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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
4. Example-4 `x=root(3)(48)`
(Previous example)		5. Secant method
(Next method)

5. Example-5 `f(x)=x^3+2x^2+x-1`


Find a root of an equation `f(x)=x^3+2x^2+x-1` using Fixed Point Iteration method

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

Here `x^3+2x^2+x-1=0`

`:.x^3+2x^2+x=1`

`:.x(x^2+2x+1)=1`

`:.x=(1)/(x^2+2x+1)`

`:.phi(x)=(1)/(x^2+2x+1)`

Here
`x`01
`f(x)`-13



Here `f(0) = -1 < 0` and `f(1) = 3 > 0`

`:.` Root lies between `0` and `1`

`x_0 = (0 + 1)/2 = 0.5`


`x_1 = phi(x_0) = phi(0.5) = 0.44444`

`x_2 = phi(x_1) = phi(0.44444) = 0.47929`

`x_3 = phi(x_2) = phi(0.47929) = 0.45698`

`x_4 = phi(x_3) = phi(0.45698) = 0.47108`

`x_5 = phi(x_4) = phi(0.47108) = 0.46209`

`x_6 = phi(x_5) = phi(0.46209) = 0.46779`

`x_7 = phi(x_6) = phi(0.46779) = 0.46416`

`x_8 = phi(x_7) = phi(0.46416) = 0.46647`

`x_9 = phi(x_8) = phi(0.46647) = 0.465`

`x_10 = phi(x_9) = phi(0.465) = 0.46593`

`x_11 = phi(x_10) = phi(0.46593) = 0.46534`

`x_12 = phi(x_11) = phi(0.46534) = 0.46572`


Approximate root of the equation `x^3+2x^2+x-1` using Iteration method is `0.46572` (After 12 iterations)

`n``x_0``x_1=phi(x_0)`UpdateDifference
`|x_1-x_0|`
20.50.44444`x_0 = x_1`0.05556
30.444440.47929`x_0 = x_1`0.03485
40.479290.45698`x_0 = x_1`0.02231
50.456980.47108`x_0 = x_1`0.0141
60.471080.46209`x_0 = x_1`0.00899
70.462090.46779`x_0 = x_1`0.0057
80.467790.46416`x_0 = x_1`0.00363
90.464160.46647`x_0 = x_1`0.0023
100.466470.465`x_0 = x_1`0.00146
110.4650.46593`x_0 = x_1`0.00093
120.465930.46534`x_0 = x_1`0.00059
130.465340.46572`x_0 = x_1`0.00038


This material is intended as a summary. Use your textbook for detail explanation.
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4. Example-4 `x=root(3)(48)`
(Previous example)		5. Secant method
(Next method)


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