# N-th root of a number

Given two numbers N and A, find N-th root of A. In mathematics, Nth root of a number A is a real number that gives A, when we raise it to integer power N. These roots are used in Number Theory and other advanced branches of mathematics.
Examples:

```Input : A = 81
N = 4
Output : 3
3^4 = 81```
Recommended Practice

As this problem involves a real valued function A^(1/N) we can solve this using Newtonâ€™s method, which starts with an initial guess and iteratively shift towards the result.
We can derive a relation between two consecutive values of iteration using Newton’s method as follows,

```according to newtonâ€™s method
x(K+1) = x(K) â€“ f(x) / fâ€™(x)
here    f(x)  = x^(N) â€“ A
so    fâ€™(x) = N*x^(N - 1)
and     x(K) denoted the value of x at Kth iteration
putting the values and simplifying we get,
x(K + 1) = (1 / N) * ((N - 1) * x(K) + A / x(K) ^ (N - 1))```

Using above relation, we can solve the given problem. In below code we iterate over values of x, until difference between two consecutive values of x become lower than desired accuracy.
Below is the implementation of above approach:

## C++

 `// C++ program to calculate Nth root of a number``#include ``using` `namespace` `std;` `//  method returns Nth power of A``double` `nthRoot(``int` `A, ``int` `N)``{``    ``// initially guessing a random number between``    ``// 0 and 9``    ``double` `xPre = ``rand``() % 10;` `    ``//  smaller eps, denotes more accuracy``    ``double` `eps = 1e-3;` `    ``// initializing difference between two``    ``// roots by INT_MAX``    ``double` `delX = INT_MAX;` `    ``//  xK denotes current value of x``    ``double` `xK;` `    ``//  loop until we reach desired accuracy``    ``while` `(delX > eps)``    ``{``        ``//  calculating current value from previous``        ``// value by newton's method``        ``xK = ((N - 1.0) * xPre +``              ``(``double``)A/``pow``(xPre, N-1)) / (``double``)N;``        ``delX = ``abs``(xK - xPre);``        ``xPre = xK;``    ``}` `    ``return` `xK;``}` `//    Driver code to test above methods``int` `main()``{``    ``int` `N = 4;``    ``int` `A = 81;` `    ``double` `nthRootValue = nthRoot(A, N);``    ``cout << ``"Nth root is "` `<< nthRootValue << endl;` `    ``/*``        ``double Acalc = pow(nthRootValue, N);``        ``cout << "Error in difference of powers "``             ``<< abs(A - Acalc) << endl;``    ``*/` `    ``return` `0;``}`

