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I want to transform \begin{equation} {_3F_1}(-\frac n2,-\frac{n-1}2,\frac{n+1}2;\frac12;z^{-1}) \end{equation} into ${_2F_2}$ using equation (2.2.3.2) of the book by Lucy Joan Slater "Generalized hypergeometric Functions". The equation (2.2.3.2) is \begin{equation} {_{A+1}F_B}[(a),-m;(b);z]=\frac{((a))_m(-z)^m}{((b))_m}{_{B+1}F_A}[1-(b),1-m;1-(a);(-1)^{A+B}/z]. \end{equation} To avoid any misunderstanding, I keep the original form of the equation in the book. If you need more info, I put the section of the book in the end. page 47 page 48

I tired this way \begin{align} {_3F_1}&(-\frac n2,-\frac{n-1}2,\frac{n+1}2;\frac12;z^{-1})\\ =&\frac{(1+\frac n2)_{\frac{1-n}2}(\frac{n+1}2)_{\frac{1-n}2}}{(\frac12)_{\frac{1-n}2}}z^{\frac{n-1}2}{_2F_2}(\frac 12,\frac{n-1}2;\frac{n+1}2,\frac n2+1;-z)\\ =&\frac{\Gamma(1.5)\Gamma(0.5)}{\Gamma(\frac n2+1)\Gamma(\frac{n+1}2)\Gamma(1-\frac n2)}z^{\frac{n-1}2}{_2F_2}(\frac 12,\frac{n-1}2;\frac{n+1}2,\frac n2+1;-z), \end{align} which is not correct with numerical testing. In this case, I chose $1-m=\frac{n+1}{2}$, so that $m=\frac{1-n}{2}$. I've also tried choosing other terms to make $m$ positive, but the result still is still incorrect.

I think I misunderstand the transform. Can anyone help me?

Jobs Adam
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  • What is $m$ in your case? – Gary Jun 14 '24 at 07:38
  • @Gary I choose $1-m=\frac{n+1}{2}$, so that $m=\frac{1-n}{2}$. – Jobs Adam Jun 14 '24 at 07:45
  • What is the condition on $m$ in the book? – Gary Jun 14 '24 at 07:49
  • @Gary I've included the pages of the book, where I do not find any condition restriction on any variable. – Jobs Adam Jun 14 '24 at 08:00
  • Since it talks about the number of terms, $m$ must be a non-negative integer. – Gary Jun 14 '24 at 09:55
  • $-m$ is the numerator parameter that fixes/guarantees the terminating condition of the ${A+1}F_B$. For even $n$ we need $m=n/2$ because $-n/2+1/2$ does not run through 0, and for odd $n$ we need $m=(n-1)/2$ because $-n/2$ does not run through 0. $m=0$ is a border case because in the ${B+1}F_A$ the numerator $1-m$ does not yield a terminating series. – R. J. Mathar Jun 14 '24 at 10:17
  • @R.J.Mathar Thank you for your reply. So, what's the problem with my transform? I have also tried both even and odd $n$, the results are all incorrect. – Jobs Adam Jun 14 '24 at 11:29
  • Note that in your case, $-m=\frac{n+1}{2}$, right before $(b) =\frac{1}{2}$. Fix your calculations accordingly. – Gary Jun 15 '24 at 01:20
  • Note also the negative sign in $(-z)^m$ in the formula from the book. It seems you are missing that as well. – Gary Jun 15 '24 at 01:25
  • @Gary I take the reverse way from the right side of the equation to the left side. I will check if the other way works. – Jobs Adam Jun 15 '24 at 03:40

1 Answers1

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Note that the full-fledged version of the theorem is also in https://arxiv.org/abs/2012.13822, in our case $$ _{A+1}F_B[(a),-m; (b);z] = \frac{((a))_m }{((b))_m} (-z)^m {}_{B+1}F_A[-m,(1-b-m);(1-a-m); (-1)^{A+B}/z] $$ Formula (2.2.3.2) of the book seems to be wrong. [Obviously a formula for a polynomial of order $m$ needs to be transformed into a polynomial of order $m$ (not $m-1$) by reversal.] For even $n$, $m=n/2$, $a_1=-n/2+1/2$, $a_2=n/2+1/2$, $b=1/2$, $$ _3F_2(-n/2, -n/2+1/2,n/2+1/2;1/2; 1/z) $$ $$ % = \frac{(-n/2+1/2)_{n/2}(n/2+1/2)_{n/2}}{(1/2)_{n/2}} (-1/z)^{n/2}{}_2F_2(1/2, 1-n/2; 1/2+n/2 ; 1/2-n/2 ; -z) = \frac{(-n/2+1/2)_{n/2}(n/2+1/2)_{n/2}}{(1/2)_{n/2}} (-1/z)^{n/2}{}_2F_2(-n/2, 1/2-n/2; 1/2; 1/2-n ; -z) $$ $$ = \frac{\Gamma(1/2)\Gamma(n+1/2)\Gamma(1/2)}{\Gamma(1/2-n/2)\Gamma(n/2+1/2)\Gamma(n/2+1/2)} (-1/z)^{n/2}{}_2F_2(-n/2, 1/2-n/2; 1/2 ; 1/2-n ; -z). $$ For odd $n$ we'll leave it as a home work.

R. J. Mathar
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  • Thank you for your answer, that's exactly the problem with the book. But the form of the ${_2F_2}$ of the correct answer seems difficult for further manipulation, which is unfortunate. – Jobs Adam Jun 17 '24 at 02:21
  • This all concerns terminating hypergeometric series, i.e., polynomials of degree n/2 or n/2-1/2. We don't expect polynomials to allow ''further manipulations'' in general. – R. J. Mathar Jul 02 '24 at 09:34