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Program to find root of an equations using secant method
  • Difficulty Level : Easy
  • Last Updated : 26 Apr, 2021

The secant method is used to find the root of an equation f(x) = 0. It is started from two distinct estimates x1 and x2 for the root. It is an iterative procedure involving linear interpolation to a root. The iteration stops if the difference between two intermediate values is less than convergence factor.
Examples : 
 

Input : equation = x3 + x - 1 
        x1 = 0, x2 = 1, E = 0.0001
Output : Root of the given equation = 0.682326
         No. of iteration=5

Algorithm 
 

Initialize: x1, x2, E, n         // E = convergence indicator
calculate f(x1),f(x2)

if(f(x1) * f(x2) = E); //repeat the loop until the convergence
    print 'x0' //value of the root
    print 'n' //number of iteration
}
else
    print "can not found a root in the given interval"

 

 

C++




// C++ Program to find root of an
// equations using secant method
#include <bits/stdc++.h>
using namespace std;
// function takes value of x and returns f(x)
float f(float x)
{
    // we are taking equation as x^3+x-1
    float f = pow(x, 3) + x - 1;
    return f;
}
 
void secant(float x1, float x2, float E)
{
    float n = 0, xm, x0, c;
    if (f(x1) * f(x2) < 0) {
        do {
            // calculate the intermediate value
            x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));
 
            // check if x0 is root of equation or not
            c = f(x1) * f(x0);
 
            // update the value of interval
            x1 = x2;
            x2 = x0;
 
            // update number of iteration
            n++;
 
            // if x0 is the root of equation then break the loop
            if (c == 0)
                break;
            xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));
        } while (fabs(xm - x0) >= E); // repeat the loop
                                // until the convergence
 
        cout << "Root of the given equation=" << x0 << endl;
        cout << "No. of iterations = " << n << endl;
    } else
        cout << "Can not find a root in the given inteval";
}
 
// Driver code
int main()
{
    // initializing the values
    float x1 = 0, x2 = 1, E = 0.0001;
    secant(x1, x2, E);
    return 0;
}

Java




// Java Program to find root of an
// equations using secant method
class GFG {
     
    // function takes value of x and
    // returns f(x)
    static float f(float x) {
         
        // we are taking equation
        // as x^3+x-1
        float f = (float)Math.pow(x, 3)
                               + x - 1;
                                
        return f;
    }
     
    static void secant(float x1, float x2,
                                float E) {
         
        float n = 0, xm, x0, c;
        if (f(x1) * f(x2) < 0)
        {
            do {
                 
                // calculate the intermediate
                // value
                x0 = (x1 * f(x2) - x2 * f(x1))
                            / (f(x2) - f(x1));
         
                // check if x0 is root of
                // equation or not
                c = f(x1) * f(x0);
         
                // update the value of interval
                x1 = x2;
                x2 = x0;
         
                // update number of iteration
                n++;
         
                // if x0 is the root of equation
                // then break the loop
                if (c == 0)
                    break;
                xm = (x1 * f(x2) - x2 * f(x1))
                            / (f(x2) - f(x1));
                             
                // repeat the loop until the
                // convergence
            } while (Math.abs(xm - x0) >= E);
                                                 
            System.out.println("Root of the" +
                    " given equation=" + x0);
                     
            System.out.println("No. of "
                      + "iterations = " + n);
        }
         
        else
            System.out.print("Can not find a"
              + " root in the given inteval");
    }
     
    // Driver code
    public static void main(String[] args) {
         
        // initializing the values
        float x1 = 0, x2 = 1, E = 0.0001f;
        secant(x1, x2, E);
    }
}
 
// This code is contributed by Anant Agarwal.

Python3




# Python3 Program to find root of an
# equations using secant method
 
# function takes value of x
# and returns f(x)
def f(x):
     
    # we are taking equation
    # as x^3+x-1
    f = pow(x, 3) + x - 1;
    return f;
 
def secant(x1, x2, E):
    n = 0; xm = 0; x0 = 0; c = 0;
    if (f(x1) * f(x2) < 0):
        while True:
             
            # calculate the intermediate value
            x0 = ((x1 * f(x2) - x2 * f(x1)) /
                            (f(x2) - f(x1)));
 
            # check if x0 is root of
            # equation or not
            c = f(x1) * f(x0);
 
            # update the value of interval
            x1 = x2;
            x2 = x0;
 
            # update number of iteration
            n += 1;
 
            # if x0 is the root of equation
            # then break the loop
            if (c == 0):
                break;
            xm = ((x1 * f(x2) - x2 * f(x1)) /
                            (f(x2) - f(x1)));
             
            if(abs(xm - x0) < E):
                break;
         
        print("Root of the given equation =",
                               round(x0, 6));
        print("No. of iterations = ", n);
         
    else:
        print("Can not find a root in ",
                   "the given inteval");
 
