# Find root of a number using Newton’s method

• Difficulty Level : Hard
• Last Updated : 06 Feb, 2022

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:
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 ``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

 ``

Output:

`18.0831`

Time Complexity: O(log N)

Auxiliary Space: O(1)

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