Reading time: 35 minutes  Coding time: 10 minutes
Newton's Method, also known as NewtonRaphson method, named after Isaac Newton and Joseph Raphson, is a popular iterative method to find a good approximation for the root of a realvalued function f(x) = 0. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.
Theory
For a continous and differentiable function f(x), let x_{0} be a good approximation for root r and let r = x_{0} + h. The number h measures how far is x_{0} is from true root.
Since h = r  x_{0} is small we can use tangent line approximation as follow:
0 = f(r) = f(x_{0} + h) â‰ˆ f(x_{0}) + hfÂ´(x_{0})
And unless hfÂ´(x_{0}) is close to zero we have,
h â‰ˆ  f(x_{0}) / fÂ´(x_{0})
So this gives us the root r as,
r = x_{0} + h â‰ˆ x_{0}  f(x_{0}) / fÂ´(x_{0})
Now we put x_{1} = x_{0} + h, where x_{1} is our new improved estimate, and we get the following equation
x_{1} = x_{0}  f(x_{0}) / fÂ´(x_{0})
The next estimate x_{2} is obtained from x_{1} in exactly the same way as x_{1} was obtained from x_{0}:
x_{2} = x_{1}  f(x_{1}) / fÂ´(x_{1})
So we continue in this way, If x_{n} is the current estimate, then the next estimate x_{n+1} is given by :
We can see graphically how Newton's Method works as follow:
Newton Raphson Method.Source: Openstax
We draw a tangent line to the graph of function f(x) at point x = x_{0}. This tangent will have the equation as y= fÂ´(x_{0})(x  x_{0}) + f(x_{0}). Now, we find the x intercept of this tangent by putting y = 0. The new x that we find is the new estimate. This continues till we reach the root of the function.
Algorithm
For a given function f(x),the Newton Raphson Method algorithm works as follows:
1. Start
2. Define function as f(x)
3. Define derivative of function as g(x)
4. Input:
a. Initial guess x0
b. Tolerable Error e
c. Maximum Iteration N
5. Do
If g(x0) = 0
Print "Mathematical Error"
Stop
End If
x(i+1) = x(i)  f(x) / g(x)
step = step + 1
If step > N
Print "Not Convergent"
Stop
End If
While abs f(x1) > e
7. Print root as x(i+1)
8. Stop
Sample Problem
Now let's work with an example:
Find the root of function f(x) = x^{2}  4x  7 taking initial guess as x = 5 using the Newton's Method to determine an approximation to the root that is accurate to at least within 10^{9}.
Now, the information required to perform the Newton Raphson Method is as follow:
 f(x) = x^{2}  4x  7,
 Initial Guess x0 = 5,
 fÂ´(x) = g(x) = 2x  4,
 And tolerance e = 10^{9}
Below we show the iterative process described in the algortihm above and show the values in each iteration:
Inputs
 f(x) = x^{2}  4x  7,
 Initial Guess x0 = 5,
 And tolerance e = 10^{9}
We will calculate the fÂ´(x) :
fÂ´(x) = g(x) = 2x  4,
Iteration 1
x0 = 5
f(x0) = f(5) = 2
 Check if g(x0) is not 0
g(x0) = g(5) = 2(5)  4 = 6 which is not 0 âœ…  We then proceed to calculate x1 :
x1 = x0  f(x0)/g(x0)
x1 = 5  (2/6) = 5.333333333
Now we check the loop condition i.e. fabs(f(x1)) > e
f(x1) = f(5.333333333) = 0.1111111111
fabs(f(c)) = 0.1111111111 > e = 10^{9} âœ…
The loop condition is true so we will perform the next iteration.
Iteration 2
x1 = 5.333333333
f(x1) = f(5.333333333) = 0.1111111111
 Check if g(x1) is not 0
g(x1) = g(5.333333333) = 2(5.333333333)  4 = 6.666666666 which is not 0 âœ…  We then proceed to calculate x2 :
x2 = x1  f(x1)/g(x1)
x1 = 5.333333333  (0.1111111111/6) = 5.316666667
Now we check the loop condition i.e. fabs(f(x1)) > e
f(x2) = f(5.316666667) = 0.0002777777778
fabs(f(x2)) = 0.0002777777778 > e = 10^{9} âœ…
The loop condition is true so we will perform the next iteration.
As you can see, it converges to a solution which depends on the tolerance and number of iteration the algorithm performs.
Newton Raphson method performed on the function f(x) = x^2  4x  7C++ Implementation
#include <iostream>
#include <math.h>
#include<iomanip>
#include<chrono>
using namespace std::chrono;
using namespace std;
int N= 1000; // max iterations
static double function(double x);
double derivFunc(double x);
void newtonRaphson( double x, double precision);
int main() {
double x0;
double c;
double precision;
cout << "function(x) = x^2  4x  7 "<<endl;
cout << "Enter initial guess: ";
cin >> x0;
cout << "\nEnter precision of method: ";
cin >> precision;
newtonRaphson(x0,precision);
return 0;
}
static double function(double x) // f(x)
{
return pow(x,2)  4*x 7;
}
double derivFunc(double x) // f'(x) = g(x)
{
return 2*x  4 ;
}
void newtonRaphson(double x, double precision)
{
int iter=0;
cout<<setw(3)<<"\niterations"<<setw(8)<<"x"<<setw(30)<<"function(x)"<<endl;
auto start = high_resolution_clock::now();
if(derivFunc(x)== 0 )
{
cout<<"Error";
return;
}
double h = function(x) / derivFunc(x);
do
{
h = function(x)/derivFunc(x);
// x(i+1) = x(i)  f(x) / f'(x)
x = x  h;
iter++;
cout<<setprecision(10)<<setw(3)<<iter<<setw(25)<<x<<setw(20)<<function(x)<<endl;
if (iter > N)
{
cout<<" Not Convergent";
break;
}
}while (fabs(function(x))>=precision);
auto stop = high_resolution_clock::now();
auto duration = duration_cast<microseconds>(stop  start);
cout<<"\nRoot = "<<x<<endl;
cout<<"f(x)="<<function(x)<<endl;
cout << duration.count()<<" microseconds"<< endl;
}
Limitations
Newton Raphson Method is one of the fastest method which converges to root quickly ( as it has quadratic convergence)
but there are also some drawbacks when this algorithm is used.

We need to calculate the derivative for the function we are performing the iterations for which adds in more computation than simple methods like Bisection or Regula Falsi method.

Since it is dependent on initial guess it may encounter a point where derivative may become zero and then not continue.

Newton's method may not work if there are points of inflection, local maxima or minima around initial guess or the root.
Let us see an examples below demonstrating the limitation
Visualisation for Newton Raphson methodAs we can see in the above graph the sequence of root estimates produces by Newton Raphson Method do not converge to the true root but rather osillate.