# Hermite Polynomial Reccurence Relation

• Scootertaj
In summary, the homework statement is that Hermite polynomials can be solved using the Rodrigues form, and that the attempt at a solution is that Hn+1 = (-1)n+1ex2\frac{d^(n+1)}{dx^(n+1)}(e-x2).

## Homework Statement

Prove, using Rodrigues form, that Hn+1=2xHn -2nHn-1

## Homework Equations

The Rodrigues form for Hermite polynomials is the following:

Hn = (-1)nex2$\frac{d^n}{dx^n}$(e-x2)

## The Attempt at a Solution

Hn+1 = (-1)n+1ex2$\frac{d^(n+1)}{dx^(n+1)}$(e-x2)
where $\frac{d^(n+1)}{dx^(n+1)}$(e-x2) means the (n+1)th derivative of e-x2
2xHn = 2x*(-1)nex2$\frac{d^n}{dx^n}$(e-x2)
2nHn-1 = 2n*(-1)n-1ex2$\frac{d^(n-1)}{dx^(n-1)}$(e-x2)

So, Hn+1 - 2xHn + 2nHn-1 should be = 0.

Well,

Hn+1 - 2xHn + 2nHn-1 = (-1)n-1ex2 * [ $\frac{d^(n+1)}{dx^(n+1)}$(e-x2) + 2x * $\frac{d^n}{dx^n}$(e-x2) + 2n * $\frac{d^(n-1)}{dx^(n-1)}$(e-x2) ]

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

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If I were you, I would try induction...

susskind_leon said:
If I were you, I would try induction...
I've tried, but it doesn't seem to get me anywhere.

Base case n=1: H1+1 = 2xH1 - 2(1)H0 ?
Yep since H2 = 4x2-2 = 2x(2x) - 2

Assume Hk+1 = 2xHk - 2kHk-1
Try to prove Hk+2 = 2xHk+1 - 2(k+1)Hk

Well, we know Hk+2 = (-1)k+2ex2(dk+2/dxk+2)[e-x2]

So, does Hk+2 = 2xHk+1 - 2(k+1)Hk = (-1)k+2ex2(dk+2/dxk+2)[e-x2] ?
Well, 2xHk+1 - 2(k+1)Hk = 2x(2xHk - 2kHk-1) - 2(k+1)Hk = 4x2Hk - 4xkHk-1 - 2kHk - 2Hk

I don't see how I finish it.

Try to express $H_{k+2}$ in terms of $H_{k+1}$, without using any of the assumption. What do you need to do to get from $H_{k+1}$ to $H_{k+2}$?

That's what I've been trying to figure out, what the relationship is between Hk+2 and Hk+1 using only Rodrigues form.
The main thing I can see is that Hk+2 * (n+1)th derivative of e-x2 = -Hk+1 * (n+2)nd derivative of e-x2

How about $H_{n+1}=- e^{x^2} \frac{d}{dx}[e^{-x^2}H_n]$