I desperately with hypergeometric functions.

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SUMMARY

The discussion focuses on the transformation of the hypergeometric differential equation using the substitutions \( z = \zeta^{-1} \) and \( u = \zeta^{\alpha} v(\zeta) \). The original equation is shown to be reducible to a new form, which is essential for finding solutions in terms of hypergeometric functions. The user struggles with the conversion from \( u \) to \( v \) and seeks clarification on the correct approach to substitute \( u = \zeta^{\alpha} v \) into the differential equation. The hypergeometric function \( _{2}F_{1} \) is also referenced as part of the solution process.

PREREQUISITES
  • Understanding of hypergeometric functions, specifically \( _{2}F_{1} \)
  • Familiarity with differential equations and their transformations
  • Knowledge of variable substitution techniques in calculus
  • Basic proficiency in mathematical notation and operations
NEXT STEPS
  • Study the derivation and properties of hypergeometric functions, particularly \( _{2}F_{1} \)
  • Learn about the method of substitution in differential equations
  • Explore the implications of variable transformations in solving differential equations
  • Investigate the specific solutions of hypergeometric equations around singular points
USEFUL FOR

Students and researchers in mathematics, particularly those focusing on differential equations, hypergeometric functions, and mathematical analysis. This discussion is beneficial for anyone tackling complex transformations in their studies.

Snoofleglax
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Homework Statement


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} + <br /> \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} + <br /> \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].

Homework Equations



Hypergeometric function:

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

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.
 
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Just stick [tex]u = \zeta^{\alpha} v[/tex] into the differential equation? :)
 

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