How to solve function

Question

Ruta Dranginyte 2020-3-18

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/511595-how-to-solve-function

评论： Walter Roberson 2020-3-21

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?

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Answer 1

Ameer Hamza 2020-3-18

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/511595-how-to-solve-function#answer_420779

编辑：Ameer Hamza 2020-3-18

在 MATLAB Online 中打开

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))

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John D'Errico 2020-3-18

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. :)