If x is very small, directly computing exp(x) - 1 can be inaccurate. Numerical libraries often include a function expm1 to compute this function. The need for such a function is easiest to see when x is extremely small. If x is small enough, exp(x) = 1 in machine arithmetic and so exp(x) - 1 returns 0 even though the correct result is positive. All precision is lost. If x is small but not so extremely small, direct computation still loses precision, just not as much.
We can avoid the loss of precision by using a Taylor series to evaluate exp(x):
exp(x) = 1 + x + x2/2 + x36 + …
If |x| < 10-5, the error in approximating exp(x) – 1 by x + x2/2 is on the order of 10-15 and so the relative error is on the order of 10-10 or better. If we compute exp(10-5) – 1 directly, the absolute error is about 10-16 and so the relative error is about 10-11. So by using the two-term Taylor approximation for |x| less than 10-5 and the direct method for |x| larger than 10-5, we obtain at least 10 significant figures for all inputs.
#include <cmath>
#include <iostream>
// Compute exp(x) - 1 without loss of precision for small values of x.
double expm1(double x)
{
if (fabs(x) < 1e-5)
return x + 0.5*x*x;
else
return exp(x) - 1.0;
}
void testExpm1()
{
// Select a few input values
double x[] =
{
-1,
0.0,
1e-5 - 1e-8,
1e-5 + 1e-8,
0.5
};
// Output computed by Mathematica
// y = Exp[x] - 1
double y[] =
{
-0.632120558828558,
0.0,
0.000009990049900216168,
0.00001001005010021717,
0.6487212707001282
};
int numTests = sizeof(x)/sizeof(double);
double maxError = 0.0;
for (int i = 0; i < numTests; ++i)
{
double error = fabs(y[i] - expm1(x[i]));
if (error> maxError)
maxError = error;
}
std::cout << "Maximum error: " << maxError << "\n";
}
This code is in the public domain. Do whatever you want to with it, no strings attached.