- #1

- 97

- 0

## Homework Statement

Prove,

__using Rodrigues form__, that H

_{n+1}=2xH

_{n}-2nH

_{n-1}

## Homework Equations

The Rodrigues form for Hermite polynomials is the following:

H

_{n}= (-1)

^{n}e

^{x2}[itex]\frac{d^n}{dx^n}[/itex](e

^{-x2})

## The Attempt at a Solution

H

_{n+1}= (-1)

^{n+1}e

^{x2}[itex]\frac{d^(n+1)}{dx^(n+1)}[/itex](e

^{-x2})

where [itex]\frac{d^(n+1)}{dx^(n+1)}[/itex](e

^{-x2}) means the (n+1)th derivative of e

^{-x2}

2xH

_{n}= 2x*(-1)

^{n}e

^{x2}[itex]\frac{d^n}{dx^n}[/itex](e

^{-x2})

2nH

_{n-1}= 2n*(-1)

^{n-1}e

^{x2}[itex]\frac{d^(n-1)}{dx^(n-1)}[/itex](e

^{-x2})

So, H

_{n+1}- 2xH

_{n}+ 2nH

_{n-1}should be = 0.

Well,

H

_{n+1}- 2xH

_{n}+ 2nH

_{n-1}= (-1)

^{n-1}e

^{x2}* [ [itex]\frac{d^(n+1)}{dx^(n+1)}[/itex](e

^{-x2}) + 2x * [itex]\frac{d^n}{dx^n}[/itex](e

^{-x2}) + 2n * [itex]\frac{d^(n-1)}{dx^(n-1)}[/itex](e

^{-x2}) ]

but, I don't really know where to go from there.

Last edited: