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Legendre Polynomial (anti)symmetry proof

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]P_{n}(x)[/itex] denote the Legendre polynomial of degree n, n = 0, 1, 2, ... . Using the formula for the generating function for the sequence of Legendre polynomials, show that:

    [itex]P_{n}(-x) = (-1)^{n}P_{n}(x)[/itex]

    for any x [itex]\in[/itex] [-1, 1], n = 0, 1, 2, ... .

    2. Relevant equations

    Generating function for the sequence of Legendre polynomials:

    [itex]\sum P_{n}(x)r^{n} = (1 - 2rx + r^{2})^{-\frac{1}{2}}[/itex]

    3. The attempt at a solution

    I guess I don't really know where to begin with this. I tried differentiating both sides of the generating function with respect to r to obtain:

    [itex]\sum nP_{n}(x)r^{n-1} = (x - r)\sum P_{n}(x)r^{n}[/itex]

    but I don't see how this might move closer to the desired result. I imagine differentiating with respect to x would lead to similar difficulties.


    I then tried just directly substituting -x into the generating function, giving

    [itex](1 + 2rx + r^{2})^{-\frac{1}{2}}[/itex]

    as a generating function for [itex]P_{n}(-x)[/itex], but again I don't really see where this might lead me.

    Perhaps there's a general method I'm missing; I'm still very new to the concept of Legendre polynomials and generating functions in general. My attempts at scouring the Internet for a proof of this specific identity have turned up a sole reference to a textbook I have no access to. I'm not really looking for a complete proof to be given to me on a plate, but a nudge in the right direction would certainly be appreciated.
     
  2. jcsd
  3. Mar 5, 2012 #2

    Dick

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    Try substituting both -x for x and -r for r into the generating function expression. What does that tell you?
     
  4. Mar 5, 2012 #3
    Ahh, figures it'd be something as simple as that.

    [itex]\sum P_{n}(-x)(-r)^{n} = (1 + 2rx + r^{2})^{\frac{1}{2}} = \sum P_{n}(x)r^{n}[/itex]

    [itex]\Rightarrow \sum P_{n}(x)r^{n} = \sum P_{n}(-x)(-1)^{n}r^{n}[/itex]

    and then by applying the identity theorem for power series we get [itex]P_{n}(x) = (-1)^{n}P_{n}(x)[/itex] as required.

    Thanks for your help. :)
     
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