Find Root using Newton-Raphson method in C++

Newton-Raphson Method:

                   x_n+1 = x_n-f(x_n)/f ‘(x_n)

How it works:

Suppose you need to find the root of a continuous, differentiable function $f(x)$, and you know the root you are looking for is near the point x = x_0x=x0​. Then Newton’s method tells us that a better approximation for the root is

x_1 = x_0 – \frac{f(x_0)}{f'(x_0)}.x1​=x0​−f′(x0​)f(x0​)​.

This process may be repeated as many times as necessary to get the desired accuracy. In general, for any $x$-value x_nxn​, the next value is given by

x_{n+1} = x_n – \frac{f(x_n)}{f'(x_n)}.xn+1​=xn​−f′(xn​)f(xn​)​.

Note: the term “near” is used loosely because it does not need a precise definition in this context. However, x_0x0​ should be closer to the root you need than to any other root (if the function has multiple roots)

C++ program to find root using Newton-Raphson method:

You can also execute this code on our online compiler.

#include <iostream>
#include <math.h>
using namespace std;
float fn(float x)
{   
     return  pow(x,2)+(3*x)+1 ;
}
float de(float x)
{   
     return  2*x + 3 ;
}
int main()
{   float a,e=0,z;
    cout<<"Enter Number ";
    cin>>a;
    do
    {   e++;
        z=a-(fn(a)/de(a));
        cout<<"The iterative "<<e<<" root is "<<z;    
        a=z;
        cout<<endl;
    }while(abs(fn(z))>0.001);
    return 0;
}

Output:

$g++ -o main *.cpp
$main
Enter Number The iterative 1 root is -0.333333
The iterative 2 root is -0.380952
The iterative 3 root is -0.381966

Please comment in case of any query, issues or concerns.