# fsolve result is not desirable even giving a close starting point

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I have three nonlinear equations with three unknowns and I used fsolve function.

I know that x=[0.3173;0.3173;3.6590] is a good enough solution for the three equations. However, using fsolve, I could not get to this solution even for a very close starting point as shown below. Instead the result is (upon plugging these results into the 3 equations, the values are not close to 0):

x =

0.999999999999936

1.000000000000067

1.000000000000036

How can this be solved? Thanks!

CODE:

x0=[0.31; 0.31; 3.5];

options=optimset('Display','iter');

[x,fval]=fsolve(@myfun1,x0,options)

function F=myfun(x)

F=[13*cos(x(3)) - 5*x(1)*x(3)^2 - 5*x(2)*x(3)^2 + 13*x(1)*x(3)*sin(x(3)) + 13*x(2)*x(3)*sin(x(3)) + 10*x(1)*x(2)*x(3)^2 - 13*x(1)*x(2)*x(3)^2*cos(x(3)) + 40;

5*x(3) - 13*sin(x(3)) + 30*x(1)*x(3) + 13*x(1)*x(3)*cos(x(3)) + 13*x(2)*x(3)*cos(x(3)) - 5*x(1)*x(2)*x(3)^3 + 13*x(1)*x(2)*x(3)^2*sin(x(3));

390*x(3)^2*sin(x(3)) + 300*x(1)*x(3)^3 - 25*x(1)*x(3)^5 + 25*x(2)*x(3)^5 - 150*x(3)^3 - 260*x(1)*x(3)^3*cos(x(3)) - 130*x(2)*x(3)^3*cos(x(3)) + 130*x(1)*x(3)^4*sin(x(3)) - 130*x(2)*x(3)^4*sin(x(3)) - 169*x(1)*x(3)^3*cos(x(3))^2 + 169*x(2)*x(3)^3*cos(x(3))^2 - 169*x(1)*x(3)^3*sin(x(3))^2 + 169*x(2)*x(3)^3*sin(x(3))^2

];

##### 2 Comments

### Accepted Answer

Matt J
on 31 Jan 2014

Edited: Matt J
on 31 Jan 2014

##### 2 Comments

Matt J
on 31 Jan 2014

### More Answers (1)

Alex Sha
on 3 Sep 2023

For Qingbin's equations, although it is a problem that has passed a long time, it is worth and interesting to have a try, there are multi-solutions:

The first one:

x1: -12.7613758329431

x2: -12.7613758329431

x3: -0.755039591199301

The second one:

x1: 0.101047394240414

x2: -3.71598808815167

x3: -1.63192533266968

The third one:

x1: -0.282297628581163

x2: 0.369875309676247

x3: 3.09470113469319

The fourth one:

x1: 0.317262018077202

x2: 0.317262018077202

x3: 3.65895757009226

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