The Bisection Method and Regula Falsi Method are two fundamental numerical techniques used to find the roots of a function. While both methods work by narrowing down an interval where the root lies, the Bisection Method divides the interval equally, ensuring steady but slow convergence, whereas the Regula Falsi Method uses a secant-like approach for faster estimation. Understanding their differences helps in choosing the right method for efficient root-finding in computational mathematics.
Bisection Method
The Bisection Method repeatedly bisects an interval and then selects a subinterval in which a root must lie. It is based on the Intermediate Value Theorem, which states that if a continuous function f(x) has values of opposite signs at the endpoints of an interval, then there is at least one root within that interval.
Advantages of Bisection Method
- Simple and robust.
- It always converges if the initial interval contains a root.
Disadvantages of Bisection Method
- Convergence is relatively slow (linear convergence rate).
- Does not exploit the function's values within the interval for faster convergence.
Formula
X_2=\frac{ (X_0 + X_1)} {2}
Please refer Bisection Method for more details.
Regula Falsi Method
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.
The Regula Falsi Method improves upon the Bisection Method by using a linear interpolation to find a better approximation of the root. It calculates the root of the line segment connecting f(a) and f(b), which tends to be closer to the actual root than the midpoint used in the Bisection Method.
Advantages of Regula Falsi Method
- Faster convergence than the Bisection Method for functions that are approximately linear in the interval.
- Uses the function's values to better estimate the root.
Disadvantages of Regula Falsi Method
- May converge slowly if the function is not approximately linear within the interval.
- Can get stuck if the same subinterval is repeatedly chosen without significant improvement.
Formula
X_3=X_2−\frac{f(X_2)−f(X_1)}{f(X_2)(X_2−X_1)}
Please refer Regula Falsi Method for more details.
Differences between Bisection Method and Regula False Method
| Basis | Bisection Method | Regula Falsi Method |
|---|---|---|
| Definition | In mathematics, the bisection method is a root-finding method that applies to continuous function for which knows two values with opposite signs. | In mathematics, the false position method is a very old method for solving equations with one unknown this method is modified form is still in use. |
| Simplicity | it is simple to use and easy to implement. | Simple to use as compared to Bisection Method |
| Computational Efforts | Less as compared to Regula Falsi Method | More as compared to Bisection Method |
| Iteration required | In the bisection method, if one of the initial guesses is closer to the root, it will take a large number of iterations to reach the root. | Less as compared to Bisection Method. This method can be less precise than bisection - no strict precision is guaranteed. |
| Convergence | The order of convergence of the bisection method is slow and linear. | This method faster order of convergence than the bisection method. |
| General Iterative Formula | Formula is : | Formula is : |
| Other Names | It is also known as the Bolzano method, Binary chopping method, half Interval method. | It is also known as the False Position method. |
Examples
Problem 1 : Find a root of an equation f(x)=x3-x-1 .(Bisection method)
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
Problem 2 : 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=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=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
Practice Problems - Bisection Method and Regula Falsi Method
- Problem 1: Use the Bisection Method to find the root of the equation f of f(x) = x2 -4 in the interval [1, 3].
- Problem 2: Apply the Bisection Method to find the root of f of f(x)= sin x -1/2 in the interval
[0, \pi] . - Problem 3: Determine the root of f (x) = ln x- 2 using the Bisection Method in the interval [5, 10].
- Problem 4: Find the root of f(x)= e x - 3x2 using the Bisection Method in the interval [0, 1].
- Problem 5: Use the Regular falsi Method to approximate the root of f ( x) = x 3-x + 1 in the interval [1, 2].
- Problem 6: Apply the Regular falsi Method to find the root of f ( x) = x 3-2x -5 in the interval [2, 3].
- Problem 7: Use the Regular falsi Method to determine the root of f (x)= cos x - x in the interval
[0, \pi/ 2]. - Problem 8: Find the root of f ( x) = x 4-16 using the regular falsi Method in the interval [1, 3].
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The Difference Between Regula Falsi and Bisection Method - FAQs
Which method is better for numerical root-finding?
If you need guaranteed convergence, Bisection is more reliable but slower. If speed is important, Regula Falsi is preferable, though it may sometimes slow down in certain cases.
Can both methods be used for all types of functions?
Yes, but Bisection is safer for tricky functions, while Regula Falsi works better for smooth functions with well-behaved roots.
Is Regula Falsi an improvement over Bisection?
Regula Falsi improves speed by using a weighted approach instead of equal splitting, but it doesn’t always guarantee steady convergence like Bisection.