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

My Personal Notes

C++

Java

Python3

C#

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