Question

help with C++ please

Substituting these formulas, the Newton-Raphson update rule for computing the n' root of any value K is: 2+1 = - na"-T Write a program that includes a user-defined function named root that will calculate and return the nth root of any value using the Newton-Raphson method described above: double rootN(double x, int n); The function takes two arguments: a (type double) value > 0 and an integer root > 1. This function should work exactly as your Babylonian square root, but using the more generalized update rule. Do not use any math library functions in your solution. [Hint: it will be very helpful if you also construct a separate function to computex"] Your main program should take as input positive values for K and n, then call the rootN function to determine and display the root of the input value. Include a continuation loop that will continue this process as long as the user wishes. Examples: enter value and root: 92 the root is: 3.00002 continue? (y/n): y enter value and root: 93 the root is: 2.08008 continue? (y/n): y enter value and root: 94 the root is: 1.73207 continue? (y/n): n

Answer 1

`Hey,

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#include <iostream>

using namespace std;
double power(double x,int n)
{
if(n==0)
return 1;
return x*power(x,n-1);
}
double rootN(double x,int n)
{
double x0=x/2.0,t;
for(int i=0;i<20;i++)
{
x0=1.0/(double)n*((double)(n-1)*x0+x/power(x0,n-1));
}
return x0;
}
int main()
{
char ch='y';
while(ch=='y')
{
cout<<"enter value and root: ";
double x;
int n;
cin>>x>>n;
cout<<"root is "<<rootN(x,n)<<endl;
cout<<"continue? (y/n): ";
cin>>ch;
}
return 0;
}

Kindly revert for any queries

Thanks.