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Recurrence relation for the Legendre functions

  1. May 31, 2012 #1
    My book wants to find solutions to Legendre's equation:

    (1-x2)y'' - 2xy' 0 l(l+1)y = 0 (1)

    By assuming a solution of the form:

    y = Ʃanxn , the sum going from 0->∞ (2)

    Now by plugging (2) into (1) one finds:

    Ʃ[n(n-1)anxn-2-n(n-1)anxn - 2nanxn +
    l(l+1)anxn = 0

    Rearranging my book writes this as:

    Ʃ[(n+2)(n+1)an+2 - (n(n+1)-l(l+1))an]xn = 0

    I can't understand the emphasized bit. How did we go from xn-2 to xn using an+2?
     
    Last edited: May 31, 2012
  2. jcsd
  3. May 31, 2012 #2

    Ray Vickson

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    Until you get the hang of these things the best policy is to write it out in a bit more detail: from [itex] y = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n + \cdots, [/itex] we have
    [tex] y' = a_1 + 2 a_2 x + 3a_3 x^2 + \cdots + (n+1) a_{n+1} x^n + \cdots, \\
    y'' = 2a_2 + 3\cdot 2 a_3 x + 4 \cdot 3 a_4 x^2 + \cdots + (n+2)(n+1) a_{n+2} x^n + \cdots.[/tex] So now you can write out the terms of
    [tex] (1-x^2) y'' - 2xy' + l(l+1) y, [/tex] collect all the terms in [itex]x^0, \: x^2, \: x^2, \ldots, x^n, \ldots [/itex] and require them all to be zero.

    RGV
     
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