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) This isn't quite Mathematica's DedekindEta function, which takes the τ argument instead of q = e^(2 i π τ). For numerics and series use η[q]:=DedekindEta[Log[q]/2/I/π]==EllipticTheta[1, Pi/3, q^(1/6)]/Sqrt[3]== q^(1/24)*QPochhammer[q,q] Also, Mathematica's EllipticK conflicts with Wikipedia's Elliptic integral K: See Caution, below. Now suppose we wish to exactly find EllipticK[(2 + Sqrt[3])/4]. First, get a good numerical solution q to m == (16 η[q]^8 η[q^4]^16)/η[q^2]^24 == (2 + Sqrt[3])/4 In[849]:=FindRoot[%840[[1, 1]] == (2 + Sqrt[3])/4 /. η[q_] :> q^(1/24)*QPochhammer[q, q], {q, 0.01}, WorkingPrecision -> 34] Out[849]= {q -> 0.1630335348215804648616403561098358} See if this q is of the form e^(π*quadratic surd): In[850]:= Log[q /. %]/π Out[850]= -0.5773502691896257645091487805019575 In[851]:= %^2 Out[851]= 0.3333333333333333333333333333333333 So q=e^(-π/√3)

From my collection of special values of η:

Out[844]= {η[1/E^((4*π)/Sqrt[3])] == (3^(3/8)*(Sqrt[3] + 1)^(1/4)* Sqrt[Gamma[1/3]]^3)/(2*2^(1/8)*8^(1/4)*π), η[1/E^(π/Sqrt[3])] == (3^(3/8)* Sqrt[Gamma[1/3]]^3)/(2*2^(1/8)*(Sqrt[3] + 1)^(1/4)*π), η[1/E^((2*π)/Sqrt[3])] == (3^(3/8)*Sqrt[Gamma[1/3]]^3)/(2*2^(1/3)*π)} Plugging into the rhs of the K equation: In[846]:= FullSimplify /@ %840 /. q -> 1/E^(π/Sqrt[3]) /. ToRules[And @@ %844] Out[846]= EllipticK[1/8 (1 + Sqrt[3])^2] == (3^(3/4) Gamma[1/3]^3)/(4 2^(1/3) π) Note that we can crank out such K valuations for any q where η[q], η[q^2], and η[q^4] all appear in the collection. Actually we need only two of the three, since η[q^2]^24 == 16*η[q]^8*η[q^4]^16 + η[q]^16*η[q^4]^8 . This argues for tabulating or otherwise facilitating valuations of η(e^(π surd)). I believe there's an algorithm for half the cases of η(e^(-π rational)). Jörg? Caution: http://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of... (#Special_values, if only the hash tags worked this way) effectively gives the above EllipticK identity as K[Cos[π/12]] == (3^(3/4) Gamma[1/3]^3)/(4 2^(1/3) π) where K[k] := EllipticK[k^2], reflecting the infernal "modulus" vs "parameter" notational conflict. --rwg On Tue, Nov 6, 2012 at 3:53 PM, Bill Gosper <billgosper@gmail.com> wrote:

In[835]:= EllipticK[0]

Out[835]=π/2

so small arguments give π approximations:

From Im[EllipticK[(((1 - 5^(1/4))^8*(Sqrt[5] + 1)^16)/524288)]== (((2*5^(1/4) + 3)*(1/2 - I/10)*(Sqrt[Pi])^3)/((Gamma[3/4])^2*Sqrt[2]))]

In[832]:= N[(50*Gamma[3/4]^4)/(3 + 2*5^(1/4))^2]

Out[832]= 3.1415964411401665

or In[834]:= EllipticK[(((2^(1/4) - 1)^4)/((2^(1/4) + 1)^4))] == (((2^(1/4) + 1)^2*(Sqrt[Pi])^3)/(8*(Gamma[3/4])^2*Sqrt[2]))

or EllipticK[-(((1 - 5^(1/4))^16*(Sqrt[5] + 1)^8)/65536)]== (((2*5^(1/4) + 3)*(Sqrt[Pi])^3)/(10*(Gamma[3/4])^2*Sqrt[2]))

(Aproximating π in terms of π^(3/2) isn't cheating!)

Stay tuned for an interesting way to find similar K valuations. --rwg