On Wed, Nov 7, 2012 at 3:05 PM, Bill Gosper <billgosper@gmail.com> wrote:
The ("complete") first kind elliptic integral K(m) and parameter m are both expressible in 𝜗 constants, and those three+ 𝜗 constants are all expressible in terms of Dedekind η. So
Out[840]= EllipticK[(16 η[q]^8 η[q^4]^16)/η[q^2]^24] == (π η[q^2]^10)/(2 η[q]^4 η[q^4]^4)
Crude but effective: d/dq and eliminate EllipticK to get EllipticE[(16 \[Eta][q^4]^8)/(\[Eta][q]^8 + 16 \[Eta][q^4]^8)] == -( 1/(\[Eta][q]^8 + 16 \[Eta][q^4]^8))(-((\[Pi] \[Eta][q]^4 \[Eta][q^2]^10)/( 2 \[Eta][ q^4]^4)) - (\[Pi] \[Eta][ q^2]^9 (6 Sqrt[3] q^(3/2) logderiveta[q] \[Eta][q] \[Eta][q^2] \[Eta][q^4] - 30 Sqrt[3] q^(5/2) logderiveta[q^2] \[Eta][q] \[Eta][q^2] \[Eta][q^4] + 24 Sqrt[3] q^(9/2) logderiveta[q^4] \[Eta][q] \[Eta][q^2] \[Eta][ q^4]) (\[Eta][q]^8 + 16 \[Eta][q^4]^8))/(2 q^( 1/3) \[Eta][q]^4 \[Eta][ q^4]^4 (6 Sqrt[3] q^(7/6) logderiveta[q] \[Eta][q] \[Eta][q^4] - 24 Sqrt[3] q^(25/6) logderiveta[q^4] \[Eta][q] \[Eta][q^4])) The remaining problem is those logderivetas, but lo, ((1/(4*x) - ((logderiveta[(1/(E^x))]))/(E^x))/((DedekindEta[((x*I)/(2*Pi))])^4)) is algebraic for (x/π)^2 rational. And for such x, we have many DedekindEta[((x*I)/(2*Pi)) in our collection. In fact, people.math.carleton.ca/~*williams*/papers/*pdf*/*299*.*pdf claims to give a complete solution. Early on, it appears to omit half the odd denominators, but I assume this is remedied later in the derivation. Anyway, as a fairly strenuous example: EllipticE[1/(1 + (-1 + GoldenRatio^(3/2))^8/(16 GoldenRatio^4))] == ( 2 5^(3/4) GoldenRatio^4 π^(3/2))/( Sqrt[1 + Sqrt[5]] (1525 + 682 Sqrt[5])^(1/4) Gamma[1/20] Gamma[9/ 20]) + ((-7 - 3 Sqrt[5] + Sqrt[190 + 85 Sqrt[5]]) Gamma[1/ 20] Gamma[9/20])/( 8 (1525 + 682 Sqrt[5])^(1/4) Sqrt[5 (1 + Sqrt[5]) π]) I'm having no luck finding a table of special values of EllipticE. This may be due to a simpler formula than above to produce them from EllipticK. It looks like they're something K + something/K. At the other extreme, has anyone seen *EllipticK[(-1)^(1/3)] == ((-1)^(1/12) 3^(1/4) Gamma[1/3]^3)/( 4 2^(1/3) π) ? --rwg