sorting the roots from the solution of an equation

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Jack_111
Jack_111 el 16 de Oct. de 2013
Respondida: Jack_111 el 16 de Oct. de 2013
I have the following equation and I solve it using the Function "Solve" and the answer is as follows
syms t px py pz dx dy dz A21 A20 A19 A18 A17 A16 A15 A14 A13 A12 A11 A10 A9 A8 A7 A6 A5 A4 A3 A2 A1 Thickness;
Equation = A21.*((py +t*dy).^5) + (A20.*((py +t*dy).^4)).*(px +t*dx) + A19.*((py +t*dy).^4) + (A18.*((py +t*dy).^3)).*((px +t*dx).^2) + (A17.*((py +t*dy).^3)).*(px +t*dx) + A16.*((py +t*dy).^3) + (A15.*((py +t*dy).^2)).*((px +t*dx).^3) + (A14.*((py +t*dy).^2)).*((px +t*dx).^2) + (A13.*((py +t*dy).^2)).*(px +t*dx) + A12.*((py +t*dy).^2) + (A11.*((py +t*dy))).*((px +t*dx).^4) + (A10.*(py +t*dy)).*((px +t*dx).^3) + (A9.*(py +t*dy)).*((px +t*dx).^2) + (A8.*(py +t*dy)).*((px +t*dx)) + (A7.*(py +t*dy)) + A6.*((px +t*dx).^5) + A5.*((px +t*dx).^4) + A4.*((px +t*dx).^3) + A3.*((px +t*dx).^2) + A2.*(px +t*dx) + A1 - (pz +t*dz)
Answer = solve(S,t)
Answer = RootOf(A20*dx*dy^4*z^5 + A11*dx^4*dy*z^5 + A18*dx^2*dy^3*z^5 + A15*dx^3*dy^2*z^5 + A6*dx^5*z^5 + A21*dy^5*z^5 + 3*A18*dx^2*dy^2*py*z^4 + 3*A15*dx^2*dy^2*px*z^4 + 4*A20*dx*dy^3*py*z^4 + 2*A15*dx^3*dy*py*z^4 + 2*A18*dx*dy^3*px*z^4 + 4*A11*dx^3*dy*px*z^4 + 5*A6*dx^4*px*z^4 + 5*A21*dy^4*py*z^4 + A20*dy^4*px*z^4 + A11*dx^4*py*z^4 + A17*dx*dy^3*z^4 + A10*dx^3*dy*z^4 + A14*dx^2*dy^2*z^4 + A5*dx^4*z^4 + A19*dy^4*z^4 + 6*A18*dx*dy^2*px*py*z^3 + 6*A15*dx^2*dy*px*py*z^3 + 6*A20*dx*dy^2*py^2*z^3 + 3*A18*dx^2*dy*py^2*z^3 + 3*A15*dx*dy^2*px^2*z^3 + 6*A11*dx^2*dy*px^2*z^3 + 4*A20*dy^3*px*py*z^3 + 4*A11*dx^3*px*py*z^3 + 3*A17*dx*dy^2*py*z^3 + 2*A14*dx^2*dy*py*z^3 + 2*A14*dx*dy^2*px*z^3 + 3*A10*dx^2*dy*px*z^3 + 4*A5*dx^3*px*z^3 + 4*A19*dy^3*py*z^3 + 10*A6*dx^3*px^2*z^3 + 10*A21*dy^3*py^2*z^3 + A17*dy^3*px*z^3 + A10*dx^3*py*z^3 + A9*dx^2*dy*z^3 + A13*dx*dy^2*z^3 + A18*dy^3*px^2*z^3 + A15*dx^3*py^2*z^3 + A4*dx^3*z^3 + A16*dy^3*z^3 + 6*A18*dx*dy*px*py^2*z^2 + 6*A15*dx*dy*px^2*py*z^2 + 4*A14*dx*dy*px*py*z^2 + 2*A9*dx*dy*px*z^2 + 2*A13*dx*dy*py*z^2 + 6*A20*dy^2*px*py^2*z^2 + 3*A18*dy^2*px^2*py*z^2 + 3*A15*dx^2*px*py^2*z^2 + 6*A11*dx^2*px^2*py*z^2 + 3*A17*dy^2*px*py*z^2 + 3*A10*dx^2*px*py*z^2 + 4*A20*dx*dy*py^3*z^2 + 3*A17*dx*dy*py^2*z^2 + 4*A11*dx*dy*px^3*z^2 + 3*A10*dx*dy*px^2*z^2 + 3*A4*dx^2*px*z^2 + 3*A16*dy^2*py*z^2 + 10*A6*dx^2*px^3*z^2 + 6*A5*dx^2*px^2*z^2 + 10*A21*dy^2*py^3*z^2 + 6*A19*dy^2*py^2*z^2 + A9*dx^2*py*z^2 + A13*dy^2*px*z^2 + A8*dx*dy*z^2 + A18*dx^2*py^3*z^2 + A15*dy^2*px^3*z^2 + A14*dy^2*px^2*z^2 + A14*dx^2*py^2*z^2 + A3*dx^2*z^2 + A12*dy^2*z^2 + 4*A20*dy*px*py^3*z + 3*A17*dy*px*py^2*z + 2*A15*dy*px^3*py*z + 2*A14*dy*px^2*py*z + 2*A18*dx*px*py^3*z + 2*A14*dx*px*py^2*z + 4*A11*dx*px^3*py*z + 3*A10*dx*px^2*py*z + 2*A9*dx*px*py*z + 2*A13*dy*px*py*z + 3*A18*dy*px^2*py^2*z + 3*A15*dx*px^2*py^2*z + 2*A12*dy*py*z + 2*A3*dx*px*z + 5*A6*dx*px^4*z + 4*A5*dx*px^3*z + 5*A21*dy*py^4*z + 3*A4*dx*px^2*z + 4*A19*dy*py^3*z + 3*A16*dy*py^2*z + A8*dy*px*z + A8*dx*py*z + A9*dy*px^2*z + A20*dx*py^4*z + A17*dx*py^3*z + A13*dx*py^2*z + A11*dy*px^4*z + A10*dy*px^3*z + A7*dy*z + A2*dx*z - dz*z + A9*px^2*py + A20*px*py^4 + A17*px*py^3 + A13*px*py^2 + A11*px^4*py + A10*px^3*py + A8*px*py + A18*px^2*py^3 + A15*px^3*py^2 + A14*px^2*py^2 - pz + A6*px^5 + A5*px^4 + A4*px^3 + A21*py^5 + A3*px^2 + A19*py^4 + A16*py^3 + A12*py^2 + A7*py + A2*px + A1, z)
Is there any way to sort the roots in order to apply later the function "roots"? Or is there anyway to get the roots out of this Answer = RootOf(.......
I have all the variables at the end but they are quite a lot that's why I have to solve it generally before and then apply for loop to get the roots.
Many thanks in advance

Respuesta aceptada

Jack_111
Jack_111 el 16 de Oct. de 2013
This is the Answer
a = char(Answer); % Convert from symbolic to String
R = strrep(a,'RootOf(',''); % Preparing the Expression to be ready to get the Coefficients
R1 = strrep(R,',z)',''); % Preparing the Expression to be ready to get the Coefficients
b = sym(R1); % converting back to Symbolic
Coeff = coeffs(b,z); % I got the Coefficients

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