log1p, log1pf, log1pl - cppreference.com (original) (raw)

Defined in header <math.h>
float log1pf( float arg ); (1) (since C99)
double log1p( double arg ); (2) (since C99)
long double log1pl( long double arg ); (3) (since C99)
Defined in header <tgmath.h>
#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 arg is close to zero.

  1. 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.

Contents

[edit] Parameters

arg - floating-point value

[edit] 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.

[edit] 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),

[edit] 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.

[edit] Example

#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

[edit] See also