Given a number, n, and an approximation for its square root, a closer approximat

Question

Given a number, n, and an approximation for its square root, a closer approximation to the actual square root can be obtained by using this formula: new approximation=((n/ previous approximation)+ previous approximation)/2 Using the information, write a C++ program with Visual Studio 2015 that promts the user for a number and an initial guess as its square root. Using this input data, your program should be able to calculate to 0.00001. (HINT: Stop the loop when the difference between the two approximations is less than 0.00001.)

Explanation / Answer

// C++ code
#include <iostream>
#include <string.h>
#include <fstream>
#include <math.h>       /* asin */
#include <limits.h>
#include <cmath>        // std::abs

using namespace std;

int main()
{
   double n, new_approximation, previous_approximation, approximation;

   cout << "Enter n: ";
   cin >> n;

   cout << "Enter approximation for its square root: ";
   cin >> previous_approximation;

   new_approximation=((n/ previous_approximation)+ previous_approximation)/2;

   while(abs(new_approximation- previous_approximation ) > 0.00001)
   {
       previous_approximation = new_approximation;
       new_approximation=((n/ previous_approximation)+ previous_approximation)/2;
   }

   cout << "Approximated Square root: " << new_approximation << endl;

   return 0;
}

/*
output:

Enter n: 1000
Enter approximation for its square root: 29
Approximated Square root: 31.6228

*/

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