Difference Between Bisection Method and Regula Falsi Method

16 Dec, 2021
The bisection method is used for finding the roots of equations of non-linear equations of the form f(x) = 0 is based on the repeated application of the intermediate value property. Let f(x) is continuous function in the closed interval [x1,x2], if f(x1), f(x2) are of opposite signs , then there is at least one root α in the interval (x1,x2), such that f(α) = 0.
X2= (X0 + X1) / 2
Problem: Find a root of an equation f(x)=x3-x-1 
Solution:
Given equation  f(x)=x3-x-1
let x  = 0, 1, 2
In 1st iteration :
f(1)=-1<0 and f(2)=5>0
Root lies between these two points 1 and 2
x0=1+2/2 = 1.5
f(x0)=f(1.5)=0.875>0
In 2nd iteration :
f(1)=-1<0 and f(1.5)=0.875>0
Root lies between these two points 1 and 1.5
x1=1+1.5/2 =1.25
f(x1)=f(1.25)=-0.29688<0
In 3rd iteration :
f(1.25)=-0.29688<0 and f(1.5)=0.875>0
Root lies between these two points 1.25 and 1.5
x2=1.25+1.5/2 = 1.375
f(x2)=f(1.375)=0.22461>0
In 4th iteration :
f(1.25)=-0.29688<0 and f(1.375)=0.22461>0
Root lies between these two points 1.25 and 1.375
x3=1.25+1.375/2=1.3125
f(x3)=f(1.3125)=-0.05151<0
In 5th iteration :
f(1.3125)=-0.05151<0 and f(1.375)=0.22461>0
Root lies between these two points 1.3125 and 1.375
x4=1.3125+1.375/2=1.34375
f(x4)=f(1.34375)=0.08261>0
In 6th iteration :
f(1.3125)=-0.05151<0 and f(1.34375)=0.08261>0
Root lies between these two points 1.3125 and 1.34375
x5=1.3125+1.34375/2=1.32812
f(x5)=f(1.32812)=0.01458>0
In 7th iteration :
f(1.3125)=-0.05151<0 and f(1.32812)=0.01458>0
Root lies between these two points 1.3125 and 1.32812
x6=1.3125+1.32812/2 =1.32031
f(x6)=f(1.32031)=-0.01871<0
In 8th iteration :
f(1.32031)=-0.01871<0 and f(1.32812)=0.01458>0
Root lies between these two points 1.32031 and 1.32812
x7=1.32031+1.32812/2=1.32422
f(x7)=f(1.32422)=-0.00213<0
In 9th iteration :
 f(1.32422)=-0.00213<0 and f(1.32812)=0.01458>0
Root lies between these two points 1.32422 and 1.32812
x8=1.32422+1.32812/2=1.32617
f(x8)=f(1.32617)=0.00621>0
In 10th iteration :
f(1.32422)=-0.00213<0 and f(1.32617)=0.00621>0
Root lies between these two points 1.32422 and 1.32617
x9=1.32422+1.32617/2=1.3252
f(x9)=f(1.3252)=0.00204>0
In 11th iteration :
f(1.32422)=-0.00213<0 and f(1.3252)=0.00204>0
Root lies between these two points 1.32422 and 1.3252
x10=1.32422+1.3252/2=1.32471
f(x10)=f(1.32471)=-0.00005<0
The approximate root of the equation x3-x-1=0 using the Bisection method is 1.32471
Regula Falsi is one of the oldest methods to find the real root of an equation f(x) = 0 and closely resembles with Bisection method. It requires less computational effort as we need to evaluate only one function per iteration.
X3 = X1(fX2) - X2(fX1)/ f(X2) -f(X1)
Problem: Find a root of an equation f(x)=x3-x-1 
Solution:
Given equation, x3-x-1=0
let x  = 0, 1, 2
In 1st iteration :
 f(1)=-1<0 and f(2)=5>0
Root lies between these two points x0=1 and x1=2
x2=x0-f(x0)
= x1-x0
f(x1)-f(x0)
 x2=1-(-1)⋅
= 2-1
= 5-(-1)
 x2=1.16667
f(x2)=f(1.16667)=-0.5787<0
In 2nd iteration :
 f(1.16667)=-0.5787<0 and f(2)=5>0
Root lies between these two points x0=1.16667 and x1=2
x3=x0-f(x0)
x1-x0
f(x1)-f(x0)
x3=1.16667-(-0.5787)
2-1.16667
5-(-0.5787)
x3=1.25311
f(x3)=f(1.25311)=-0.28536<0
In 3rd iteration :
 f(1.25311)=-0.28536<0 and f(2)=5>0
Root lies between these two points x0=1.25311 and x1=2
x4=x0-f(x0)⋅
x1-x0
f(x1)-f(x0)
x4=1.25311-(-0.28536)⋅
2-1.25311
5-(-0.28536)
x4=1.29344
f(x4)=f(1.29344)=-0.12954<0
In 4th iteration :
f(1.29344)=-0.12954<0 and f(2)=5>0
Root lies between these two points x0=1.29344 and x1=2
x5=x0-f(x0)⋅
x1-x0
f(x1)-f(x0)
x5=1.29344-(-0.12954)⋅
2-1.29344
5-(-0.12954)
x5=1.31128
f(x5)=f(1.31128)=-0.05659<0
In 5th iteration :
 f(1.31128)=-0.05659<0 and f(2)=5>0
Root lies between these two points x0=1.31128 and x1=2
x6=x0-f(x0)⋅
x1-x0
f(x1)-f(x0)
x6=1.31128-(-0.05659)⋅
2-1.31128
5-(-0.05659)
x6=1.31899
f(x6)=f(1.31899)=-0.0243<0
In 6th iteration :
 f(1.31899)=-0.0243<0 and f(2)=5>0
Root lies between these two points x0=1.31899 and x1=2
x7=x0-f(x0)⋅
x1-x0
f(x1)-f(x0)
x7=1.31899-(-0.0243)⋅
2-1.31899
5-(-0.0243)
x7=1.32228
f(x7)=f(1.32228)=-0.01036<0
In 7th iteration :
 f(1.32228)=-0.01036<0 and f(2)=5>0
Root lies between these two points x0=1.32228 and x1=2
x8=x0-f(x0)⋅
x1-x0
f(x1)-f(x0)
x8=1.32228-(-0.01036)⋅
2-1.32228
5-(-0.01036)
x8=1.32368
The approximate root of the equation x3-x-1=0 using the Regula Falsi method is 1.32368
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