#include<iostream>
using namespace std;

namespace proper
{
double inverse(double x)
{
    return 1.0/x;
}// inverse

}

namespace quick
{
double inverse(double x)
{
    return 2-x; // why?!
}// inverse

}

int main()
{

    cout << proper::inverse(1.005) << endl;

    cout << quick::inverse(1.005) << endl;

    return(0);
}// main()


/* 
How is 1/x ~= 2-x?

(1+dx)*(1-dx) = 1-(dx*dx)
For SMALL dx, (dx*dx)~0, can be omitted.
So
(1+dx)*(1-dx) ~ 1, which means

1/(1+dx) ~ (1-dx)

How would we return (1-dx), in terms of our input x (which is 1+dx)?
x = 1+dx [by defn]
dx= x-1 (eg. 1.005-1, ie. .005)
1-dx = 1-(x-1) = 2-x, this is what we return :)

Eg. 1/1.004 ~= 2-1.004, ie. 0.996 [same as 1-0.004]
*/