var a; var b; var c; var d; var e; subject to cons1: a + b + c + d + e = 0; cons2: a*b + b*c + c*d + d*e + e*a = 0; cons3: a*b*c + b*c*d + c*d*e + d*e*a + e*a*b = 0; cons4: a*b*c*d + b*c*d*e + c*d*e*a + d*e*a*b + e*a*b*c = 0; cons5: a*b*c*d*e - 1 = 0; solve; display a, b, c, d, e; # TITLE : cyclic 5-roots problem # ROOT COUNTS : # total degree : 120 # 5-homogeneous Bezout number : 120 # with partition : {a }{b }{c }{d }{e } # generalized Bezout number : 106 # based on the set structure : # {a b c d e } # {a c e }{b d e } # {a d }{b d e }{c e } # {a e }{b e }{c e }{d e } # {a }{b }{c }{d }{e } # mixed volume : 70 = 14*5 # SYMMETRY GROUP : # b c d e a # e d c b a # SYMMETRIC SET STRUCTURE : # { a b c d e } # { a } { b } { c } { d } { e } # { a } { b } { c } { d } { e } # { a } { b } { c } { d } { e } # { a } { b } { c } { d } { e } # with generalized Bezout bound : 120, leading to 12 generating solutions. # REFERENCES : # See G\"oran Bj\"orck and Ralf Fr\"oberg: # `A faster way to count the solutions of inhomogeneous systems # of algebraic equations, with applications to cyclic n-roots', # in J. Symbolic Computation (1991) 12, pp 329--336. # THE GENERATING SOLUTIONS : # 7 5 # =========================================================== # solution 1 : # t : 1.00000000000000E+00 0.00000000000000E+00 # m : 10 # the solution for t : # a : 3.09016994374947E-01 -9.51056516295154E-01 # b : 3.09016994374947E-01 -9.51056516295154E-01 # c : -8.09016994374948E-01 2.48989828488278E+00 # d : -1.18033988749895E-01 3.63271264002680E-01 # e : 3.09016994374948E-01 -9.51056516295154E-01 # == err : 8.556E-16 = rco : 6.220E-02 = res : 7.022E-16 == # solution 2 : # t : 1.00000000000000E+00 0.00000000000000E+00 # m : 10 # the solution for t : # a : 1.00000000000000E+00 -3.31628872515627E-75 # b : 1.00000000000000E+00 -8.84343660041671E-75 # c : -2.61803398874990E+00 3.31628872515627E-75 # d : -3.81966011250105E-01 1.65814436257813E-75 # e : 1.00000000000000E+00 6.90893484407556E-75 # == err : 4.713E-15 = rco : 6.850E-02 = res : 4.441E-16 == # solution 3 : # t : 1.00000000000000E+00 0.00000000000000E+00 # m : 10 # the solution for t : # a : 3.09016994374948E-01 9.51056516295154E-01 # b : 3.09016994374947E-01 9.51056516295154E-01 # c : -8.09016994374948E-01 -2.48989828488278E+00 # d : -1.18033988749895E-01 -3.63271264002680E-01 # e : 3.09016994374947E-01 9.51056516295154E-01 # == err : 6.582E-16 = rco : 6.220E-02 = res : 4.965E-16 == # solution 4 : # t : 1.00000000000000E+00 0.00000000000000E+00 # m : 10 # the solution for t : # a : -8.09016994374947E-01 5.87785252292473E-01 # b : -8.09016994374947E-01 5.87785252292473E-01 # c : 2.11803398874990E+00 -1.53884176858763E+00 # d : 3.09016994374947E-01 -2.24513988289793E-01 # e : -8.09016994374948E-01 5.87785252292473E-01 # == err : 5.945E-15 = rco : 6.765E-02 = res : 4.003E-16 == # solution 5 : # t : 1.00000000000000E+00 0.00000000000000E+00 # m : 10 # the solution for t : # a : -8.09016994374947E-01 -5.87785252292473E-01 # b : -8.09016994374947E-01 -5.87785252292473E-01 # c : 2.11803398874990E+00 1.53884176858763E+00 # d : 3.09016994374947E-01 2.24513988289793E-01 # e : -8.09016994374948E-01 -5.87785252292473E-01 # == err : 5.945E-15 = rco : 6.765E-02 = res : 4.003E-16 == # solution 6 : # t : 1.00000000000000E+00 0.00000000000000E+00 # m : 10 # the solution for t : # a : 1.00000000000000E+00 -7.24393703353565E-18 # b : -8.09016994374947E-01 -5.87785252292473E-01 # c : 3.09016994374947E-01 9.51056516295154E-01 # d : 3.09016994374947E-01 -9.51056516295154E-01 # e : -8.09016994374948E-01 5.87785252292473E-01 # == err : 7.269E-16 = rco : 2.571E-01 = res : 4.442E-16 == # solution 7 : # t : 1.00000000000000E+00 0.00000000000000E+00 # m : 10 # the solution for t : # a : -8.09016994374947E-01 5.87785252292473E-01 # b : -8.09016994374947E-01 -5.87785252292473E-01 # c : 3.09016994374947E-01 -9.51056516295153E-01 # d : 1.00000000000000E+00 2.82553319327192E-17 # e : 3.09016994374947E-01 9.51056516295153E-01 # == err : 6.769E-16 = rco : 2.221E-01 = res : 7.022E-16 ==