Fixed Point Iteration Method Can anyone explain or prove the Fixed Point Iteration method? 
I know the conditions of fixed point existence.
A fixed point is a point of a function ${f}$ on a continuous interval ${(a,b)}$ which satisfies this condition : ${x=f(x)}$, But i don't know how to conclude this method :${{x}_{n+1}=f(x_n)}$
 A: The Fixed Point Iteration Method takes an equation
$$f(x)=0$$
and converts it into the form
$$x=g(x)$$
You then make an initial guess, say $x_0$, and recursively compute $$x_{n+1}= g(x_n)$$
Continue this process until one of the following criteria is met:


*

*A specific number of iterations are done (which you define yourself)

*The error $E= | x_{n+1}-g(x_n)| \leq \epsilon$, where $\epsilon$ is the degree of accuracy which you prefer.

A: If $f$ is a continuous function and a sequence $x_n$ defined by the recurrence $x_{n+1} = f(x_n)$ happens to have a limit $X$, then 
$$ X = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} f(x_n) = f(X)$$
so $X$ is a fixed point.  In general there is no guarantee that the limit will exist, but in some cases it will.  In particular, suppose there is a solution $p$
and $f$ is continuously differentiable with $|f'(p)| < 1$,  and take $\epsilon > 0$ such that $|f'(x)| < 1$ for $p-\epsilon < x < p + \epsilon$.  Then by the 
Mean Value Theorem $x_{n+1} - p = f(x_n) - f(p) = f'(\xi) (x_n - p)$ for some
$\xi$ between $x_n$ and $p$.   Then if $p -\epsilon < x_0 < p + \epsilon$ the
sequence $x_n$ will converge to $p$, with $|x_n - p| < |x_0 - p| r^n$ for some 
$r < 1$.
