Saalehorizontale

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@rcorless That is an interesting ap...

Answered: Saalehorizontale 35

December 19 2024

0 0

@rcorless That is an interesting approach. Thank you. So this process transform my system of equations and inequalities in an easier system? Because it does not give the solution immediately. So I know would like to take the new system and use solve, is that right? However I dont know how to exctract the equations and inequalities from the solutions of regularchains. How can I do that, because there are multiple lines. Do you have advice? RegularChains.mw

$restart; with(RegularChains); equations := {-y*(m-p) = 0, ((-x-y+1)*k+x)*n+s*y-t = 0, (k-x-y)*t-k*p+y = 0, (-m+n+y)*x+m-1 = 0, -(x+y-1)*(p-t)*k+(-x-y+1)*t+x*p = 0, y^2+(-m-1)*y+1+x*(p-1) = 0, (-x-y+1)*t+(-m+1)*x+y*n+m-1 = 0, -k*n+s*x = 0}; restrictions := {0 < k, 0 < m, 0 < s, 0 < x, 0 < y, 0 < n+(t-1)*p, 0 < (m*y-1)*n+(1-p)*(m*x-m+1), 0 < (m*x-m-t+1)*p+m*y*(t-n), 1 < x+y, k < 1, m < 1, s < t, t < 1}; sys := `union`(equations, restrictions); SuggestVariableOrder(sys); R := PolynomialRing(%); dec := RealTriangularize(sys, R); ZZZ := Display(dec, R)$

$sys := {-y*(m-p) = 0, ((-x-y+1)*k+x)*n+s*y-t = 0, (k-x-y)*t-k*p+y = 0, (-m+n+y)*x+m-1 = 0, -(x+y-1)*(p-t)*k+(-x-y+1)*t+x*p = 0, y^2+(-m-1)*y+1+x*(p-1) = 0, (-x-y+1)*t+(-m+1)*x+y*n+m-1 = 0, -k*n+s*x = 0, 0 < k, 0 < m, 0 < s, 0 < x, 0 < y, 0 < n+(t-1)*p, 0 < (m*y-1)*n+(1-p)*(m*x-m+1), 0 < (m*x-m-t+1)*p+m*y*(t-n), 1 < x+y, k < 1, m < 1, s < t, t < 1}$

(1)

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Sol_traditional and Sol_backsolve should...

Answered: Saalehorizontale 35

December 19 2024

0 0

@rcorless Yes you are right. But when that solutions fullfills the equations under the restrictions, than it should appear if I solve the system without restrictions as well, since with restrictions you get a subset of solutions. So why is the marked solution you gave not part of the solutions of Sol_traditinal and Sol_backsolve? Thats my main concern. THey should be there, right? Since PolynomialSystem (under some engignes) does not find a solution which has to be there I am loosing trust into the function. Do you see my point?

Missing Solutions and strange Output for...

Answered: Saalehorizontale 35

December 09 2024

0 0

Thank you for your resonse @Scot Gould . I see your point. But why does adding 0<s makes the problem and not 0<x and 0<y? And shouldnt Maple find out at itself, that although I ask for 8 variables and it is a parametric solution that it can give me just 7 variables back?

I would be glad if you have time to look at my second approach. In the attached file you can see the comparison of three solving approaches. In the first case I solve equations and inequalities together and get a nice solution, this is the one I want to replicate. In the second and thirsd case you can see the two stepped approch. If I take the solutions found while solving 8 equations and put it together with the inequalities it does not find any solution. But as we know from sol_1 there is one. Why is that?

In the third case I asked for allvalues first and now he finds the solution which is missing in case 2. However the way the solution is shown is weird. The conditions are really complicated and not necesarry. You could summarisze them since they are all depend from x. furthermore it is given in two rows, but the second row is useless. I ask for solutions and it tells me the area where it cant find a solution. How can I change the way Maple gives back solutions?

By the way sol_1 and sol_3 are equal, you can show that k and t in both cases have the same value although they look diffrent.

