Question

9. If we define f(r)= r"-c, then solving f(x) = 0 is equivalent to finding e/n for n > 0. Given an approximate value of r, the Newton-Raphson method consists of calculating a approximation rnew, by using new f(x) Inew r In our case, this reduces to: (n-1)r c Inew= nrn-1 I 2 Write & program to find cn for positive n and c. The program should prompt for n (as a positive double) and c (as a positive double). positive integer), the initial value of r (as a Incorrect entries should be trapped and a prompt for a new value issued. Use the above formula for Inew to iterate towards an approximate value of r, and hence, c. The program should terminate if the number of iterations exceeds a reasonable limit. difference between two successive iterations is in some sense small. For instance, you might or the try a limit of twenty iterations and a difference between two successive iterations that is close to machine precision. Try different values of n and c, and list the approximation to each root.

we are currently using C++ through reply.it thanks for your help.

Answer 1

NOTE:: The following code takes input from the user and calculates the root of the aforesaid mentioned equation.

NOTE:: Lines starting with // are actually comments, will be ignored by the compiler

//including the iostream for basic input and output operations
#include <iostream>
//including math library for the mathematical calculation like taking absolute
#include <cmath>
//using default std namespace
#define MAX_COUNT 20
//defined the maximum number of iteration. Please change it to your need
#define EPSILON 0.001
//defined the lowest precision value. Please change it to suit your need
using namespace std;

//function to compute the new value of x given the previous ones

double compute_new_x(int n, double x_old, double c){
    //implementing the formula to calculate the new value of x
    double x_new=(((n-1)*x_old)/n)+(c/(n*(pow(x_old,n-1))));
    //returning the calculated value
    return x_new;
}

//working the main function around

int main() {
   //taking the input for the value of n
   int n;
   double x_ini,c;
   //taking input from user as n
   cout<<"Please enter the value of n"<<endl;
   cin>>n;
   //Asking the value of x
   cout<<"Please enter the initial value of x"<<endl;
   cin>>x_ini;
   //Asking the value of c
   cout<<"Please enter the value of c"<<endl;
   cin>>c;
   //checking whether all the inputs are positive if not then asking again
   while(n<0||x_ini<0||c<0){
        cout<<"You entered something negativive:: Not Allowed";
        cout<<"Please enter the value of n"<<endl;
        cin>>n;
        cout<<"Please enter the initial value of x"<<endl;
        cin>>x_ini;
        cout<<"Please enter the value of c"<<endl;
        cin>>c;
   }
   //computing the value of new root for the first time
   double x_new=compute_new_x(n,x_ini,c);
   //Printing the value of root at first iteration
   cout<<"Value at iteration 1 "<<x_new<<endl;
   //Starting the root convergence
   int count=0;
   //Looping till the max count or reach the precision
   while((abs(x_ini-x_new)>EPSILON)&&count<MAX_COUNT){
        x_ini=x_new;
        x_new=compute_new_x(n,x_ini,c);
        cout<<"Value at iteration "<<count+2<<" "<<x_new<<endl;
        count++;
   }
   //printing so that you can cross check the result
   double result=(1/(double)n);
   //typecasted n to double to put it in pow function
   cout<<"Value of c^(1/n) "<<pow(c,result)<<endl;
   return 0;
}

The above code is perfectly running. I tested it at the codechef online IDE.

If you face any issues please revert back, I will be happy to assist you further.

Hope this helps!!!

Thanks and regards