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I need a numerically stable way to compute the following ratio:

$$\frac{_{2}F_{1}(a+1,b;c;x)}{_{2}F_{1}(a,b;c;x)},$$

All the parameters are real numbers, with $a< 0$,$\ $ $b,c > 0$ and $0<x<1$.

Right now I am using GSL's implementation of the hypergeometric function, but I keep getting underflows/overflows.

Is there a simplification that I can use?

Update: Following @Leiucippus comment, is there a continued fraction expansion for $\frac{_{2}F_{1}(a+1,b;c;x)}{_{2}F_{1}(a,b;c;x)}$ and/or $\frac{_{2}F_{1}(a+2,b;c;x)}{_{2}F_{1}(a,b;c;x)}$? Hopefully one that converges quickly so it is numerically useful.

a06e
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    Gauss has a method for a continued fraction method http://en.wikipedia.org/wiki/Gauss_continued_fraction – Leucippus Aug 05 '14 at 19:56
  • @Leucippus I don't see how that helps. There you have equations relating $_2F_1(a+1,b;c+1;x)$ to $_2F_1(a,b;c;x)$, but not $_2F_1(a+1,b;c;x)$ to $_2F_1(a,b;c;x)$. How can I use that? – a06e Aug 06 '14 at 14:48

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I know this is a little late to the game but you can work with the contigious relations of the hypergeometric function to get the ratio into a more useful form for using continued fraction expansions. Using the notation of the linked paper we may write $$ \begin{aligned} F\left({a+1,b\atop c};z\right)% &=\mathcal A_1 F\left({a,b\atop c};z\right)\\ &=\left(\mathcal I+\frac{bz}{c}\mathcal A_1\mathcal A_2\mathcal A_3\right) F\left({a,b\atop c};z\right). \end{aligned} $$ Hence, $$ \begin{aligned} \frac{F(a+1,b;c;z)}{F(a,b;c;z)}% &=1+\frac{b z}{c}\frac{F(a+1,b+1;c+1;z)}{F(a,b;c;z)}\\ &=1+bz\frac{\mathbf F(a+1,b+1;c+1;z)}{\mathbf F(a,b;c;z)}. \end{aligned} $$ Then, DLMF $\S 15.7$ provides continued fraction expansions for the remaining ratio of hypergeometric functions for certain values of $z$. If this does not help you can further work with these contiguous relations to turn your expression into a ratio of hypergeometric functions that will match one of the continued fraction expansions.

Aaron Hendrickson
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