Hermite Polynomial Reccurence Relation

  • Thread starter Scootertaj
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  • #1
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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[itex]\frac{d^n}{dx^n}[/itex](e-x2)


The Attempt at a Solution


Hn+1 = (-1)n+1ex2[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
2xHn = 2x*(-1)nex2[itex]\frac{d^n}{dx^n}[/itex](e-x2)
2nHn-1 = 2n*(-1)n-1ex2[itex]\frac{d^(n-1)}{dx^(n-1)}[/itex](e-x2)



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



Well,

Hn+1 - 2xHn + 2nHn-1 = (-1)n-1ex2 * [ [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:

Answers and Replies

  • #2
If I were you, I would try induction...
 
  • #3
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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.
 
  • #4
Try to express [itex]H_{k+2}[/itex] in terms of [itex]H_{k+1}[/itex], without using any of the assumption. What do you need to do to get from [itex]H_{k+1}[/itex] to [itex]H_{k+2}[/itex]?
 
  • #5
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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

I'm sadly lost :(
 
  • #6
How about [itex]H_{n+1}=- e^{x^2} \frac{d}{dx}[e^{-x^2}H_n][/itex]
Can you use this relation to make the induction step?
 
  • #7
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Ahh I got it I think, thanks!
 

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