Numerical methods are the set of tasks by applying arithmetic operations to numerical equations. We can formulate mathematical problems to find the approximate result. This formulation is called the numerical implementation of the problem. In this, there is no need for algorithms. Programming logic is then developed for numerical implementation. The programming is usually done with some high-level languages like Fortran, Basic, etc.
Bisection Method
This method 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.
Bisection Method
Formula is: x2 = (x0 + x1) / 2
Newton Raphson Method
The Newton Raphson Method is the process for the determination of a real root of an equation f(x)=0 given just one point close to the desired root.
Newton Raphson method
Formula is: x1 = x0 – f(x0)/f'(x0)
Comparison between Bisection Method and Newton Raphson Method
Sr. No. |
Bisection Method |
Newton Raphson Method |
1. |
In the Bisection Method, the rate of convergence is linear thus it is slow. |
In the Newton Raphson method, the rate of convergence is second-order or quadratic. |
2. |
In Bisection Method we used following formula
x2= (x0 + x1) / 2
|
In Newton Raphson method we used following formula
x1 = x0 – f(x0)/f'(x0)
|
3. |
In this method, we take two initial approximations of the root in which the root is expected to lie. |
In this method, we take one initial approximation of the root. |
4. |
The computation of function per iteration is 1. |
The computation of function per iteration is 2. |
5. |
The initial approximation is less sensitive. |
The initial approximation is very sensitive. |
6. |
In the Bisection Method, there is no need to find derivatives. |
In the Newton Raphson method, there is a need to find derivatives. |
7. |
This method is not applicable for finding complex, multiple, and nearly equal two roots. |
This method is applicable for finding complex, multiple, and nearly equal two roots. |
Question 1: Find a root of an equation f(x) = x3 – x – 1
Solution:
Given equation f(x) = x3 – x – 1
let x = 0, 1, 2
f(1) = -1 < 0 and f(2) = 5 > 0
The root lies between these two points 1 and 2
x0 = 1 + 2/2 = 1.5
f(x0) = f(1.5) = 0.875 > 0
f(1) = -1 < 0 and f(1.5) = 0.875 > 0
The 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
f(1.25) = -0.29688 < 0 and f(1.5) = 0.875 > 0
The 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
f(1.25) = -0.29688 < 0 and f(1.375) = 0.22461 > 0
The 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
f(1.3125) = -0.05151 < 0 and f(1.375) = 0.22461 > 0
The 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
f(1.3125) = -0.05151 < 0 and f(1.34375) = 0.08261 > 0
The 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
f(1.3125) = -0.05151 < 0 and f(1.32812) = 0.01458 > 0
The 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
f(1.32031) = -0.01871 < 0 and f(1.32812) = 0.01458 > 0
The 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
f(1.32422) = -0.00213 < 0 and f(1.32812) = 0.01458 > 0
The 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
f(1.32422) = -0.00213 < 0 and f(1.32617) = 0.00621 > 0
The 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
f(1.32422) = -0.00213 < 0 and f(1.3252) = 0.00204 > 0
The 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
Question 2: Find a root of an equation f(x) = 2x3 – 2x – 5
Solution:
Given Equation f(x) = 2x3 – 2x – 5
f(1) = -5 < 0 and f(2) = 7 > 0
The root lies between these two points 1 and 2
x0 = 1 + 2/2 = 1.5
f(x0) = f(1.5) = 2 × 1.53 – 2 × 1.5 – 5 = -1.25 < 0
f(1.5) = -1.25 < 0 and f(2) = 7 > 0
The root lies between these two points 1.5 and 2
