But note the very cheap recurrence %pi k 2 sin(a - -----) = y , n + 1 k which can be computed ultrastably from the coupled recurrences x = x - y , y = e x + y , with k + 1 k k k + 1 k + 1 k 2 %pi e := 4 sin -------, and 2 n + 2 %pi cos(------- + a) 2 n + 2 x = - ----------------, y = 2 sin(a). 0 %pi 0 sin ------- 2 n + 2 This gives you all n+1 sins for the cost of two sins and a cos. Note that if you algebraically eliminate x[k] in favor of a second order recurrence for y[k], you lose the reversibility and hence the stability. --rwg Bill Gosper <rwmgosper@yahoo.com> wrote: n n ==== /===\ \ k | | %pi k > (- 1) | | 2 sin(a - -----) n / | | j n + 1 ==== ==== j = 0 \ k = 0 sin( > a ) = ------------------------------------, / k 2 (n + 1) ==== k = 1 where %pi n a := -----. 0 2 There are n^2 sins on the right, vs n2^(n-1) for the usual addition formula. __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com