Fixed point :  A point, say, s is called a fixed point if it satisfies the equation x = g(x). 

Fixed point Iteration : The transcendental equation f(x) = 0 can be converted algebraically into the form x = g(x) and then using the iterative scheme with the recursive relation 

x_i+1= g(x_i),           i = 0, 1, 2, . . ., 

with some initial guess x₀  is called the fixed point iterative scheme.

C1. Fixing apriori the total number of iterations N . 

C2. By testing the condition  | x_i+1 - g(x_i) | (where i is the iteration number) less than some tolerance limit, say epsilon, fixed apriori.

 

 
 
 

Numerical Example :

Find a root of  x⁴-x-10 = 0                                [ Graph] 
Consider g1(x) = 10 / (x³-1) and the fixed point iterative scheme x_i+1=10 / (x_i³ -1),      i = 0, 1, 2, . . .let the initial guess x₀ be 2.0

So the iterative process with g1 gone into an infinite loop without converging. 

Consider another function g2(x) = (x + 10)^1/4   and the fixed point iterative scheme 
x_i+1= (x_i + 10)^1/4,     i = 0, 1, 2, . . .

let the initial guess x₀ be 1.0, 2.0 and 4.0

That is for g2  the iterative process is converging to 1.85558 with any initial guess. 

Consider g3(x) =(x+10)^1/2/x  and the fixed point iterative scheme 

x_i+1=( x_i+10)^1/2 /x_i,     i = 0, 1, 2, . . .

let the initial guess x₀ be 1.8, 

That is for g3 with any initial guess the iterative process is converging but very slowly to 

Geometric interpretation of convergence with g1, g2 and g3 

Fig g1	Fig g2	Fig g3

The graphs Figures Fig g1, Fig g2 and Fig g3 demonstrates the Fixed point Iterative Scheme with g1, g2 and g3 respectively for some initial approximations. It's clear from the 
 

Fig g1, the iterative process does not converge for any initial approximation.

Fig g2, the iterative process converges very quickly to the root which is the intersection point of y = x and y = g2(x) as shown in the figure.

Fig g3, the iterative process converges but very slowly.

 

Example 2 :The equation  x⁴ +  x = Î, where Î is a small number , has a root which is close to Î. Computation of this root is done by the expression x = Î - Î ⁴ + 4Î⁷  Then find an iterative formula of the form x_n+1 = g(x_n ), if we start with x₀ = 0 for the computation then show that we get the expression given above as a solution. Also find the error in the approximation in the interval [0, 0.2] .

Proof

Given  x⁴ +  x = Î

x(x³ +  1) = Î

x = Î/(1 +  x³)    or      x_i = Î/(1 + x_i ³)         i = 0, 1, 2, . . .

x₀ = 0

x₁ = Î

x₂ = Î/ (1 + Î_i³) = Î(1 + Î_i³ )^-1
     = Î(1 - γ + Î⁶ + . . .)
     = Î - Î⁴ + Î⁷ + . . .

x₃ = Î/( 1 + (Î - Î ⁴ + Î⁷)³) = Î[1 + ( Î - Î⁴ + Î⁷)^-3] = Î - Î⁴ + 4Î ⁷

Now taking x = Î - Î⁴ + 4Î⁷

error = x ⁴ + x - Î
        = (Î - Î ⁴ + 4Î⁷)⁴ + (Î - Î ⁴ + 4Î⁷) - Î
        = 22ι⁰ + higher order power of Î

Condition for Convergence :

If g(x) and g'(x) are continuous on an interval J about their root s of the equation x = g(x), and if |g'(x)|<1 for all x in the interval J then the fixed point iterative process x_i+1=g( x_i), i = 0, 1, 2, . . ., will converge to the root x = s for any  initial approximation x₀ belongs to the interval J .