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log1p, log1pf, log1pl

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.

Parameters

- -
arg - floating point value

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),

-

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");
-}

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

References

See also

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

- © cppreference.com
Licensed under the Creative Commons Attribution-ShareAlike Unported License v3.0.
- https://en.cppreference.com/w/c/numeric/math/log1p -

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