Hermite Polynomial Reccurence Relation

[Scootertaj]

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)ⁿe^x2$\frac{d^n}{dx^n}$(e^-x2)

The Attempt at a Solution

H_n+1 = (-1)ⁿ⁺¹e^x2$\frac{d^(n+1)}{dx^(n+1)}$(e^-x2)
where $\frac{d^(n+1)}{dx^(n+1)}$(e^-x2) means the (n+1)th derivative of e^-x2
2xH_n = 2x*(-1)ⁿe^x2$\frac{d^n}{dx^n}$(e^-x2)
2nH_n-1 = 2n*(-1)^n-1e^x2$\frac{d^(n-1)}{dx^(n-1)}$(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-1e^x2 * [ $\frac{d^(n+1)}{dx^(n+1)}$(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.

 

[susskind_leon]

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

 

[Scootertaj]

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 H_k+1 = 2xH_k - 2kH_k-1
Try to prove H_k+2 = 2xH_k+1 - 2(k+1)H_k


Well, we know H_k+2 = (-1)^k+2e^x2(d^k+2/dx^k+2)[e^-x2]

So, does H_k+2 = 2xH_k+1 - 2(k+1)H_k = (-1)^k+2e^x2(d^k+2/dx_k+2)[e^-x2] ?
Well, 2xH_k+1 - 2(k+1)H_k = 2x(2xH_k - 2kH_k-1) - 2(k+1)H_k = 4x²H_k - 4xkH_k-1 - 2kH_k - 2H_k

I don't see how I finish it.

 

[susskind_leon]

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 $H_{k+2}$?

 

[Scootertaj]

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

I'm sadly lost :(

 

[susskind_leon]

How about $H_{n+1}=- e^{x^2} \frac{d}{dx}[e^{-x^2}H_n]$
Can you use this relation to make the induction step?

 

[Scootertaj]

Ahh I got it I think, thanks!