On Wed, Sep 23, 2015 at 6:03 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Sun, Aug 9, 2015 at 2:42 AM, Bill Gosper <billgosper@gmail.com> wrote:
Oops, of course I meant parameters mod 1. Can we say for sure whether or not 2F1[1/3,2/3,1,z] or 2F1[1/6,2/3,1,z] or 2F1[1/6,5/6,1,z] is algebraic K[algebraic]? (Mod some factors of π.)
Almost certainly, at least for the 1st & 3rd. https://en.wikipedia.org/wiki/J-invariant#Inverse_function Gives identities of the form tau = c K'/K[algebraic1[[J]] = d F'/F[algebraic2[J]], where F is 2F1[1/3,2/3,1,z] or 2F1[1/6,5/6,1] and F'[z]:=F[1-z]. Repeating with J' s.t. algebraic1[J'] = 1 - algebraic1[J] should permit eliminating F' to give F in terms of K and K', at worst. "In theory there is no difference between theory and practice. In practice there is." So far, the algebra has proved intractable. But the Wiki gives the impression that Ramanujan had figured out 2F1[n/d,1-n/d,1,z] in terms of K. --rwg
Bruce Berndt says no, only n/d = 1/2, 1/3, 1/4, and 1/6. He calls these elliptic integrals of signatures 2,3,4, and 6 respectively. Eqn 5.17 of chap 33 of Part 5 of Berndt's magnum opus is Ramanujan's amazing connection between signature 2 and signature 3, which is what I was looking for: (2 EllipticK[(p^3 (2 + p))/(1 + 2 p)])/(Sqrt[1 + 2 p] \[Pi]) == Hypergeometric2F1[1/3, 2/3, 1, (27 p^2 (1 + p)^2)/ (4 (1 + p + p^2)^3)]/(1 + p + p^2) But expressing the K argument in terms of the 2F1 argument (or vice versa) leads to breath-taking radical-storms, from which I extracted Hypergeometric2F1[1/3, 2/3, 1, 1 - GoldenRatio] == (2 Pi Sqrt[GoldenRatio])/ (5^(7/12) Gamma[1/3] Gamma[11/15] Gamma[14/15]) == 3^(3/4) EllipticK[1/64 (47 - 17 I Sqrt[3] - 21 Sqrt[5] + 7 I Sqrt[15])]* E^(I ArcCot[1/8 (1+Sqrt[3])^4 (Sqrt[3]+Sqrt[5])^2])/(Pi Sqrt[GoldenRatio]) and Hypergeometric2F1[1/3, 2/3, 1, -1] == -(((-1)^(11/12) 3^( 3/4) (2 (674484539 - 581632 Sqrt[2])^(2/3) - 1734 (-674484539 + 581632 Sqrt[2])^(1/3) + 769097 I (I + Sqrt[3]))^(1/4) EllipticK[1/32 (16 - \[Sqrt](6 (-14867 + 17 (-674484539 - 581632 Sqrt[2])^(1/3) - 17 (-1)^(2/3) (674484539 - 581632 Sqrt[2])^( 1/3))))])/(2 Sqrt[2] (674484539 - 581632 Sqrt[2])^( 1/12) \[Pi])) and just now Hypergeometric2F1[1/3, 2/3, 1, 9/250 (7 + 19 2^(1/3) - 2 2^(2/3))] == (3 (1 + 2^(2/3)) Gamma[7/6])/(Sqrt[\[Pi]] Gamma[2/3]) (mapping simple to hairy and vice versa) and Hypergeometric2F1[1/3, 2/3, 1, -(27/500) (-16 + Sqrt[6])] == (Sqrt[-1 + 2 Sqrt[3] + Sqrt[6] ] (-10 + 7 Sqrt[2] - 6 Sqrt[3] + 5 Sqrt[6]) Gamma[1/8] Gamma[3/8])/ (8 2^(1/4) \[Pi]^(3/2)) . I wonder if signature 6 will be this weird. --rwg Can that Hypergeometric2F1[1/3,2/3,1,-1] really be in simplest terms?
Glad to see Mathematica is learning about contiguity! --rwg
To do this, Mathematica must know at least two contiguous identities. If there's a symbolic parameter a, say, you can manufacture a contiguous identity just by incrementing a, but the contiguity recurrences may only provide a sublattice of valuations due to division by 0, so you need a pair of "independent" valuations. Below are two such pairs, of which Mathematica and Functions.wolfram.com seem completely unaware. I can supply a number of others which can be hard to find with or without a library.
