Program to find root of an equations using secant method - GeeksforGeeks

08 Oct, 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 the 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++
Java
Python3
C#
PHP
Javascript
// 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 interval";} // Driver codeint main(){    // initializing the values    float x1 = 0, x2 = 1, E = 0.0001;    secant(x1, x2, E);    return 0;}
// Java Program to find root of an// equations using secant methodclass 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 interval");    }         // 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 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 interval"); # Driver code # initializing the valuesx1 = 0;x2 = 1;E = 0.0001;secant(x1, x2, E); # This code is contributed by mits
// C# Program to find root of an// equations using secant methodusing 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 interval");    }         // 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 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 interval";} // 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.?>
<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 interval");} // 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_methodThis article is contributed by Niteesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@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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nitin mittal
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