products. To get prod(trig(t/k^n),n), choose the undetermined coefficients in a small polynomial p(x) to annihilate remainder(p(x^k),p(x)). E.g., for a cubic and k=2, (c1) (x^3+a*x^2+b*x+c,remainder(subst(x^2,x,%%),%%,x)) (d1) (2 a c + b + (- 3 a - a + 1) b + a + a ) x 2 2 3 2 + ((2 b - a - a) c - 2 a b + (a + a ) b) x 2 3 2 + c + (- 2 a b + a + a + 1) c giving three eqns and 3 unknowns, with numerous solutions, [[b = 0, c = 0, a = - 1], [c = 0, b = 1, a = - 2], [c = 0, b = 1, a = 1], sqrt(3) %i - 1 sqrt(3) %i - 1 [a = 0, b = 0, c = 0], [c = --------------, b = --------------, 2 sqrt(3) %i + 1 sqrt(3) %i 1 sqrt(3) %i + 1 sqrt(3) %i + 1 a = ---------- + -], [c = - --------------, b = --------------, 2 2 2 sqrt(3) %i - 1 1 sqrt(3) %i a = - - ----------], [c = 1, b = - 1, a = - 1], 2 2 2 3 6 5 4 3 b - b [c = - 1, b = root_of(b - 2 b - b - 6 b , b), a = -------], 2 b [a = 0, c = 0, b = - 1], [a = 0, b = 0, c = - 1], [c = %i, b = - 1, a = - %i], [c = - %i, b = - 1, a = %i]], each yielding a product identity. The most interesting are from the root_of sextic which could (should?) have reduced to quadratic surds. One of these gives 2 2 p(x ) 3 (- sqrt(7) %i - 1) x (sqrt(7) %i - 1) x ----- = x + --------------------- + ------------------ + 1. p(x) 2 2 Now substitute x= exp(i t/2^n) and the lhs takes the form f(n-1)/f(n). After some simplification, product(2*cos(3*t/2^n)+sqrt(7)*sin(t/2^n)-cos(t/2^n),n,1,inf) = (2*sin(3*t)+sin(t))/sqrt(7)+cos(t) inf /===\ | | 3 t t t 2 sin(3 t) + sin(t) | | (2 cos(---) + sqrt(7) sin(--) - cos(--)) = ------------------- + cos(t) | | n n n sqrt(7) n = 1 2 2 2 Likewise for -sqrt(7). Tayloring a few terms, closedform(taylor(%,t,0,6)) 2 3 4 5 6 t (55 sqrt(7)) t t (487 sqrt(7)) t t 1 + sqrt(7) t - -- - --------------- + -- + ---------------- - --- + . . . 2 42 24 840 720 2 3 4 5 6 t (55 sqrt(7)) t t (487 sqrt(7)) t t = 1 + sqrt(7) t - -- - --------------- + -- + ---------------- - --- + . . . 2 42 24 840 720 (As of 7.0, Mathematica still won't Series expand Sums and Products!) In vain hope of brevity, we skip the more routine solutions and try for prod(trig(t/5^n),n), which I don't recall seeing. (x^2+a*x+b,remainder(subst(x^5,x,%%),%%,x)) gives the two equations 2 2 4 2 2 4 0 = a (b - 3 a b + a ) (5 b - 5 a b + a - 1) 4 2 3 4 2 6 2 8 4 = b (b - 10 a b + 15 a b - 7 a b + 2 a b + a - a - 1) The very simple case a=2, b=1 gives the square of the identity inf /===\ t | | 2 t t cos(-) = | | (4 cos (--) - 2 cos(--) - 1) 2 | | n n n = 1 5 5 The case a=2i, b=-1 gives the square of the identity inf /===\ t t | | t 2 t cos(-) + sin(-) = | | (1 + 2 sin(--) - 4 sin (--)) 2 2 | | n n n = 1 5 5 a=-2, b=1 gives the square of 2 t t t inf 4 cos (--) + 2 cos(--) - 1 2 sin(-) /===\ n n 2 | | 5 5 -------- = | | -------------------------- t | | 5 n = 1 The peculiar solution a=sqrt(3i), b=1 gives t pi t 4 (sqrt(3) - sqrt(2)) cos(- + --) sin(-) + 1 = 2 4 2 inf /===\ | | 2 t t pi 4 t | | (4 sin(---) (sqrt(3) cos(-- - --) - 1) + 2 cos(---) - 1) | | n n 4 n n = 1 5 5 5 Likewise for -sqrt(3), but not -sqrt(2). There are two root_of quartics saying b can be any 10th root of 1. Choosing the first led, via almost two days of simplifying and debugging, to product(2*cos(t/5^n+3*%pi/20)*(2*cos(3*t/5^n-3*%pi/20)-(sqrt(5)-1)*cos(t/5^n-%pi/20)),n,1,inf) = sin(t)+cos(t) inf /===\ | | t 3 pi 3 t 3 pi t pi | | 2 cos(-- + ----) (2 cos(--- - ----) - (sqrt(5) - 1) cos(-- - --)) | | n 20 n 20 n 20 n = 1 5 5 5 = sin(t) + cos(t), a puzzling result, given that the individual factors, product(cos(t/5^n+3*%pi/20)/cos(3*%pi/20),n,1,inf) t 3 pi inf cos(-- + ----) /===\ n 20 | | 5 | | -------------- and <cofactor> | | 3 pi n = 1 cos(----) 20 appear not to telescope, even though there were hints during the derivation that they would. One nice simplifier subproblem was %e^-(%i*t)*((sqrt(5)-1)*%e^(%i*t+%i*%pi/10)/2+%e^(2*%i*t)+%e^(%i*%pi/5))/(%e^(%i*%pi/5)+(sqrt(5)-1)*%e^(%i*%pi/10)/2+1) = sqrt(2)*cos(t/2-%pi/4)*cos(t/2+3*%pi/20)/cos(3*%pi/20) i pi i t + ---- i pi 10 ---- (sqrt(5) - 1) e 2 i t 5 t pi t 3 pi ------------------------- + e + e sqrt(2) cos(- - --) cos(- + ----) 2 2 4 2 20 ------------------------------------------ = --------------------------------- i pi 3 pi i pi ---- cos(----) ---- 10 20 i t 5 (sqrt(5) - 1) e e (e + ------------------- + 1) 2 which FullSimplify cannot even verify for special cases, let alone derive. --rwg