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Normalization of Orthogonal Polynomials?

  1. Nov 9, 2013 #1
    The generalized Rodrigues formula is of the form

    [tex]K_n\frac{1}{w}(\frac{d}{dx})^n(wp^n)[/tex]

    The constant [itex]K_n[/itex] is seemingly chosen completely arbitrarily, & I really need to be able to figure out a quick way to derive whether it should be [itex]K_n = \tfrac{(-1)^n}{2^nn!}[/itex] in the case of Jacobi polynomials (reducable to Legendre, Chebyshev or Gegenbauer), [itex]K_n = \tfrac{1}{n!}[/itex] for Laguerre polynomials & [itex]K_n = (-1)^n[/itex] for Hermite polynomials. The best I have so far is actually working out the n'th derivative of [itex](wp^n)[/itex] in the case of Legendre polynomials, but that method becomes crazy with any of the other polynomials & as Hassani says the choices are arbitrary so they probably don't work. My question is, how do I get derive constants without any memorization, whether by some nice trick or by the method one uses to arbitrarily choose their values - I'd really appreciate it.
     
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  3. Nov 9, 2013 #2

    UltrafastPED

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  4. Nov 9, 2013 #3
    Unfortunately not, it just states what the constants are equal to without any explanation as to why they take those values, at least from my reading of that page.
     
  5. Nov 9, 2013 #4

    UltrafastPED

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    The normalization constants are chosen so that when you integrate the polynomial over the standard range ... you get 1. Hence orthonormal: orthogonal and normalized.
     
  6. Nov 9, 2013 #5
    I just do not see how that idea works when you try it on the Jacobi polynomials, I spent half a day trying to figure out some way of ending up with [itex]\tfrac{(-1)^n}{2^nn!}[/itex] by starting with [itex](1 - z)^{-a}(1 + z)^{-b} (\tfrac{d}{dx})^n[(1-z)^{n+a}(1+z)^{n+b}][/itex], whether by working out the differentiation in Rodrigues formula explicitly, or by trying that orthogonality idea, & nothing works - I just cannot do it, I'm not even sure if it gives the right answer, at most I think it might give a different normalization constant (as there are 3 different conventions), but the calculation leads nowhere but circles & pages of paper for me thus I think I'm missing the fundamental point underlying this seemingly arbitrary convention (one of 3 arbitrary conventions... :frown: ) & hoping somebody knows how to do this & doesn't mind showing me. Everything else makes perfect sense, why one can only have 3 classical weights & how to derive that fact, why Rodrigues works in the first place, all that stuff - it's just that this random convention is ruining everything :cry:
     
  7. Nov 9, 2013 #6

    UltrafastPED

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    Try it with a polynomial with weight=1. Then there are no choices ... you should get the standard normalization factor.

    I looked at your reference link; its been too many years since I did one of these, so I don't have any definite advice.
     
  8. Nov 9, 2013 #7

    AlephZero

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    I don't think there is any "nice trick" here, because the usual definitions of the polynomials (e.g. see Wolfram Alpha) are not orthonormal, but only orthogonal.

    Historically, the various sets of polynomials were invented/discovered independently of each other, for getting series solutions of ODEs etc, and the "normalizations" were probably whatever was most convenient in the days before pocket calculators and symbolic algebra software.

    When somebody later discovered all the sets of polynomials were variations on a more general theme, the conventional scaling factors turned out to be "arbitrary." It doesn't matter much what you use from a pure math point of view, except that changing the standard definitions would invalidate a lot of existing published work, or lead to even more confusion about what convention any particular book or paper had used.
     
  9. Nov 9, 2013 #8
    Thanks a lot guys, through some furious googling I found something that might help, on the bottom of page 67 of Szego he's apparently claiming to derive it, where the (4.1.1) he refers to is to [itex]P_n^{(\alpha, \beta)} (1) = {n+\alpha\choose n}[/itex], & it even seems as though this crazy convention is motivated when you look at page 68 where he apparently derives the [itex]{n+\alpha\choose n}[/itex] out of differentiating n times - even though it looks like a bit of an arbitrary step in the calculation it's at least partially motivated. I do not see how to actually derive it all yet, I can't properly follow the argument he makes, but at least this is progress - if anyone grok's it I'd sincerely appreciate help with this stuff :cool:
     
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