Find root of a number using Newton’s method

Given an integer N and a tolerance level L, the task is to find the square root of that number using Newton’s Method.

Examples:

Input: N = 16, L = 0.0001
Output: 4
42 = 16

Input: N = 327, L = 0.00001
Output: 18.0831

Newton’s Method:
Let N be any number then the square root of N can be given by the formula:



root = 0.5 * (X + (N / X)) where X is any guess which can be assumed to be N or 1.

  • In the above formula, X is any assumed square root of N and root is the correct square root of N.
  • Tolerance limit is the maximum difference between X and root allowed.

Approach: The following steps can be followed to compute the answer:

  1. Assign X to the N itself.
  2. Now, start a loop and keep calculating the root which will surely move towards the correct square root of N.
  3. Check for the difference between the assumed X and calculated root, if not yet inside tolerance then update root and continue.
  4. If the calculated root comes inside the tolerance allowed then break out of the loop.
  5. Print the root.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the square root of
// a number using Newtons method
double squareRoot(double n, float l)
{
    // Assuming the sqrt of n as n only
    double x = n;
  
    // The closed guess will be stored in the root
    double root;
  
    // To count the number of iterations
    int count = 0;
  
    while (1) {
        count++;
  
        // Calculate more closed x
        root = 0.5 * (x + (n / x));
  
        // Check for closeness
        if (abs(root - x) < l)
            break;
  
        // Update root
        x = root;
    }
  
    return root;
}
  
// Driver code
int main()
{
    double n = 327;
    float l = 0.00001;
  
    cout << squareRoot(n, l);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach 
class GFG 
{
      
    // Function to return the square root of 
    // a number using Newtons method 
    static double squareRoot(double n, double l) 
    
        // Assuming the sqrt of n as n only 
        double x = n; 
      
        // The closed guess will be stored in the root 
        double root; 
      
        // To count the number of iterations 
        int count = 0
      
        while (true)
        
            count++; 
      
            // Calculate more closed x 
            root = 0.5 * (x + (n / x)); 
      
            // Check for closeness 
            if (Math.abs(root - x) < l) 
                break
      
            // Update root 
            x = root; 
        
      
        return root; 
    
      
    // Driver code 
    public static void main (String[] args) 
    
        double n = 327
        double l = 0.00001
      
        System.out.println(squareRoot(n, l)); 
    
}
  
// This code is contributed by AnkitRai01

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function to return the square root of 
# a number using Newtons method 
def squareRoot(n, l) :
  
    # Assuming the sqrt of n as n only 
    x =
  
    # To count the number of iterations 
    count = 0 
  
    while (1) :
        count += 1 
  
        # Calculate more closed x 
        root = 0.5 * (x + (n / x)) 
  
        # Check for closeness 
        if (abs(root - x) < l) :
            break 
  
        # Update root 
        x = root
  
    return root 
  
# Driver code 
if __name__ == "__main__"
  
    n = 327
    l = 0.00001 
  
    print(squareRoot(n, l)) 
  
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach 
using System;
  
class GFG 
{
      
    // Function to return the square root of 
    // a number using Newtons method 
    static double squareRoot(double n, double l) 
    
        // Assuming the sqrt of n as n only 
        double x = n; 
      
        // The closed guess will be stored in the root 
        double root; 
      
        // To count the number of iterations 
        int count = 0; 
      
        while (true)
        
            count++; 
      
            // Calculate more closed x 
            root = 0.5 * (x + (n / x)); 
      
            // Check for closeness 
            if (Math.Abs(root - x) < l) 
                break
      
            // Update root 
            x = root; 
        
      
        return root; 
    
      
    // Driver code 
    public static void Main() 
    
        double n = 327; 
        double l = 0.00001; 
      
        Console.WriteLine(squareRoot(n, l)); 
    
}
  
// This code is contributed by AnkitRai01

chevron_right


Output:

18.0831

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.