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:

1. 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.
2. 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:

1. 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.
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 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 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 ";         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.

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