Program for Muller Method


Given a function f(x) on floating number x and three initial distinct guesses for root of the function, find the root of function. Here, f(x) can be an algebraic or transcendental function.

Examples:

Input : A function f(x) = x^3 + 2x^2 + 10x - 20 
        and three initial guesses - 0, 1 and 2 .
Output : The value of the root is 1.3688 or 
         any other value within permittable deviation 
         from the root.         

Input : A function f(x) = x^5 - 5x + 2 
        and three initial guesses - 0, 1 and 2 .
Output :The value of the root is 0.4021 or 
        any other value within permittable deviation 
        from the root. 

Muller Method

Muller Method is a root-finding algorithm for finding the root of a equation of the form, f(x)=0. It was discovered by David E. Muller in 1956.

It begins with three initial assumptions of the root, and then constructing a parabola through these three points, and takes the intersection of the x-axis with the parabola to be the next approximation. This process continues until a root with the desired level of accuracy is found .



Why to learn Muller’s Method?

Muller Method, being one of the root-finding method along with the other ones like Bisection Method, Regula – Falsi Method, Secant Method and Newton – Raphson Method. But, it offers certain advantages over these methods, as follows –

  1. The rate of convergence, i.e., how much closer we move to the root at each step, is approximately 1.84 in Muller Method, whereas it is 1.62 for secant method, and linear, i.e., 1 for both Regula – falsi Method and bisection method . So, Muller Method is faster than Bisection, Regula – Falsi and Secant method.
  2. Although, it is slower than Newton – Raphson’s Method, which has a rate of convergence of 2, but it overcomes one of the biggest drawbacks of Newton-Raphson Method, i.e., computation of derivative at each step.

So, this shows that Muller Method is an efficient method in calculating root of the function.

Algorithm And Its Working

  1. Assume any three distinct initial roots of the function, let it be x0, x1 and x2.
  2. Now, draw a second degree polynomial, i.e., a parabola, through the values of function f(x) for these points – x0, x1 and x2.
    The equation of the parabola, p(x), through these points is as follows-
    p(x) = c + b(x – x_2) + a(x – x_2)^2, where a, b and c are constants.

  3. After drawing the parabola, then find the intersection of this parabola with the x-axis, let us say x3 .
  4. Finding the intersection of parabola with the x-axis, i.e., x3:
    • To find x_3, the root of p(x), where p(x) = c + b(x – x_2) + a(x – x_2)^2, such that p(x_3) = c + b(x_3 – x_2) + a(x_3 – x_2)^2 = 0, apply the quadratic formula to p(x).Since, there will be two roots, but we have to take that one which is closer to x_2.To avoid round-off errors due to subtraction of nearby equal numbers, use the following equation:

      x_3 - x_2 = \frac{-2c}{b\pm \sqrt{b^{2}-4ac}}
      Now, since, root of p(x) has to be closer to x_2, so we have to take that value which has a greater denominator out of the two values possible from the above equation.

    • To find a, b and c for the above equation, put x in p(x) as x_0, x_1 and x_2, and let these values be p(x_0), p(x_1) and p(x_2), which are as follows-

      p(x_0) = c + b(x_0 – x_2) + a(x_0 – x_2)^2 = f(x_0).
      p(x_1) = c + b(x_1 – x_2) + a(x_1 – x_2)^2 = f(x_1).
      p(x_2) = c + b(x_2 – x_2) + a(x_2 – x_2)^2 = c = f(x_2).

    • So, we have three equations and three variables – a, b, c. After solving them to found out the values of these variables, we get the following values of a, b and c-
      c = p(x_2) = f(x_2) .
      b = (d_2*(h_1)^2 - d_1*(h_2)^2 ) / ( h_1h_2 * (h_1 - h_2)) .
      a = (d_1*(h_2) - d_2*(h_1)) / ( h_1h_2 * (h_1 - h_2)).
      

      where,
      d_1 = p(x_0) – p(x_2) = f(x_0) – f(x_2)
      d_2 = p(x_1) – p(x_2) = f(x_1) – f(x_2)
      h_1 = x_0 – x_2
      h_2 = x_1 – x_2

    • Now, put these values in the expression for x_3 – x_2, and obtain x_3.
      This is how root of p(x) = x_3 is obtained.
  5. If x_3 is very close to x_2 within the permittable error, then x_3 becomes the root of f(x), otherwise, keep repeating the process of finding the next x_3, with previous x_1, x_2 and x_3 as the new x_0, x_1 and x_2 .

