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Convergence of a hypergeometric

  1. Jul 20, 2015 #1

    CAF123

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    The hypergeometric function, ##{}_{2}F_1(a,b,c;z)## can be written in terms of a power series in ##z## as follows, $${}_{2}F_1(a,b,c;z) = \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}\,\,\,\,\,\text{provided}\,\,\,\,|z|<1$$

    So we may reexpress any hypergeometric function as a power series like this as long as the last argument modulus is less than 1.

    My question is, I also have an expansion of a hypergeometric of the form ##{}_2 F_1(\alpha,\beta+\epsilon, \gamma- \epsilon, z)##, where ##\alpha, \beta, \gamma## are real numbers, by using the HypExp package on Mathematica (expansion in ##\epsilon##) and was wondering if I use this expansion do I also require ##|z|<1##? When I write the code in Mathematica, the last argument is replaced by simply ##x## say and Mathematica gives me an expansion regardless of the size of ##x## so I am thinking ##{\it this}## expansion (and not the power series one) is maybe valid independent of ##|x|## but would be nice to confirm.

    Thanks!
     
  2. jcsd
  3. Jul 20, 2015 #2

    micromass

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    This specific form of the hypergeometric function is valid for ##|z|<1##. The hypergeometric function can be extended beyond ##|z|<1## by analytic continuation. So this analytic continuation is likely what mathematica is calculating.
     
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