Notice the big gap between 344 and 51109. Once again, pi has certain excellent approximations. 'rapprox' uses the greedy algorithm in an attempt to minimize the largest power; the 2nd argument is an epsilon. (%i1) rapprox(%pi,0.0000001); (%o1) [468856, 364508, 260515, 52529, 51109, 344, 77, 33, 30, 14, 11, 5, 1, 0] (%i2) rexpand(%); 468856 364508 260515 52529 51109 344 77 33 30 (%o2) r + r + r + r + r + r + r + r + r 14 11 5 1 1 1 1 1 1 1 1 1 + r + r + r + r + - + -- + --- + --- + --- + --- + --- + ---- + ------ r 5 11 14 30 33 77 344 51109 r r r r r r r r 1 1 1 1 + ------ + ------- + ------- + ------- + 1 52529 260515 364508 468856 r r r r (%i3) %,r=exp(%i); 468856 %i 364508 %i 260515 %i 52529 %i 51109 %i (%o3) %e + %e + %e + %e + %e 344 %i 77 %i 33 %i 30 %i 14 %i 11 %i 5 %i %i + %e + %e + %e + %e + %e + %e + %e + %e - %i - 5 %i - 11 %i - 14 %i - 30 %i - 33 %i + %e + %e + %e + %e + %e + %e - 77 %i - 344 %i - 51109 %i - 52529 %i - 260515 %i + %e + %e + %e + %e + %e - 364508 %i - 468856 %i + %e + %e + 1 (%i4) %,rectform; (%o4) 2 cos(468856) + 2 cos(364508) + 2 cos(260515) + 2 cos(52529) + 2 cos(51109) + 2 cos(344) + 2 cos(77) + 2 cos(33) + 2 cos(30) + 2 cos(14) + 2 cos(11) + 2 cos(5) + 2 cos(1) + 1 (%i5) bfloat(%); (%o5) 3.141588840097392b0