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I Legendre polynomials and Rodrigues' formula

  1. Sep 3, 2016 #1
    Source: http://www.nbi.dk/~polesen/borel/node4.html#1
    Differentiating this equation we get the second order differential eq. for fn,
    [itex](1-x^2)f''_n-2(n-1)xf'_n+2nf_n=0[/itex] .............(22)
    But when I differentiate to 2nd order, I get this instead,
    [itex](1-x^2)f''_n+2(n-1)xf'_n+2nf_n=0[/itex]


    Applying General Leibniz formula to (22) we easily get,
    [itex](1-x^2)f^{(n+2)}_n-2xf^{(n+1)}_n+n(n+1)f^{(n)}_n=0[/itex] ,................(24)
    How did the author get to (24)? I thought he get that by subbing the Leibniz formula into (22) by replacing the f with fn but it doesn't get to (24) and it doesn't make sense at all.
    What I get,
    [itex](1-x^2)f^{(n+2)}_n+2(n-1)xf^{(n+1)}_n+2nf^{(n)}_n=0[/itex]

    which is exactly Legendre's differential equation (1-49). This equation is therefore satisfied by the polynomials
    $$y=\frac{d^n}{dx^n}(x^2-1)^n$$......................(25)
    The Legendre polynomials Pn(x) are normalized by the requirement Pn(1)=1. Using y=2nn! for x=1,
    Is equation (25) supposed to be the Legendre polynomial? And why do we normalized with x=1?Besides, attempt to normalize the Legendre polynomial with x=1 doesn't get me anywhere.
    $$P_l(1)=\sum^n_{k=0} (-1)^k \frac{(2n-2k)!}{2^nk!(n-k)!(n-2k)!}*1$$
     
    Last edited: Sep 3, 2016
  2. jcsd
  3. Sep 3, 2016 #2

    mathman

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    Science Advisor
    Gold Member

    Your initial statement, that the text is wrong, is correct. I haven't gone further.
     
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