log1p, log1pf, log1pl

float       log1pf( float arg );

double      log1p( double arg );

long double log1pl( long double arg );

#define log1p( arg )

Parameters

Return value

If no errors occur ln(1+arg) is returned.

If a domain error occurs, an implementation-defined value is returned (NaN where supported).

If a pole error 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.

Domain error occurs if arg is less than -1.

Pole error may occur if arg is -1.

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 and FE_DIVBYZERO is raised.
• If the argument is less than -1, NaN is returned and FE_INVALID is raised.
• 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.

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("log1p(0) = %f\n", log1p(0));
    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("log(1+1e-16) = %g, but log1p(1e-16) = %g\n",
           log(1+1e-16), log1p(1e-16));
    // special values
    printf("log1p(-0) = %f\n", log1p(-0.0));
    printf("log1p(+Inf) = %f\n", log1p(INFINITY));
    //error handling
    errno = 0; feclearexcept(FE_ALL_EXCEPT);
    printf("log1p(-1) = %f\n", log1p(-1));
    if(errno == ERANGE) perror("    errno == ERANGE");
    if(fetestexcept(FE_DIVBYZERO)) puts("    FE_DIVBYZERO raised");
}

log1p(0) = 0.000000
Interest earned in 2 days on $100, compounded daily at 1%
 on a 30/360 calendar = 0.005556
log(1+1e-16) = 0, but log1p(1e-16) = 1e-16
log1p(-0) = -0.000000
log1p(+Inf) = Inf
log1p(-1) = -Inf
    errno == ERANGE: Result too large
    FE_DIVBYZERO raised

References

• C11 standard (ISO/IEC 9899:2011):
• 7.12.6.9 The log1p functions (p: 245)
• 7.25 Type-generic math <tgmath.h> (p: 373-375)
• F.10.3.9 The log1p functions (p: 522)
• C99 standard (ISO/IEC 9899:1999):
• 7.12.6.9 The log1p functions (p: 226)
• 7.22 Type-generic math <tgmath.h> (p: 335-337)
• F.9.3.9 The log1p functions (p: 459)

See also