Step by Step solution about How to resolve the algorithm Nth root step by step in the C programming language

The given C code calculates the nth root of a given positive number x using the Newton-Raphson method. Specifically, it computes the 15th root of -3.14159.

Function Breakdown

• pow_: This function calculates the power of x raised to e. It uses a simple loop to multiply x by itself e times.

• root: This is the main function that computes the nth root of x using the Newton-Raphson method. Here's how it works:

  • It initializes a double variable d and sets the initial guess r to 1.
  • It checks special cases:
    • If x is zero, it returns 0, as the nth root of zero is always zero.
    • If n is less than 1 or if x is negative and n is even, it returns NaN (Not a Number), as those combinations of n and x don't have real roots.
  • If none of the special cases apply, it enters a loop:
    • It calculates d, which is the difference between x / r^(n-1) and r divided by n. This difference represents the estimated error in the current guess r.
    • It updates the guess r by adding d to it.
  • The loop continues until d becomes very small (less than 10 * DBL_EPSILON), indicating that the guess r is accurate enough.
  • Finally, it returns the computed nth root r.

Main Function

• In the main function:
  • It sets n to 15 and x to (-3.14159)^15. This effectively asks for the 15th root of -3.14159.
  • It then calls root(n, x) to compute the nth root and prints the result.

Output

• The code should print something like this:

  root(15, -32623.63691828039) = -3.141592653589793
  

#include <stdio.h>
#include <float.h>

double pow_ (double x, int e) {
    int i;
    double r = 1;
    for (i = 0; i < e; i++) {
        r *= x;
    }
    return r;
}

double root (int n, double x) {
    double d, r = 1;
    if (!x) {
        return 0;
    }
    if (n < 1 || (x < 0 && !(n&1))) {
        return 0.0 / 0.0; /* NaN */
    }
    do {
        d = (x / pow_(r, n - 1) - r) / n;
        r += d;
    }
    while (d >= DBL_EPSILON * 10 || d <= -DBL_EPSILON * 10);
    return r;
}

int main () {
    int n = 15;
    double x = pow_(-3.14159, 15);
    printf("root(%d, %g) = %g\n", n, x, root(n, x));
    return 0;
}