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. [Added 2012-11-16 07:03 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 No one seemed to recognize it, so, grotesque though it is, NeilB and I decided it's time to write this up. So Neil bashed on it with Mathematica, and it got a simpler. Then I bashed on it so hard that it got unnervingly ungrotesque: (*) EllipticK[1/4 (2 + (-2 + t)/Sqrt[1 - t])] == (1 - t)^(1/4) EllipticK[t] How could anything this simple be new? It looks like a routine 2F1 quadratic transformation. But which? As before, it makes the strange little results EllipticK[1/4] == E^(-((I \[Pi])/6)) EllipticK[Sqrt[3] E^((I \[Pi])/6)] and EllipticK[I] == (E^((I \[Pi])/16) EllipticK[1/2 - 1/4 Sqrt[7/2 - I/2]])/2^(1/8) Are these familiar? And somehow, (*) has miraculously lost its Re[t]<1/2 restriction:. E.g. for t=2, EllipticK[1/2] == (-1)^(1/4) EllipticK[2], as one might hope. For t=(-1)^(1/6), EllipticK[1/4 (2 + (-2 + (-1)^(1/6))/Sqrt[1 - (-1)^(1/6)])] == (2 - Sqrt[3])^(1/8) E^(-((5 I \[Pi])/48)) EllipticK[E^((I \[Pi])/6)] Angling for a K'/K identity, solve 1/4 (2 + (-2 + t)/Sqrt[1 - t]) == 1 - t : t -> 1/32 (31 + 3 I Sqrt[7]) So, Log[EllipticNomeQ[1/32 (31 - 3 I Sqrt[7])]] == -(1/2) Sqrt[1/2 (-3 I + Sqrt[7])] E^((I \[Pi])/4) \[Pi] along with its formal conjugate. Stop me if you've heard this one. --rwg (Is there a (dlmf?) page of special values of nome? Or just garden varieties:-)