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Legendre's associated function - number of zeros?

  1. Jul 3, 2008 #1
    Hi.

    It is stated that the associated Legendre functions change their sign n-m times in the interval -1 <= t <= 1, where t = cos(theta)...


    Pnm(t) = {1/(2nn!)}(1 - t2)m/2Dn+m(t2 - 1)n ... Associated Legendre function


    I can see how this number arises having differentiated (t2 - 1)n, n+m times. But this is then multiplied by a factor of (1 - t2)m/2, which is a polynomial in t of degree m.

    So multiplying both polynomials you have a polynomial of degree [2n - (n+m)] + [m] = n

    Where have I gone wrong in my understanding of this?

    Thanks
     
  2. jcsd
  3. Jul 3, 2008 #2
    Found the reason why...

    The (1-t2)m/2 term of the function has no effect on the zero's of the function as it will always be positive (|t| = |cos(theta)| =< 1)

    Therefore the number of crossings will be n-m as stated.
     
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