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2. False Position method (regula falsi method) example ( 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

3. Example-3 `x=sqrt(12)`
(Previous example)
5. Example-5 `f(x)=x^3+2x^2+x-1`
(Next example)

4. Example-4 `x=root(3)(48)`





Find `root(3)(48)` using False Position method (regula falsi method)

Solution:
Let `x=48^(1/3)`

`:.x^3=48`

`:.x^3-48=0`

i.e. `f(x)=x^3-48`

Here
`x`01234
`f(x)`-48-47-40-2116



`1^(st)` iteration :

Here `f(3) = -21 < 0` and `f(4) = 16 > 0`

`:.` Now, Root lies between `x_0 = 3` and `x_1 = 4`

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

`x_2=3 - (-21) * (4 - 3)/(16 - (-21))`

`x_2=3.5676`

`f(x_2)=f(3.5676)=3.5676^3-48=-2.5936 < 0`


`2^(nd)` iteration :

Here `f(3.5676) = -2.5936 < 0` and `f(4) = 16 > 0`

`:.` Now, Root lies between `x_0 = 3.5676` and `x_1 = 4`

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

`x_3=3.5676 - (-2.5936) * (4 - 3.5676)/(16 - (-2.5936))`

`x_3=3.6279`

`f(x_3)=f(3.6279)=3.6279^3-48=-0.2513 < 0`


`3^(rd)` iteration :

Here `f(3.6279) = -0.2513 < 0` and `f(4) = 16 > 0`

`:.` Now, Root lies between `x_0 = 3.6279` and `x_1 = 4`

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

`x_4=3.6279 - (-0.2513) * (4 - 3.6279)/(16 - (-0.2513))`

`x_4=3.6336`

`f(x_4)=f(3.6336)=3.6336^3-48=-0.0237 < 0`


`4^(th)` iteration :

Here `f(3.6336) = -0.0237 < 0` and `f(4) = 16 > 0`

`:.` Now, Root lies between `x_0 = 3.6336` and `x_1 = 4`

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

`x_5=3.6336 - (-0.0237) * (4 - 3.6336)/(16 - (-0.0237))`

`x_5=3.6342`

`f(x_5)=f(3.6342)=3.6342^3-48=-0.0022 < 0`


`5^(th)` iteration :

Here `f(3.6342) = -0.0022 < 0` and `f(4) = 16 > 0`

`:.` Now, Root lies between `x_0 = 3.6342` and `x_1 = 4`

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

`x_6=3.6342 - (-0.0022) * (4 - 3.6342)/(16 - (-0.0022))`

`x_6=3.6342`

`f(x_6)=f(3.6342)=3.6342^3-48=-0.0002 < 0`


Approximate root of the equation `x^3-48=0` using False Position method is `3.6342` (After 5 iterations)

`n``x_0``f(x_0)``x_1``f(x_1)``x_2``f(x_2)`Update
13-214163.5676-2.5936`x_0 = x_2`
23.5676-2.59364163.6279-0.2513`x_0 = x_2`
33.6279-0.25134163.6336-0.0237`x_0 = x_2`
43.6336-0.02374163.6342-0.0022`x_0 = x_2`
53.6342-0.00224163.6342-0.0002`x_0 = x_2`





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3. Example-3 `x=sqrt(12)`
(Previous example)
5. Example-5 `f(x)=x^3+2x^2+x-1`
(Next example)





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