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The Gauss theorem $${_2F_1}(a,b;c;1)=\frac{\Gamma(c-a)\Gamma(c-b)}{\Gamma(c)\Gamma(c-a-b)}$$ allows to compute the analytic continuation of ${_2F_1}(a,b;c;1)$ for $a+b>c$ when the series definition diverges. The same can be done for $_3F_2(1)$ via Thomae relations. My question is how to find analytic continuation for $_4F_3(1)$ to the subdomains of $\mathbb{C}^7$ where the sum of the upper parameters is greater than the sum of the lower parameters, so that the series diverges. I know that no direct analogues of Thomae relations exit in this case so the formulas may be more complicated. I have seen some work by Allen Miller and other authors giving transformations for $_4F_3(1)$, but these transformations leave the excess (total upper parameters minus total lower parameters) invariant, so that divergent series is transformed into divergent series. The same question, of course, pertains to $_pF_{p-1}(1)$ with $p>4$...

Any help is highly appreciated.

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Anton, thank you for correction! –  Dmitri Jan 7 '11 at 1:42

2 Answers 2

The documentation to Christian Krattenthaler's HYP package contains many contiguous relations for arbitrary $_pF_q$s. In particular, C20 on page 17 of http://www.mat.univie.ac.at/~kratt/hyp_hypq/hypm.pdf looks as though it might help.

The URL for the whole HYP package is http://www.mat.univie.ac.at/~kratt/hyp_hypq/hyp.html.

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Ira, thanks a lot for the links! Contiguous relations do provide a transition from divergent series to convergent, however in my case the excess is some indeterminate positive $n$ so that I get as finite sum on the right hand side which I was trying to avoid by finiding a compact transformation a la Thomae relations... –  Dmitri Jan 7 '11 at 1:48

Dmitri, I am confused by your way to continue analytically to a point rather than to a domain. First of all, the Gauss summation formula is valid only if $\operatorname{Re}(c-a-b)>0$, so that your value at 1 is only a formal quantity assigned to the right-hand side (the ratio of gamma functions) when $a+b>c$. "The same can be done for ${}_3F_2$ via Thomae's transformations" is not correct by the same reason. So, your question is not about analytic continuation: you cannot continue the Gauss series at 1, viewed as an analytic function of three variables $a$, $b$ and $c$, through the barrier $\operatorname{Re}(c-a-b)=0$.

If you look for a former way of assigning some reasonable values to the hypergeometric functions $ {}_p F _{\substack{p-1}} $ at 1, it is naturally to play with the classical integral representations: the $(p-1)$-fold integral due to Euler or the complex Barnes integral (expressing the hypergeometric functions as Meijer's $G$-functions). All these can be found in the monographs of W.N. Bailey or L. Slater on hypergeometric functions; alternative sources are Andrews--Askey--Roy and Whittaker--Watson. However, this can never be used in deriving identities/transformations of the hypergeometric functions because of validity issues.

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Wadim, thank you for clarification! I admit I have misunderstood analytic continuation due to very poor knowledge of several complex variables. Nevertheless, looking at each parameter separately is the following argument wrong? Fix some complex $c\ne{0,-1,-2,\ldots}$ and $b$. Take $$f_1(a)={_2F_1}(a,b;c;1),~~\Re(a)<\Re(c-b)$$ and $$f_2(a)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$. The function $f_2(a)$ is meromorphic in the whole complex $a$-plane and we have $f_1(a)=f_2(a)$ in the half-plane $\Re(a)<\Re(c-b)$. Is this not called analytic continuation by definition? –  Dmitri Jan 7 '11 at 2:00
    
Dmirti, thanks for clarification. It's indeed an analytic continuation to a meromorhpic function. And if you divide your $f_1(a)$ by $\Gamma(c-a-b)$, then it can be continued analytically to an entire function... So, your question is about how to "see" the poles of hypergeometric functions at 1 using the summation and transformation theorems. Because the latter are quite "rare", using the integral representations I mention is an efficient way to do the job. Another way is to introduce a dummy integer parameter $-k$ and compensate it in the denominator and then take the limit $k\to+\infty$. –  Wadim Zudilin Jan 7 '11 at 3:15

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