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Home > Numerical methods calculators > Fixed Point Iteration method example
4. Fixed Point Iteration method example ( Enter your problem )
( Enter your problem )
  1. Algorithm & Example-1 `f(x)=x^3-x-1`
  2. Example-2 `f(x)=2x^3-2x-5`
  3. Example-3 `x=sqrt(12)`
  4. Example-4 `x=root(3)(48)`
  5. Example-5 `f(x)=x^3+2x^2+x-1`
 		Other related methods
  1. Bisection method
  2. False Position method (regula falsi method)
  3. Newton Raphson method
  4. Fixed Point Iteration method
  5. Secant method
  6. Muller method
  7. Halley's method
  8. Steffensen's method
  9. Ridder's method
1. Algorithm & Example-1 `f(x)=x^3-x-1`
(Previous example)		3. Example-3 `x=sqrt(12)`
(Next example)

2. Example-2 `f(x)=2x^3-2x-5`


Find a root of an equation `f(x)=2x^3-2x-5` using Fixed Point Iteration method

Solution:
Method-1
Let `f(x) = 2x^3-2x-5`

Here `2x^3-2x-5=0`

`:.2x^3=2x+5`

`:.x^3=(2x+5)/2`

`:.x=root (3)((2x+5)/2)`

`:.phi(x)=root (3)((2x+5)/2)`

Here
`x`012
`f(x)`-5-57



Here `f(1) = -5 < 0` and `f(2) = 7 > 0`

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

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


`x_1 = phi(x_0) = phi(1.5) = 1.5874`

`x_2 = phi(x_1) = phi(1.5874) = 1.59888`

`x_3 = phi(x_2) = phi(1.59888) = 1.60037`

`x_4 = phi(x_3) = phi(1.60037) = 1.60057`


Approximate root of the equation `2x^3-2x-5` using Iteration method is `1.60057` (After 4 iterations)

`n``x_0``x_1=phi(x_0)`UpdateDifference
`|x_1-x_0|`
21.51.5874`x_0 = x_1`0.0874
31.58741.59888`x_0 = x_1`0.01148
41.598881.60037`x_0 = x_1`0.0015
51.600371.60057`x_0 = x_1`0.00019


This material is intended as a summary. Use your textbook for detail explanation.
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1. Algorithm & Example-1 `f(x)=x^3-x-1`
(Previous example)		3. Example-3 `x=sqrt(12)`
(Next example)


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