Question

(C++)

Solve the following polynomial using Newton’s method.

            X⁴+ 2x³-9x²-2x +8 = 0

For a root that lies between x=-2 and x= -10.

Print out for each iteration the current x, stop when the solution is within 10^-5

Answer 1

//                     f( x(n) )

// x(n+1) = x(n) - -----------------

//                    f'( x(n) )

// find root of f(x) = x^4 + 2x^3 -9x^2 -2x +8 upto 3 decimal places

// So, f'(x) = 4x^3 + 6x^2 - 18x

// f(-10) = 7128

// f(-2) = -24

// as the sign is different, so the root lies in between [ -10 , -2 ]

#include<iostream>

#include<cmath>

using namespace std;

int main()

{

    int n = 5;

   

    // initially set the x(n) root to right bound

    double x_n = -10;

   

    // calculate f(x(n))

    double f_x = pow( x_n, 4 ) + 2 * pow( x_n, 3 ) - 9 * pow( x_n, 2 ) - 2 * x_n + 8;

   

    // calculate f'(x(n))

    double f_der_x = 4 * pow( x_n, 3 ) + 6 * pow( x_n, 2 ) - 18 * x_n;

   

    // calculate x(n+1)

    double x_n_1 = x_n - ( f_x / f_der_x );

   

    while(true)

    {

        // if the root is correct upto n decimal places

       if( abs( x_n_1 - x_n ) < pow( 0.1, n ) )

           break;

       

        // set x(n+1) as the x(n) th root

        x_n = x_n_1;

       

        cout<<"x : "<<x_n<<endl;

       

        // calculate f(x(n))

        f_x = pow( x_n, 4 ) + 2 * pow( x_n, 3 ) - 9 * pow( x_n, 2 ) - 2 * x_n + 8;

   

        // calculate f'(x(n))

        f_der_x = 4 * pow( x_n, 3 ) + 6 * pow( x_n, 2 ) - 18 * x_n;

   

        // calculate x(n+1)

        x_n_1 = x_n - ( f_x / f_der_x );

    }

   

    cout<<"Root lies at x = "<<x_n;

   

    return 0;

}

Sample Output