d/dx) Hn(x) = 2n H(n-1(x)).
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2) We want to show that (d/dx) H₂(x) = 4 H₁(x) with Rodrigues' formula.

(d/dx) H₂(x)
= (d/dx) [exp(x^2) * d^2/dx^2 (exp(−x^2))], by Rodrigues'
= (d/dx) [exp(x^2) * (d/dx) -2x exp(−x^2))]
= (d/dx) [exp(x^2) * (-2 exp(−x^2) + 4x^2 exp(−x^2))]
= (d/dx) (-2 + 4x^2)
= 8x.

4 * H₁(x) = 4 * (−1) exp(x^2) d/dx exp(−x^2), by Rodrigues'
...............= 4 * (−1) exp(x^2) * -2x exp(−x^2)
...............= 8x.

So, (d/dx) H₂(x) = 4 H₁(x).

I hope this helps!