expm1, expm1f, expm1l

Parameters

Return value

If no errors occur earg
-1 is returned.

If a range error due to overflow occurs, +HUGE_VAL, +HUGE_VALF, or +HUGE_VALL is returned.

If a range error occurs due to underflow, the correct result (after rounding) is returned.

Error handling

Errors are reported as specified in math_errhandling.

If the implementation supports IEEE floating-point arithmetic (IEC 60559),

• If the argument is ±0, it is returned, unmodified
• If the argument is -∞, -1 is returned
• If the argument is +∞, +∞ is returned
• If the argument is NaN, NaN is returned

Notes

The functions expm1 and log1p are useful for financial calculations, for example, when calculating small daily interest rates: (1+x)n
-1 can be expressed as expm1(n * log1p(x)). These functions also simplify writing accurate inverse hyperbolic functions.

For IEEE-compatible type double, overflow is guaranteed if 709.8 < arg.

Example

#include <stdio.h>
#include <math.h>
#include <float.h>
#include <errno.h>
#include <fenv.h>
#pragma STDC FENV_ACCESS ON
int main(void)
{
    printf("expm1(1) = %f\n", expm1(1));
    printf("Interest earned in 2 days on $100, compounded daily at 1%%\n"
           " on a 30/360 calendar = %f\n",
           100*expm1(2*log1p(0.01/360)));
    printf("exp(1e-16)-1 = %g, but expm1(1e-16) = %g\n",
           exp(1e-16)-1, expm1(1e-16));
    // special values
    printf("expm1(-0) = %f\n", expm1(-0.0));
    printf("expm1(-Inf) = %f\n", expm1(-INFINITY));
    //error handling
    errno = 0; feclearexcept(FE_ALL_EXCEPT);
    printf("expm1(710) = %f\n", expm1(710));
    if(errno == ERANGE) perror("    errno == ERANGE");
    if(fetestexcept(FE_OVERFLOW)) puts("    FE_OVERFLOW raised");
}

expm1(1) = 1.718282
Interest earned in 2 days on $100, compounded daily at 1%
 on a 30/360 calendar = 0.005556
exp(1e-16)-1 = 0, but expm1(1e-16) = 1e-16
expm1(-0) = -0.000000
expm1(-Inf) = -1.000000
expm1(710) = inf
    errno == ERANGE: Result too large
    FE_OVERFLOW raised

References