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Laguerre polynomials

  1. Oct 10, 2006 #1
    What differential equation does

    [tex]\phi_n (x) := e^{-x/2} L_n (x)[/tex]

    solve? [tex]L_n[/tex] is a Laguerre polynomial.

    Please give me a hint on this one. I haven't got a clue where to start.
     
  2. jcsd
  3. Oct 10, 2006 #2

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    What differential equation do the Laguerre polynomials solve? If you don't know, look it up.
     
  4. Oct 11, 2006 #3
    The Laguerre differential equation is

    [tex]xy'' + (1 - x)y' + ny = 0[/tex]

    and [tex]L_n (x)[/tex] is a solution to this but my [tex]\phi_n (x)[/tex] is not a solution to the Laguerre equation, is it?
    I know that [tex]\phi_n (x)[/tex] should solve a self adjoint differential equation but I don't think the Laguerre eq. is?
     
  5. Oct 11, 2006 #4

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    Find [itex]\phi',\phi''[/itex] in terms of the derivatives of L and use the differential equation relating the derivatives of L to get a DE relating the derivatives of [itex]\phi[/itex].
     
  6. Oct 11, 2006 #5
    Ok, so then I get

    [tex]x \phi_n^{''} (x) + (1-x) \phi_n^{'} + (n + \frac{1}{2} - \frac{x}{4}) \phi_n (x) = 0[/tex]

    but this isn't self-adjoint?
    In my case, self-adjoint means it can be written in the form

    [tex]\frac{d}{dx} (p(x) \frac{d}{dx} \phi_n (x) ) + q(x) \phi_n (x)[/tex]
     
  7. Oct 12, 2006 #6
    I made a mistake. The equation I get is

    [tex]x \phi_n^{''} + \phi_n^{'} + (n + \frac{1}{2} - \frac{x}{4}) \phi_n = 0[/tex]

    and this is indeed self-adjoint. Thanks for your help!
     
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