3
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I've found that this identity seems to be true for all positive integers $n$: $$1= \,_4F_3\left(\frac{1}{6},\frac{5}{6},\frac{1}{2}-\frac{n}{2},1-\frac{n}{2};\frac{3}{2},\frac{5}{6}-\frac{n}{2},\frac{7 }{6}-\frac{n}{2};1\right)2\frac{\left(\frac{n+1}{2}\right)_{n-1} }{\left(\frac{n}{2}+1\right)_{n-1}}\\ -\,_4F_3\left(-\frac{1}{3},\frac{1}{3},\frac{1}{2}-\frac{n}{2},-\frac{n}{2};\frac{1}{2},\frac{1}{3}-\frac{n}{2},\frac{2 }{3}-\frac{n}{2};1\right) $$ I've searched for identities involving the $_4F_3$ hypergeometric function but have not found anything matching this set of parameters.

Note that both hypergeometric functions are Saalschützian and their series are terminating for $n$ a positive integer.

Do you have any ideas of how to prove this is true or find an applicable identity?

The identity shows up in trying to prove that the solution found here: https://mathoverflow.net/questions/291546/second-order-recurrence-relation-for-third-order-polynomial-root is correct using induction. That relation in turn showed up in a quantum field theory calculation.

Best regards, Petter

Mathematica expression for convenience:

0==-1 - HypergeometricPFQ[{-(1/3), 1/3, 1/2 - n/2, -(n/2)}, {1/2, 
   1/3 - n/2, 2/3 - n/2}, 1] + (
 2 HypergeometricPFQ[{1/6, 5/6, 1/2 - n/2, 1 - n/2}, {3/2, 5/6 - n/2, 
    7/6 - n/2}, 1] Pochhammer[(1 + n)/2, -1 + n])/
 Pochhammer[1 + n/2, -1 + n]
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  • $\begingroup$ First guess would be Induction, did you try that? $\endgroup$ – Rudi_Birnbaum Mar 28 '18 at 12:13
  • $\begingroup$ I suppose $(a)_n$ is supposed to be the Pochhammer symbol $\Gamma(a+n)/\Gamma(a)$? $\endgroup$ – Rudi_Birnbaum Mar 28 '18 at 12:22
  • $\begingroup$ Do you mean induction in n? Proving the identity for n=k by assuming it works for n=k-1. That could be a route but I would then need to find an identity between the 4F3 functions relating different n. $\endgroup$ – Petter Mar 28 '18 at 12:28
  • $\begingroup$ Indeed, $(a)_n$ is the Pochhammer symbol, denoting the rising factorial. $\endgroup$ – Petter Mar 28 '18 at 12:29
  • $\begingroup$ Yes, to get a feeling for the thing it would probably be good to start with $n=1$ "manually". $\endgroup$ – Rudi_Birnbaum Mar 28 '18 at 12:30

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