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Interesting proof

  1. Feb 6, 2005 #1
    How can it be proved that Rnl(r) has (n-l-1) zeros (not counting those at r-0 and r= infinity?

    I tried doing this inductively but found it hard to get anywhere since the equations for Rnl(r) aren't too pretty. For instance, Rnl(r)=Anlunl

    where Anl= sqrt((n-l-1)!/(2n((n+l)!)^3)

    this last equation seems to bear a similarity with what I am supposed to prove but I am having trouble with this. Help anyone?
     
  2. jcsd
  3. Feb 6, 2005 #2

    dextercioby

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    I'm sorry for not sharing your enthusiasm,but i don't find that proof to be interesting... :tongue2:
    Yu can find the general formula for [itex] R_{nl}(r) [/itex] here

    I think you can build the proof really easily.

    Good luck!! :smile:

    Daniel.
     
  4. Feb 6, 2005 #3
    I don't think you need to worry about [tex]A_{nl}[/tex] in your proof, since that is just a constant for a given n and l, and not dependent on r. I think what you want to look at is the Laguerre Polynomials, (specifically their order), and remember what the fundamental theorem of algebra says about how many zeros a polynomial will have, and somehow come up with an argument that those zeros will be real.
     
  5. Feb 6, 2005 #4
    The link Daniel provided states that
    [tex]R_{n\ell}(r) = r^\ell \exp\left(\frac{zr}{na}\right) \sum_{j=0}^{n-\ell-1} b_jr^j.[/tex]
    The sum is simply the definition of a polynomial of degree [tex]n-\ell-1[/tex], which the fundamental theorem of algebra guarantees will have exactly [tex]n-\ell-1[/tex] complex roots (not necessarily distinct). The coefficients in front of the sum provide the zeros at [tex]r=0[/tex] and [tex] z\to-\infty[/tex], and the others are provided by the polynomial.

    If you need to show that there are [tex]n-\ell-1[/tex] distinct zeros then there will obviously be something more to the proof (I would guess it involves looking at the definition of the [tex]b_j[/tex]), but this might be a good starting point at least....
     
    Last edited: Feb 6, 2005
  6. Feb 6, 2005 #5
    Ha ha, I just didn't know what to title it. Thanks for help.
     
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