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Decompositoin of f(x) in Legendre polynomials

  1. Sep 6, 2008 #1

    In Wikipedia it's stated that
    Legendre polynomials are useful in expanding functions like

    \frac{1}{\sqrt{1 + \eta^{2} - 2\eta x}} = \sum_{k=0}^{\infty} \eta^{k} P_{k}(x)[/tex]

    Unfortunately, I am failing to see how this can be true. Is there a way of showing this?

    I know that Legendre polynomials form an orthonormal set, and so given any function, we should be able to decompose it into a 'linear combination' of these polynomials. But what form does this decomposition take?
  2. jcsd
  3. Sep 6, 2008 #2


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    If ui is an orthonormal basis for a vector space and v is any vector in that space, then v can be written as [itex]v= \sum a_i u_i[/itex]. Taking the inner product of v with any member of the basis, say uk gives [itex]<u_k, v>= <u_k, \sum a_iu_i>= \sum a_i<u_k, u_i>= a_k[/itex] because <uk, uk>= 1 while <u_k, u_i>=0 for any value of i other than k. That is, you can find the coefficient of Pk by taking the innerproduct (defined as an integral) of the function with Pk. According to Wikipedia (yes, I had to look it up!) the coefficients of the function f(x) for the Legendre polynomials is
    [tex]<f, P_k>= \frac{1}{2n+1}\int_0^1 f(x)P_k(x)dx[/tex]
  4. Sep 6, 2008 #3
    Thank you very much for your explanation. I understand now, the reasoning behind the equation.
    On a technical note,
    1. where does the term [tex] \frac{1}{2n+1}[tex]in front of the integral come from?
    2. why do we integrate from 0-1?

    and lastly,
    3. would the value of [tex]/eta[tex] in our function [tex] \frac{1}{\sqrt{1 + \eta^{2} - 2\eta x}} [tex] play any role in our calculation of Pn's coefficients?
  5. Sep 6, 2008 #4
    Ok, so I think I have the answer to my first two questions and for anyone reading this thread, I'm going to try to answer them.
    The term [tex]
    in front of the integral comes from the fact that the Lagendre polynomials are only orthogonal and not orthonormal.
    By normalizing the polynomials, we can follow HallsofIvy's reasoning.
    The integral I believe should be taken from -1 to 1, since the polynomials Pn are orthogonal on -1[tex]\leq[/tex]x[tex]\leq[/tex]1.
    Having said that, the term that is multiplied by the integral would be [tex]
    rather than [tex]

    What HallsofIvy has done is that he's taken the parity (even/odd) of the functions into account. Since for any function f, the product of f and Pn is even.

    I'm still working on 3. Hope this helps.
  6. Sep 6, 2008 #5
    Actually I think the terms
    [tex] \frac{1}{2n+1}[/tex]
    should in fact be
    [tex] \frac{2n+1}{1}[/tex]
    [tex] \frac{2n+1}{2}[/tex]
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