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Good evening

Say we are working with a hypergeometric $3F3(a,b,c; d,e,f; z)$ function. I know that $\frac{d}{dz}3F3(a,b,c; d,e,f; z)=\frac{abc}{def}3F3(a+1,b+1,c+1; d+1,e+1,f+1; z)$. There are numerous references for why this is so.

For my specific parameters, $3F3(1,1,1; 2,2,2;z)$ Mathematica gives a different solution. The result I get is:

$$ \frac{d}{dz}3F3(1,1,1; 2,2,2; z)=2F2(1,1; 2,2; z)-3F3(1,1,1; 2,2,2; z)$$

Can anyone explain why this would be the case, or provide a reference to this statement? I have been through Gradstein and Ryzhik and a few other books, but none of them give an identity that will result in a decrease in the number of hypergeometric parameters!

Martin Sleziak
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Albert
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  • This post discusses the solution https://math.stackexchange.com/questions/3084146/an-identity-on-small-pf-q-left-left-beginarrayc-a-11-a-21-dots-a-p?rq=1 – Albert Aug 30 '19 at 14:13

1 Answers1

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There is something strange.

Working on Wolfram Cloud, I got $$\frac d {dz}\Big[\, _3F_3(1,1,1;2,2,2;z)\Big]=-\frac{z \, _3F_3(1,1,1;2,2,2;z)+\log (-z)+\Gamma (0,-z)+\gamma }{z^2}$$

Computing for $z=\frac 1 \pi$ using central differences $(\Delta z=10^{-6})$, the result is $0.137625891463763$ while the above formula gives $0.1376258914637604$; so, at least, this can be considered as correct.

If I use your formula $$\,_2F_2(1,1; 2,2; z)-\,_3F_3(1,1,1; 2,2,2; z)$$ this would give $0.0438076818477723$ and the ratio of the numbers is $\pi$ !

Just entering $\,_2F_2(1,1; 2,2; z)$ gives as output $$-\frac{\log (-z)+\Gamma (0,-z)+\gamma }{z}$$ Apparently, you lost a factor somewhere

Claude Leibovici
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  • I see what you mean. After rerunning the code I get the same solution you do. However, for more parameters, say 4F4 or 5F5 I do get the results that I mentioned in the original question. – Albert Aug 29 '19 at 13:57