log1p, log1pf, log1pl

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Common mathematical functions
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log1p
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Classification
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Macro constants
 
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.
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.

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),

  • 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

[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 <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 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 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] See also

(C99)(C99)
computes natural (base-e) logarithm (ln(x))
(function)
(C99)(C99)
computes common (base-10) logarithm (log10(x))
(function)
(C99)(C99)(C99)
computes base-2 logarithm (log2(x))
(function)
(C99)(C99)(C99)
computes e raised to the given power, minus one (ex-1)
(function)
C++ documentation for log1p