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jncarter
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Homework Statement
Show that by letting [itex]z = \zeta^-1 [/itex] and [itex] u = \zeta^{\alpha}v(\zeta)[/itex] that the differential equation,
[itex]z(1-z)\frac{d^{2}u(z)}{d^{2}z}+{\gamma - (\alpha+\beta+1)z}\frac{du(z)}{dz}-\alpha \beta u(z) = 0 [/itex]
can be reduced to[itex]\zeta(1-\zeta)\frac{d^{2}v(\zeta)}{d\zeta^{2}} + {\alpha-\beta+1-(2\alpha-\gamma+2)\zeta}\frac{dv(\zeta)}{d\zeta}-\alpha(\alpha-\gamma+1)v(\zeta) = 0[/itex]
and find the solutions of the first equation about z = ∞ in therms of hypergeometric functions of z.Homework Equations
The first solution to the top equation about z=0 is given by
[itex] u_{1} = 1 + \Sigma^{\infty}_{n=1}\frac{(\alpha)_{n}(\beta)_{n}}{(\gamma)_{n}n!}z^{n} = F(\alpha,\beta,\beta;z)[/itex]
The Attempt at a Solution
First, I'm trying to find expressions for all of the derivatives.
[itex]z=\zeta^{-1}
dz=-\zeta^{-2}d\zeta
\frac{d\zeta}{dz}=-\zeta^{2}[/itex]
First question, is this the correct expression for d2z? I am concerned that I may have to use the chain rule in some manner that I am missing.
[itex]\frac{d^{2}\zeta}{dz^{2}} = 2\zeta^{-3} [/itex] or [itex]d^{2}z = d(-\zeta^{-2}d\zeta) = 2 \zeta^{-3}d\zeta - \zeta^{-2}d^{2}\zeta[/itex]
which gives some funny expression for [itex]\frac{d^{2}\zeta}{dz^{2}}[/itex]
[itex]\frac{d}{dz}=\frac{d\zeta}{dz}\frac{d}{d\zeta} = -\zeta^{2}\frac{d}{d\zeta}[/itex]
[itex]\frac{d^{2}}{dz^{2}} = \frac{d^{2}\zeta}{dz^{2}}\frac{d^{2}}{d\zeta^{2}} = \frac{\zeta^{3}}{2}\frac{d^{2}}{d\zeta^{2}}[/itex]
Now the next part can be done in a different order, but I have decided to go with a method that seems to be more careful.
[itex]\frac{du}{dz} = \frac{d}{dz}(\zeta^{\alpha}v) [/itex]
[itex] = \frac{d}{dz} \zeta^{\alpha} v+\zeta^{\alpha} \frac{dv}{dz}[/itex]
Now use [itex]\frac{d}{dz}(\zeta^{\alpha}) = \frac{d}{dz}(z^{-\alpha}) = -\alpha \zeta^{\alpha - 1}[/itex] to get
[itex]\frac{du}{dz} = -\alpha \zeta^{\alpha - 1} v + \zeta^{\alpha}\frac{dv}{dz}[/itex]
[itex] = -\alpha \zeta^{\alpha - 1} v - \zeta^{\alpha+2}\frac{dv}{d \zeta}[/itex]
Similarly,
[itex]\frac{d^{2}u}{dz^2} = \frac{d^2}{dz^2}(\zeta^{\alpha})v + 2 \frac{d}{dz}(\zeta^{\alpha}) \frac{dv}{dz} + \zeta^{\alpha} \frac{d^2 v}{dz^2}[/itex]
[itex]= - \alpha (\alpha - 1)\zeta^{\alpha -2} v +2 \alpha \zeta^{\alpha-1} \frac{dv}{d \zeta} +\frac{1}{2} \zeta^{\alpha +3} \frac{d^{2}v}{d \zeta^{2}}[/itex]
I've tried plugging this all in a couple of different times and keep getting weird answers. More specifically, I can't seem to get the [itex]\zeta (1- \zeta)\frac{d^{2}v}{d\zeta^{2}}[/itex]. So I tried finding common factors that I could divide out of the equation. So far, it's been no good. This got me thinking that maybe I had the incorrect expression for the second derivatives. Please take a look at my work and point out any problems. I always find that a fresh pair of eyes is enlightening in these problems. Thanks for any help!
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