Off-list Rich asked if maybe the identities in my crankout message weren't fairly obvious when converted back to finite products of polynomials. Well, obvious modulo solving for some undetermined coefficients. While hand simulating an algorithm that ought to evaluate most of these products, I constructed the following screw case: prod(8*cos(t/2^n-5*%pi/12)*cos(t/2^n+%pi/12)*cos(t/2^n+%pi/3),n,1,inf) = (sqrt(2)*sin(2*t-%pi/12)+1)*(sqrt(2)*sin(4*t-%pi/12)+1)/(6*cos(t-5*%pi/12)*cos(2*t-5*%pi/12)) inf /===\ | | t 5 pi t pi t pi | | 8 cos(-- - ----) cos(-- + --) cos(-- + --) = | | n 12 n 12 n 3 n = 1 2 2 2 pi pi (sqrt(2) sin(2 t - --) + 1) (sqrt(2) sin(4 t - --) + 1) 12 12 ------------------------------------------------------- 5 pi 5 pi 6 cos(t - ----) cos(2 t - ----) 12 12 The prodand can *not* (I think) be written as trig(2^(n+1))/trig(2^n). But the rhs, which is of the form f(t) f(2t), reveals the trick*: Do the odd and even terms separately, making the lhs *two* products which actually *can* be put in the form trig(4^(n+1))/trig(4^n). So we have to try bisecting and trisecting (etc., 'til when?) products that resist telescopy. This might even handle such grotesqueries as prod((cos(t/2^n)+cos(%pi*2^n/5))/(cos(%pi*2^n/5)+1),n,1,inf) = (1-1/sqrt(5))*cos(t)+1/sqrt(5) n t 2 pi inf cos(--) + cos(-----) /===\ n 5 | | 2 1 1 | | -------------------- = (1 - -------) cos(t) + ------- | | n sqrt(5) sqrt(5) n = 1 2 pi cos(-----) + 1 5 And speaking of cos(pi/5), can anyone prove this little goodie?: sum(k*fib(k+1)*binom(-n,n-k),k,1,n) = n n ==== \ > k fib(k + 1) binomial(- n, n - k) = n / ==== k = 1 This is large integers summing to a small one. E.g., for n=9, 495 - 2448 + 6615 - 12870 + 19800 - 25740 + 27027 - 25740 + 12870 = 9 Note all the suggestive coincidences. But they don't last! 1584 - 9790 + 32670 - 77792 + 147147 - 234234 + 320320 - 388960 + 393822 - 369512 + 184756 = 11 Replacing fib(k) with q^k gave no hint of a degenerate bibasic sum. --rwg INTERMORAINIC RECRIMINATION (feuding glaciologists) *(divulgent infinite product)