Question

Hello !!


Make an algorithm that calculates the nth root of a number in C language.

Example :

$\mathsf{\sqrt[n]{x}}$

Thank you !​

Answer 1

Answer:

Following code also gives a good approximation:

1. #include<stdio.h>
2. #include<math.h>
3. int main(){
4. double b=0.0;
5. double a=0.001;
6. double r=0;
7. printf("Type the number and root value (e.g.2 for square root)\n");
8. scanf("%lf %lf",&b,&r);

thank you for your opportunity

Explanation:

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Answer 2

Answer:

Hey mate here is your answer...

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.

Below is the implementation of above approach:

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;

hope it's helpful for you