I am trying to find solution for this system of nonlinear equations.
$f(x) = 0$
- a) Calculate all the zero`s in the system (mannualy)
- b) Find the Jacobioan matrix, J(x). (Notice that J(x) is singular for x(3) = 0)
- c) Consider independent two starting solutions:
1) x(0) = {-0.01, -0.01, -0.01}T
2) x(0) = {-0.1, -0.1, -0.1}T
I stucked here: $$\begin{cases} x_1^5 \pm x_2^3 =0 \\[2ex] x_1^2 * x_2 * (\pm 1) = 0 \\[2ex] x_3 = \pm 1 \end{cases} $$
Alright! What I am going to do is give you an example of a problem similar to the one you are doing. That way you can understand what exactly you are trying to do, but without actually doing your homework for you. Without further ado, let $f: \mathbb R^3 \to \mathbb R^3$ such that $$f(\vec x):= \begin{bmatrix} 2x_1^2-x_2x_3-8\\ x^2_1x_2+2x_1x_2x_3+x_2x_3^2\\ x_3^2-4 \end{bmatrix}$$
a) find $\vec x$ such that $f(\vec x)=\vec 0$.
b) find the Jacobian matrix of $f$
a) we have a set of solutions where $$\begin{bmatrix} 2x_1^2-x_2x_3-8\\ x^2_1x_2+2x_1x_2x_3+x_2x_3^2\\ x_3^2-4 \end{bmatrix}= \begin{bmatrix}0\\0\\0\end{bmatrix}$$ Solving independently we have that $$x_3=\pm 2,$$ $$x_2(x_1+x_3)^2=0,$$ and $$2x_1^2-x_2x_3=8.$$
If $x_2 =0$. Then from the third equation we have $x_1^2=8/2=4$. So $x_1=\pm 2$.
If $(x_1+x_3)=0$. Then $x_3=-x_1=\pm 2$, and from the third equation we have $$2x_1^2+x_1x_2=8$$ $\Rightarrow$ $$x_1(2x_1+x_2)=8.$$ As $x_1=\pm 2$ this implies that $$x_2=\pm 4-\pm 4=0$$ So for the zero's we have $$\begin{bmatrix}2\\0\\2\end{bmatrix},\:\begin{bmatrix}2\\0\\-2\end{bmatrix},\:\begin{bmatrix}-2\\0\\2\end{bmatrix},\:\mathrm{and}\;\:\begin{bmatrix}-2\\0\\-2\end{bmatrix}$$
b) The Jacobian matrix is the matrix of partial derivatives. Hence, $$J_f=\begin{bmatrix} \frac{\partial f_1}{\partial x_1}&\frac{\partial f_1}{\partial x_2}&\frac{\partial f_1}{\partial x_3}\\ \frac{\partial f_2}{\partial x_1}&\frac{\partial f_2}{\partial x_2}&\frac{\partial f_2}{\partial x_3}\\ \frac{\partial f_3}{\partial x_1}&\frac{\partial f_3}{\partial x_2}&\frac{\partial f_3}{\partial x_3} \end{bmatrix}$$
Since $f_1=2x_1^2-x_2x_3-8$, $f_2=x^2_1x_2+2x_1x_2x_3+x_2x_3^2$, and $f_3=x_3^2-4$, this implies that $$J_f=\begin{bmatrix} 4x_1&-x_3&-x_2\\ x_2x_3^2&(x_1+x_3)^2&x_1^2x_2\\ 0&0&2x_3 \end{bmatrix}$$
As for c) all you've asked to do is consider those vector points. There is literally no question. This is like asking me to "Observe your keyboard for a minute". Okay, I've observed my keyboard for a minute now what?