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Instead of solving the recursion relation (154) for the coefficients
in the Legendre polynomials, it is easier to use the following trick:
Consider the function
f_{n}(x)=(x^{2}1)^{n}.

(20) 
By differentiation we see that it satisfies the following first order
differential equation,
(1x^{2})f'_{n}+2nxf_{n}=0.

(21) 
Differentiating this equation we get the second order differential eq. for
f_{n},
(1x^{2})f_{n}''2(n1)xf_{n}'+2nf_{n}=0.

(22) 
We wish to differentiate this n times by use of Leibniz's formula,

(23) 
Applying this to (22) we easily get
(1x^{2})f_{n}^{(n+2)}2xf_{n}^{(n+1)}+n(n+1)f_{n}^{(n)}=0,

(24) 
which is exactly Lergendre's differential equation (149). This equation is
therefore satisfied by the polynomials

(25) 
The Legendre polynomials P_{n}(x) are normalized by the requirement P_{n}(1)=1.
Using

(26) 
we get

(27) 
This is Rodrigues' formula for the Legendre function. By means of the
binominal formula we get

(28) 
The summation starts at r=0 and end when n2r is 0 (n=even) or
1 (n=odd). In evaluating P_{n} it is most easy to use (27) directly.
Thus,
.
Next: The Gamma function and
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Previous: The Green's function
Mette Lund
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