How to shorten my output? Currently getting 'inf'

Question

I'm writing a program to calculate the 'Taylor Series' and I'm getting 'inf' as my output. I'm not done with the program and everything isn't working properly yet, but I want to be able to see my output. Is there a way to shorten the output or make it visible? Thanks for the help.

#include <iostream>
#include <math.h>
using namespace std;

long double taylorSeries(long double input, int degree);
long int factorial(int input);
long double derivative(long double input);

int main(int argc, const char * argv[])
{

    long double taylorInput = cos(0);    

    cout << taylorSeries(taylorInput, 3) << endl;

    return 0;
}

// taylor series function
long double taylorSeries(long double input, int degree){

    long double i = 0;
    long double accumulation = 0;


    while (i < degree) {
        derivative(input);
        accumulation += (pow(input, i))/factorial(i); 
        i++;
    }

    return accumulation;
}

// function to calculate factorial
long int factorial(int input){

    long int factorial = 0;

    if (input == 1) {
        factorial = 0;
    }

    while (input > 0) {
        if (factorial == 0) {
            factorial = input * (input - 1);
            input -= 2;
        }

        else if (input > 0) {
            factorial *= input;
            input--;
        }


    }
    return factorial;
}

long double derivative(long double input){
     long double derivativeResult = 0;

    derivativeResult = (((input + .001) - (input)) / .001);

    return derivativeResult;
}

Answer 1

In accumulation += (pow(input, i))/factorial(i); you are dividing by 0.

On the factorial function you are missing the case for 0, factorial of 0 is 1.

EDIT: Factorial of 1 is also not 0, you should take a look at that factorial function first before going forward.

Answer 2

Your factorial function is a bit screwy. Try this:

long factorial( long input ) {
    long output = 1;
    for( long i = 2; i <= input; ++ i ) {
        output *= i;
    }
    return output;
}

Answer 3

The double value inf is a special value to avoid losing information about an overflowing intermediate calculation when doing a subsequent math operation. Once a chain of operations produces inf it will "stick" rather than producing a valid number that is an incorrect value.

A common way to get inf is to divide by zero in an environment where that's not a floating point exception. So x = 1.0 / 0.0 ... (any operations here) will be inf because of the initial inf. You have to find the problem calculation and eliminate it before you can see the rest of your result.