1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relationship between Legendre polynomials and Hypergeometric functions

  1. May 4, 2013 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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]

    3. 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.
     
  2. jcsd
  3. May 6, 2013 #2
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Relationship between Legendre polynomials and Hypergeometric functions
Loading...