How to solve function

Hi,

I am trying to solve the fuction according to x1:

syms x1 x2

Eq2 = (1*(-0.0296505547))+(x1*(-1.4168313955))+(x2*(-0.4039704255))+(x1^2*(-0.0746402042))+((x2^2)*(-1.6203228042))+(x1^3*0.0000356471)+(x2^3*0.0400047846)+(x1*x2*0.7706539390)+((x1^2)*x2*0.0015188307)+(x1*(x2^2)*(-0.0173642989))==-log(1/0.4 - 1);

t_sol1 = solve(Eq2, x1)

The answer i get is this:

t_sol1 =

root(z^3 + (73786976294838206464*z^2*((3502185151774141*x2)/2305843009213693952 - 1344598383287911/18014398509481984))/2630291722679727 - (73786976294838206464*z*(- (3470716792512005*x2)/4503599627370496 + (2502459201778877*x2^2)/144115188075855872 + 1595210336205155/1125899906842624))/2630291722679727 + (2951832092960308736*x2^3)/2630291722679727 - (119558720343491166208*x2^2)/2630291722679727 - (9935918736728068096*x2)/876763907559909 + 9243406514527810816/876763907559909, z, 1)

root(z^3 + (73786976294838206464*z^2*((3502185151774141*x2)/2305843009213693952 - 1344598383287911/18014398509481984))/2630291722679727 - (73786976294838206464*z*(- (3470716792512005*x2)/4503599627370496 + (2502459201778877*x2^2)/144115188075855872 + 1595210336205155/1125899906842624))/2630291722679727 + (2951832092960308736*x2^3)/2630291722679727 - (119558720343491166208*x2^2)/2630291722679727 - (9935918736728068096*x2)/876763907559909 + 9243406514527810816/876763907559909, z, 2)

root(z^3 + (73786976294838206464*z^2*((3502185151774141*x2)/2305843009213693952 - 1344598383287911/18014398509481984))/2630291722679727 - (73786976294838206464*z*(- (3470716792512005*x2)/4503599627370496 + (2502459201778877*x2^2)/144115188075855872 + 1595210336205155/1125899906842624))/2630291722679727 + (2951832092960308736*x2^3)/2630291722679727 - (119558720343491166208*x2^2)/2630291722679727 - (9935918736728068096*x2)/876763907559909 + 9243406514527810816/876763907559909, z, 3)

Could You explain why the z occurs in solved function?

Answer 1

[Ameer Hamza]

The solution given by MATLAB shows that for a given value of x2, x1 has 3 solutions. However, MATLAB is not able to express those solutions analytically; therefore, it gave the output in terms of another function root(). If you look carefully, you can see that the polynomials in three solutions are the same, only the last number is different, indicating which root will be output by root() function. See: https://www.mathworks.com/help/symbolic/sym.root.html. To get a numeric answer for a specific value of x2, try

syms x1 x2
Eq2 = (1*(-0.0296505547))+(x1*(-1.4168313955))+(x2*(-0.4039704255))+(x1^2*(-0.0746402042))+((x2^2)*(-1.6203228042))+(x1^3*0.0000356471)+(x2^3*0.0400047846)+(x1*x2*0.7706539390)+((x1^2)*x2*0.0015188307)+(x1*(x2^2)*(-0.0173642989))==-log(1/0.4 - 1);
x2_val = 1;
double(subs(t_sol1, x2, x2_val))

[John D'Errico]

If you want the results in a symbolic floating point form, then use vpa instead of double. Of course, since the coefficients in the original equation were only provided with 10 significant digits, then computing an answer that is accurate to 40 digits is arguably a bit silly. :)

[Walter Roberson]

MATLAB is able to express the solution analytically: it is just abbreviating it.

Answer 2

[Ruta Dranginyte]

Thank You guys for the help :)

Answer 3

[Walter Roberson]

Change

t_sol1 = solve(Eq2, x1)

To

t_sol1 = solve(Eq2, x1, 'MaxDegree', 3)

To get the solutions.

They will be long and difficult to read, but they will be the solutions.

MATLAB knows that the exact solutions to degree 3 and 4 are very long, and automatically abbreviates them into the root() form.