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Lengendre Polynomials of cos(theta)

  1. Mar 20, 2009 #1
    Hey guys,

    I've been working on a quantum related problem in my math physics class and I've run into a snag. When dealing with Legendre Polynomials ( specifically : [tex] P_{lm} (x) [/tex] ), I can find the general expression that can be used to derive the polynomial for any sets of l and m (wolfram math has that explicit equation on their website on this page, eq 65: http://mathworld.wolfram.com/LegendrePolynomial.html" [Broken].)

    However, when trying to find the general expression for the Legendre polynomial of cos: [tex] P_{lm}(cos(\theta)) [/tex], I find nothing. I tried to come up with the expression on my own by substituting for [tex] x = cos(\theta) [/tex] and [tex] dx = -sin(\theta)d\theta [/tex] , but I do not know how to handle the term with the derivative which goes like: [tex] \frac{d^{l+m}}{dx^{l+m}} [/tex].

    My first guess was to try and replace the differential [tex] dx^{l+m} [/tex] with [tex] (-sin(\theta)d\theta)^{l+m} [/tex], but this didn't seem to give the correct final result.

    If any body knows of a website where the [tex]P_{lm}(cos(\theta)) [/tex] formula is explicity given, or if someone knows how to actually derive the general form for these polynomials, let me know. I would really appreciate it. Thanks for your time.

    PCalhoun

    P.S. If more information is needed about the problem, I would be glad to elaborate.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 21, 2009 #2

    tiny-tim

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    Hey pcalhoun! :smile:

    (have a theta: θ :wink:)

    d/dx = d/dθ dθ/dx = (-1/sinθ)d/dθ

    d2/dx2 = (-1/sinθ)d/dθ [(-1/sinθ)d/dθ] = … :smile:
     
  4. Mar 21, 2009 #3

    clem

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    Just use (64) with cos theta=x. Then at the end go back to cos theta.
     
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