计算未输出预期答案

Question

I'm trying to work out the arctan of a number using the formula:

arctan(x) = x - x^3/3 + x^5/5 - x^7/7...

I have to calculate it to 20 decimal places. The answer should be 0.78539....

This is the code I have written, including some debugging statements. The problem is in the calculation I think but I just can't see it. Could someone point me in the right direction please?

EDIT : Can't use the atan function, has to be manually calculated using a double variable from user input.

#include <iomanip>
#include <cstdlib>
#include <iostream>
#include <cmath>
using namespace std;

int main(void)
{
 double x;
 int i;
 int j;
 int y=3;

  cout<<"Please enter the number you wish to calculate the arctan of:"<<endl;
  cin>>x;

   //Calculate arctan of this number
   cout<<x;
   cout<<"\n";
   cout<<y;
   cout<<"\n";

   cout<<"Start\n";

   x=x-(pow(x,y)/y);
   y=y+2;
   cout <<  setprecision (20) << x;
   cout<<"=x before loop\n";
   cout<<y;
   cout<<"=y before loop\n";

   for(i=0;i<9;i++)
    {
     x=x+(pow(x,y)/y);
      cout<<x;
      cout<<"=x1 in loop\n";
     y=y+2;
      cout<<y;
      cout<<"=y1 in loop\n";

     x-(pow(x,y)/y);
      cout<<x;
      cout<<"=x2 in loop\n";
     y=y+2; 
      cout<<y;
      cout<<"=y2 in loop\n";
    }
return 0;

}

Answer 1

Well, your x is changing! You probably want to use a different variable to store the value computed so far and the argument to your function. That said, don't expect to precise outputs because all those computations involve rounding.

Answer 2

This line:

 x-(pow(x,y)/y);

might have something to do with your problem.

Answer 3

I would strong advise you use the inbuilt atan function, it is more than likely been well optimised for you architecture, as well as being a standard function recognised by most C++ programmers.

#include <cmath>
#include <iostream>    

int main()
{
    double d;
    std::cout << "enter number" << std::endl;
    std::cin  >> d;
    std::cout << "atan of: " << d 
              << " is "      << std::atan(d) 
              << std::endl;
    return 0;
}

Answer 4

I agree with @Mystical. I don't think you're going to get 20 digits of precision out of a double. I think you need a long double (if that exists on your system) or, perhaps you need to implement your own big-num class...