If you factor the equations before passing them to solve then the solution is ``` >>> fsol [{a2: 0, a1: 0}, {a2: 0, a1: 0}, {a2: 0, b2: 0}, {a2: 0, b1: b2/2, a1: 0}, {a2: (64 + (-8 + (1 - sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))*(1 - sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))**2/(36*(1 - sqrt(3)*I)**2*(37 + 3*sqrt(303)*I)**(2/3)), b1: b2/2, a1: (-64 + (1 - sqrt(3)*I)*(8 + (-1 + sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))*(37 + 3*sqrt(303)*I)**(1/3))/(6*(1 - sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))}, {a2: (64 + (-8 + (1 + sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))*(1 + sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))**2/(36*(1 + sqrt(3)*I)**2*(37 + 3*sqrt(303)*I)**(2/3)), b1: b2/2, a1: (-64 + (1 + sqrt(3)*I)*(8 - (1 + sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))*(37 + 3*sqrt(303)*I)**(1/3))/(6*(1 + sqrt(3)*I)*(37 + 3*sqrt(303)*I)**(1/3))}, {a2: (16 + (4 + (37 + 3*sqrt(303)*I)**(1/3))*(37 + 3*sqrt(303)*I)**(1/3))**2/(9*(37 + 3*sqrt(303)*I)**(2/3)), b1: b2/2, a1: 4/3 + 16/(3*(37 + 3*sqrt(303)*I)**(1/3)) + (37 + 3*sqrt(303)*I)**(1/3)/3}, {a2: a1**2, b1: 0, b2: 0}] ``` and all of them are true solutions: ``` >>> [[e.subs(s).simplify() for e in Sys] for s in fsol] [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]] ```
On Sunday, December 5, 2021 at 8:35:28 AM UTC-6 Chris Smith wrote: > `nonlinsolve(Sys, Unk)` gives > ``` > {(0, 0, b1, b2), (0, 0, b2/2, b2), (a1, 0, b1, 0), (-sqrt(a2), a2, 0, 0), > (sqrt(a2), a2, 0, 0)} > ``` > On Saturday, December 4, 2021 at 4:36:04 PM UTC-6 emanuel.c...@gmail.com > wrote: > >> A bug in Sympy’s solve >> >> Inspired by this ask.sagemath.orgquestion >> <https://ask.sagemath.org/question/59063/weird-c-values-from-solving-system-of-equations/> >> . >> >> This question was about the names of the symbolic variables created by >> Maxima’s solve to denote arbitrary constants. Exploring the example used >> in this questions revealed a non-trivial problem with sympy’s solve. >> Problem >> >> Solve the system : >> $$ >> \begin{align*} >> >> - a*{1}^{3} a*{2} + a*{1} a*{2}^{2} &= 0 \ >> - 3 a*{1}^{2} a*{2} b*{1} + 2 a*{1} a*{2} b*{2} - a*{1} b*{2} + a*{2}^{2} >> b*{2} &= 0 \ >> - a*{1}^{2} a*{2}^{2} + a_{2}^{3} &= 0 \ >> - 2 a*{1}^{2} a*{2} b*{2} - 2 a*{2}^{2} b*{1} + 3 a*{2}^{2} b_{2} &= 0 >> \end{align*} >> $$ >> >> # Set up sympy (brutal version)import sympyfrom sympy import * >> init_session() >> init_printing(pretty_print=False)# System to solve >> a1, a2, b1, b2 = symbols('a1 a2 b1 b2') >> Unk = [a1, a2, b1, b2] >> eq1 = a1 * a2**2 - a2 * a1**3 >> eq2 = 2*a1*a2*b2 + b2*a2**2 - 3*a2*a1**2*b1 - a1*b2 >> eq3 = a2**3 - a2**2*a1**2 >> eq4 = 3*a2**2*b2 - 2*a2*a1**2*b2 - 2*a2**2*b1 >> Sys = [eq1, eq2, eq3, eq4] >> >> IPython console for SymPy 1.9 (Python 3.9.9-64-bit) (ground types: gmpy) >> >> These commands were executed: >> >>> from __future__ import division >> >>> from sympy import * >> >>> x, y, z, t = symbols('x y z t') >> >>> k, m, n = symbols('k m n', integer=True) >> >>> f, g, h = symbols('f g h', cls=Function) >> >>> init_printing() >> >> Documentation can be found at https://docs.sympy.org/1.9/ >> >> Attempt to use the “automatic” Sympy solver: >> >> Sol = solve(Sys, Unk) >> DSol = [dict(zip(Unk, u)) for u in Sol] >> DSol >> >> [{a1: 0, a2: 0, b1: b1, b2: b2}, {a1: 0, a2: 0, b1: b1, b2: b2}, {a1: 0, a2: >> 0, b1: b1, b2: b2}, {a1: 0, a2: 0, b1: b2/2, b2: b2}, {a1: a1, a2: 0, b1: >> b1, b2: 0}, {a1: -sqrt(a2), a2: a2, b1: 0, b2: 0}, {a1: >> -2**(5/6)*sqrt(48*2**(1/3) + 624/(1499 + 3*sqrt(303)*I)**(1/3) + >> 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/6, a2: 16/3 + >> 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 + >> 3*sqrt(303)*I)**(1/3)/3, b1: b2/2, b2: b2}, {a1: -sqrt(6)*sqrt(32 - >> 2**(1/3)*(1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + >> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)))/6, a2: 