Thank you

$restart; equations_3042 := {((m*y-p)*k+p)*x-m*y = 0, (-x-y+1)*s+y*n = 0, (-k*x-y+1)*p+k*m*x*y = 0, -1-y^2*m+((1-x)*m+p+x)*y = 0, x*(m*x-m-n+1)*k+(1-y)*n+y*s-t = 0, -(x-1)*(x+y-1)*m+x^2-x+1+(n-1)*y = 0, k*m*x^2+((-m-n+1)*k-m+n-t)*x+m+(1-y)*t+y-1 = 0, p*y-1 = 0}; inequalities := {0 < k, 0 < m, 0 < s, 0 < x, 0 < y, 0 < n+(p-1)*s, 0 < (m*y-1)*n+(m*x-m+1)*(1-p), 0 < (m*x-m-s+1)*p+m*y*(s-n), 1 < x+y, k < 1, m < 1, s < t, t < 1}; sol_1 := solve(`union`(equations_3042, inequalities))$

${k = (1+x^(1/2))/(x^(1/2)*(x^(3/2)+2*x-1)), m = x/(x^(1/2)+x), n = x^(1/2)/(1+x^(1/2)), p = 1/(1+x^(1/2)), s = x^(1/2)/(x^(1/2)+x), t = (5*x^(3/2)+2*x^2-x^(1/2)+3*x-1)/(x^(5/2)+5*x^(3/2)+4*x^2-2*x^(1/2)+x-1), y = 1+x^(1/2), 1 < x}$

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>	$sol_eq := SolveTools:-PolynomialSystem(equations_3042, indets(equations_3042), engine = triade); for expr in sol_eq do sol_2 := solve(`union`(expr, inequalities), {k, m, n, p, s, t, y}) end do$

${k = k, m = 0, n = 0, p = 1, s = s, t = s, x = 0, y = 1}, {k = (x*RootOf(_Z^2-2*_Z-x+1)-RootOf(_Z^2-2*_Z-x+1)-2*x+1)/((x^2-3*x+1)*x), m = (-RootOf(_Z^2-2*_Z-x+1)+x+1)/(x-1), n = (-RootOf(_Z^2-2*_Z-x+1)+x+1)/(x-1), p = (RootOf(_Z^2-2*_Z-x+1)-2)/(x-1), s = (RootOf(_Z^2-2*_Z-x+1)-2)/(x-1), t = (2*x*RootOf(_Z^2-2*_Z-x+1)-RootOf(_Z^2-2*_Z-x+1)-5*x+2)/(x^2-3*x+1), x = x, y = RootOf(_Z^2-2*_Z-x+1)}, {k = 1, m = m, n = 1-m, p = m, s = 1-m, t = 1, x = 1, y = 1/m}, {k = k, m = 1, n = 0, p = 1, s = 0, t = k, x = 1, y = 1}$

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>	$sol_all := NULL; for expr in sol_eq do sol_av := allvalues(expr); sol_all := sol_all, sol_av end do; print("sol_all", sol_all); for expr in sol_all do sol_3 := solve(`union`(expr, inequalities), {k, m, n, p, s, t, y}) end do$

$"sol_all", {k = k, m = 0, n = 0, p = 1, s = s, t = s, x = 0, y = 1}, {k = ((1+sqrt(x))*x-sqrt(x)-2*x)/((x^2-3*x+1)*x), m = (-sqrt(x)+x)/(x-1), n = (-sqrt(x)+x)/(x-1), p = (-1+sqrt(x))/(x-1), s = (-1+sqrt(x))/(x-1), t = ((2*(1+sqrt(x)))*x+1-sqrt(x)-5*x)/(x^2-3*x+1), x = x, y = 1+sqrt(x)}, {k = (x*(1-sqrt(x))+sqrt(x)-2*x)/((x^2-3*x+1)*x), m = (sqrt(x)+x)/(x-1), n = (sqrt(x)+x)/(x-1), p = (-1-sqrt(x))/(x-1), s = (-1-sqrt(x))/(x-1), t = (2*x*(1-sqrt(x))+1+sqrt(x)-5*x)/(x^2-3*x+1), x = x, y = 1-sqrt(x)}, {k = 1, m = m, n = 1-m, p = m, s = 1-m, t = 1, x = 1, y = 1/m}, {k = k, m = 1, n = 0, p = 1, s = 0, t = k, x = 1, y = 1}$

