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++

// 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;
}

Java

// 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

Python3

# 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 = n

    # 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

C#

// 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

Javascript

<script>
    // Javascript implementation of the approach

    // Function to return the square root of
    // a number using Newtons method
    function squareRoot(n, l)
    {
        // Assuming the sqrt of n as n only
        let x = n;

        // The closed guess will be stored in the root
        let root;

        // To count the number of iterations
        let 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.toFixed(4);
    }

    let n = 327;
    let l = 0.00001;

    document.write(squareRoot(n, l));

 // This code is contributed by divyesh072019.
</script>

Output

18.0831

Time Complexity: O(log N)

Auxiliary Space: O(1)

Recursive Approach:

• Start by defining the function findSqrt that takes three arguments – the number whose square root is to be found N, the current guess guess, and the tolerance level tolerance.
• Compute the next guess using the Newton’s formula next_guess = (guess + N/guess) / 2.
• Check if the difference between the current guess and the next guess is <= tolerance level tolerance using the abs() function. If the condition is satisfied, return the next guess.
• Otherwise, recursively call the findSqrt function with the new guess.
• Last print the result

Below is the implementation of the above approach: 

C++

#include <iostream>
#include <cmath>

using namespace std;

double findSqrt(double N, double guess, double tolerance)
{
    double next_guess = (guess + N/guess) / 2;
    if (abs(guess - next_guess) <= tolerance) {
        return next_guess;
    }
    else {
        return findSqrt(N, next_guess, tolerance);
    }
}

int main()
{
    double N=327, L=0.00001;
    double guess = N/2; // Initialize the guess to N/2
    double sqrt = findSqrt(N, guess, L);
    cout << sqrt << endl;
    return 0;
}

Java

public class Main {
    public static double findSqrt(double N, double guess,
                                  double tolerance)
    {
        double nextGuess = (guess + N / guess) / 2;
        if (Math.abs(guess - nextGuess) <= tolerance) {
            return nextGuess;
        }
        else {
            return findSqrt(N, nextGuess, tolerance);
        }
    }

    public static void main(String[] args)
    {
        double N = 327, L = 0.00001;
        double guess = N / 2; // Initialize the guess to N/2
        double sqrt = findSqrt(N, guess, L);
        System.out.printf("%.4f%n", sqrt);
    }
}

// This code is contributed by Samim Hossain Mondal.

Python3

def find_sqrt(N, guess, tolerance):
    next_guess = (guess + N / guess) / 2
    if abs(guess - next_guess) <= tolerance:
        return next_guess
    else:
        return find_sqrt(N, next_guess, tolerance)

if __name__ == "__main__":
    N = 327
    tolerance = 0.00001
    guess = N / 2  # Initialize the guess to N/2
    sqrt = find_sqrt(N, guess, tolerance)
    sqrt = round(sqrt, 4)
    print(sqrt)

C#

using System;

class Program
{
    static double FindSqrt(double N, double guess, double tolerance)
    {
        double nextGuess = (guess + N / guess) / 2;
        if (Math.Abs(guess - nextGuess) <= tolerance)
        {
            return nextGuess;
        }
        else
        {
            return FindSqrt(N, nextGuess, tolerance);
        }
    }

    static void Main()
    {
        double N = 327;
        double tolerance = 0.00001;
        double guess = N / 2; // Initialize the guess to N/2
        double sqrt = FindSqrt(N, guess, tolerance);
        sqrt = Math.Round(sqrt, 4);
        Console.WriteLine(sqrt);
    }
}

Javascript

function findSqrt(N, guess, tolerance) {
    let nextGuess = (guess + N / guess) / 2;
    if (Math.abs(guess - nextGuess) <= tolerance) {
        return nextGuess;
    } else {
        return findSqrt(N, nextGuess, tolerance);
    }
}

let N = 327;
let tolerance = 0.00001;
let guess = N / 2; // Initialize the guess to N/2
let sqrt = findSqrt(N, guess, tolerance);
console.log(sqrt.toFixed(4));

Output

18.0831

Time Complexity: O(log N), where N is the input number.

Auxiliary Space: O(log N)