[math-fun] EllipticK[1/3]?
(Question from Julian.) Apparently, K(k) has a closed form at and only at "singular values", i.e. when K(1-k)/K(k) = sqrt(rational), (periodic CF). There's no period in the first 10^4 terms of K(2/3)/K(1/3), but, e.g., EllipticK[4 Sqrt[2] (-1 + Sqrt[2])^2] -> ((2 + Sqrt[2]) Pi^(3/2))/(4 Gamma[3/4]^2) EllipticK[(-1 + Sqrt[2])^4] == ((2 + Sqrt[2]) Pi^(3/2))/(8 Gamma[3/4]^2) EllipticK[(-Sqrt[2] + Sqrt[3])^4/(-1 + Sqrt[2])^4] == (3^(3/4) (-1 + Sqrt[2]) (1 + Sqrt[3]) Gamma[1/3]^3)/( 16 2^(5/6) (-Sqrt[2] + Sqrt[3]) Pi) EllipticK[( 16 Sqrt[2] (-Sqrt[2] + Sqrt[3])^2)/((-1 + Sqrt[2])^2 (1 + Sqrt[ 3])^3)] == (3^(1/4) (-1 + Sqrt[2]) (1 + Sqrt[3]) Gamma[1/3]^3)/( 8 2^(5/6) (-Sqrt[2] + Sqrt[3]) Pi) EllipticK[(-1 + 2^(1/4))^4/(1 + 2^(1/4))^4] == ((-1 + Sqrt[ 2])^2 Pi^(3/2))/(8 Sqrt[2] (-1 + 2^(1/4))^2 Gamma[3/4]^2) EllipticK[(8 2^(1/4) (1 + Sqrt[2]))/(1 + 2^(1/4))^4] == ((1 + 2^(1/4))^2 Pi^(3/2))/(2 Sqrt[2] Gamma[3/4]^2) I don't recall seeing the last four (which are equivalent in pairs). I have to check if K(1-k)/K(k) = sqrt(rational1) + i sqrt(rational2) can also work. Also, that unlikely conjecture that denestable <-> factorable into binomials has survived a minor complexes test: DedekindEta[-((I*((I*Pi)/2 - Pi/4))/(2*Pi))] == (Gamma[1/4]*Sqrt[Sqrt[2] + 1]* I^(13/24))/(2^(25/32)*(Sqrt[2] - 1)^(1/8)*Sqrt[2^(1/4)*I - 1]*Pi^(3/4)) (A virtue of binomial factorization is that it tends to minimize the most "deeply rooted" subexpressions, and we know how unwelcome radicalism is around here.) --rwg
I computed the first N terms of the continued fraction of K(2/3)/K(1/3), for N given below: N | Size | Link ---- | -------- | ---------------------------------------------------------- 10^6 | 1.2 MiB | http://dl.dropbox.com/u/734346/elliptic23_13__10e6.tar.bz2 10^7 | 10.9 MiB | http://dl.dropbox.com/u/734346/elliptic23_13__10e7.tar.bz2 There is one term per line. I haven't done any analysis on the terms yet, but thought you might want to play with them. -Robert On Sun, Mar 27, 2011 at 6:15 PM, Bill Gosper <billgosper@gmail.com> wrote:
(Question from Julian.) Apparently, K(k) has a closed form at and only at "singular values", i.e. when K(1-k)/K(k) = sqrt(rational), (periodic CF). There's no period in the first 10^4 terms of K(2/3)/K(1/3), but, e.g.,
EllipticK[4 Sqrt[2] (-1 + Sqrt[2])^2] -> ((2 + Sqrt[2]) Pi^(3/2))/(4 Gamma[3/4]^2)
EllipticK[(-1 + Sqrt[2])^4] == ((2 + Sqrt[2]) Pi^(3/2))/(8 Gamma[3/4]^2)
EllipticK[(-Sqrt[2] + Sqrt[3])^4/(-1 + Sqrt[2])^4] == (3^(3/4) (-1 + Sqrt[2]) (1 + Sqrt[3]) Gamma[1/3]^3)/( 16 2^(5/6) (-Sqrt[2] + Sqrt[3]) Pi)
EllipticK[( 16 Sqrt[2] (-Sqrt[2] + Sqrt[3])^2)/((-1 + Sqrt[2])^2 (1 + Sqrt[ 3])^3)] == (3^(1/4) (-1 + Sqrt[2]) (1 + Sqrt[3]) Gamma[1/3]^3)/( 8 2^(5/6) (-Sqrt[2] + Sqrt[3]) Pi)
EllipticK[(-1 + 2^(1/4))^4/(1 + 2^(1/4))^4] == ((-1 + Sqrt[ 2])^2 Pi^(3/2))/(8 Sqrt[2] (-1 + 2^(1/4))^2 Gamma[3/4]^2)
EllipticK[(8 2^(1/4) (1 + Sqrt[2]))/(1 + 2^(1/4))^4] == ((1 + 2^(1/4))^2 Pi^(3/2))/(2 Sqrt[2] Gamma[3/4]^2)
I don't recall seeing the last four (which are equivalent in pairs). I have to check if K(1-k)/K(k) = sqrt(rational1) + i sqrt(rational2) can also work.
