- #1
cephron
- 124
- 0
Hi guys, I'm dealing with a function whose integral (via Wolfram integrator) carries a hypergeometric function term: 2F1([itex]\frac{1}{4}, \frac{1}{2}, \frac{3}{2},[/itex] 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([itex]\frac{1}{4}, \frac{1}{2}, \frac{-3}{2},[/itex] 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.
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([itex]\frac{1}{4}, \frac{1}{2}, \frac{-3}{2},[/itex] 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.