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Hermite Polynomial Reccurence Relation

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove, using Rodrigues form, that Hn+1=2xHn -2nHn-1



    2. Relevant equations
    The Rodrigues form for Hermite polynomials is the following:

    Hn = (-1)nex2[itex]\frac{d^n}{dx^n}[/itex](e-x2)


    3. 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: Nov 1, 2011
  2. jcsd
  3. Nov 2, 2011 #2
    If I were you, I would try induction...
     
  4. Nov 2, 2011 #3
    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.
     
  5. Nov 2, 2011 #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]?
     
  6. Nov 2, 2011 #5
    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 :(
     
  7. Nov 2, 2011 #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?
     
  8. Nov 2, 2011 #7
    Ahh I got it I think, thanks!
     
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