Algorithm

Fixed Point Iteration method Steps (Rule)
Step-1:	First write the equation `x = phi(x)`
Step-2:	Find points `a` and `b` such that `a < b` and `f(a) * f(b) < 0`.
Step-3:	If `f(a)` is more closer to 0 then `f(b)` then `x_0=a` else `x_0=b`
Step-4:	`x_1 = phi(x_0)` `x_2 = phi(x_1)` `x_3 = phi(x_2)` ... Repeat until `|f(x_i)-f(x_(i-1))| ~~ 0`

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Example-1
1. Find a root of an equation `f(x)=x^3-x-1` using Fixed Point Iteration method

Solution:
Method-1
Let `f(x) = x^3-x-1`

Here `x^3-x-1=0`

`:.x^3=x+1`

`:.x=root (3)(x+1)`

`:.phi(x)=root (3)(x+1)`

Here

`x`	0	1	2
`f(x)`	-1	-1	5

Here `f(1) = -1 < 0` and `f(2) = 5 > 0`

`:.` Root lies between `1` and `2`

`x_0 = (1 + 2)/2 = 1.5`


`x_1 = phi(x_0) = phi(1.5) = 1.35721`

`x_2 = phi(x_1) = phi(1.35721) = 1.33086`

`x_3 = phi(x_2) = phi(1.33086) = 1.32588`

`x_4 = phi(x_3) = phi(1.32588) = 1.32494`

`x_5 = phi(x_4) = phi(1.32494) = 1.32476`


Approximate root of the equation `x^3-x-1` using Iteration method is `1.32476` (After 5 iterations)

`n`	`x_0`	`x_1=phi(x_0)`	Update	Difference `|x_1-x_0|`
2	1.5	1.35721	`x_0 = x_1`	0.14279
3	1.35721	1.33086	`x_0 = x_1`	0.02635
4	1.33086	1.32588	`x_0 = x_1`	0.00498
5	1.32588	1.32494	`x_0 = x_1`	0.00094
6	1.32494	1.32476	`x_0 = x_1`	0.00018

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