FIXED POINT ITERATION METHOD
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 x_{0} is called the fixed
point iterative scheme.
Algorithm  Fixed Point Iteration Scheme
Given an equation f(x) = 0
Convert f(x) = 0 into the form x = g(x)
Let the initial guess be x_{0 }
Do
x_{i+1}= g(x_{i})
while (none of the convergence criterion C1 or C2 is met) 
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 x^{4}x10 = 0
[
Graph]
Consider g1(x) = 10 / (x^{3}1) and the fixed point
iterative scheme x_{i+1}=10 / (x_{i}^{3} 1),
i = 0, 1, 2, . . .let the initial guess x_{0} be 2.0
i

0

1

2

3

4

5

6

7

8

x_{i}

2

1.429

5.214

0.071

10.004

9.978E3

10

9.99E3

10

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 x_{0} be 1.0, 2.0 and 4.0
1ffb
i

0

1

2

3

4

5

6

x_{i}

1.0

1.82116

1.85424

1.85553

1.85558

1.85558


x_{i}

2.0

1.861

1.8558

1.85559

1.85558

1.85558
 
x_{i}

4.0

1.93434

1.85866

1.8557

1.85559

1.85558

1.85558

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 x_{0} be 1.8,
i

0

1

2

3

4

5

6

. . .

98

x_{i}

1.8

1.9084

1.80825

1.90035

1.81529

1,89355

1.82129

. . .

1.8555

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
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 x^{4} + 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} + 4Î^{7} Then
find an iterative formula of the form x_{n+1} = g(x_{n}
), if we start with x_{0} = 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 x^{4} + x = Î
x(x^{3} + 1) = Î
x = Î/(1 + x^{3})
or x_{i} = Î/(1
+ x_{i}^{ 3})
i = 0, 1, 2, . . .
x_{0} = 0
x_{1} = Î
x_{2} = Î/ (1 + Î_{i}^{3})
= Î(1 + Î_{i}^{3}
)^{1}
= Î(1
 Î^{3} + Î^{6}
+ . . .)
= Î 
Î^{4}
+ Î^{7} + . . .
x_{3} = Î/( 1 + (Î

Î^{
4} + Î^{7})^{3})
= Î[1 + ( Î

Î^{4}
+ Î^{7})^{3}] = Î
 Î^{4} + 4Î^{
7}
Now taking x = Î

Î^{4}
+ 4Î^{7}
error = x^{ 4} + x
 Î
= (Î

Î^{
4} + 4Î^{7})^{4}
+ (Î  Î^{
4} + 4Î^{7})  Î
= 22Î^{10}
+ 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 x_{0} belongs to the interval J
.
Worked out problems
Exapmple 1 
Find a root of cos(x)  x * exp(x) = 0 
Solution 
Exapmple 2 
Find a root of x^{4}x10 = 0 
Solution 
Exapmple 3 
Find a root of xexp(x) = 0 
&nb
c35
sp;Solution 
Exapmple 4 
Find a root of exp(x) * (x^{2}5x+2)
+ 1= 0 
Solution 
Exapmple 5 
Find a root of xsin(x)(1/2)= 0 
Solution 
Exapmple 6 
Find a root of exp(x) = 3log(x) 
Solution 
Problems
to workout

Work out with the
Fixed Point Iteration method here
Note :Few
examples of how to enter equations are given below . . . (i)
exp[x]*(x^2+5x+2)+1
(ii) x^4x10
(iii) xsin[x](1/2)
(iv)
exp[(x+212+1)]*(x^2+5x+2)+1 (v)
(x+10) ^ (1/4) 
