Fixed point iteration:

Let $\xi$ be a root of $f(x)=0$ and $g(x)$ be an associated iteration function. Say, $x_0$ is the given starting point. Then one can generate a sequence of successive approximations of $\xi$ as:

$$x_{1}=g(x_{0})$$

$$x_{2}=g(x_{1})$$

$$x_{n}=g(x_{n-1})$$

This sequence ${x_{0},x_{1}...x_{n}...}$ is said to converge to $\xi$ iff $\vert x_{n}-\xi\vert\rightarrow 0$ as $n\rightarrow\infty$.
Now the natural question that would arise is what are the conditions on $g(x)$ s.t. the sequence $\{x_{n}\}\rightarrow\xi$   as $\quad n \rightarrow \infty$
Here, we state few important comments on such a convergence:
(i)Suppose on an interval $I=[a,b], \forall \,\, x \,\epsilon \, I$ $g(x)$ is defined and $g(x)\, \epsilon \, I$. i.e. g(x) maps I into itself.
(ii) The iteration function $g(x)$ is continuous on I=[a,b].
(iii)The iteration function g(x) is differentiable on $I=[a,b]$ and $\exists\,\, 0<k< 1$ s.t. $\forall \,\,x \, \epsilon \, I \quad \vert g'(x)\vert\leq K$

Theorem :
Let g(x) be an iteration function satisfying (i), (ii) and (iii) then g(x) has exactly one fixed point $\xi$ in I and starting with any $x_{0}\, \epsilon \, I$, the sequence $\{x_n\}$ generated by fixed point iteration function converges to $\xi$. (you may refer to [ ] for proof).

(iv) If $e_{n}=\xi-x_{n},\quad e_{n+1}=\xi -x_{n+1}$ then $e_{n+1}\approx g'(\xi)e_{n}$. For rapid convergence it is desirable that $g'(\xi)=0$. Under this condition for the Newton Raphson method one can show that $${e_{n+1}\approx\frac{-1}{2}g''(\xi)e^{2}_{n}}$$ (i.e. quadratic Convergence).

Remark: =1in =1    One can generalize all the iterative methods for a system of nonlinear equations. For instance, if we have two non-linear equations $f(x_{1},\, x_{2})=0,\quad g(x_{1},\, x_{2})=0$ then given a suitable starting point $(x_{0},y_{0})$ the Newton-Raphson algorithm may be written as follows:
For i=1,2... until satisfied do

$$x_{i+1}=x_i-\frac{\left\vert%% \begin{array}{cc} f & g \\ ... ...artial g}{\partial x_2} \\ \end{array}%% \right\vert(x_i,y_i)}$$

$$y_{i+1}=y_i-\frac{\left\vert%% \begin{array}{cc} \frac{\part... ...artial g}{\partial x_2} \\ \end{array}%% \right\vert(x_i,y_i)}$$