Re: reduction/simplification of hypergeometric-function-related formula

*To*: mathgroup at smc.vnet.net*Subject*: [mg127897] Re: reduction/simplification of hypergeometric-function-related formula*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>*Date*: Thu, 30 Aug 2012 04:11:27 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <k1k8np$90f$1@smc.vnet.net>

On 29 Aug., 07:19, Paul Slater <sla... at kitp.ucsb.edu> wrote: > Note: In the previous posting of this query, the variable "alpha" was employed, which led to now-apparent problems in the mailing with the presentation of the formula in plaintext. I've replaced "alpha" by "a", so hopefully now the formula below will be usable as intended. > > > > > > > I posted a short preprint > > >http://arxiv.org/pdf/1203.4498v2.pdf > > > a few months ago. > > > The central object in it is the formula in Figure 3--given in plain text at the bottom of this email. (It can be copied-and-pasted into a Mathematica notebook.). > > > The formula contains a "family" of six 7F6 hypergeometric functions. It seems to have a number of very interesting (quantum-information-theoretic) properties--as indicated in the preprint. (The upper and lower parameters form intriguing sequences, and the argument of all the six functions is (27/64) = (3/4)^3. For non-negative integers and half-integers, it appears to yield rational values.) > > > It took considerable work to get the formula as "concise" as it is now (LeafCount=530). The original form, generated using the Mathematica FindSequenceFunction command on a sequence of length 32, extended over several pages of output--and also had six (different) hypergeometric functions embedded in it. > > > I have devoted a considerable amount of effort, unsuccessfully, to see if it can be made more concise/digestible. In particular, I have never been able to derive an equivalent form in which fewer than six independent hypergeometric formulas are present. > > > Any thoughts? > > > Thanks! > > > Paul B. Slater > > > Formula in question: > > (4^(-3 - 2 a) > Gamma[5/2 + 3 a] Gamma[ > 2 + 5 a] ((-54 + > a (39 + > 5 a (628 + 25 a (161 + 2 a (-581 + 740 a))))) HypergeometricPFQ[{1, > 2/5 + a, 3/5 + a, 4/5 + a, 5/6 + a, 7/6 + a, 6/5 + a}, {13/10 + a, > 3/2 + a, 17/10 + a, 19/10 + a, 2 + a, 21/10 + a}, 27/ > 64] + (347274 + > 5 a (-312019 + > 25 a (22255 + 8 a (-2431 + 925 a)))) HypergeometricPFQ[{2, 2/5 + a, > 3/5 + a, 4/5 + a, 5/6 + a, 7/6 + a, 6/5 + a}, {13/10 + a, 3/2 + a, > 17/10 + a, 19/10 + a, 2 + a, 21/10 + a}, 27/64] + > 10 ((-769797 + > 25 a (66227 + 4 a (-12843 + 3700 a))) HypergeometricPFQ[{3, > 2/5 + a, 3/5 + a, 4/5 + a, 5/6 + a, 7/6 + a, 6/5 + a}, {13/10 + a, > 3/2 + a, 17/10 + a, 19/10 + a, 2 + a, 21/10 + a}, 27/64] + > 75 ((44133 + 8 a (-6131 + 1850 a)) HypergeometricPFQ[{4, 2/5 + a, > 3/5 + a, 4/5 + a, 5/6 + a, 7/6 + a, 6/5 + a}, {13/10 + a, > 3/2 + a, 17/10 + a, 19/10 + a, 2 + a, 21/10 + a}, 27/64] + > 8 ((-7981 + 3700 a) HypergeometricPFQ[{5, 2/5 + a,3/5 + a, > 4/5 + a, 5/6 + a, 7/6 + a, 6/5 + a}, {13/10 + a, 3/2 + a, > 17/10 + a, 19/10 + a, 2 + a, 21/10 + a}, 27/64] + > 3700 HypergeometricPFQ[{6, 2/5 + a, 3/5 + a,4/5 + a, 5/6 + a, > 7/6 + a, 6/5 + a}, {13/10 + a, 3/2 + a, 17/10 + a, 19/10 + a, > 2 + a, 21/10 + a}, 27/64])))))/(3 Gamma[1 + a] Gamma[ > 3 + 2 a] Gamma[13/2 + 5 a]) I don't see any easy way to simplify your expression but here are two suggestions to understand it better. 1) Structure Let us first look at the structure of your complete expression which I shall call f[a]: Decomposing as f[a] = g[a] h[a] where g[a_] := 4^(-3 - 2*a)*Gamma[5/2 + 3*a]*(Gamma[2 + 5*a]/(3*Gamma[1 + a]*Gamma[3 + 2*a]*Gamma[13/2 + 5*a])) and h[a] = Sum[ h1[m,a] p[m,a],{m,1,5}] h1[ m_,a_] := HypergeometricPFQ[{m, 2/5 + a, 3/5 + a, 4/5 + a, 5/6 + a, 7/6 + a, 6/5 + a}, {13/10 + a, 3/2 + a, 17/10 + a, 19/10 + a, 2 + a, 21/10 + a}, 27/64] where the arguments can be simplified to t1 = Join[{m},{2/5 + a, 3/5 + a, 4/5 + a, 5/6 + a, 7/6 + a, 6/5 + a}] t2 = Table[(2k+1)/10 + a,{k,6,10}] If there was a typo in t1 (is it?) you can even reduce it to t11 = Join[{m},Table[k + a,{k,2,7}]] The degree of the polynomials p[m,a] is 6-m. The coefficients do not seem follow any simple pattern (you might know better, because you have calculated them). 2) Graphical and numerical study Mathematica can help you a lot in this aspect of study. Use Plot[{1 - f[a], 1 - 1/(1 + Pi*a^2)}, {a, 0, 2}, PlotRange -> {0, 1.1}] to see that f[a] has a very simple smmooth and bounded behaviour for a>=0. Also I have plottet a suggested approximate formula fa[a] = 1/(1 + Pi*a^2) You can look at compley a using e.g. Plot3D[Abs[h1[3, x + I*y]], {x, -3, 0}, {y, -1, 1}] Next we can let Mathematica calculate some numerical values f[0] 1 f[1]; N[%, 30] 0.242424242424242424242424242424 i.e. f[1] = 24/99 = 8/33 No periodicity is seen in the next example f[2]; N[%,30] 0.0804953560371517027863777089783281733746130030959752321981424148606811145= 5108359133126934984520123839 but again a rational number is abtained at a=1/2 f[1/2]; N[%,30] 0.4531250000000000000000000000000000000000000000000000000000000000000000000= 000000000000000000000000000 Interesting study (at least for me) which makes use of some of Mathematica's features. Regards, Wolfgang

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