x1 = 1.5 + 2/2 = 1.75
f(x1) = f(1.75) = 2 × 1.753 – 2 × 1.75 – 5 = 2.21875 > 0
f(1.5) = -1.25 < 0 and f(1.75) = 2.21875 > 0
The root lies between these two points1.5 and 1.75
x2 = 1.5 + 1.75/2 = 1.625
f(x2) = f(1.625) = 2 × 1.6253 – 2 × 1.625 – 5 = 0.33203 > 0
f(1.5) = -1.25 < 0 and f(1.625) = 0.33203 > 0
The root lies between these two points 1.5 and 1.625
x3 = 1.5 + 1.625/2 = 1.5625
f(x3) = f(1.5625) = 2 × 1.56253 – 2 × 1.5625 – 5 = -0.49561 < 0
f(1.5625) = -0.49561 < 0 and f(1.625) = 0.33203 > 0
The root lies between these two points 1.5625 and 1.625
x4 = 1.5625 + 1.625/2 = 1.59375
f(x4) = f(1.59375) = 2 × 1.593753 – 2 × 1.59375 – 5 = -0.09113 < 0
f(1.59375) = -0.09113 < 0 and f(1.625) = 0.33203 > 0
The root lies between these two points 1.59375 and 1.625
x5 = 1.59375 + 1.625/2 = 1.60938
f(x5) = f(1.60938) = 2 × 1.609383 – 2 × 1.60938 – 5 = 0.1181 > 0
f(1.59375) = -0.09113 < 0 and f(1.60938) = 0.1181 > 0
The root lies between these two points 1.59375 and 1.60938
x6 = 1.59375 + 1.60938/2 = 1.60156
f(x6) = f(1.60156) = 2 × 1.601563 – 2 × 1.60156 – 5 = 0.0129 > 0
f(1.59375) = -0.09113 < 0 and f(1.60156) = 0.0129 > 0
The root lies between these two points 1.59375 and 1.60156
x7 = 1.59375 + 1.60156/2 = 1.59766
f(x7) = f(1.59766) = 2 × 1.597663 – 2 × 1.59766 – 5 = -0.03926 < 0
f(1.59766) = -0.03926 < 0 and f(1.60156) = 0.0129 > 0
The root lies between these two points 1.59766 and 1.60156
x8 = 1.59766 + 1.60156/2 = 1.59961
f(x8) = f(1.59961) = 2 × 1.599613 – 2 × 1.59961 – 5 = -0.01322 < 0
Here f(1.59961) = -0.01322 < 0 and f(1.60156) = 0.0129 > 0
The root lies between these two points 1.59961 and 1.60156
x9 = 1.59961 + 1.60156/2 = 1.60059
f(x9) = f(1.60059) = 2 × 1.600593 – 2 × 1.60059 – 5 = -0.00017 < 0
The Approximate root of the equation 2x3 – 2x – 5 = 0 using Bisection method is 1.60059
Question 3: Find a root of an equation f(x) = x3 – x – 1
Solution:
Given equation x3 – x – 1 = 0
Using differentiate method the equation is
∴ f′(x) = 3x2 – 1
Here f(1) = -1 < 0 and f(2) = 5 > 0
∴ Root lies between 1 and 2
x0 = 1 + 2/ 2 = 1.5
f(x0) = f(1.5) = 0.875
f′(x0) = f′(1.5) = 5.75
x1 = x0 – f(x0) / f′(x0)
x1 = 1.5 – 0.875/ 5.75
x1 = 1.34783
f(x1) = f(1.34783) = 0.10068
f′(x1) = f′(1.34783) = 4.44991
x2 = x1 – f(x1)/f′(x1)
x2 = 1.34783 – 0.10068/4.44991
x2 = 1.3252
f(x2) = f(1.3252) = 0.00206
f′(x2) = f′(1.3252) = 4.26847
x3 = x2 – f(x2)/f′(x2)
x3 = 1.3252 – 0.00206/4.26847
x3 = 1.32472
f(x3) = f(1.32472) = 0
f′(x3) = f′(1.32472) = 4.26463
x4 = x3 – f(x3)/f′(x3)
x4 = 1.32472 – 0/ 4.26463
x4 = 1.32472
The Approximate root of the equation x3 – x – 1 = 0 using the Newton Raphson method is 1.32472
Question 4: Find a root of an equation f(x) = 2x3 – 2x – 5
Solution:
Given equation 2x3 – 2x – 5 = 0
Using differentiate method the equation is
∴ f′(x) = 6x2 – 2
Here f(1) = -5 < 0 and f(2) = 7 > 0
∴ Root lies between 1 and 2
x0 = 1 + 2/ 2 = 1.5
f(x0) = f(1.5) = 2 × 1.53 – 2 × 1.5 – 5 = -1.25
f′(x0) = f′(1.5) = 6 × 1.52 – 2 = 11.5
x1 = x0 – f(x0)/f′(x0)
x1 = 1.5 – (-1.25)/11.5
x1 = 1.6087
f(x1) = f(1.6087) = 2 × 1.60873 – 2 × 1.6087 – 5 = 0.1089
f′(x1) = f′(1.6087) = 6 × 1.60872 – 2 = 13.52741
x2 = x1 – f(x1)/f′(x1)
x2 = 1.6087 – 0.1089/13.52741
x2 = 1.60065
f(x2) = f(1.60065) = 2 × 1.600653 – 2 × 1.60065 – 5 = 0.00062
f′(x2) = f′(1.60065) = 6 × 1.600652 – 2 = 13.37239
x3 = x2 – f(x2)/f′(x2)
x3 = 1.60065 – 0.00062/13.37239
x3 = 1.6006
The Approximate root of the equation 2x3 – 2x – 5 = 0 using the Newton Raphson method is 1.6006
Last Updated :
28 Jan, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...