Newton method to solve nonlinear system How can I solve this problem:
${e^x}^y +{x}^2+y-1.2=0 \\
{x}^2 +{y}^2+x-0.55=0$
for various loaded inputs with start points $(x_0 , y_0) = 0.6, 0.5$
Could anybody help me. Thank you in advance 
 A: Using Solving a set of equations with Newton-Raphson, I am going to jump start you, but you need to show some work.
The regular Newton-Raphson method is initialized with a starting point $x_0$ and then you iterate $\tag 1x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$ 
In higher dimensions, there is an exact analog. We define:
$$F\left(\begin{bmatrix}x_1\\x_2\end{bmatrix}\right) = \begin{bmatrix}f_1(x_1,x_2) \\  f_2(x_1,x_2)  \end{bmatrix} = \begin{bmatrix}e^{x_1 x_2}+x_1^2+x_2-1.20 \\   x_1^2+x_2^2 + x_1 -0.55  \end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$$
The derivative of this system is the $2x2$ Jacobian given by:
$$J(x_1,x_2) = \begin{bmatrix} \dfrac{\partial f_1}{\partial x_1} & \dfrac{\partial f_1}{\partial x_2} \\ \dfrac{\partial f_2}{\partial x_1} & \dfrac{\partial f_2}{\partial x_2}  \end{bmatrix} = \begin{bmatrix} x_2~ e^{ x_1 x_2}+2 x_1 & x_1~ e^{ x_1 x_2}+1  \\ 2 x_1 + 1 & 2x_2   \end{bmatrix}$$
The function $G$ is defined as:
$$G(x) = x - J(x)^{-1}F(x)$$
and the functional Newton-Raphson method for nonlinear systems is given by the iteration procedure that evolves from selecting an initial $x^{(0)}$ and generating for $k \ge 1$ (compare this to $(1)$),
$$x^{(k)} = G(x^{(k-1)}) = x^{(k-1)} - J(x^{(k-1)})^{-1}F(x^{(k-1)}).$$
We can write this as:
$$\begin{bmatrix}x_1^{(k)}\\x_2^{(k)}\end{bmatrix} = \begin{bmatrix}x_1^{(k-1)}\\x_2^{(k-1)}\end{bmatrix} + \begin{bmatrix}y_1^{(k-1)}\\y_2^{(k-1)}\end{bmatrix}$$
where:
$$\begin{bmatrix}y_1^{(k-1)}\\y_2^{(k-1)}\end{bmatrix}= -\left(J\left(x_1^{(k-1)},x_2^{(k-1)}\right)\right)^{-1}F\left(x_1^{(k-1)},x_2^{(k-1)}\right)$$
Here is a starting vector:
$$x^{(0)} = \begin{bmatrix}x_1^{(0)}\\x_2^{(0)}\end{bmatrix} = \begin{bmatrix}0.6\\0.5\end{bmatrix}$$
You should end up with a solution that looks something like (you can do the iteration formula steps as it converges very quickly):
$$\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix} 0.393849452834869\\0.0321427389437406 \end{bmatrix}$$
Of course, for a different starting vector you are going to get a different solution and perhaps no solution at all.
