Legendre's associated function - number of zeros?

[JP O'Donnell]

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)...


P_nm(t) = {1/(2ⁿn!)}(1 - t²)^m/2D^n+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

 

[JP O'Donnell]

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.