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Legendre Polynomials - how to find P0(u) and P2(u)?

  1. May 8, 2014 #1
    Pl(u) is normalized such that Pl(1) = 1. Find P0(u) and P2(u)

    note: l, 0 and 2 are subscript




    recursion relation

    an+2 = [n(n+1) - l (l+1) / (n+2)(n+1)] an


    n is subscript

    substituted λ = l(l+1) and put n=0 for P0(u) and n=2 for P2(u), didnt get very far

    please could someone give me a point in the right direction or a youtube video to explain the process used to solve this? really lost.

    thanks for any advice
     
  2. jcsd
  3. May 8, 2014 #2

    Ray Vickson

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    You wrote
    [tex] a_n + 2 = \left[n(n+1) - \frac{l(l+1)}{(n+2)(n+2)} \right] a_n [/tex]
    Is that what you really mean, or did you want
    [tex] a_{n+2} = \left[ \frac{n(n+1) - l(l+1)}{(n+2)(n+1)}\right] a_n ?[/tex]
    If so, please use parentheses properly.

    Anyway, I don't see what this approach will get you, since you have the recursion for a fixed value of ##l##---that is, the recursion that tells you how to get ##P_l(x)## for fixed ##l##, given some initial conditions. There is nothing there that tells you how to relate ##P_{l-1}(x)## and ##P_{l+1}(x)## to ##P_l(x)##. Normalization comes from outside the differential equation and outside the recursion. Look up Legendre functions in a textbook or via Google.

    Note added in edit: I see that you might have meant ##P_l##; the font you used made it look like ##P_1##. Anyway, you might find it useful to look at http://www.phys.ufl.edu/~fry/6346/legendre.pdf .
     
    Last edited: May 8, 2014
  4. May 8, 2014 #3
    apologies, yes it's this:

    i think the idea is to put l=0 for P0(u) and l=2 for P2(u)

    but not really sure what to do after that.

    will have a look at the link, thank-you.
     
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