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.

Refer Wiki page for more information.

Examples:

Input : A = 81 N = 4 Output : 3 3^4 = 81

As this problem involves a real valued function A^(1/N) we can solve this using Newton’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 newton’s method x(K+1) = x(K) – f(x) / f’(x) here f(x) = x^(N) – A so f’(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.

## C++

// C++ program to calculate Nth root of a number #include <bits/stdc++.h> using namespace std; // method returns Nth power of A double nthRoot(int A, int N) { // intially 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 untill 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) { // intially 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 untill 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 untill 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.

Output:

Nth root is 3

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