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Home > Numerical methods calculators > False Position method example
2. False Position method (regula falsi 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 False Position method (regula falsi method)

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

Let `f(x) = 2x^3-2x-5`

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



`1^(st)` iteration :

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

`:.` Now, Root lies between `x_0 = 1` and `x_1 = 2`

`x_2 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_2 = 1 - (-5) * (2 - 1)/(7 - (-5))`

`x_2 = 1.41667`

`f(x_2)=f(1.41667)=2*1.41667^(3)-2*1.41667-5=-2.14699 < 0`


`2^(nd)` iteration :

Here `f(1.41667) = -2.14699 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.41667` and `x_1 = 2`

`x_3 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_3 = 1.41667 - (-2.14699) * (2 - 1.41667)/(7 - (-2.14699))`

`x_3 = 1.55359`

`f(x_3)=f(1.55359)=2*1.55359^(3)-2*1.55359-5=-0.60759 < 0`


`3^(rd)` iteration :

Here `f(1.55359) = -0.60759 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.55359` and `x_1 = 2`

`x_4 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_4 = 1.55359 - (-0.60759) * (2 - 1.55359)/(7 - (-0.60759))`

`x_4 = 1.58924`

`f(x_4)=f(1.58924)=2*1.58924^(3)-2*1.58924-5=-0.15063 < 0`


`4^(th)` iteration :

Here `f(1.58924) = -0.15063 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.58924` and `x_1 = 2`

`x_5 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_5 = 1.58924 - (-0.15063) * (2 - 1.58924)/(7 - (-0.15063))`

`x_5 = 1.59789`

`f(x_5)=f(1.59789)=2*1.59789^(3)-2*1.59789-5=-0.0361 < 0`


`5^(th)` iteration :

Here `f(1.59789) = -0.0361 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.59789` and `x_1 = 2`

`x_6 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_6 = 1.59789 - (-0.0361) * (2 - 1.59789)/(7 - (-0.0361))`

`x_6 = 1.59996`

`f(x_6)=f(1.59996)=2*1.59996^(3)-2*1.59996-5=-0.00858 < 0`


`6^(th)` iteration :

Here `f(1.59996) = -0.00858 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.59996` and `x_1 = 2`

`x_7 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_7 = 1.59996 - (-0.00858) * (2 - 1.59996)/(7 - (-0.00858))`

`x_7 = 1.60045`

`f(x_7)=f(1.60045)=2*1.60045^(3)-2*1.60045-5=-0.00203 < 0`


`7^(th)` iteration :

Here `f(1.60045) = -0.00203 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.60045` and `x_1 = 2`

`x_8 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_8 = 1.60045 - (-0.00203) * (2 - 1.60045)/(7 - (-0.00203))`

`x_8 = 1.60056`

`f(x_8)=f(1.60056)=2*1.60056^(3)-2*1.60056-5=-0.00048 < 0`


Approximate root of the equation `2x^3-2x-5=0` using False Position method is `1.60056`

`n``x_0``f(x_0)``x_1``f(x_1)``x_2``f(x_2)`Update
11-5271.41667-2.14699`x_0 = x_2`
21.41667-2.14699271.55359-0.60759`x_0 = x_2`
31.55359-0.60759271.58924-0.15063`x_0 = x_2`
41.58924-0.15063271.59789-0.0361`x_0 = x_2`
51.59789-0.0361271.59996-0.00858`x_0 = x_2`
61.59996-0.00858271.60045-0.00203`x_0 = x_2`
71.60045-0.00203271.60056-0.00048`x_0 = x_2`


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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