On Fri, Apr 4, 2014 at 4:26 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Thu, Apr 3, 2014 at 4:33 PM, Bill Gosper <billgosper@gmail.com> wrote [...]
WDS> --These questions are just new questions...
If you were to take the attitude you are not even going to permit elliptic/modular functions to appear, then it might be the reflection and ntupling is all there is and Chowla & Selberg might be irrelevant.
rwg>A tool like "ToElliptic" would be valuable, and might synergize with this duction hack, but the user should keep control. If my memory is right about eta(e^-(pi sqrt(n/d))) always coming out in Gammas, there remains a chance of finding an "inaccessible" identity. --rwg
I finally see what Warren was driving at. My "ductions" list has, for (n/24)!,
(5/24)! -> (5 Sqrt[1/2 (-1 + Sqrt[2]) (-1 + Sqrt[3])] (1/24)! (1/3)!)/(2 (1/6)!), (7/24)! -> (7 (1/24)! Sqrt[( 5 (-1 + Sqrt[3]) \[Pi] Sin[\[Pi]/24] Sin[(5 \[Pi])/24])/((1/12)! (5/ 12)!)])/(6 3^(1/4)), (11/24)! -> (11 Sqrt[5 (1 + Sqrt[3]) \[Pi]] (1/24)! (1/3)! Sin[\[Pi]/24])/( 4 2^(1/4) 3^(3/4) (1/6)! Sqrt[(1/12)! (5/12)!]),
I.e., reduction to 1/24, but no further. But I find in my notes \[Eta][(1/(E^((Sqrt[2] * Pi)/(Sqrt[3]))))]== ((Gamma[1/24] * (Tan[Pi/24])^(1/4) * (Sin[Pi/8])^(1/6))/(2 * 2^(1/12) * 3^(1/8) * Sqrt[Gamma[1/12]] * Sqrt[Pi])) i.e., we can reduce to (1/12)! by allowing Dedekind eta.
and then rest of my list takes us all the way down to (1/3)! and (1/4)!. And I have etas for those, so (n/24)! = <algebraic>*etas. Recall also my conjecture that (n/prime)! is "irreducible" if prime>2n. Yet (1/7)! (2/7)!)/(3/7)! == 16 π Cos[π/14] DedekindEta[I/Sqrt[7]]^2/21/7^(1/4) gets (n/7)! down to at worst (2/7)! . People should probably love etas more. Meanwhile, a 13.7 hour computation just failed to find an algebraic relation among (1/7)!, (2/7)!, (3/7)!, and pi. If there is one, it's a bear. --rwg Warren, I miss the Maple I had years ago. It wasn't stupid. You probably need some simple fix. OtOH, Mma's Reduce is pretty impressive, in case someone is up for translating your formula.
eta satisfies infinitely many relations like
0==27 * (\[Eta][q])^3 * (\[Eta][q^9])^9 + 9 * (\[Eta][q])^6 * (\[Eta][q^9])^6 + (\[Eta][q])^9 * (\[Eta][q^9])^3 - (\[Eta][q^3])^12] but as far as I can tell, the Gammas just cancel out. --rwg BtW, the more I understand http://en.wikipedia.org/wiki/Dedekind_eta_function , the more it sucks.