# Relationship between Legendre polynomials and Hypergeometric functions

1. May 4, 2013

### Rulonegger

1. The problem statement, all variables and given/known data
If we define $\xi=\mu+\sqrt{\mu^2-1}$, show that
$$P_{n}(\mu)=\frac{\Gamma(n+\frac{1}{2})}{n!\Gamma(\frac{1}{2})}\xi^{n}\: _2F_1(\frac{1}{2},-n;\frac{1}{2}-n;\xi^{-2})$$ where $P_n$ is the n-th Legendre polynomial, and $_2F_1(a,b;c;x)$ is the ordinary hypergeometric function.

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

3. The attempt at a solution
I just tried to write down how $_2F_1(\frac{1}{2},-n;\frac{1}{2}-n;\xi^{-2})$ is, and expand $\xi^{-2}$ with the binomial theorem in terms of $\mu$, 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.

2. May 6, 2013

### 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.

Last edited: May 6, 2013