float log1pf( float arg );
(1) (since C99)double log1p( double arg );
(2) (since C99)long double log1pl( long double arg );
(3) (since C99)#define log1p( arg )
(4) (since C99) 1-3)Computes the natural (base
e) logarithm of
1 + arg. This function is more precise than the expression
log(1 + arg)if
argis close to zero.
4) Type-generic macro: If arg has type long double, log1pl
is called. Otherwise, if arg has integer type or the type double, log1p
is called. Otherwise, log1pf
is called.
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.
[edit] Error handlingErrors 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),
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.
#include <errno.h> #include <fenv.h> #include <float.h> #include <math.h> #include <stdio.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"); }
Possible output:
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[edit] References
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