Program for Method Of False Position

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 : -1.00
        OR any other value close to root.

We strongly recommend to refer below post as a prerequisite of this post.
Solution of Algebraic and Transcendental Equations | Set 1 (The Bisection Method)

In this post The Method Of False Position is discussed. This method is also known as Regula Falsi or The Method of Chords.

Similarities with Bisection Method:

Same Assumptions: This method also assumes that function is continuous in [a, b] and given two numbers 'a' and 'b' are such that f(a) * f(b) < 0.
Always Converges: like Bisection, it always converges, usually considerably faster than Bisection--but sometimes very much more slowly than Bisection.

Differences with Bisection Method:
It differs in the fact that we make a chord joining the two points [a, f(a)] and [b, f(b)]. We consider the point at which the chord touches the x axis and named it as c.

Steps:

Write equation of the line connecting the two points.

y – f(a) =  ( (f(b)-f(a))/(b-a) )*(x-a)

Now we have to find the point which touches x axis. 
For that we put y = 0.

so x = a - (f(a)/(f(b)-f(a))) * (b-a)
   x = (a*f(b) - b*f(a)) / (f(b)-f(a)) 

This will be our c that is c = x.

If f(c) == 0, then c is the root of the solution.
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 and may converge very slow in worst case, we iterate for a very large number of times such that the answer becomes closer to the root.

Following is the implementation.

C++

// C++ program for implementation of Bisection Method for 
// solving equations 
#include<bits/stdc++.h> 
using namespace std; 
#define MAX_ITER 1000000 
  
// 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) in interval [a, b] 
void regulaFalsi(double a, double b) 
{ 
    if (func(a) * func(b) >= 0) 
    { 
        cout << "You have not assumed right a and b\n"; 
        return; 
    } 
  
    double c = a;  // Initialize result 
  
    for (int i=0; i < MAX_ITER; i++) 
    { 
        // Find the point that touches x axis 
        c = (a*func(b) - b*func(a))/ (func(b) - func(a)); 
  
        // Check if the above found point is root 
        if (func(c)==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; 
    regulaFalsi(a, b); 
    return 0; 
} 

Java

// java program for implementation  
// of Bisection Method for 
// solving equations 
import java.io.*; 
  
class GFG { 
  
    static int MAX_ITER = 1000000; 
  
    // 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)  
    // in interval [a, b] 
    static void regulaFalsi(double a, double b) 
    { 
        if (func(a) * func(b) >= 0)  
        { 
            System.out.println("You have not assumed right a and b"); 
        } 
        // Initialize result 
        double c = a;  
  
        for (int i = 0; i < MAX_ITER; i++)  
        { 
            // Find the point that touches x axis 
            c = (a * func(b) - b * func(a))  
                 / (func(b) - func(a)); 
  
            // Check if the above found point is root 
            if (func(c) == 0) 
                break; 
  
            // Decide the side to repeat the steps 
            else if (func(c) * func(a) < 0) 
                b = c; 
            else
                a = c; 
        } 
        System.out.println("The value of root is : " + (int)c); 
    } 
  
    // Driver program  
    public static void main(String[] args) 
    { 
        // Initial values assumed 
        double a = -200, b = 300; 
        regulaFalsi(a, b); 
    } 
} 
  
// This article is contributed by vt_m 

Python3

# Python3 implementation of Bisection 
# Method for solving equations 
  
MAX_ITER = 1000000
  
# 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) in interval [a, b] 
def regulaFalsi( a , b): 
    if func(a) * func(b) >= 0: 
        print("You have not assumed right a and b") 
        return -1
      
    c = a # Initialize result 
      
    for i in range(MAX_ITER): 
          
        # Find the point that touches x axis 
        c = (a * func(b) - b * func(a))/ (func(b) - func(a)) 
          
        # Check if the above found point is root 
        if func(c) == 0: 
            break
          
        # Decide the side to repeat the steps 
        elif func(c) * func(a) < 0: 
            b = c 
        else: 
            a = c 
    print("The value of root is : " , '%.4f' %c) 
  
# Driver code to test above function 
# Initial values assumed 
a =-200
b = 300
regulaFalsi(a, b) 
  
# This code is contributed by "Sharad_Bhardwaj". 

C#

// C# program for implementation  
// of Bisection Method for 
// solving equations 
using System; 
  
class GFG { 
  
    static int MAX_ITER = 1000000; 
  
    // 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)  
    // in interval [a, b] 
    static void regulaFalsi(double a, double b) 
    { 
        if (func(a) * func(b) >= 0)  
        { 
            Console.WriteLine("You have not assumed right a and b"); 
        } 
        // Initialize result 
        double c = a;  
  
        for (int i = 0; i < MAX_ITER; i++)  
        { 
            // Find the point that touches x axis 
            c = (a * func(b) - b * func(a))  
                / (func(b) - func(a)); 
  
            // Check if the above found point is root 
            if (func(c) == 0) 
                break; 
  
            // Decide the side to repeat the steps 
            else if (func(c) * func(a) < 0) 
                b = c; 
            else
                a = c; 
        } 
  
        Console.WriteLine("The value of root is : " + (int)c); 
    } 
  
    // Driver program  
    public static void Main(String []args) 
    { 
        // Initial values assumed 
        double a = -200, b = 300; 
        regulaFalsi(a, b); 
    } 
} 
  
// This code is contributed by Sam007. 

Output:

The value of root is : -1

This method always converges, usually considerably faster than Bisection. But worst case can be very slow.

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/False_position_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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