Some background theory: By Taylor's theorem we know that an approximation of $f(x)$ with two terms around an approximation $x_k\in \mathbb{R}^n$ is given by $f(x)\approx f(x_k)+J(x_k)(x-x_k),$ where $J$ is the jacobian. So a linear model of our problem $f(x)=0$ becomes
$$f(x_k)+J(x_k)(x-x_k)=0. \tag{1}$$
The solution to the system (1), is taken as our new approximation $x_{k+1}.$ If we let $s_k=x_{k+1}-x_k,$ one iteration with Newtons method can be written as
$$\left\{ \begin{array}{rcr} J(x_k)s_k & = & -f(x_k) \\ x_{k+1} & = & x_k+s_k \\ \end{array} \right. \tag{2}$$
Problem: Formulate Newtons method for the system
$$\left\{ \begin{array}{rcr} x_1^2+x_2^2 & = & 1 \\ x_1^2-x_2 & = & 0 \\ \end{array} \right.$$
First of all, I don't see how we can have the jacobian $J(x_k)$ of one variable. We are still working with a single variable function $f$ as I understand but somehow we can start using Jacobians. I feel I need a better explanation of the background theory.
Second, the anser in the book is
$$\begin{pmatrix} x_1 \\ x_2\end{pmatrix}^{(k+1)}=\begin{pmatrix} x_1 \\ x_2\end{pmatrix}^{(k)}+\begin{pmatrix} d_1 \\ d_2\end{pmatrix}^{(k)}.$$
I see that this is related to the second row in (2) but what is the last matrix in terms of $d_1, \ d_2?$
You're supposed to reformulate Newton's method assuming everything is now a vector -- you have a vector of coordinates, $\vec{x}$, a vector of functions, $\vec{f}$, and a matrix of derivatives, $J$.
You have a vector of functions receiving a vector of $x$s. Written as vectors, $$ \vec{f}(\vec{x}) = \dots \text{.} $$ Written with explicit components, $$ \begin{pmatrix} f_1 \\ f_2 \end{pmatrix}\left( \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \right) = \begin{pmatrix} x_1^2 + x_2^2 \\ x_1 - x_2 \end{pmatrix} \text{.} $$
Then the Jacobian is $$ J(\vec{x}) = \left. \begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{pmatrix} \right|_{\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \vec{x}} $$ where, for example, $\frac{\partial f_1}{\partial x_1} = 2x_1$.