How to calculate exponential of a function

[hasan_researc]

I have to calculate the exponetial of a large (on the order of 10^5) negative number. I tried using exp(), the exponential function from the cmath library, but I get an answer of 0. So, I tried to use the power series for the exponential function. For numbers from -1 to -14, I get answers which are accurate (within the percentage error set in the while statement). But for any number above -14, the answer diverges from the true value. For a number as small as 10^-5, the answer is a large positive number (which is nonsenical).

Please help me understand what's wrong with the code, if anything, and how it can be improved. (Or is there another better way to calculate the exp of a large -ve number?)

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
	double t = - 16;  // t is the argument of the exponential function.
	double fsum = 1.0; // fsum is the cumulative sum in the calc of the answer.
	double fbefore = fsum; // to be used for the while statement
	int n = 1;             // n is the denominator of the power series
	double comparing;      // to be used in the while statement
	int iterations = 0;     
	double term = 1;        // each additional term in the series.
	do 
	{ 
		iterations = iterations + 1;
		cout << iterations << endl;
		term = term * ( t/n );
		fsum = fsum + term;
		n = n + 1;
		double fafter = fsum;
		comparing = (fbefore - fafter)/fbefore;
		fbefore = fafter;
		cout << fsum << endl;
	}
	while ( abs(comparing) > 0.0000000001);

 

[Coto]

It seems like this is just a rounding problem.

Why not try calculating 1/e^x , where x is very small (i.e. -16). Taking n steps in the sum you would obtain

\frac{1}{e^x} = \frac{1}{x^n} \cdot \left( 1/\left[\frac{1}{x^n} + \frac{1}{x^{n-1}} + ... + 1\right]\right). The idea being that \frac{1}{x^n} is very small for sufficiently large n, and so has little effect on\left[\frac{1}{x^n} + \frac{1}{x^{n-1}} + ... + 1\right].

 

[zzzoak]

For example you want to find
y=e^{-16}
lg(y)=-16lg(e)=-16*0.4343=-6.949=-7+0.051
y=10^{-7}10^{0.051}=1.125*10^{-7}
You may calculate with any accuracy.

 

[hasan_researc]

I understand that very well. Thank you! But I am wondering whether it would be possible to calculate the exponent of -10^5.