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I am trying to find a relation between hypergeometrics $${}_2F_1(a,b,c;z)\,\,\text{and}\,\,{}_2F_1(a+1,b+1,c+1;z)$$

I can see that $$\frac{\partial}{\partial z}{}_2F_1(a,b,c;z) = \frac{ab}{c}{}_2F_1(a+1,b+1,c+1;z)$$ but I was wondering if I can write a relation between them that admits the form $${}_2F_1(a+1,b+1,c+1;z) = f(a,b,c,z) \cdot {}_2F_1(a,b,c;z) + \,\,\text{some other hypergeometric} $$

I know there is a hypergeometric identities page on Wolfram with a plethora of identities, but I couldn't see such an identity for the case I am considering. I am not sure, however, if this list is exhaustive.

Thanks!

CAF
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  • What parameters you would like "some other hypergeometric" to have? – Start wearing purple Jan 21 '16 at 21:24
  • @Startwearingpurple: I was hoping the 'some other hypergeometric' would actually be a trivial hypergeometric in the sense that its arguments are such that the hypergeometric turns out to be just $z$ to some power or something at least simpler than a hypergeometric. – CAF Jan 23 '16 at 12:41
  • Did you look at: http://dlmf.nist.gov/15.5.E19 ? Which is basically the discrete extension of the differential equation. – rrogers May 19 '16 at 15:56
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    In addition if a,b,c are integer you can almost certainly transform the series into a combinatorial summation. – rrogers May 19 '16 at 16:00

2 Answers2

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Little late to the game here but your request is possible by using the contigious relations of the hypergeometric function. Using the notation in the linked paper $$ F\left({a_1+1,a_2+1\atop a_3+1};x\right)=\mathcal A_1\mathcal A_2\mathcal A_3 F\left({a_1,a_2\atop a_3};x\right). $$ There are a number of operators you can derive from the contigious relations that will be equivalent to $\mathcal A_1\mathcal A_2\mathcal A_3$. For example, equation $(17)$ gives $\mathcal A_1\mathcal A_2\mathcal A_3=-\frac{a_3}{a_2 z}(\mathcal I-\mathcal A_1)$ so that $$ \begin{aligned} F\left({a_1+1,a_2+1\atop a_3+1};x\right)% &=-\frac{a_3}{a_2 z}(\mathcal I-\mathcal A_1) F\left({a_1,a_2\atop a_3};x\right)\\ &=-\frac{a_3}{a_2 z}F\left({a_1,a_2\atop a_3};x\right)+\frac{a_3}{a_2 z}F\left({a_1+1,a_2\atop a_3};x\right), \end{aligned} $$ which is in the form presented in the question.

If any of the parameters $a_1$, $a_2$, or $a_3$ are integers it is likely that repeated application of these contigious relations can reduce the second hypergeometric term above to some "simpler" product of elementary functions.

Aaron Hendrickson
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I do not have the answer for the general form you asked. However, I found out one special case which was useful for my work: a = b = c -1. Here is the relation:

$_2F_1(K+1, K+1; K+2; x) = \frac{K+1}{Kx}\big[\frac{1}{(1-x)^K}-_2F_1(K,K;K+1;x)\big]$

Proof can be found here.

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