Program for Bisection Method

3.1

Given a function f(x) on floating number x and two numbers ‘a’ and ‘b’ such that f(a)*f(b) < 0 and f(x) is continuous in [a, b]. Here f(x) represents algebraic or transcendental equation. Find root of function in interval [a, b] (Or find a value of x such that f(x) is 0).

Input: A function of x, for example x3 - x2 + 2.     
       And two values: a = -200 and b = 300 such that
       f(a)*f(b) < 0, i.e., f(a) and f(b) have
       opposite signs.
Output: The value of root is : -1.0025
        OR any other value with allowed
        deviation from root.

What are Algebraic and Transcendental functions?
Algebraic function are the one which can be represented in the form of polynomials like f(x) = a1x3 + a2x2 + ….. + e where aa1, a2, … are constants and x is a variable.
Transcendental function are non algebraic functions, for example f(x) = sin(x)*x – 3 or f(x) = ex + x2 or f(x) = ln(x) + x ….

What is Bisection Method?
The method is also called the interval halving method, the binary search method or the dichotomy method. This method is used to find root of an equation in a given interval that is value of ‘x’ for which f(x) = 0 .

The method is based on The Intermediate Value Theorem which states that if f(x) is a continuous function and there are two real numbers a and b such that f(a)*f(b) 0 and f(b) < 0), then it is guaranteed that it has at least one root between them.

Assumptions:

  1. f(x) is a continuous function in interval [a, b]
  2. f(a) * f(b) < 0

Steps:

  1. Find middle point c= (a + b)/2 .
  2. If f(c) == 0, then c is the root of the solution.
  3. Else f(c) != 0
    1. If value f(a)*f(c) < 0 then root lies between a and c. So we recur for a and c
    2. Else If f(b)*f(c) < 0 then root lies between b and c. So we recur b and c.
    3. Else given function doesn’t follow one of assumptions.

Since root may be a floating point number, we repeat above steps while difference between a and b is less than a value ? (A very small value).

bisection

Below is C++ implementation of above steps.

C++

// C++ program for implementation of Bisection Method for
// solving equations
#include<bits/stdc++.h>
using namespace std;
#define EPSILON 0.01

// An example function whose solution is determined using
// Bisection Method. The function is x^3 - x^2  + 2
double func(double x)
{
    return x*x*x - x*x + 2;
}

// Prints root of func(x) with error of EPSILON
void bisection(double a, double b)
{
    if (func(a) * func(b) >= 0)
    {
        cout << "You have not assumed right a and b\n";
        return;
    }

    double c = a;
    while ((b-a) >= EPSILON)
    {
        // Find middle point
        c = (a+b)/2;

        // Check if middle point is root
        if (func(c) == 0.0)
            break;

        // Decide the side to repeat the steps
        else if (func(c)*func(a) < 0)
            b = c;
        else
            a = c;
    }
    cout << "The value of root is : " << c;
}

// Driver program to test above function
int main()
{
    // Initial values assumed
    double a =-200, b = 300;
    bisection(a, b);
    return 0;
}

Java


// Java program for implementation of Bisection Method 
// for solving equations
class GFG{
    static final float EPSILON = (float)0.01;

    // An example function whose solution is determined using
    // Bisection Method. The function is x^3 - x^2  + 2
    static double func(double x)
    {
        return x*x*x - x*x + 2;
    }

    // Prints root of func(x) with error of EPSILON
    static void bisection(double a, double b)
    {
        if (func(a) * func(b) >= 0)
        {
            System.out.println("You have not assumed"
                        + " right a and b");
            return;
        }

        double c = a;
        while ((b-a) >= EPSILON)
        {
            // Find middle point
            c = (a+b)/2;

            // Check if middle point is root
            if (func(c) == 0.0)
                break;

            // Decide the side to repeat the steps
            else if (func(c)*func(a) < 0)
                b = c;
            else
                a = c;
        }
                //prints value of c upto 4 decimal places
        System.out.printf("The value of root is : %.4f"
                        ,c);
    }

    // Driver program to test above function
    public static void main(String[] args)
    {
        // Initial values assumed
        double a =-200, b = 300;
        bisection(a, b);
    }
    // This code is contributed by Nirmal Patel
}

Python3

# Python program for implementation
# of Bisection Method for
# solving equations

 
# An example function whose
# solution is determined using
# Bisection Method.
# The function is x^3 - x^2  + 2
def func(x):
    return x*x*x - x*x + 2
 
# Prints root of func(x)
# with error of EPSILON
def bisection(a,b):

    if (func(a) * func(b) >= 0):
        print("You have not assumed right a and b\n")
        return
 
    c = a
    while ((b-a) >= 0.01):

        # Find middle point
        c = (a+b)/2
 
        # Check if middle point is root
        if (func(c) == 0.0):
            break
 
        # Decide the side to repeat the steps
        if (func(c)*func(a) < 0):
            b = c
        else:
            a = c
            
    print("The value of root is : ","%.4f"%c)
    
# Driver code
# Initial values assumed
a =-200
b = 300
bisection(a, b)

# This code is contributed
# by Anant Agarwal.

Output:

The value of root is : -1.0025

Time complexity :- Time complexity of this method depends on the assumed values and the function.

What are pros and cons?
Advantage of the bisection method is that it is guaranteed to be converged. Disadvantage of bisection method is that it cannot detect multiple roots.

In general, Bisection method is used to get an initial rough approximation of solution. Then faster converging methods are used to find the solution.

We will soon be discussing other methods to solve algebraic and transcendental equations

References:
Introductory Methods of Numerical Analysis by S.S. Sastry
https://en.wikipedia.org/wiki/Bisection_method

This article is contributed by Abhiraj Smit. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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