The sum in Watson's identity can also be done with Jacobi's triple product, giving product = product - product. http://mathworld.wolfram.com/QuintupleProductIdentity.html gives this in terms of "Ramanujan theta functions", but with enough screwing around, it's equivalent to a strangish but sightly identity relating ordinary thetas: (EllipticTheta[1,x,q^6]*EllipticTheta[2,x,q^3])/EllipticTheta[1,Pi/3,q]== (1/3)*(-EllipticTheta[3,Pi/6+x,q^2]+EllipticTheta[4,Pi/3+x,q^2]) In:= FullSimplify[Series[List @@ %, {q, 0, 68}].{1, -1}] 69 Out= O[q ] theta[2](x,q^3)*theta[1](x,q^6)/theta[1](%pi/3,q) = (theta[4](x+%pi/3,q^2)-theta[3](x+%pi/6,q^2))/3 3 6 pi 2 pi 2 theta (x, q ) theta (x, q ) theta (x + --, q ) - theta (x + --, q ) 2 1 4 3 3 6 --------------------------- = --------------------------------------- . pi 3 theta (--, q) 1 3 Or, as a five card straight, theta[2](x,q^3)*theta[1](x,q^6)/eta(q^6) = (theta[4](x+%pi/3,q^2)-theta[3](x+%pi/6,q^2))/sqrt(3) 3 6 pi 2 pi 2 theta (x, q ) theta (x, q ) theta (x + --, q ) - theta (x + --, q ) 2 1 4 3 3 6 --------------------------- = ---------------------------------------, 6 sqrt(3) eta(q ) with one each theta_1, theta_2, theta_3, theta_4, and eta. --rwg