Question

Using Rodrigue’s formula, obtain expression for the Hermite polynomial
H3(x)
and show that
H3'(x) = 6H2(x)

Answer 1

The Rodrigues formula for the Hermite Polynomial can be written as

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

For $H_2(x)$ we obtain

$H_2(x)=(-1)^xe^{x^2}\frac{d^2}{dx^2}e^{-x^2}=4x^2-2$

For $H_3(x)$

$H_3(x)=(-1)^xe^{x^2}\frac{d^3}{dx^3}e^{-x^2}=8x^3-12x$

$H'_3(x)=24x^2-12=6(4x^2-2)=6H_2(x)$