Escaping the unit disk

Hypergeometric functions are defined in terms of their power series representation. This power series converges inside the unit circle and diverges outside. The functions extend to most [1] of the complex plane, and do so in a computationally nice way.

Analytic continuation is often sort of existential: we prove that it can be done, but we may not be able to say much explicitly about the result. But for hypergeometric functions, there is an explicit formula.

This formula is so complicated that it’s possible to miss its basic structure. Here’s an outline version:

F(\ldots z) = \ldots F(\ldots z^{-1}) + \ldots F(\ldots z^{-1})

This says that F evaluated at a point z outside of the unit circle is equal to the sum of two terms involving F evaluated at 1/z, a point inside the unit circle. This means you can evaluate F everywhere [1] if you can evaluate it inside the unit circle.

The parameters in each of the F‘s above are different, so the formula doesn’t relate a single function’s value at z to the same function’s values at 1/z. Instead, it relates the value of one instance of a class of functions at z to values of other instances of the class at 1/z.

And now without further ado, here is the analytic continuation formula in all its glory:

\begin{align*}  \frac{\Gamma(a)\Gamma(b)}{\Gamma(c)}F(a, b;c; z) &=  \frac{\Gamma(a)\Gamma(b-a)}{\Gamma(c-a)} \,\,(-z)^{-a}\,\,F(a, 1-c+a; 1-b+a; z^{-1}) \\ &+ \,\frac{\Gamma(b)\Gamma(a-b)}{\Gamma(c-b)} \,\,(-z)^{-b}\,\,F(b,\, 1-c+b; 1-a+b; z^{-1}) \end{align*}

[1] There may be a singularity at 1, and there may be a branch cut along the real axis from 1 to infinity.

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