CS-111: Floating Point Arithmetic

float.cpp

/*
 * Copyright 2019 (c) Ralph Wilson Aguilar
 * For CS-111
 * Do not copy or redistribute without permission.
 */
#include <iostream>
#include <cmath>

using namespace std;

int main ()
{
    float x, y;
    const float pi = 3.14159265359;
    

    cout << "Enter the value for x: ";
    cin >> x;

    /*
    * There's nothing really special about this, we just translated the algebraic equation from here.
    * So how this works is basically, we put x to the power of zero,
    * put pi in the power of zero, then solve the first part of the equation
    * which is x^2 / pi ^2 (x^2 + 1/2 or 0.5). Once we have a value of the first part,
    * We then move on the the second one, with the same principles we used in the first one.
    * This is very confusing by its own as the way we should put the parentheses to denote proper PEMDAS
    * really threw everyone off. Basically what we did here is to wrap the divisor in another sub-expression
    * so it solves properly.
    */
    y = ( (pow(x, 2.0)) / ( (pow(pi , 2.0)) * ( (pow(x, 2.0)) + 0.5 ) ) ) * ( 1 + (pow(x, 2.0)) / ( (pow(pi, 2.0))  * pow(( ((pow(x, 2.0)) - 0.5 )), 2.0) ) );

    cout << "y = " << y << endl;

    return 0;
}