I have post once here about this question but i had some problems and i stopped with solving it. Now i started again and i think i`ve made it better. Here is where i need help.

This is the system of nonlinear equations.

f(x) = 0 where:

$$ \left\{ \begin{array}{c} x_1^5+x_2^3+x_3^4+1 \\ x_1^2*x_2* x_3 \\ x_3^4-1 \end{array} \right. $$

The right side of the equation is 0.

A) Find manually all the zeroes of the system.

B) Calculate the Jacobian J(X). (Notice that J(x) is singular for x3 = 0)

C) Check this two starting solutions:

1) X0 = {-0.01, -0.01, -0.01}

2) X0 = {-0.1, -0.1, -0.1}

D) Calculate the determinants |J(X0)| and |J^-1(X0)| for the two starting solutions. Notice that Jacobians are almost singular, altough the starting solutions are not so far from the real solutions.


This is the task that i need to solve. I started from here.

From this system we can see that $$ X_3 = \pm 1 $$ So going from this we can make 4 solutions.

$$I:x_3=1, x_2=0, x_1=-1$$ $$II:x_3=1, x_2=-1, x_1=0$$ $$III:x_3=-1, x_2=0, x_1=-1$$ $$IV:x_3=-1, x_2=-1, x_1=0$$

After this i started with the Jacobian. I have made this for the Jacobian matrix.

$$ \left[ \begin{array}{ccc} 5x_1^4 & 3x_2^2x_3^4 &4x_3^3x_2^3\\ 2x_1 x_2 x_3 & 1*x_1^2 x_3 & 1*x_1^2 x_2\\ 0 & 0 & 4x_3^3 \end{array} \right] $$

After the jacobian i calculate the determinants for all 4 solutions and I have:

I: the determinant is: 20.

II: the determinant is: 0.

III: the determinant is: 20.

IV: the determinant is: 0.

I stucked on this: When i need to see the two starting solutions C). And when i need to calculate their determinant. The first number that i calculate is 0.00000005 and i said okay its enough i am something wrong.

  • $\begingroup$ i think your Solutions are not complete or wrong $\endgroup$ – Dr. Sonnhard Graubner May 22 '17 at 14:16
  • $\begingroup$ Can you help me to find my mistake? $\endgroup$ – Bambus May 22 '17 at 14:18
  • $\begingroup$ Your solutions $I-IV$ are wrong. You can check by direct substitution. $\endgroup$ – mattos May 22 '17 at 14:48
  • $\begingroup$ Okay, My I solution is x1 = -1 , x2 = 0, x3 = 1. You can check like this: X_3^4 -1 = 0 so by default x_3 must be +- 1. In the first solution i go with 1. Going in the middle equation you can notice that some of X1, X2 and X3 must be 0. So i decided to go with x2. And if X3=1 and x2=0 you can notice from first equation that x1 = -1. So the first solution is correct. $\endgroup$ – Bambus May 22 '17 at 16:11
  • $\begingroup$ If you are assuming that you want all real solutions, then please state that explicitly in your question. $\endgroup$ – Somos May 22 '17 at 19:44

HINT: eliminating $$x_2,x_3$$ from your System we get for $x_1$ $$2x_1^2+x_1^7=0$$ you can start with: one variable of your System must be Zero. If we have $x_2=0$ then we get$$x_1^5+2=0$$ If we have $$x_1=0$$ then $$x_2^3+2=0$$ from my equation you will get $$x_1=0$$ or $$x_1^5+2=0$$ the whole number of solutions for $(x_1, x_2, x_3)$ are $$ (0, -\sqrt[3]2, -1), \quad (0, -\sqrt[3]2, 1), \quad (-\sqrt[5]2, 0, -1), \quad (-\sqrt[5]2, 0, 1)$$

  • $\begingroup$ I didnt get it why you go with x1 when is simple with x3? And how you get that solution for x1? $\endgroup$ – Bambus May 22 '17 at 14:36
  • $\begingroup$ Yes and in every solution i have variable that is zero. $\endgroup$ – Bambus May 22 '17 at 14:46
  • $\begingroup$ no $x_3$ can not be Zero, thus you must consider $$x_1,x_2$$ $\endgroup$ – Dr. Sonnhard Graubner May 22 '17 at 14:48
  • $\begingroup$ Yes i know that. In my solutions X3 is 1 and -1. $\endgroup$ – Bambus May 22 '17 at 14:49
  • $\begingroup$ I still cant get it what you want to say. You think that i have mistake in my 4 solutions? And that i cant go first with x3? I cant get it from where you have 2X1^2 + x1^7 =0 and what is the point of it. $\endgroup$ – Bambus May 22 '17 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.