## Java

 `// Java program to calculate Nth root of a number``class` `GFG``{``    ` `    ``// method returns Nth power of A``    ``static` `double` `nthRoot(``int` `A, ``int` `N)``    ``{``        ` `        ``// initially guessing a random number between``        ``// 0 and 9``        ``double` `xPre = Math.random() % ``10``;``    ` `        ``// smaller eps, denotes more accuracy``        ``double` `eps = ``0.001``;``    ` `        ``// initializing difference between two``        ``// roots by INT_MAX``        ``double` `delX = ``2147483647``;``    ` `        ``// xK denotes current value of x``        ``double` `xK = ``0.0``;``    ` `        ``// loop until we reach desired accuracy``        ``while` `(delX > eps)``        ``{``            ``// calculating current value from previous``            ``// value by newton's method``            ``xK = ((N - ``1.0``) * xPre +``            ``(``double``)A / Math.pow(xPre, N - ``1``)) / (``double``)N;``            ``delX = Math.abs(xK - xPre);``            ``xPre = xK;``        ``}``    ` `        ``return` `xK;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `N = ``4``;``        ``int` `A = ``81``;``    ` `        ``double` `nthRootValue = nthRoot(A, N);``        ``System.out.println(``"Nth root is "``        ``+ Math.round(nthRootValue*``1000.0``)/``1000.0``);``    ` `        ``/*``            ``double Acalc = pow(nthRootValue, N);``            ``cout << "Error in difference of powers "``                ``<< abs(A - Acalc) << endl;``        ``*/``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 program to calculate``# Nth root of a number``import` `math``import` `random` `# method returns Nth power of A``def` `nthRoot(A,N):` `    ``# initially guessing a random number between``    ``# 0 and 9``    ``xPre ``=` `random.randint(``1``,``101``) ``%` `10`` ` `    ``#  smaller eps, denotes more accuracy``    ``eps ``=` `0.001`` ` `    ``# initializing difference between two``    ``# roots by INT_MAX``    ``delX ``=` `2147483647`` ` `    ``#  xK denotes current value of x``    ``xK``=``0.0`` ` `    ``#  loop until we reach desired accuracy``    ``while` `(delX > eps):` `        ``# calculating current value from previous``        ``# value by newton's method``        ``xK ``=` `((N ``-` `1.0``) ``*` `xPre ``+``              ``A``/``pow``(xPre, N``-``1``)) ``/``N``        ``delX ``=` `abs``(xK ``-` `xPre)``        ``xPre ``=` `xK;``        ` `    ``return` `xK` `# Driver code``N ``=` `4``A ``=` `81``nthRootValue ``=` `nthRoot(A, N)` `print``(``"Nth root is "``, nthRootValue)` `## Acalc = pow(nthRootValue, N);``## print("Error in difference of powers ",``##             abs(A - Acalc))` `# This code is contributed``# by Anant Agarwal.`

## C#

 `// C# program to calculate Nth root of a number``using` `System;``class` `GFG``{``    ` `    ``// method returns Nth power of A``    ``static` `double` `nthRoot(``int` `A, ``int` `N)``    ``{``        ``Random rand = ``new` `Random();``        ``// initially guessing a random number between``        ``// 0 and 9``        ``double` `xPre = rand.Next(10);;``    ` `        ``// smaller eps, denotes more accuracy``        ``double` `eps = 0.001;``    ` `        ``// initializing difference between two``        ``// roots by INT_MAX``        ``double` `delX = 2147483647;``    ` `        ``// xK denotes current value of x``        ``double` `xK = 0.0;``    ` `        ``// loop until we reach desired accuracy``        ``while` `(delX > eps)``        ``{``            ``// calculating current value from previous``            ``// value by newton's method``            ``xK = ((N - 1.0) * xPre +``            ``(``double``)A / Math.Pow(xPre, N - 1)) / (``double``)N;``            ``delX = Math.Abs(xK - xPre);``            ``xPre = xK;``        ``}``    ` `        ``return` `xK;``    ``}``    ` `    ``// Driver code``    ``static` `void` `Main()``    ``{``        ``int` `N = 4;``        ``int` `A = 81;``    ` `        ``double` `nthRootValue = nthRoot(A, N);``        ``Console.WriteLine(``"Nth root is "``+Math.Round(nthRootValue*1000.0)/1000.0);``    ``}``}` `// This code is contributed by mits`

## PHP

 ` ``\$eps``)``    ``{``        ``// calculating current``        ``// value from previous``        ``// value by newton's method``        ``\$xK` `= ((int)(``\$N` `- 1.0) * ``                     ``\$xPre` `+ ``\$A` `/ ``                     ``(int)pow(``\$xPre``, ``                              ``\$N` `- 1)) / ``\$N``;``        ``\$delX` `= ``abs``(``\$xK` `- ``\$xPre``);``        ``\$xPre` `= ``\$xK``;``    ``}` `    ``return` `floor``(``\$xK``);``}` `// Driver code ``\$N` `= 4;``\$A` `= 81;` `\$nthRootValue` `= nthRoot(``\$A``, ``\$N``);``echo` `"Nth root is "` `, ``      ``\$nthRootValue` `,``"\n"``;` `// This code is contributed by akt_mit``?>`

## Javascript

 ``

Output
```Nth root is 3
```

Time Complexity: O(log(eps)), where eps is the desired accuracy.
Space Complexity: O(1)

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