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Lengendre polynomials

  1. Apr 8, 2004 #1
    I know that legendre polynomials are solutions of the differential equation is (1-x^2)d^2y/dx^2 - 2x dy/dx+l(l+1)y=0 where l is an integer. The first five solutions are P0(x)=1, P1(x)=x, P2(x)=3/2x^2-1/2, P3(x)=5/2x^3-3/2x, P4(x)=35/8x^4-15/4x^2+3/8

    The problem is that I don't understand what the problem is telling me to do. It says to show that each of the polynomials Pl(x) solves the differentil equation with its particular value l. Do I just plug in l? For example, for P0(x)=1, would I plug in 1 for x and 0 for l? I'm really confused.

    Another problem is that I have to show by doing 10 integrals that if l is not equal to m, that integral from -1 to 1 dxPl(x)Pm(x)=0 so that these polynomials are orthogonal on the interva1 [-1,1].

    Do I just take a value for l and one for m 10 times. So for the first integral, m=1 and n=2?
  2. jcsd
  3. Apr 9, 2004 #2
    [tex]y_l = P_l(x)[/tex]

    is a solution to the differential equation

    [tex](1-x^2)\frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + l(l+1)y = 0[/tex]

    They make up pairs. When l = 2, we have y_2 = P_2(x) is a solution to the differential equation given if 2 is plugged in for l (meaning the last term is 6).

    If l = 15, then we have to plug in l = 15 into y_l and l = 15 into the differential equation. Then the solution y_15 will solve the differential equation made when we substitute 15 for l.

  4. Apr 9, 2004 #3

    matt grime

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    z solves an equation, f, if f(z)=0, o yes you just plug in the z which is P_l into the equation defining the l^th legendre polynomial.

    secondly you must do the integrals for every pair of numbers (l,m) where l and m are one of 1,2,3,4
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