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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? :)
     
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