Newton-Raphson Root Finding Method
Algorithm:
xi+1 = xi - f(xi)/f'(xi)
Example: find the roots of f(x) = 2x2 -x -21
% cat newt3.cpp
// Newton-Raphson method for finding the
// roots of f(x) = 0 for a polynomial function
#include <iostream>
using namespace std;
#include <cmath>
double newton(double a, double b, double c, double x0, double crit);
int main()
{
double a, b, c, root, guess, conv_crit;
cout << "Enter values for a, b, and c :";
cin >> a >> b >> c;
cout << "Enter a guess for the root :";
cin >> guess;
cout << "Enter the convergence criterion: ";
cin >> conv_crit;
// call function newton
root = newton(a, b, c, guess, conv_crit);
// display result
cout.setf(ios::fixed | ios::showpoint);
cout.precision(4);
cout << "One root is " << root << endl;
return 0;
}
double newton(double a, double b, double c, double x0, double crit)
{
double x1, delta;
int iters = 0;
// use a do-while loop
do
{
x1 = x0 - (a*x0*x0 + b*x0 + c) / ( 2*a*x0 + b);
delta = fabs ( x1 - x0 );
x0 = x1; // new approximation becomes
the old
// approximation for the next iteration
iters++; // count the number of
iterations
} while (delta > crit && iters <= 30);
cout << iters << " iterations" << endl;
return x1;
}
% a.out
Enter values for a, b, and c :2 -1 -21
Enter a guess for the root :10
Enter the convergence criterion: .0001
5 iterations
One root is 3.5000
% a.out
Enter values for a, b, and c :2 -1 -21
Enter a guess for the root :-9
Enter the convergence criterion: .0001
5 iterations
One root is -3.0000
