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Legendre polynomials and Rodrigues' formula

Instead of solving the recursion relation (1-54) for the coefficients in the Legendre polynomials, it is easier to use the following trick:

Consider the function

fn(x)=(x2-1)n. (20)

By differentiation we see that it satisfies the following first order differential equation,

(1-x2)f'n+2nxfn=0. (21)

Differentiating this equation we get the second order differential eq. for fn,

 
(1-x2)fn''-2(n-1)xfn'+2nfn=0. (22)

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

\begin{displaymath}\frac{d^n}{dx^n}A(x)B(x)=\sum_{k=0}^n\frac{n!}{k!(n-k)!}~\frac{d^kA}{dx^k}~
\frac{d^{n-k}B}{dx^{n-k}}.
\end{displaymath} (23)

Applying this to (22) we easily get

(1-x2)fn(n+2)-2xfn(n+1)+n(n+1)fn(n)=0, (24)

which is exactly Lergendre's differential equation (1-49). This equation is therefore satisfied by the polynomials

\begin{displaymath}y=\frac{d^n}{dx^n}(x^2-1)^n.
\end{displaymath} (25)

The Legendre polynomials Pn(x) are normalized by the requirement Pn(1)=1. Using

\begin{displaymath}y=2^nn!~~{\rm for}~~x=1,
\end{displaymath} (26)

we get

 \begin{displaymath}P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n.
\end{displaymath} (27)

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

\begin{displaymath}P_n(x)=\frac{1}{2^n}\sum_r(-1)^r\frac{(2n-2r)!}{r!(n-r)!(n-2r)!}~x^{n-2r},
\end{displaymath} (28)

The summation starts at r=0 and end when n-2r is 0 (n=even) or 1 (n=odd). In evaluating Pn it is most easy to use (27) directly. Thus, $P_0(x)=1, P_1(x)=x, P_2(x)=\frac{1}{2}(3x^2-1),...$ .


next up previous
Next: The Gamma function and Up: No Title Previous: The Green's function
Mette Lund
2000-04-27