Also, that unlikely conjecture that denestable <-> factorable into binomials has survived a minor complexes test:
DedekindEta[-((I*((I*Pi)/2 - Pi/4))/(2*Pi))] == (Gamma[1/4]*Sqrt[Sqrt[2] + 1]* I^(13/24))/(2^(25/32)*(Sqrt[2] - 1)^(1/8)*Sqrt[2^(1/4)*I - 1]*Pi^(3/4))
(A virtue of binomial factorization is that it tends to minimize the most "deeply rooted" subexpressions, and we know how unwelcome radicalism is around here.) --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
rwg>I have to check if K(1-k)/K(k) = sqrt(rational1) + i sqrt(rational2) can also work. Going through the motions of singular value (1-i)/10 (squared), I got instead the one for (5+i)/26 (squared) ! ? EllipticK[((1 - 5^(1/4))^8*(1 + Sqrt[5])^16)/524288] == ((1/2 - I/10)*(3 + 2*5^(1/4))*Pi^(3/2))/(Sqrt[2]*Gamma[3/4]^2) EllipticK[-(((1 - 5^(1/4))^16*(1 + Sqrt[5])^8)/65536)] == ((3 + 2*5^(1/4))*Pi^(3/2))/(10*Sqrt[2]*Gamma[3/4]^2) Lots of interesting spinoffs, e.g. DedekindEta[1/20 + I/20] == -(((-1 + 5^(1/4))*E^((7*I*Pi)/24)*Gamma[1/4])/ (2^(5/8)*Sqrt[1 + Sqrt[5]]*Pi^(3/4))) --rwg On Sun, Mar 27, 2011 at 4:15 PM, Bill Gosper <billgosper@gmail.com> wrote:
(Question from Julian.) Apparently, K(k) has a closed form at and only at "singular values", i.e. when K(1-k)/K(k) = sqrt(rational), (periodic CF). There's no period in the first 10^4 terms of K(2/3)/K(1/3), but, e.g.,
EllipticK[4 Sqrt[2] (-1 + Sqrt[2])^2] -> ((2 + Sqrt[2]) Pi^(3/2))/(4 Gamma[3/4]^2)
EllipticK[(-1 + Sqrt[2])^4] == ((2 + Sqrt[2]) Pi^(3/2))/(8 Gamma[3/4]^2)
EllipticK[(-Sqrt[2] + Sqrt[3])^4/(-1 + Sqrt[2])^4] == (3^(3/4) (-1 + Sqrt[2]) (1 + Sqrt[3]) Gamma[1/3]^3)/( 16 2^(5/6) (-Sqrt[2] + Sqrt[3]) Pi)
EllipticK[( 16 Sqrt[2] (-Sqrt[2] + Sqrt[3])^2)/((-1 + Sqrt[2])^2 (1 + Sqrt[ 3])^3)] == (3^(1/4) (-1 + Sqrt[2]) (1 + Sqrt[3]) Gamma[1/3]^3)/( 8 2^(5/6) (-Sqrt[2] + Sqrt[3]) Pi)
EllipticK[(-1 + 2^(1/4))^4/(1 + 2^(1/4))^4] == ((-1 + Sqrt[ 2])^2 Pi^(3/2))/(8 Sqrt[2] (-1 + 2^(1/4))^2 Gamma[3/4]^2)
EllipticK[(8 2^(1/4) (1 + Sqrt[2]))/(1 + 2^(1/4))^4] == ((1 + 2^(1/4))^2 Pi^(3/2))/(2 Sqrt[2] Gamma[3/4]^2)
I don't recall seeing the last four (which are equivalent in pairs). I have to check if K(1-k)/K(k) = sqrt(rational1) + i sqrt(rational2) can also work.
Also, that unlikely conjecture that denestable <-> factorable into binomials has survived a minor complexes test:
DedekindEta[-((I*((I*Pi)/2 - Pi/4))/(2*Pi))] == (Gamma[1/4]*Sqrt[Sqrt[2] + 1]* I^(13/24))/(2^(25/32)*(Sqrt[2] - 1)^(1/8)*Sqrt[2^(1/4)*I - 1]*Pi^(3/4))
(A virtue of binomial factorization is that it tends to minimize the most "deeply rooted" subexpressions, and we know how unwelcome radicalism is around here.) --rwg
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