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# Program for Bisection Method

• Difficulty Level : Medium
• Last Updated : 06 Apr, 2021

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).
Example:

```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). Below is implementation of above steps.

## C++

 `// C++ program for implementation of Bisection Method for``// solving equations``#include``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.`

## C#

 `// C# program for implementation``// of Bisection Method for``// solving equations``using` `System;` `class` `GFG``{``static` `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)``    ``{``        ``Console.WriteLine(``"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``    ``Console.WriteLine(``"The value of "` `+``                      ``"root is : "``+ c);``}` `// Driver Code``static` `public` `void` `Main ()``{``    ``// Initial values assumed``    ``double` `a = -200, b = 300;``    ``bisection(a, b);``}``}` `// This code is contributed by ajit`

## PHP

 `= 0)``    ``{``        ``echo` `"You have not assumed "` `.``                 ``"right a and b"``,``"\n"``;``        ``return``;``    ``}` `    ``\$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``;``    ``}``    ``echo` `"The value of root is : "` `, ``\$c``;``}` `// Driver Code` `// Initial values assumed``\$a` `=-200;``\$b` `= 300;``bisection(``\$a``, ``\$b``);` `// This code is contributed by ajit``?>`

## Javascript

 ``

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