C++ – dealing with infinitesimal numbers

March 3, 2022

I need to find some way to deal with infinitesimial double values.
For example:

exp(-0.00000000000000000000000000000100000000000000000003)= 0.99999999999999999999999999999899999999999999999997

But exp function produce result = 1.000000000000000000000000000000

So my first thought was to make my own exp function. Unfortunately I am getting same output.


double my_exp(double x)
{
    bool minus = x < 0;
    x = abs(x);
    double exp = (double)1 + x;
    double temp = x;
    for (int i = 2; i < 100000; i++)
    {
        temp *= x / (double)i;
        exp = exp + temp;
    }
    return minus ? exp : (double)1 / exp;
}

I found that issue is when such small numbers like 1.00000000000000000003e-030 doesn’t work well when we try to subtract it, neither both if we subtracting or adding such a small number the result always is equal to 1.

Have U any idea how to manage with this?

>Solution :

Try using std::expm1

Computes the e (Euler’s number, 2.7182818) raised to the given power
arg, minus 1.0. This function is more accurate than the expression
std::exp(arg)-1.0 if arg is close to zero.

#include <iostream>
#include <cmath>

int main()
{
    std::cout << "expm1(-0.00000000000000000000000000000100000000000000000003) = " << std::expm1(-0.00000000000000000000000000000100000000000000000003) << '\n';
}

Run the example in the below source by changing the arguments to your very small numbers.

Source: https://en.cppreference.com/w/cpp/numeric/math/expm1