Relationship between Legendre polynomials and Hypergeometric functions

[Rulonegger]

Homework Statement

If we define [itex]\xi=\mu+\sqrt{\mu^2-1}[/itex], show that
[tex]P_{n}(\mu)=\frac{\Gamma(n+\frac{1}{2})}{n!\Gamma(\frac{1}{2})}\xi^{n}\: _2F_1(\frac{1}{2},-n;\frac{1}{2}-n;\xi^{-2})[/tex] where [itex]P_n[/itex] is the n-th Legendre polynomial, and [itex]_2F_1(a,b;c;x)[/itex] is the ordinary hypergeometric function.

Homework Equations

[tex]\frac{1}{\sqrt{1-2\mu t+t^2}}=\sum_{n=0}^{\infty}{t^n P_{n}(\mu)}[/tex]
[tex]_2F_1(a,b;c;x)=\sum_{n=0}^{\infty}{\frac{(a)_n (b)_n}{(c)_n}\frac{x^n}{n!}}[/tex]
[tex](\alpha)_n=\alpha(\alpha+1) _\cdots (\alpha+n-1)[/tex]

The Attempt at a Solution

I just tried to write down how [itex]_2F_1(\frac{1}{2},-n;\frac{1}{2}-n;\xi^{-2})[/itex] is, and expand [itex]\xi^{-2}[/itex] with the binomial theorem in terms of [itex]\mu[/itex], but it results in a little complicated double infinite sum, so i feel that there is another way to prove it, but i cannot find it.

 

[andrien]

It will be better if you deal with eqn itself.Just try to convert hypergeometric differential eqn to legendre one by change of variable.Also see what those a,b and c are by comparison.make the change as t=1/2(1-u),u is of legendre and t for hypergeometric.