# Driver code
 
# initializing the values
x1 = 0;
x2 = 1;
E = 0.0001;
secant(x1, x2, E);
 
# This code is contributed by mits

C#




// C# Program to find root of an
// equations using secant method
using System;
 
class GFG {
     
    // function takes value of
    // x and returns f(x)
    static float f(float x)
    {
         
        // we are taking equation
        // as x^3+x-1
        float f = (float)Math.Pow(x, 3)
                                + x - 1;
        return f;
    }
     
    static void secant(float x1, float x2,
                    float E)                
                     
    {
         
        float n = 0, xm, x0, c;
        if (f(x1) * f(x2) < 0)
        {
            do {
                 
                // calculate the intermediate
                // value
                x0 = (x1 * f(x2) - x2 * f(x1))
                    / (f(x2) - f(x1));
         
                // check if x0 is root of
                // equation or not
                c = f(x1) * f(x0);
         
                // update the value of interval
                x1 = x2;
                x2 = x0;
         
                // update number of iteration
                n++;
         
                // if x0 is the root of equation
                // then break the loop
                if (c == 0)
                    break;
                xm = (x1 * f(x2) - x2 * f(x1))
                    / (f(x2) - f(x1));
                             
                // repeat the loop until
                // the convergence
            } while (Math.Abs(xm - x0) >= E);
                                                 
            Console.WriteLine("Root of the" +
                    " given equation=" + x0);
                     
            Console.WriteLine("No. of " +
                              "iterations = " + n);
        }
         
        else
            Console.WriteLine("Can not find a" +
                              " root in the given inteval");
    }
     
    // Driver code
    public static void Main(String []args)
    {
         
        // initializing the values
        float x1 = 0, x2 = 1, E = 0.0001f;
        secant(x1, x2, E);
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP Program to find root of an
// equations using secant method
 
// function takes value of x
// and returns f(x)
function f( $x)
{
     
    // we are taking equation
    // as x^3+x-1
    $f = pow($x, 3) + $x - 1;
    return $f;
}
 
function secant($x1, $x2, $E)
{
    $n = 0; $xm;
    $x0; $c;
    if (f($x1) * f($x2) < 0)
    {
        do {
             
            // calculate the intermediate value
            $x0 = ($x1 * f($x2) - $x2 *
                  f($x1)) / (f($x2) - f($x1));
 
            // check if x0 is root
            // of equation or not
            $c = f($x1) * f($x0);
 
            // update the value of interval
            $x1 = $x2;
            $x2 = $x0;
 
            // update number of iteration
            $n++;
 
            // if x0 is the root of equation
            // then break the loop
            if ($c == 0)
                break;
            $xm = ($x1 * f($x2) - $x2 * f($x1)) /
                              (f($x2) - f($x1));
                               
        // repeat the loop
        // until the convergence
        } while (abs($xm - $x0) >= $E);
         
        echo "Root of the given equation=". $x0."\n" ;
        echo "No. of iterations = ". $n ;
         
    } else
        echo "Can not find a root in the given inteval";
}
 
// Driver code
{
     
    // initializing the values
    $x1 = 0; $x2 = 1;
    $E = 0.0001;
    secant($x1, $x2, $E);
    return 0;
}
 
// This code is contributed by nitin mittal.
?>

Javascript




<script>
// JavaScript Program to find root of an
// equations using secant method
 
// function takes value of x and returns f(x)
function f(x)
{
    // we are taking equation as x^3+x-1
    let f = Math.pow(x, 3) + x - 1;
    return f;
}
 
function secant(x1, x2, E)
{
    let n = 0, xm, x0, c;
    if (f(x1) * f(x2) < 0) {
        do {
            // calculate the intermediate value
            x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));
 
            // check if x0 is root of equation or not
            c = f(x1) * f(x0);
 
            // update the value of interval
            x1 = x2;
            x2 = x0;
 
            // update number of iteration
            n++;
 
            // if x0 is the root of equation then break the loop
            if (c == 0)
                break;
            xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));
        } while (Math.abs(xm - x0) >= E); // repeat the loop
                                // until the convergence
 
        document.write("Root of the given equation=" + x0.toFixed(6) + "<br>");
        document.write("No. of iterations = " + n + "<br>");
    } else
        document.write("Can not find a root in the given inteval");
}
 
// Driver code
    // initializing the values
    let x1 = 0, x2 = 1, E = 0.0001;
    secant(x1, x2, E);
 
 
// This code is contributed by Surbhi Tyagi.
</script>

Output : 
 

Root of the given equation = 0.682326
No. of iterations = 5

Time Complexity = O(1)
Reference 
https://en.wikipedia.org/wiki/Secant_method
This article is contributed by Niteesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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