C++

// C++ Program to find root of a function, f(x)
#include<bits/stdc++.h>
using namespace std;
  
const int MAX_ITERATIONS = 10000;
  
// Function to calculate f(x)
float f(float x)
{
    // Taking f(x) = x ^ 3 + 2x ^ 2 + 10x - 20
    return 1*pow(x, 3) + 2*x*x + 10*x - 20;
}
  
void Muller(float a, float b, float c)
{
    int i;
    float res;
  
    for (i = 0;;++i)
    {
        // Calculating various constants required
        // to calculate x3
        float f1 = f(a);
        float f2 = f(b);
        float f3 = f(c);
        float d1 = f1 - f3;
        float d2 = f2 - f3;
        float h1 = a - c;
        float h2 = b - c;
        float a0 = f3;
        float a1 = (((d2*pow(h1, 2)) - (d1*pow(h2, 2)))
                    / ((h1*h2) * (h1-h2)));
        float a2 = (((d1*h2) - (d2*h1))/((h1*h2) * (h1-h2)));
        float x = ((-2*a0) / (a1 + abs(sqrt(a1*a1-4*a0*a2))));
        float y = ((-2*a0) / (a1-abs(sqrt(a1*a1-4*a0*a2))));
  
        // Taking the root which is closer to x2
        if (x >= y)
            res = x + c;
        else
            res = y + c;
  
        // checking for resemblance of x3 with x2 till
        // two decimal places
        float m = res*100;
        float n = c*100;
        m = floor(m);
        n = floor(n);
        if (m == n)
            break;
        a = b;
        b = c;
        c = res;
        if (i > MAX_ITERATIONS)
        {
            cout << "Root cannot be found using"
                 << " Muller's method";
            break;
        }
    }
    if (i <= MAX_ITERATIONS)
         cout << "The value of the root is " << res;
}
  
// Driver main function
int main()
{
    float a = 0, b = 1, c = 2;
    Muller(a, b, c);
    return 0;
}

Java

// Java Program to find root of a function, f(x)
import java.io.*;
import static java.lang.Math.*;
  
class Muller
{
    static final int MAX_ITERATIONS = 10000;
  
    // function to calculate f(x)
    static double f(double x)
    {
        // Taking f(x) = x ^ 3 + 2x ^ 2 + 10x - 20
        return 1*pow(x, 3) + 2*x*x + 10*x - 20;
    }
  
    static void Muller(double a, double b, double c)
    {
        int i;
        double res;
  
        for (i = 0;; ++i)
        {
            // Calculating various constants required
            // to calculate x3
            double f1 = f(a);
            double f2 = f(b);
            double f3 = f(c);
            double d1 = f1 - f3;
            double d2 = f2 - f3;
            double h1 = a - c;
            double h2 = b - c;
            double a0 = f3;
            double a1 = (((d2*pow(h1, 2)) - (d1*pow(h2, 2)))
                        / ((h1*h2) * (h1-h2)));
            double a2 = (((d1*h2) - (d2*h1))/((h1*h2) * (h1-h2)));
            double x = ((-2*a0)/(a1 + abs(sqrt(a1*a1-4*a0*a2))));
            double y = ((-2*a0)/(a1-abs(sqrt(a1*a1-4*a0*a2))));
  
            // Taking the root which is closer to x2
            if (x >= y)
                res = x + c;
            else
                res = y + c;
  
            // checking for resemblance of x3 with x2 till
            // two decimal places
            double m = res*100;
            double n = c*100;
            m = floor(m);
            n = floor(n);
            if (m == n)
                break;
            a = b;
            b = c;
            c = res;
            if (i > MAX_ITERATIONS)
            {
                System.out.println("Root cannot be found using" +
                                   " Muller's method");
                break;
            }
        }
        if (i <= MAX_ITERATIONS)
            System.out.println("The value of the root is " + res);
    }
  
    // Driver main function
    public static void main(String args[])
    {
        double a = 0, b = 1, c = 2;
        Muller(a, b, c);
    }
}


Output:

The value of the root is 1.3688

Advantages

  • Can find imaginary roots.
  • No need to find derivatives.

Disadvantages

  • Long to do by hand, more room for error.
  • Extraneous roots can be found.

Reference-

  1. Higher Engineer Mathematics by B.S. Grewal.



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