Question

Using Rodrigue’s formula, obtain the expression for the Hermite polynomial H₄(x)

Answer 1

Answer on Question #84507 – Math – Differential Equations

Question

Using Rodrigue’s formula, obtain the expression for the Hermite polynomial $H_4(x)$.

Solution

Hermite polynomials, named after the French mathematician Charles Hermite, are orthogonal polynomials, in a sense to be described below, of the form

$$H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2}$$

for $n = 0, 1, 2, 3, \ldots$

This is the Rodrigues formula for the Hermite polynomial.

for $n = 0$: $H_0(x) = (-1)^0 e^{x^2} \left( e^{-x^2} \right) = 1$

for $n = 1$: $H_1(x) = (-1)^1 e^{x^2} \frac{d}{dx} \left( e^{-x^2} \right) = -1(-2x) e^{x^2} \left( e^{-x^2} \right) = 2x$

for $n = 2$: $H_2(x) = (-1)^2 e^{x^2} \frac{d^2}{dx^2} \left( e^{-x^2} \right) = e^{x^2} \frac{d}{dx} \left( -2x e^{-x^2} \right) =$

$$= e^{x^2} \left( -2e^{-x^2} - 2x(-2x) e^{-x^2} \right) = 4x^2 - 2$$

for $n = 3$: $H_3(x) = (-1)^3 e^{x^2} \frac{d^3}{dx^3} \left( e^{-x^2} \right) = -e^{x^2} \frac{d^2}{dx^2} \left( -2x e^{-x^2} \right) =$

$$= -e^{x^2} \frac{d}{dx} \left( -2e^{-x^2} - 2x(-2x) e^{-x^2} \right) = 8x^3 - 12x$$

for $n = 4$: $H_4(x) = (-1)^4 e^{x^2} \frac{d^4}{dx^4} \left( e^{-x^2} \right) = e^{x^2} \frac{d^3}{dx^3} \left( -2x e^{-x^2} \right) =$

$$= e^{x^2} \frac{d^2}{dx^2} \left( -2e^{-x^2} - 2x(-2x) e^{-x^2} \right) = 6x^2 \frac{d}{dx} \left( -2(-2x) e^{-x^2} + 8x e^{-x^2} + 4x^2(-2x) e^{-x^2} \right) = 16x^4 - 48x^2 + 12$$

for $n = 4$: $H_6(x) = (-1)^6 e^{x^2} \frac{d^4}{dx^4} \left( e^{-x^2} \right) = e^{x^2} \frac{d^3}{dx^3} \left( -2x e^{-x^2} \right) =$

$$= e^{x^2} \frac{d^2}{dx^2} \left( -2e^{-x^2} - 2x(-2x) e^{-x^2} \right) = 2x$$

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