$piecewise(Or(And(1 < x, -1 < x^(1/2), 0 < x^(1/2)+x, 0 < x/(1+x^(1/2))^2, 0 < x^(1/2)/(1+x^(1/2)), 0 < x^(1/2)/(1+x^(1/2))^2, 0 < 1/(x^(1/2)*(x+x^(1/2)-1)), (2*x^(1/2)-1)/(x+x^(1/2)-1) < 1, 1/(x^(1/2)*(x+x^(1/2)-1)) < 1, 0 < (x^2+x^(3/2))/(1+x^(1/2))^2, 1/(1+x^(1/2)) < (2*x^(1/2)-1)/(x+x^(1/2)-1), -x^2+3*x < 1, 0 < 1/(1+x^(1/2))), And(1 < x, -1 < x^(1/2), 0 < x^(1/2)+x, 0 < x/(1+x^(1/2))^2, 0 < x^(1/2)/(1+x^(1/2)), 0 < x^(1/2)/(1+x^(1/2))^2, 0 < 1/(x^(1/2)*(x+x^(1/2)-1)), (2*x^(1/2)-1)/(x+x^(1/2)-1) < 1, 1/(x^(1/2)*(x+x^(1/2)-1)) < 1, 0 < (x^2+x^(3/2))/(1+x^(1/2))^2, 1/(1+x^(1/2)) < (2*x^(1/2)-1)/(x+x^(1/2)-1), x^2-3*x < -1, 0 < 1/(1+x^(1/2))), And(0 < x, x < 1, -1 < x^(1/2), 0 < x^(1/2)+x, 0 < x/(1+x^(1/2))^2, 0 < x^(1/2)/(1+x^(1/2)), 0 < x^(1/2)/(1+x^(1/2))^2, 0 < 1/(x^(1/2)*(x+x^(1/2)-1)), (2*x^(1/2)-1)/(x+x^(1/2)-1) < 1, 1/(x^(1/2)*(x+x^(1/2)-1)) < 1, 0 < (x^2+x^(3/2))/(1+x^(1/2))^2, 1/(1+x^(1/2)) < (2*x^(1/2)-1)/(x+x^(1/2)-1), -x^2+3*x < 1, 0 < 1/(1+x^(1/2))), And(0 < x, x < 1, -1 < x^(1/2), 0 < x^(1/2)+x, 0 < x/(1+x^(1/2))^2, 0 < x^(1/2)/(1+x^(1/2)), 0 < x^(1/2)/(1+x^(1/2))^2, 0 < 1/(x^(1/2)*(x+x^(1/2)-1)), (2*x^(1/2)-1)/(x+x^(1/2)-1) < 1, 1/(x^(1/2)*(x+x^(1/2)-1)) < 1, 0 < (x^2+x^(3/2))/(1+x^(1/2))^2, 1/(1+x^(1/2)) < (2*x^(1/2)-1)/(x+x^(1/2)-1), x^2-3*x < -1, 0 < 1/(1+x^(1/2)))), [{k = 1/(x^(1/2)*(x+x^(1/2)-1)), m = x^(1/2)/(1+x^(1/2)), n = x^(1/2)/(1+x^(1/2)), p = 1/(1+x^(1/2)), s = 1/(1+x^(1/2)), t = (2*x^(1/2)-1)/(x+x^(1/2)-1), y = 1+x^(1/2)}], [])$

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Download Strange_Output_form_and_missing_solutions.mw

@dharr Thank you for the fast respo...

Answered: Saalehorizontale 35

October 15 2024

0 0

@dharr Thank you for the fast response. That is a good explination. I didnt thought of that. Does that mean, that in first place alle the solutions of the RootOf expression are possible solutions? If Maple would mean one specifc RootOf solution the programm would make that clear either with an index or with a numeric expression like 0,355...0,356 behind the RootOf expression right?

And does this also explain, why solving the second step with the symbolic expression of RootOf does take so much more time? Because there is more the the on value you get with evalf? I just changed to the numeric expression via evalf because it took to much time and I already saw the "error" ,you just explained, inbetween.

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