Algebraic Manipulation of Hypergeometric F'n Parameters

Click For Summary
SUMMARY

The discussion focuses on evaluating the hypergeometric function 2F1(1/4, 1/2, 3/2, z) for |z| > 1, where traditional series expansion fails due to convergence issues. The user discovered a special case of the hypergeometric function, 2F1(1/4, 1/2, -3/2, z), which simplifies to a polynomial but is hindered by a negative sign. The conversation emphasizes the need for algebraic manipulations or alternative evaluation techniques to overcome this obstacle and efficiently compute the function for given values of z.

PREREQUISITES
  • Understanding of hypergeometric functions, specifically 2F1 parameters.
  • Familiarity with convergence criteria for power series.
  • Basic knowledge of algebraic manipulation techniques.
  • Experience with computational tools like Wolfram Integrator.
NEXT STEPS
  • Research algebraic manipulations applicable to hypergeometric functions.
  • Explore alternative evaluation methods for hypergeometric functions when |z| > 1.
  • Study the properties and applications of special cases of hypergeometric functions.
  • Learn about convergence tricks for logarithmic functions and their extensions to other functions.
USEFUL FOR

Mathematicians, physicists, and computational scientists dealing with hypergeometric functions, particularly those needing efficient evaluation methods for complex integrals.

cephron
Messages
123
Reaction score
0
Hi guys, I'm dealing with a function whose integral (via Wolfram integrator) carries a hypergeometric function term: 2F1(\frac{1}{4}, \frac{1}{2}, \frac{3}{2}, z).

I need to evaluate this function twice for every integral, but |z| will often be greater than 1, so I can't use the hypergeometric series to evaluate this term because it won't converge.

I looked in the database of special cases for the hypergeometric function on wolfram, and found this one, for parameters 2F1(\frac{1}{4}, \frac{1}{2}, \frac{-3}{2}, z), which equates it to a nice, explicit polynomial expression. It would be perfect except for that darned negative sign!

So before I go playing a violin for myself, I was wondering if anyone knew any algebraic-ish manipulations one could do that might make this work; any way to work around the negative sign? I'll post more detail about the whole function on demand.

Failing that, I'm open to input on how to best (ie. most speedily on computer) evaluate this term for a given z.
 
Physics news on Phys.org
There is a trick for the natural logarithm. We have a power series for ##\ln(1+x)## which converges for ##|x|<1##. To obtain a result for ##x>1## we can use ##\ln(1+x)=-\ln\left(1+\left(- \dfrac{x}{1+x}\right)\right)##

I think you have to look for similar tricks.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K