This transformation is decidedly not an involution. Iterating just seems to make ever more exotic identities, provided Re[m] remains ≤ 1/2. E.g., starting with m=(-1)^(1/3), I got to EllipticK[(-2 - Sqrt[2] + Sqrt[3] + Sqrt[6])^4] == (3^(3/4) (-1 + 2 Sqrt[2] + Sqrt[3]) Gamma[1/3]^3)/(16 2^(5/6) π) and EllipticK[(2 - Sqrt[2] - Sqrt[3] + Sqrt[6])^4] == (3^(1/4) Sqrt[7] (1 + 2 Sqrt[2] - Sqrt[3]) E^(-I ArcTan[2/Sqrt[3]]) Gamma[1/3]^3)/(16 2^(5/6) π) in three steps. This actually led to a profusion of results, because typically two of the eight sign patterns were confirmed. Whenever Re(m) exceeded 1/2, RootApproximant gave an identity for the "konjugate" because, empirically, K'/K was always algebraic. Is there a converse of Chowla-Selberg: K "nice" ⇒ K/'K algebraic? How can we automate FunctionExpand[EllipticK[m]] to produce algebraic*Gammas, where possible? Earlier I proposed finding q to make the lhs of 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) be EllipticK[m], and then table-lookup the η[q]. Can we automate the construction of such an η table, or find η in real time? I gather this is only solved for q=exp(- √rational π). Perhaps better would be to tabulate thousands of K identities as above. Given Mma's vagaries, I'm not sure how to automatically crank out either K or η identities. Perhaps if I do enough manually... Elsewhere I wondered rwg> where one might find tables much more extensive than, e.g.,http://functions.wolfram.com/EllipticIntegrals/EllipticK/03/01/ or http://dlmf.nist.gov/19.6 and I cited http://people.math.carleton.ca/~williams/papers/pdf/299.pdf , which says "The values of K[λ] for X = 1,2,...,16 are given in [I, Table 9.1, p. 298]." Duh, Borwein & Borwein, of course! (My old edition has the λ=√8 case wrong. Should be EllipticK[113 + 80 Sqrt[2] - 4 Sqrt[2 (799 + 565 Sqrt[2])]] -> 1/16 (1/2 (1 + Sqrt[2]))^(1/4) Sqrt[( 2 Sqrt[2] + Sqrt[1 + 5 Sqrt[2]])/π] Gamma[1/8] Gamma[3/8] I haven't checked the last seven, √10 thru √16.) All 16 can serve to initialize the iteration of the new(?) K transformation. Here's one several steps beyond λ=√7: EllipticK[(8 + 6 Sqrt[2] + 3 Sqrt[7] + 2 Sqrt[14])^4] == (Sqrt[29] (3 - 4 Sqrt[2] + Sqrt[7]) E^(I/4 (3 π - 2 ArcTan[27/(4 Sqrt[7])])) Gamma[1/7] Gamma[2/7] Gamma[4/7])/(32 Sqrt[2] 7^(1/4) π) And since there seems to be an E identity for every K, there are scads of closed forms for 2F1[a+1/2,b+1/2,c;z], integer a,b,c. And scads of ellipses with perimeters in closed form. --rwg On Mon, Nov 5, 2012 at 11:44 PM, Bill Gosper <billgosper@gmail.com> wrote:
Purely empirically, for 0<r<1/2, EllipticK[4 E^(I ArcTan[1/2 Sqrt[-4 + 1/(1 - r) + 1/r]]) Sqrt[(1- r) r]] == E^(I ArcTan[Sqrt[r/(1 - r)]]) EllipticK[r]
E.g., for r=1/4 EllipticK[Sqrt[3] E^(I*π/6)] == E^(I*π/6) EllipticK[1/4]
For r=1/2, it gets the conjugate of the right answer, which is EllipticK[2] == -(-1)^(3/4) EllipticK[1/2] which can be found by listing all eight sign variations of the three square roots. Then we can try complex r: EllipticK[1/2 + I/2] == (-3 + 2 Sqrt[2])^(1/4) EllipticK[4 - 2 Sqrt[2]] which puts the lhs in polar form.
Namely, EllipticK[1/2 + I/2] == Sqrt[-1 + Sqrt[2]] E^(I π/4) EllipticK[4 - 2 Sqrt[2]]
And for r = I, EllipticK[(8 + 8 I) - 4 Sqrt[-1 + 7 I]] == (2^(1/4) E^(-(I*π/8)) - I E^(I*π/4)) EllipticK[I]
None of the eight sign guesses works for Re(r)>1/2.
If no one finds a proof of this stuff, imagine some future Mathematica just trying everything and choosing the numerically plausible one. --rwg