Program to find root of an equations using secant method

• Difficulty Level : Easy
• Last Updated : 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

C++

 // C++ Program to find root of an// equations using secant method#include 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

 // 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

 # 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#

 // 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

 = \$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.?>

Javascript



Output :

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

Time Complexity = O(1)

Reference
https://en.wikipedia.org/wiki/Secant_method
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