Hello!

Section 6.12 gives code on how to evaluate the hypergeometric function 2F1 outside the convergence interval of the usual series representation of the function. I have tried this code for the values a = b = c = 1 and for x ranging from -2 to 2 with all variables being real (imaginary part zero).

I then got Mathematica 5.0 to evaluate the same thing and compared the two. NR does perfectly up to x = 1 and then gets very poor results.

The whole integration machinery seems to work based on the [-2,1] results, so is it the starting point of the path integral?

Could you help me?

Thank you very much.
Patrick

Hi Pat,

I'm unable to reproduce your problem - the routine works fine, even for e.g. x=1.00001. The choice of starting point is crucial, as described in the text and implemented in the routine. In particular, you cannot try to integrate though the point x=1. You have to go around it in the complex plane.

Saul Teukolsky