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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`

1. Algorithm & Example-1 `f(x)=x^3-x-1`





Algorithm
False Position method (regula falsi method) Steps (Rule)
Step-1: Find points `x_0` and `x_1` such that `x_0 < x_1` and `f(x_0) * f(x_1) < 0`.
Step-2: Take the interval `[x_0, x_1]` and
find next value using
Formula-1 : `x_2=x_0-f(x_0)*(x_1-x_0)/(f(x_1)-f(x_0))`
or Formula-2 : `x_2=(x_0*f(x_1)-x_1*f(x_0))/(f(x_1)-f(x_0))`
or Formula-3 : `x_2=x_1-f(x_1)*(x_1-x_0)/(f(x_1)-f(x_0))`
(Using any of the formula, you will get same x2 value)
Step-3: If `f(x_2) = 0` then `x_2` is an exact root,
else if `f(x_0) * f(x_2) < 0` then `x_1 = x_2`,
else if `f(x_2) * f(x_1) < 0` then `x_0 = x_2`.
Step-4: Repeat steps 2 & 3 until `f(x_i) = 0` or `|f(x_i)| <= "Accuracy"`

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

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

Let `f(x) = x^3-x-1`

Here
`x`012
`f(x)`-1-15



`1^(st)` iteration :

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

`x_2 = 1.16667`

`f(x_2) = f(1.16667) = -0.5787 < 0`


`2^(nd)` iteration :

Here `f(1.16667) = -0.5787 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.16667` and `x_1 = 2`

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

`x_3 = 1.16667 - (-0.5787) * (2 - 1.16667)/(5 - (-0.5787))`

`x_3 = 1.25311`

`f(x_3) = f(1.25311) = -0.28536 < 0`


`3^(rd)` iteration :

Here `f(1.25311) = -0.28536 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.25311` and `x_1 = 2`

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

`x_4 = 1.25311 - (-0.28536) * (2 - 1.25311)/(5 - (-0.28536))`

`x_4 = 1.29344`

`f(x_4) = f(1.29344) = -0.12954 < 0`


`4^(th)` iteration :

Here `f(1.29344) = -0.12954 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.29344` and `x_1 = 2`

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

`x_5 = 1.29344 - (-0.12954) * (2 - 1.29344)/(5 - (-0.12954))`

`x_5 = 1.31128`

`f(x_5) = f(1.31128) = -0.05659 < 0`


`5^(th)` iteration :

Here `f(1.31128) = -0.05659 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.31128` and `x_1 = 2`

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

`x_6 = 1.31128 - (-0.05659) * (2 - 1.31128)/(5 - (-0.05659))`

`x_6 = 1.31899`

`f(x_6) = f(1.31899) = -0.0243 < 0`


`6^(th)` iteration :

Here `f(1.31899) = -0.0243 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.31899` and `x_1 = 2`

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

`x_7 = 1.31899 - (-0.0243) * (2 - 1.31899)/(5 - (-0.0243))`

`x_7 = 1.32228`

`f(x_7) = f(1.32228) = -0.01036 < 0`


`7^(th)` iteration :

Here `f(1.32228) = -0.01036 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.32228` and `x_1 = 2`

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

`x_8 = 1.32228 - (-0.01036) * (2 - 1.32228)/(5 - (-0.01036))`

`x_8 = 1.32368`

`f(x_8) = f(1.32368) = -0.0044 < 0`


`8^(th)` iteration :

Here `f(1.32368) = -0.0044 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.32368` and `x_1 = 2`

`x_9 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_9 = 1.32368 - (-0.0044) * (2 - 1.32368)/(5 - (-0.0044))`

`x_9 = 1.32428`

`f(x_9) = f(1.32428) = -0.00187 < 0`


`9^(th)` iteration :

Here `f(1.32428) = -0.00187 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.32428` and `x_1 = 2`

`x_10 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_10 = 1.32428 - (-0.00187) * (2 - 1.32428)/(5 - (-0.00187))`

`x_10 = 1.32453`

`f(x_10) = f(1.32453) = -0.00079 < 0`


`10^(th)` iteration :

Here `f(1.32453) = -0.00079 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.32453` and `x_1 = 2`

`x_11 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_11 = 1.32453 - (-0.00079) * (2 - 1.32453)/(5 - (-0.00079))`

`x_11 = 1.32464`

`f(x_11) = f(1.32464) = -0.00034 < 0`


Approximate root of the equation `x^3-x-1=0` using False Position method is `1.32464`

`n``x_0``f(x_0)``x_1``f(x_1)``x_2``f(x_2)`Update
11-1251.16667-0.5787`x_0 = x_2`
21.16667-0.5787251.25311-0.28536`x_0 = x_2`
31.25311-0.28536251.29344-0.12954`x_0 = x_2`
41.29344-0.12954251.31128-0.05659`x_0 = x_2`
51.31128-0.05659251.31899-0.0243`x_0 = x_2`
61.31899-0.0243251.32228-0.01036`x_0 = x_2`
71.32228-0.01036251.32368-0.0044`x_0 = x_2`
81.32368-0.0044251.32428-0.00187`x_0 = x_2`
91.32428-0.00187251.32453-0.00079`x_0 = x_2`
101.32453-0.00079251.32464-0.00034`x_0 = x_2`





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