Is the Expansion of Hypergeometric Function Valid for Any |z|?

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SUMMARY

The hypergeometric function ##{}_{2}F_1(a,b,c;z)## is defined by a power series that converges for ##|z|<1##. However, it can be extended beyond this domain through analytic continuation. The discussion highlights the use of the HypExp package in Mathematica for expanding the hypergeometric function ##{}_2 F_1(\alpha,\beta+\epsilon, \gamma- \epsilon, z)##, suggesting that this expansion may be valid for any value of ##|x|## as Mathematica provides results regardless of the size of ##x##. Thus, the analytic continuation is the key factor enabling this broader applicability.

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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!
 
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CAF123 said:
The hypergeometric function, ##{}_{2}F_1(a,b,c;z)## can be written in terms of a power series in ##z## as follows, $${}_{2}F_1(a,b,c;z) = \sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}\,\,\,\,\,\text{provided}\,\,\,\,|z|<1$$

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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