(Oops, sorry, Gene, you're right--I completely missed the 1/n!. Unbelievable that something with that smooth an edge could have a curtain of poles there. Even less believable with the edge hidden by your challenge continuation. Nice plot. only one nontrivial root.) eta(%i*%e^-(%pi/(2*sqrt(3)))) = 3^(3/8)*gamma(1/3)^(3/2)/(2^(5/6)*%i^(1/6)*%pi) 3/8 3/2 1 3 gamma (-) i 3 eta(----------) = ---------------- . pi 5/6 1/6 --------- 2 i pi 2 sqrt(3) e eta(-%e^-(%pi/sqrt(3))) = 3^(3/8)*%i^(1/12)*gamma(1/3)^(3/2)/(2*%pi) 3/8 3/2 1 1/12 3 gamma (-) i 1 3 eta(- --------) = ---------------------- . pi 2 pi ------- sqrt(3) e theta[4](0,%i*%e^-(%pi/(2*sqrt(3)))) = 3^(3/8)*gamma(1/3)^(3/2)/(2^(2/3)*%i^(5/12)*%pi) 3/8 3/2 1 3 gamma (-) i 3 theta (0, ----------) = ---------------- . 4 pi 2/3 5/12 --------- 2 i pi 2 sqrt(3) e eta(%e^(2*%i*%pi/3-%pi/(3*sqrt(3)))) = (sqrt(3)+1)^(1/4)*(2^(5/6)*((2^(2/3)+2*2^(1/3)+1)*sqrt(3)-3*2^(1/3)+3)+%i*(2*2^(1/6)*(2^(1/3)-1)*sqrt(3)+2^(5/6)*(sqrt(3)-3)))*gamma(1/3)^(3/2)/(8*2^(3/8)*3^(13/24)*%pi) 2 i pi pi ------ - --------- 3 3 sqrt(3) 1/4 3/2 1 eta(e ) = (sqrt(3) + 1) gamma (-) 3 1/6 1/3 5/6 ((2 2 (2 - 1) sqrt(3) + 2 (sqrt(3) - 3)) i 5/6 2/3 1/3 1/3 3/8 13/24 + 2 ((2 + 2 2 + 1) sqrt(3) - 3 2 + 3))/(8 2 3 pi) . --rwg PS, to compute the volume of a tetraroller, I need to run the prismatoid formula on a wedge cut off the base of a cone. This requires finding two saggitae. Since osage orange is most highly prized for making bows, it is therefore completely germane to math-fun. PRISMATOID DIATROPISM, CONSIDERATE DESECRATION