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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

1. Bisection method
(Previous method)
2. Example-2 `f(x)=2x^3-2x-5`
(Next example)

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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1. Bisection method
(Previous method)
2. Example-2 `f(x)=2x^3-2x-5`
(Next example)





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