Newton's method

[luuska]

Hello. This is my first C++ code (and my first post).

I've created a code, that calculates the root of function f(x) = sin(x). I use 3 as x_0. Here's the output:

Insert the number of iterations. 
5
Starting iteration for 5 rounds. Using xOld = 3
3.1425465430742778316641761193750426173210144042969
3.1415926533004769893864249752368777990341186523438
3.141592653589793115997963468544185161590576171875
3.141592653589793115997963468544185161590576171875
3.141592653589793115997963468544185161590576171875
 

So the digits don't change after the third iteration. The code goes like this:

#include <iostream>
#include <stdlib.h>
#include <math.h>
using namespace std;
 
int main()
{
        // double tol = 1.0E-300;
        int nMax;
        int n;
        double xOld;
        double xNew;
 
        cout << "Insert the number of iterations. \n";
        cin >> nMax;
        cout << "Starting iteration for " << nMax << " rounds. Using xOld = 3" << endl;
        xOld = 3;
        cout.precision(50);
 
        for( n = 1; n <= nMax ; n++ ) 
        {
                xNew = xOld - tan(xOld);
                cout << xNew << endl;
                        // if( fabs((xNew - xOld) / xOld ) < tol ) break;
                xOld = xNew;
        }
 
}
 

I've commented the tolerancy parts off.

If anyone knows what the problem is, do tell. Also improvements are welcome. Thanks in advance.

[Banfa]

You have not actually asked a question or stated a problem or you could have stated the behaviour you were expecting as well as what actually happened. If you were a customer of mine I would have told you to go away and come back when you had a proper description of the problem.

You have stated this behaviour of your code

So the digits don't change after the third iteration. The code goes like this:

Which I assume is what you are querying, you were expecting the digits to continue changing?

OK 2 points to bear in mind,

1. there is no point setting an output precision of 50 decimal places, a double is only accurate to 15 decimal places.

2. Since the result you are getting mimics PI to 15 decimal places I would guess that it has found the correct answer in 3 iterations (with-in the limits of the precision of a double).



Basically your code is working correctly, if you want more precision then on some platforms the type long double gives a few extra places of precision (but still not 50).