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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:

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

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




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

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.

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$.

(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 (i.e. quadratic convergence).

Remark 1:     One can generalize all the iterative methods for a system of nonlinear equations. For instance, if we have two non-linear equations 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


Exercises: Solve the following systems of equations by Newton Raphson Method.


Use the initial approximation


Use the initial approximation



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