Out[1026]= {Hypergeometric2F1[a, 1/2 + a, 3/2 - 4 a, 1/5] == ( 2^(3/2 - 10 a) 5^(-(3/2) + 6 a) Sqrt[(5 + Sqrt[5]) \[Pi]] Gamma[3/2 - 4 a])/(Gamma[4/5 - 2 a] Gamma[6/5 - 2 a]), Hypergeometric2F1[a, 1/2 + a, 5/2 - 4 a, 1/5] == ( 2^(7/2 - 10 a) 5^(-(5/2) + 6 a) Sqrt[(5 - Sqrt[5]) \[Pi]] Gamma[5/2 - 4 a])/(Gamma[7/5 - 2 a] Gamma[8/5 - 2 a]), Hypergeometric2F1[5 a, 1/2 + 5 a, 9/10 + 4 a, 1/5] == ( 2^(3/10 - 2 a) Sqrt[\[Pi] - \[Pi]/Sqrt[5]] Gamma[9/10 + 4 a])/( Gamma[3/5 + 2 a] Gamma[4/5 + 2 a]), Hypergeometric2F1[5 a, 1/2 + 5 a, 7/10 + 4 a, 1/5] == ( 2^(-(1/10) - 2 a) Sqrt[(1 + 1/Sqrt[5]) \[Pi]] Gamma[7/10 + 4 a])/( Gamma[2/5 + 2 a] Gamma[4/5 + 2 a])}
Specializations to make an EllipticK: In[1027]:= FunctionExpand[% /. a -> 1/8]
Out[1027]= {( 2 (2/(1 + 2/Sqrt[5]))^(1/4) EllipticK[1/2 - 1/(5^(1/4) Sqrt[1 + 2/Sqrt[5]])])/\[Pi] == ( 11 Sqrt[(5 + Sqrt[5]) \[Pi]])/( 10 10^(3/4) Gamma[19/20] Gamma[31/20]), ( 64 5^(3/4) (2 (1 + 2/Sqrt[5]))^(1/4) EllipticE[1/2 - 1/(5^(1/4) Sqrt[1 + 2/Sqrt[5]])])/(21 \[Pi]) - ( 32 (2/(1 + 2/Sqrt[5]))^(1/4) EllipticK[1/2 - 1/(5^(1/4) Sqrt[1 + 2/Sqrt[5]])])/(21 \[Pi]) - ( 32 Sqrt[5] (2/(1 + 2/Sqrt[5]))^(1/4) EllipticK[1/2 - 1/(5^(1/4) Sqrt[1 + 2/Sqrt[5]])])/(21 \[Pi]) - ( 32 5^(3/4) (2 (1 + 2/Sqrt[5]))^(1/4) EllipticK[1/2 - 1/(5^(1/4) Sqrt[1 + 2/Sqrt[5]])])/(21 \[Pi]) == ( 4 2^(1/4) Sqrt[(5 - Sqrt[5]) \[Pi]])/( 5 5^(3/4) Gamma[23/20] Gamma[27/20]), -(8/ 99) (913 Hypergeometric2F1[-(7/8), 5/8, 2/5, 1/5] + 945 Hypergeometric2F1[-(3/8), 1/8, 2/5, 1/5] - 1526 Hypergeometric2F1[1/8, 5/8, 2/5, 1/5]) == -(( 4 2^(1/20) Sqrt[5 (5 - Sqrt[5]) \[Pi]] Gamma[7/5])/( 3 Gamma[-(3/20)] Gamma[21/20])), -(4/ 51) (237 Hypergeometric2F1[-(7/8), 5/8, 1/5, 1/5] + 1445 Hypergeometric2F1[-(3/8), 1/8, 1/5, 1/5] - 1366 Hypergeometric2F1[1/8, 5/8, 1/5, 1/5]) == -(( 2 2^(13/20) Sqrt[5 (5 + Sqrt[5]) \[Pi]] Gamma[6/5])/( 7 Gamma[-(7/20)] Gamma[21/20]))}
In[1028]:= FunctionExpand[%% /. a -> 1/40]
Out[1028]= {-(16/ 5) (5 Hypergeometric2F1[-(39/40), 21/40, 2/5, 1/5] - 4 Hypergeometric2F1[1/40, 21/40, 2/5, 1/5]) == ( 2 2^(1/4) Sqrt[(5 + Sqrt[5]) \[Pi]] Gamma[7/5])/( 5 5^(7/20) Gamma[3/4] Gamma[23/20]), 256/55 (3 Hypergeometric2F1[-(39/40), 21/40, 2/5, 1/5] - 2 Hypergeometric2F1[1/40, 21/40, 2/5, 1/5]) == ( 8 2^(1/4) Sqrt[(5 - Sqrt[5]) \[Pi]] Gamma[12/5])/( 25 5^(7/20) Gamma[27/20] Gamma[31/20]), ( 2 (2/(1 + 2/Sqrt[5]))^(1/4) EllipticK[1/2 - 1/(5^(1/4) Sqrt[1 + 2/Sqrt[5]])])/\[Pi] == -(( 4 2^(1/4) Sqrt[5 (5 - Sqrt[5]) \[Pi]])/( 7 Gamma[-(7/20)] Gamma[17/20])), Hypergeometric2F1[1/8, 5/8, 4/5, 1/5] == -(( 2 2^(17/20) Sqrt[5 (5 + Sqrt[5]) \[Pi]] Gamma[4/5])/( 11 Gamma[-(11/20)] Gamma[17/20]))}
On Sat, Aug 8, 2015 at 8:37 PM, Oleksandr Pavlyk <pavlyk@wolfram.com> wrote:
On 8/8/2015 10:19 PM, Bill Gosper wrote:
2F1[rational,rational,1 or 1/2,x] ==algebraic1[x] EllipticK[algebraic2[x]] identities? Or at least any more than a,b,c =
Did you try
FunctionExpand[Hypergeometric2F1[1/4 - 2, 1/4 + 1, 1 + 3, x]]
FunctionExpand[Hypergeometric2F1[1/4 - 2, 3/4 + 1, 1/2 + 3, x]]
FunctionExpand[Hypergeometric2F1[1/8 + 4, 5/8 - 2, 1 + 1, x]]
etc.
--Sasha
1/2,1/2,1 1/4,1/2,1 1/4,3/4,1 1/4/1/4,1 1/4,1/4,1/2 1/4,3/4,1/2 (sort of) 1/8,5/8,1 1/12,7/12,1 ? This last one features seriously unwieldy cubics. Tnx, --rwg