I often cite Jacobi's "aequatio identica satis abstrusa" polynomial relating the 8th powers of eta(q), eta(q^2), and eta(q^4). Solving for eta(q)^8 (c112) solve(gimme(1,2,4),eta(q)^8) 24 4 24 2 12 4 8 sqrt(64 eta (q ) + eta (q )) + 8 eta (q ) (d112) [eta (q) = - --------------------------------------------, 4 4 eta (q ) 24 4 24 2 12 4 8 sqrt(64 eta (q ) + eta (q )) - 8 eta (q ) eta (q) = --------------------------------------------] 4 4 eta (q ) leads to the apparent absurdity that it is an even function, which is quickly refuted by numerical experiment: (c113) expand(subst(-0.5d0,q,%)) (d113) [3.17357586985114d0 %i + 1.83226488275225d0 = - 3.66452976550451d0, 3.17357586985114d0 %i + 1.83226488275225d0 = 3.83964397049972d-5] eta(<negative>) is not even real. So what is eta(-q)? Massaging the infinite product definition gives eta(q^2)^3/eta(q)/eta(q^4) times some 24th root of 1. But empirically, there is no such root that works for the whole unit disk. Rather, the power is the crazy four-valued step-function(arg q) with a point discontinuity (mainly derived by tutor Julian) in http://gosper.org/etaminus.pdf . If this is an example of Truth = Beauty, the corresponding formula for eta(i q) must be just unbearably gorgeous. And what I'm really after is eta(cis(2*pi/5) q) for nice R-R CF valuations. However, once we know that CF = algebraic/q^(1/5), we can just LatticeReduce for hugepolynomial(CF*q^(1/5)). --rwg