Solving a nonlinear system of two equations with two unknowns using Newton's method Consider the function $g : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $$g(x,y)= \begin{pmatrix} x\\ x^2 + y^2-1 \end{pmatrix}$$
and the non-linear system of equations
$$g(x,y)= \begin{pmatrix} 0\\0 \end{pmatrix}.$$ 
Newton's method takes the form
$$\begin{pmatrix} x_{n+1}\\ y_{n+1} \end{pmatrix}=\begin{pmatrix} x_{n}\\ y_{n} \end{pmatrix}-Dg(x_{n},y_{n})^{-1}g(x_{n},y_{n}).$$
Given the starting point $(x_{0},y_{0})=(\frac{1}{2},\frac{1}{2})$, how do I compute the next iterate $(x_{1},y_{1})$ via Newton's method? 
I obtain $(x_{1},y_{1})=(0,3/2)$, but I am not sure if this is correct.
 A: 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:
$$\tag 2 F\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}f_1(x,y) \\  f_2(x,y)  \end{bmatrix} = \begin{bmatrix}x \\   x^2+y^2 -1  \end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$$
The derivative of this system is the $2x2$ Jacobian given by:
$$\tag 3 J(x, y) = \begin{bmatrix} \dfrac{\partial f_1}{\partial x} & \dfrac{\partial f_1}{\partial y} \\ \dfrac{\partial f_2}{\partial x} & \dfrac{\partial f_2}{\partial y}  \end{bmatrix} = \begin{bmatrix} 1 & 0  \\ 2x & 2y   \end{bmatrix}$$
The function $G$ is defined as:
$$\tag 4 G(x) = x - J(x)^{-1}F(x)$$
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:
$$\begin{bmatrix}y_1^{(k-1)}\\y_2^{(k-1)}\end{bmatrix}= -\left(J\left(x_1^{(k-1)},x_2^{(k-1)}\right)\right)^{-1}F\left(x_1^{(k-1)},x_2^{(k-1)}\right)$$
Using the starting vector:
$$x^{(0)} = \begin{bmatrix}x^{(0)}\\y^{(0)}\end{bmatrix} = \begin{bmatrix}0.5\\0.5\end{bmatrix}$$
The iterates are


*

*$(0.5, 0.5), (0., 1.5), (0., 1.0833333333333333), (0., 1.0032051282051282), (0., 1.0000051200131073), (0., 1.000000000013107), (0., 1.)$


By inspection, we can see that the zero should be 
$$(x, y) = (0, 1)$$
In other words, for this initial condition, we have converged to the correct solution. Of course, for a different starting vector you may get a different solution and perhaps no solution at all.
A: Ok If I am not totally mistaken 
$(x_{0},y_{0})=(\frac{1}{2},\frac{1}{2})$
$\begin{pmatrix} 
x_{n+1}\\
y_{n+1}
\end{pmatrix}=\begin{pmatrix} 
x_{n}\\
y_{n}
\end{pmatrix}-(D_{g}(x_{n},y_{n}))^{-1}g(x_{n},y_{n})$
=$\begin{pmatrix} 
x_{1}\\
y_{1}
\end{pmatrix}=\begin{pmatrix} 
x_{0}\\
y_{0}
\end{pmatrix}-(D_{g}(x_{0},y_{0}))^{-1}g(x_{0},y_{0})$
=$\begin{pmatrix} 
\frac{1}{2}\\
\frac{1}{2}
\end{pmatrix}-\begin{pmatrix} 
1 & 0\\
-1 & 1
\end{pmatrix}$
$\begin{pmatrix} 
\frac{1}{2}\\
-\frac{1}{2}
\end{pmatrix}$
=(0,3/2)
and from here how do we get $(x_{1},y_{1})=(0,1)$? :i...
