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ThePerfectHacker · #7 July 29th, 2007, 04:35 PM

Here is a way to get the Rodrigues formula since it was mentioned.

Let $z=(x^2-1)^n$ .

We see that,
$(x^2-1)\frac{dz}{dx} - 2nx z = 0$

Take the derivative $n$ times to get,
$(1-x^2)\frac{dz}{dx}+n(n+1)z^{(n-1)}=0$
Take the derivative one last time,
$\left[(1-x^2)\frac{dz}{dx}\right]'+n(n+1)z^n=0$
Hence,
$\frac{d^{n}(x^2-1)^n}{dx^n}$
Is a polynomial which solves the equation.
But we write it as,
$\frac{1}{2^nn!}\cdot \frac{d^n(x^2-1)^n}{dx^n}$
Because if we solve the Legendre equation with a power series we find that the leading coefficient of the polynomial starts with that factor. So to make Rodrigues formula agree with an infinite series solution we make the leading term equal to that factor.
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Warning Complex Analysis Ahead

Here is a fabulous identity due to Laplace.
$P_n(x) = \frac{1}{\pi}\int_0^{\pi} (x+\sqrt{x^2-1}\cos \phi)^n d\phi$

We begin by noting that,
$P_n(x) = \frac{1}{2\pi i}\oint_{\gamma} \frac{1}{2^n}\frac{(z^2-1)^n}{(z-x)^{n+1}}dz$
Where $\gamma$ is any peicewise smooth simple closed curve containing $x$ .
This is immediately true by Cauchy's Integral Formula.

Let $x>1$ define $\gamma = x+\sqrt{x^2-1}e^{i\phi}$ for $0\leq \phi \leq 2\pi$ . That is a circle centered at $x$ with radius $\sqrt{x^2-1}$ .

Evaluating the above integral with this parametrization (and doing some simplification, not shown) we get Laplace's formula.