Question

Not in C++, only C code please
In class, we have studied the bisection method for finding a root of an equation. Another method for finding a root, Newton's method, usually converges to a solution even faster than the bisection method, if it converges at all. Newton's method starts with an initial guess for a root, xo, and then generates successive approximate roots X1, X2, .... Xj, Xj+1, .... using the iterative formula: f(x;) X;+1 = x; - f'(x;) Where f'(x;) is the derivative of function f evaluated at x = x;. The formula generates a new guess, Xj+1, from a previous one, xj. Sometimes Newton's method will fail to converge to a root. In this case, the program should terminate after many trials, perhaps 100. / y = f(x) Root хэ х2 X1 Xo The figure above shows the geometric interpolation of Newton's method where Xo, X1, and X2 represent successive guesses for the root. At each point X;, the derivative, f'(x;), is the slope of the tangent to the curve, f(x). The next guess for the root, Xj+1, is the point at which the tangent crosses the x-axis.

Answer 1

C code:

#include <stdio.h> 
#include <stdlib.h>
#include <math.h>
#include <limits.h>

//  this will find the n th root of a number x
double find_nthRoot(int x, int n){ 
    //count iteration
        int count=0;
        // declare the accuracy
        double accuracy=0.000001;
        
        // intially guess
        double prev = pow(x, n)/2;
    
        // it stores the difference between two roots
    double diff = INT_MAX; 
  
    //  it stores current value of x^n
    double current_x; 
  
    //  repeat until the accuracy achieved
    while (diff > accuracy && count<100){ 
        //  calculating current value using previous 
        current_x = ((n - 1.0) * prev + (double)x/pow(prev, n-1)) / (double)n; 
        diff = current_x - prev;
                // make positive
                if(diff<0)
                        diff=-diff; 
        prev = current_x; 
        count++;
    } 
    return current_x; 
} 
  
// Driver code
int main(){ 
    // declare variables
        int x, n;
        double result;
        
        //take input
        printf("Enter the value of x in nthtroot(x): ");
        scanf("%d",&x);
        printf("Enter the value of n in nthtroot(x): ");
        scanf("%d",&n);
    
    // call the function
        result = find_nthRoot(x, n); 
        
        //display the result
    printf("%d root of %d is %lf",n,x,result); 
  
    return 0; 
}

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Output:

Case 1:

Enter the value of x in nthtroot(x): 3
Enter the value of n in nthtroot(x): 2
2 root of 3 is 1.732051

Case 2:

Enter the value of x in nthtroot(x): 3
Enter the value of n in nthtroot(x): 10
10 root of 3 is 1.116123


Case 3:

Enter the value of x in nthtroot(x): 2
Enter the value of n in nthtroot(x): 3
3 root of 2 is 1.259921

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