# I desperately need help with hypergeometric functions.

1. Nov 9, 2009

### Snoofleglax

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 $$z=\zeta^{-1}$$ and $$u=\zeta^{\alpha}v(\zeta)$$ that the hypergeometric differential equation

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

can be reduced to

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

and find the solutions of the first equation about $$z=\infty$$ in terms of hypergeometric functions of $$z$$.

2. Relevant equations

Hypergeometric function:

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

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

$$\frac{du}{dz} = -\zeta^{2}\frac{du}{d\zeta}$$

$$\frac{d^{2}u}{dz^{2}} = \zeta^{4}\frac{d^{2}u}{dz^{2}}+2\zeta^{3}\frac{du}{d\zeta}$$.

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

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

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

$$\frac{du}{d\zeta} = \frac{du}{dv}\frac{dv}{d\zeta}$$

but it doesn't seem to work out, or there's a wrinkle I'm missing somewhere. Any help is appreciated.

2. Nov 9, 2009

### clamtrox

Just stick $$u = \zeta^{\alpha} v$$ into the differential equation? :)