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!

I desperately need help with hypergeometric functions.

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    I have three problems on my homework set that I can't figure out. I'll start with the longest one:

    Show that by letting [tex]z=\zeta^{-1}[/tex] and [tex]u=\zeta^{\alpha}v(\zeta)[/tex] that the hypergeometric differential equation

    [tex]z(1-z)\frac {d^2u}{dz^2} +
    \left[\gamma-(\alpha+\beta+1)z \right] \frac {du}{dz} - \alpha\beta u = 0[/tex]

    can be reduced to

    [tex]\zeta(1-\zeta)\frac {d^2v}{d\zeta^2} +
    \left[\alpha -\beta +1 -(2\alpha+\gamma+2)\zeta \right] \frac {dv}{d\zeta} - \alpha(\alpha-\gamma+1)v = 0[/tex]

    and find the solutions of the first equation about [tex]z=\infty[/tex] in terms of hypergeometric functions of [tex]z[/tex].

    2. Relevant equations

    Hypergeometric function:

    [tex] _{2}F_{1}=\sum^{\infty}_{n=0}\frac{(\alpha)_{n}(\beta)_{n}}{(\gamma)_{n}}\frac{z^{n}}{n!}[/tex]

    3. The attempt at a solution
    What I've gotten so far is finding how the differentials work out:

    [tex]\frac{du}{dz} = -\zeta^{2}\frac{du}{d\zeta}[/tex]

    [tex]\frac{d^{2}u}{dz^{2}} = \zeta^{4}\frac{d^{2}u}{dz^{2}}+2\zeta^{3}\frac{du}{d\zeta}[/tex].

    Then I've substituted into the hypergeometric equation to get:

    [tex]\frac{1}{\zeta}\left(1-\frac{1}{\zeta} \right)\left(\zeta^{4}\frac{d^{2}u}{dz^{2}}+2\zeta^{3}\frac{du}{d\zeta}\right)+\left[\gamma-(\alpha+\beta+1)\frac{1}{\zeta} \right]\left(-\zeta^{2} \frac {du}{d\zeta}\right) - \alpha\beta u = 0[/tex]

    The part where I'm having trouble is trying to convert from u into v. It seems like it should be fairly elementary, but I can't figure it out. And after that I have no idea where to go to get the form my professor wants. The best I can come up with for converting u to v is

    [tex]\frac{du}{d\zeta} = \frac{du}{dv}\frac{dv}{d\zeta}[/tex]

    but it doesn't seem to work out, or there's a wrinkle I'm missing somewhere. Any help is appreciated.
  2. jcsd
  3. Nov 9, 2009 #2
    Just stick [tex] u = \zeta^{\alpha} v [/tex] into the differential equation? :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook