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Proof that the legendre polynomials are orthogonal polynomials

  1. Oct 11, 2011 #1
    I'm now studying the application of legendre polynomials to numerical integration in the so called gaussian quadrature. There one exploits the fact that an orthogonal polynomial of degree n is orthogonal to all other polynomials of degree less than n with respect to some weight function. For legendre polynomials that must mean that

    [tex]\int_{-1}^{1} L_n(x) P_m(x) dx = 0[/tex]

    for all P(x) where m is less than n. How does one prove that the legendre polynomials are in the set of such orthogonal polynomials? It's okay that they are orthogonal among themselves, but I wonder how to show that they are orthogonal to everyone else with lower degree?
     
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  3. Oct 11, 2011 #2

    mathman

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    This is unclear. What is the set you are referring to?

    Any polynomial of degree m can be represented as a linear combination of Legendre polynomials of degree at most m.
     
  4. Oct 11, 2011 #3

    Deveno

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    show that the legendre polynomials of degree ≤ n, are linearly independent, and thus form a basis for all polynomials of degree ≤ n.

    therefore, every polynomial of degree ≤ n can be written as a linear combination of the Lj (j = 0,1,2,...,n):

    [tex]P_m(x) = a_0L_0(x) + \dots + a_nL_n(x)[/tex]

    which will make the only surviving term in the inner product the nth one:

    [tex]\int_{-1}^1L_n(x)a_nL_n(x) dx[/tex]

    but if Pm is of degree < n, the coefficent an of Ln will be 0.
     
  5. Oct 12, 2011 #4
    Ah, thanks! That was the argument I was looking for.
     
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