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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.

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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.

Bambus
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  • i think your Solutions are not complete or wrong – Dr. Sonnhard Graubner May 22 '17 at 14:16
  • Can you help me to find my mistake? – Bambus May 22 '17 at 14:18
  • Your solutions $I-IV$ are wrong. You can check by direct substitution. – Matthew Cassell May 22 '17 at 14:48
  • 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.

     – Bambus May 22 '17 at 16:11
  • If you are assuming that you want all real solutions, then please state that explicitly in your question. – Somos May 22 '17 at 19:44
  • Okay. So now the real solutions are okay. And what about jacobian? – Bambus May 23 '17 at 14:13
  • $(x_1, x_2, x_3) = (-1, 0, 1)$ is not a zero of the system. Plugging those into the very first equation results in $1$, not $0$. –  May 26 '17 at 13:19
  • Actually, none of the four solutions you currently have listed are correct. It is true that $x_3 = \pm 1$ and it is true that one of $x_1$ and $x_2$ must be zero. Beyond that the solutions are wrong. –  May 26 '17 at 13:23

1 Answers1

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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 $$ (0, -\sqrt[3]2, -1), \quad (0, -\sqrt[3]2, 1), \quad (-\sqrt[5]2, 0, -1), \quad (-\sqrt[5]2, 0, 1)$$

Dr. Sonnhard Graubner
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