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If one clicks on the last of the nine formulas on the Wolfram Functions site, one obtains Hypergeometric2F1[a, b, c, z] == HypergeometricU[a, b, a + b - c + 1, 1 - z], but Mathematica asserts that the latter function has only three (not four) arguments (according to the built-in symbol definition). So, there appears to be an inconsistency/error here, in relating the Gaussian hypergeometric function to the confluent hypergeometric function. If so, is there some suitable replacement relation?

I am interested in integrating (see eq.(37) in this preprint for some background context--with $\varepsilon= \sqrt{T}$)

(1 - T)^d T^((3 d)/2 + k)
   Hypergeometric2F1Regularized[2 + 3 d + 2 k, 2 + 3 d + 2 k, 4 + 6 d + 4 k, 1 - T]
   HypergeometricPFQRegularized[{d/2, d, -(d/2) - k}, {1 + d/2, 1 + (3 d)/2 + k}, T]

that is, \begin{equation} (1-T)^d T^{\frac{3 d}{2}+k} \, _2\tilde{F}_1(3 d+2 k+2,3 d+2 k+2;6 d+4 k+4;1-T) \, _3\tilde{F}_2\left(\frac{d}{2},d,-\frac{d}{2}-k;\frac{d}{2}+1,\frac{3 d}{2}+k+1;T\right) \end{equation} over $T \in [0,1]$, for positive integers $d$ and nonnegative integers $k$, and this seemed to be a transformation that I might investigate using, since it would render the two hypergeometric arguments, the same, that is, $T$.

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  • $\begingroup$ How does Mathematica "assert that the latter function has only three (not four) arguments"? $\endgroup$ – David G. Stork May 28 '18 at 0:45
  • $\begingroup$ David G. Stork--can you use the Find Selected Function on the Help page in your Mathematica version? $\endgroup$ – Paul B. Slater May 28 '18 at 0:54
  • $\begingroup$ Yes... You should put that matter in your question more explicitly. $\endgroup$ – David G. Stork May 28 '18 at 0:55
  • $\begingroup$ HypergeometricU in Mathematica is the confluent hypergeometric function U(a, b, z). The function in functions.wolfram.com U(a, b, c, z) is not implemented in Mathematica except through this relation as Hypergeometric2F1. $\endgroup$ – Bob Hanlon May 28 '18 at 1:26
  • $\begingroup$ Thanks, Michael E2. Along the lines you suggest I sent this link to the authors of people.maths.ox.ac.uk/porterm/papers/hypergeometric-final.pdf What's the parent company of Mathematica? $\endgroup$ – Paul B. Slater May 28 '18 at 2:11
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If you visit the Notations tab you will find, under U, the definition of the Gauss type hypergeometric function U (which is not built-in).

To answer your original question, relations including three Kummer's solutions can be used to obtain hypergeometric with the same argument. Separately, I've also sent an email to Michael Trott at Wolfram Research (the developer of Mathematica) as he's one of the driving forces behind the Functions website (or which I'm a regular user).

I note that the paper Numerical methods for the computation of the confluent and Gauss hypergeometric functions does not even mention rational minimax approximations, which is surprising and this is very widely used and is indeed used for machine precision computation of most special functions in Mathematica.

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  • $\begingroup$ Very edifying, thanks! Seems like a good deal of confusing terminology here and there. So, my interest in converting the indicated $1-T$ (2F1) hypergeometric argument to $T$ (of the 3F2) is not accomplishable via the route I hoped. Further, there also do not seem to be conversion rules of such nature for pFq's ($p>2,q>1$). But the Kummer link, TheDoctor gives, does convert $z$ to $1-z$ (at the expense of an added second function), so I'll have to try that. (It should be noted that my formulas are expressed in terms of regularized functions--but that is a matter readily dealt with.) $\endgroup$ – Paul B. Slater May 28 '18 at 15:05
  • $\begingroup$ Upon further reflection, the Kummer link does not seem applicable to my particular 2F1 function, since (in the notation of the link) $a+b-c=c-a-b=0$, so the required ratios $\frac{\Gamma(a+b-c) \Gamma(c)}{\Gamma(a) \Gamma(b)}$ and $\frac{\Gamma (c) \Gamma (-a-b+c)}{\Gamma (c-a) \Gamma (c-b)}$ are indeterminate. $\endgroup$ – Paul B. Slater May 28 '18 at 15:20
  • $\begingroup$ Since $a+b-c=0$, your particular 2FI is logarithmic in $T$, so identities such as functions.wolfram.com/07.23.17.0151.01 can be used. BTW, it's a good idea not to use capital letters for variables (e.g. N, D, C, K are all reserved). You can use other fonts, such as script capitals. $\endgroup$ – TheDoctor May 30 '18 at 2:32
  • $\begingroup$ OK, thanks again, TheDoctor. I will surely investigate this logarithmic relation. As to the use of (capital) $T$, I had an earlier variable $t$ to which I had applied the transformation $t \rightarrow \sqrt{T}$, so perhaps "my heart was somewhat in the right place". $\endgroup$ – Paul B. Slater May 30 '18 at 14:01
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As The Doctor has previously noted, the four-argument version of HypergeometricU[] is not (yet) implemented in Mathematica, tho the writers of the Wolfram Functions site seems to have seen it fit to involve it in their list of formulae. This is not the only instance of the site using an unsupported function, e.g. this previous thread.

I myself find the representation in this page for $U(a,b,c,z)$ as a sum of two Gaussian hypergeometric functions unwieldy; nevertheless, one can prove that $U(a,b,c,z)$ is actually intended to be an independent solution of the hypergeometric DE, which has a logarithmic singularity. (This is in complete analogy to the confluent case, where Kummer's Hypergeometric1F1[] is the "regular" solution, while Tricomi's HypergeometricU[] is the "singular" solution.)

Comparing the first and seventh, as well as the second and eighth formulae in this link, however, shows that you can in fact express $U(a,b,c,z)$ in terms of the Meijer $G$ function, which might be more amenable to further symbolic manipulation:

$$U(a,b,c,z)=\frac{\Gamma(1+a+b-c)}{\Gamma(a)\Gamma(1+a-c)\Gamma(b)\Gamma(1+b-c)}G_{2,2}^{2,2}\left(z\middle|{{1-a,1-b}\atop{1-c,0}} \right)$$

or in Mathematica format, Gamma[1 + a + b - c]/(Gamma[a] Gamma[1 + a - c] Gamma[b] Gamma[1 + b - c]) MeijerG[{{1 - a, 1 - b}, {}}, {{1 - c, 0}, {}}, z].

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