Massaging the infinite products gives %i %pi ------ 24 3 2 %e eta (q ) eta(- q) = -----------------. 4 eta(q) eta(q ) Enormous screwing around gave 5 %i %pi - -------- 2 8 2 2 48 4 4 eta (q ) %i eta (q ) 1/4 eta(%i q) = %e eta(q ) (---------- + -----------) , 2 2 2 8 eta (q ) eta (q ) valid in the wedge -pi/8 < arg(q) <= pi/8. This is actually amazingly simple, given that the polynomial relating eta(q), eta(q^4), and eta(q^8) is degree 64, and is apparently too hard for Macsyma and Mma to factor over cis(pi/24), let alone 48. Using our earlier eta valuations, ETA(-%E^-(%PI/SQRT(3))) = 3^(3/8)*%I^(1/12)*GAMMA(1/3)^(3/2)/(2*%PI) pi 3/8 3/2 1 1/12 - ------- 3 gamma (-) i sqrt(3) 3 eta(- %e ) = ----------------------, 2 pi (which is actually simpler than ETA(%E^-(%PI/(2*SQRT(3)))) = 3^(3/8)*(5*SQRT(3)-4*SQRT(2)-3)^(1/8)*GAMMA(1/3)^(3/2)/(2*2^(13/48)*%PI) pi 3/8 1/8 3/2 1 - --------- 3 (5 sqrt(3) - 4 sqrt(2) - 3) gamma (-) 2 sqrt(3) 3 eta(%e ) = ----------------------------------------------- 13/48 2 2 pi ) and ETA(-%I*%E^-(%PI/(2*SQRT(3)))) = 3^(3/8)*%E^(%I*%PI/12)*GAMMA(1/3)^(3/2)/(2^(5/6)*%PI) i pi ---- pi 3/8 3/2 1 12 - --------- 3 gamma (-) %e 2 sqrt(3) 3 eta(- i %e ) = ----------------------- 5/6 2 pi This came up because q can get multiplied by a root of unity when expressing special values of thetas as theta constants: THETA[1](%PI/10,Q) = (%I*SQRT(SQRT(5)/8+5/8)*(2*SQRT(5)-2)-SQRT(5)-5)*THETA[2](0,%E^(4*%I*%PI/5)*Q)/8+(SQRT(SQRT(5)/8+5/8)*(SQRT(5)+1)+%I*SQRT(5))*THETA[2](0,%E^(2*%I*%PI/5)*Q)/4-(-SQRT(5)+4*%I*SQRT(SQRT(5)/8+5/8)-3)*THETA[2](0,Q)/8 %pi sqrt(5) 5 theta (---, q) = (%i sqrt(------- + -) (2 sqrt(5) - 2) - sqrt(5) - 5) 1 10 8 8 4 %i %pi -------- 5 sqrt(5) 5 theta (0, %e q)/8 + (sqrt(------- + -) (sqrt(5) + 1) 2 8 8 2 %i %pi -------- 5 + %i sqrt(5)) theta (0, %e q)/4 2 sqrt(5) 5 (- sqrt(5) + 4 %i sqrt(------- + -) - 3) theta (0, q) 8 8 2 - -----------------------------------------------------, 8 which I got only a few months ago but have totally forgotten how. With q=e^(-pi/10), EllipticTheta[1, Pi/10, (1/(E^(Pi/10)))]==((1/4)! * ((Sqrt[5] + 1) * ((1/(Sqrt[ - 5^(3/4) + 5^(1/4) + 2])) - ((Sqrt[5 * (Sqrt[5] + 2)])/2)) - 5^(3/4) * Sqrt[(2/(Sqrt[5] + 1))] + 5^(1/4) * (Sqrt[5] - 1))/(2^(5/8) * Pi^(3/4))) THETA[1](%PI/10,%E^-(%PI/10)) = (-SQRT(2)*5^(3/4)/SQRT(SQRT(5)+1)+(SQRT(5)+1)*(1/SQRT(2-5^(1/4)*(SQRT(5)-1))-SQRT(5)*SQRT(SQRT(5)+2)/2)+5^(1/4)*(SQRT(5)-1))*(1/4)!/(2^(5/8)*%PI^(3/4)) pi - pi/10 2 3/4 theta (--, e ) = (- sqrt(-----------) 5 1 10 sqrt(5) + 1 1 sqrt(5 (sqrt(5) + 2)) + (----------------------- - ---------------------) (sqrt(5) + 1) 3/4 1/4 2 sqrt(- 5 + 5 + 2) 1/4 1 5/8 3/4 + 5 (sqrt(5) - 1)) (-)!/(2 pi ) 4 = 0.02071028405516 This was a good day's work, including six new special values of eta. It took me a couple of hours to simplify, but that's faster than Mma's FullSimplify, which is still working on it from yesterday. If you guess the transcendental factors, you could in principle determine the algebraic factor numerically, but probably the only way to solve the (degree 8*128) polynomial in radicals would be to guess most of the extensions over which it factorizes. --rwg TORT-FEASING FORETASTING GRAFTONITES ICONODULIST DISLOCUTION NOSE TO NOSE ON ONE'S TOES