2**(2/3)*(-208/3 + (1 + >> sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)/((1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)), >> b1: b2/2, b2: b2}, {a1: -sqrt(6)*sqrt(32 - 416*2**(2/3)/((1 - >> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3))/6, a2: 2**(2/3)*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + >> sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)/((1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)), >> b1: b2/2, b2: b2}] >> >> Something went sideways: the six first solutions are okay,but the last >> three use expressions, some of them being polynomials in b2. >> >> Attempt to check them formally : >> >> Chk=[[u.subs(s) for u in Sys] for s in DSol] >> Chk >> >> [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, >> 0, 0, 0], [-2**(5/6)*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) >> + 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)**2*sqrt(48*2**(1/3) + 624/(1499 >> + 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/6 + >> sqrt(2)*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + >> 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)*(48*2**(1/3) + 624/(1499 + >> 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))**(3/2)/54, >> -2**(2/3)*b2*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + >> 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)*(48*2**(1/3) + 624/(1499 + >> 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/12 - >> 2**(5/6)*b2*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + >> 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)*sqrt(48*2**(1/3) + 624/(1499 + >> 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/3 + >> b2*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 + >> 3*sqrt(303)*I)**(1/3)/3)**2 + 2**(5/6)*b2*sqrt(48*2**(1/3) + 624/(1499 + >> 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/6, >> -2**(2/3)*(16/3 + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + >> 2**(1/3)*(1499 + 3*sqrt(303)*I)**(1/3)/3)**2*(48*2**(1/3) + 624/(1499 + >> 3*sqrt(303)*I)**(1/3) + 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/18 + (16/3 >> + 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 + >> 3*sqrt(303)*I)**(1/3)/3)**3, -2**(2/3)*b2*(16/3 + 104*2**(2/3)/(3*(1499 + >> 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 + >> 3*sqrt(303)*I)**(1/3)/3)*(48*2**(1/3) + 624/(1499 + 3*sqrt(303)*I)**(1/3) + >> 3*2**(2/3)*(1499 + 3*sqrt(303)*I)**(1/3))/9 + 2*b2*(16/3 + >> 104*2**(2/3)/(3*(1499 + 3*sqrt(303)*I)**(1/3)) + 2**(1/3)*(1499 + >> 3*sqrt(303)*I)**(1/3)/3)**2], [2**(1/6)*sqrt(3)*(-208/3 + (1 + >> sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)*(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)))**(3/2)/(18*(1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)) - 2**(5/6)*sqrt(3)*(-208/3 + (1 + sqrt(3)*I)*(32 + >> (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)**2*sqrt(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)))/(3*(1 + sqrt(3)*I)**2*(1499 + >> 3*sqrt(303)*I)**(2/3)), -2*2**(1/6)*sqrt(3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 >> + (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)*sqrt(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)))/(3*(1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - >> 3*2**(2/3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + >> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)*(16/3 - 2**(1/3)*(1 >> + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)/6 - 208*2**(2/3)/(3*(1 + >> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)))/(2*(1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)) + sqrt(6)*b2*sqrt(32 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 >> + 3*sqrt(303)*I)**(1/3) - 416*2**(2/3)/((1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)))/6 + 2*2**(1/3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 + >> (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)**2/((1 + sqrt(3)*I)**2*(1499 + >> 3*sqrt(303)*I)**(2/3)), -2*2**(1/3)*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - >> sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)**2*(16/3 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)/6 - 208*2**(2/3)/(3*(1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)))/((1 + sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3)) + >> 4*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + >> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**3/((1 + >> sqrt(3)*I)**3*(1499 + 3*sqrt(303)*I)), -2*2**(2/3)*b2*(-208/3 + (1 + >> sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)*(16/3 - 2**(1/3)*(1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)/6 - 208*2**(2/3)/(3*(1 + sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)))/((1 + sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) + >> 4*2**(1/3)*b2*(-208/3 + (1 + sqrt(3)*I)*(32 + (-1 - sqrt(3)*I)*(2998 + >> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**2/((1 + >> sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3))], [-2**(5/6)*sqrt(3)*(-208/3 + >> (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 >> + 6*sqrt(303)*I)**(1/3)/12)**2*sqrt(32 - 416*2**(2/3)/((1 - sqrt(3)*I)*(1499 >> + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3))/(3*(1 - sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3)) >> + 2**(1/6)*sqrt(3)*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 + >> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)*(32 - >> 416*2**(2/3)/((1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - >> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3))**(3/2)/(18*(1 - sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)), sqrt(6)*b2*sqrt(32 - 416*2**(2/3)/((1 - >> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3))/6 + 2*2**(1/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 >> + sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)**2/((1 - sqrt(3)*I)**2*(1499 + >> 3*sqrt(303)*I)**(2/3)) - 3*2**(2/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + >> sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)*(16/3 - 208*2**(2/3)/(3*(1 - sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)/6)/(2*(1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - >> 2*2**(1/6)*sqrt(3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 >> + 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)*sqrt(32 - >> 416*2**(2/3)/((1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - >> sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3))/(3*(1 - sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)), 4*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + >> sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)**3/((1 - sqrt(3)*I)**3*(1499 + 3*sqrt(303)*I)) - >> 2*2**(1/3)*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 + >> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**2*(16/3 - >> 208*2**(2/3)/(3*(1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 >> - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)/6)/((1 - sqrt(3)*I)**2*(1499 + >> 3*sqrt(303)*I)**(2/3)), -2*2**(2/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + >> sqrt(3)*I)*(2998 + 6*sqrt(303)*I)**(1/3))*(2998 + >> 6*sqrt(303)*I)**(1/3)/12)*(16/3 - 208*2**(2/3)/(3*(1 - sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)) - 2**(1/3)*(1 - sqrt(3)*I)*(1499 + >> 3*sqrt(303)*I)**(1/3)/6)/((1 - sqrt(3)*I)*(1499 + 3*sqrt(303)*I)**(1/3)) + >> 4*2**(1/3)*b2*(-208/3 + (1 - sqrt(3)*I)*(32 + (-1 + sqrt(3)*I)*(2998 + >> 6*sqrt(303)*I)**(1/3))*(2998 + 6*sqrt(303)*I)**(1/3)/12)**2/((1 - >> sqrt(3)*I)**2*(1499 + 3*sqrt(303)*I)**(2/3))]] >> >> Some of these expressions are polynomial in b2 whose coefficients of >> not-null dregree are nort *obviously* null. >> A bit of hit-and-miss trials leads to this attempt at *numerical* check : >> >> def chz(x): >> r = x.factor().is_zero >> if r is None: return x.coeff(b2).n() >> return r >> [[chz(u) for u in v] for v in Chk] >> >> [[True, True, True, True], >> [True, True, True, True], >> [True, True, True, True], >> [True, True, True, True], >> [True, True, True, True], >> [True, True, True, True], >> [True, -223.427427555427 + 0.e-25*I, True, True], >> [True, -0.388012232966408 - 5.25038392589403e-29*I, True, True], >> [True, -0.e-137 + 1.12445821450677e-139*I, True, True]] >> >> Solution #8 may be exactr, but #6 and #7 cannot. >> Manual solution. >> >> eq1 and eq3 give us solutions for a1 and a2 : >> >> S13=solve([eq3, eq1], [a2, a1], dict=True) ; S13 >> >> [{a1: -sqrt(a2)}, {a1: sqrt(a2)}, {a2: 0}] >> >> which we prefer to rewrite as : >> >> S13 = [{a2:a1**2}, {a2:0}]; S13 >> >> [{a2: a1**2}, {a2: 0}] >> >> Substituting these values in eq4 gives us : >> >> E4 = [eq4.subs(s) for s in S13] ; E4 >> >> [-2*a1**4*b1 + a1**4*b2, 0] >> >> The second solution tells us that S13[1] is also,a solution to [eq1, >> eq3, eq4]. >> >> S134=S13[1:] ; S134 >> >> [{a2: 0}] >> >> Substituting S13[0] ineq4and solving for the variables of the resulting >> polynomial augments the setS134of solutions of[eq1, eq3`, eq4]’ : >> >> V4=E4[0].free_symbols >> S4=[solve(E4[0], v, dict=True) for v in V4] ; S4for S in S4: >> for s in S: >> S0={u:S13[0][u].subs(s) for u in S13[0].keys()} >> S134 += [S0.copy()|s] >> S134 >> >> [{a2: 0}, {a2: a1**2, b1: b2/2}, {a1: 0, a2: 0}, {a2: a1**2, b2: 2*b1}] >> >> Again, we prefer to rewrite it in a simpler (and shorter) fashion : >> >> S134=[{a2:0}, {a2:a1**2, b2:2*b1}] ; S134 >> >> [{a2: 0}, {a2: a1**2, b2: 2*b1}] >> >> Substituting in eq3 gives : >> >> [eq2.subs(s) for s in S134] >> >> [-a1*b2, -a1**4*b1 + 4*a1**3*b1 - 2*a1*b1] >> >> Solving these equations for their free symbols a,d merging with the >> previous partial solutions gives us the solutions of the full system : >> >> S1234=[]for S in S134: >> # print("S=",S) >> E=eq2.subs(S) >> # print("E=",E) >> S1=[solve(E, v, dict=True) for v in E.free_symbols] >> # print("S1=",S1) >> for s in flatten(S1): >> # print("s=",s) >> S0={u:S[u].subs(s) if "subs" in dir(S[u]) else S[u] for u in >> S.keys()} >> S1234+=[S0.copy()|s] >> S1234 >> >> [{a1: 0, a2: 0}, {a2: 0, b2: 0}, {a2: a1**2, b1: 0, b2: 0}, {a1: 0, a2: 0, >> b2: 2*b1}, {a1: 4/3 + (-1/2 - sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3) + >> 16/(9*(-1/2 - sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3)), a2: (4/3 + (-1/2 >> - sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3) + 16/(9*(-1/2 - >> sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3)))**2, b2: 2*b1}, {a1: 4/3 + >> 16/(9*(-1/2 + sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3)) + (-1/2 + >> sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3), a2: (4/3 + 16/(9*(-1/2 + >> sqrt(3)*I/2)*(37/27 + sqrt(303)*I/9)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(37/27 + >> sqrt(303)*I/9)**(1/3))**2, b2: 2*b1}, {a1: 4/3 + 16/(9*(37/27 + >> sqrt(303)*I/9)**(1/3)) + (37/27 + sqrt(303)*I/9)**(1/3), a2: (4/3 + >> 16/(9*(37/27 + sqrt(303)*I/9)**(1/3)) + (37/27 + sqrt(303)*I/9)**(1/3))**2, >> b2: 2*b1}] >> >> Again, some solutions, substituted in eq2, give first-degree monomials >> in b1 whose oefficient cannot be shownt to be null byis_zero : >> >> [[e.subs(s).is_zero for e in Sys] for s in S1234] >> >> [[True, True, True, True], >> [True, True, True, True], >> [True, True, True, True], >> [True, True, True, True], >> [True, None, True, True], >> [True, None, True, True], >> [True, None, True, True]] >> >> But, this time, the numerical check points to a probably null result : >> >> [Sys[1].subs(S1234[u]).coeff(b1).n() for u in range(3,6)] >> >> [0, 0.e-125 + 0.e-127*I, 0.e-125 - 0.e-127*I] >> >> >> > -- You received this message because you are subscribed to the Google Groups "sympy" group. 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