Cubic(v)... algo style

5ff28735-46d1-42dc-8e96-4ad62195441e 5 4 5 false false 0 decimal

&[DATE] &[TIME] - &[FILENAME]

v Cubic a v : S1 36 a 2 el * a 3 el * 108 a 1 el * - 8 a 3 el (3^* - 1212 a 2 el (3^* 3 a 2 el (2^* a 3 el (2^* - 54 a 1 el * a 2 el * a 3 el * - 81 a 1 el (2^* + 12 a 1 el * a 3 el (3^* + (12 /^* + (13 /^: q 13 / a 2 el * 19 / a 3 el (2^* - : Z 16 / S1 * 6 q S1 / * - 13 / a 3 el * - 1 - 12 / S1 * 3 q S1 / * + 13 / a 3 el * - i 3 sqrt 2 / * 16 / S1 * 6 q S1 / * + (* + 1 - 12 / S1 * 3 q S1 / * + 13 / a 3 el * - i 3 sqrt 2 / * 16 / S1 * 6 q S1 / * + (* - 31 mat : 31 line Z 31 line :

cubicroots(a,b,c,d)

d c b a cubicroots a 0 ≠ Q 3 a * c * b 2^- 9 a 2^* / eval : R 9 a * b * c * 27 a 2^* d * - 2 b 3^* - 54 a 3^* / eval : S R Q 3^R 2^+ sqrt + 3 nthroot eval : T R Q 3^R 2^+ sqrt - 3 nthroot eval : S T + b 3 a * / - S T + 2 / - b 3 a * / - i 3 sqrt * 2 / S T - (* + S T + 2 / - b 3 a * / - i 3 sqrt * 2 / S T - (* - 31 mat eval 51 line b 0 ≠ c c 2^4 b * d * - sqrt - 2 b * / - c c 2^4 b * d * - sqrt + 2 b * / - 21 mat eval c 0 ≠ d c / - eval 0 if if if 11 line :

v 04 - 6114 mat transpose :

d v 1 el : c v 2 el : b v 3 el : a v 4 el : 14 mat

v Cubic 0.6056 8.2897 1015 -^* i * - 6.6056 - 6.4121 1016 -^* i * - 3.2895 1015 -^* - 8.9309 1015 -^* i * + 31 mat

Cubic() returns complex residuals where cubicroots() does not Why is that? Also, Cubic() is structured such that a must be equal to one Doesn't that make it less flexible?

d c b a cubicroots 0.6056 6.6056 - 3.9968 1015 -^* - 31 mat

x f a x 3^* b x 2^* + c x * + d + :

x f