rwg>Maybe an atan(tan^k) hack would be better. Weird(?): using 2 3 %pi t 2 (%e - 1) atan(tan (-----)) t 2 %e (1 - -----------------------------) %pi instead of t 2 2 %e ((%e - 1) tanh(cot(%pi t)) + %e + 1) ------------------------------------------ 2 (matching only three derivatives) on [-1,0] gives a whole 'nother digit: f(t)=1.17652167574325d-4*sin(11*%pi*t)+6.08113365897434d-4*cos(11*%pi*t) +0.00100036999827d0*sin(10*%pi*t)-1.50249524904942d-4*cos(10*%pi*t)-3.81283020669115d-4 *sin(9*%pi*t)-0.00172079920634d0*cos(9*%pi*t)-0.00369552568607d0*sin(8*%pi*t) +9.47550861840517d-4*cos(8*%pi*t)+0.00173215817055d0*sin(7*%pi*t)+0.00784143653234d0 *cos(7*%pi*t)+0.0148972048649d0*sin(6*%pi*t)-0.00302598613176d0*cos(6*%pi*t) -0.00704134867511d0*sin(5*%pi*t)-0.02895637122193d0*cos(5*%pi*t)-0.0655859716264d0 *sin(4*%pi*t)+0.01897353641646d0*cos(4*%pi*t)+0.04756611866171d0*sin(3*%pi*t) +0.1616146418401d0*cos(3*%pi*t)+0.39334833081999d0*sin(2*%pi*t)-0.11547127986474d0 *cos(2*%pi*t)-0.39081697791208d0*sin(%pi*t)-0.99853697078804d0*cos(%pi*t) +1.95774972920477d0 [f(0.0d0) = 0.99987335148369d0, f(1.0d0) = 2.71817325043964d0] . For screen resolution, you could discard the five "d-4" terms. This seems in line with 11^-4 = 1/N^(derivs+1), but tan^5 is worse, and tan^7 worse yet. Presumably, with large enough N, tan^5 will win, and then tan^7, and eventually tanh(cot). ? --rwg --------------------------------- Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now.