expm1, expm1f, expm1l
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Defined in header
<math.h>
float expm1f( float arg );
(1)
(since C99)
double expm1( double arg );
(2)
(since C99)
long double expm1l( long double arg );
(3)
(since C99)
Defined in header
<tgmath.h>
#define expm1( arg )
(4)
(since C99)
1-3) Computes the e (Euler's number,
2.7182818) raised to the given power arg, minus 1.0. This function is more accurate than the expression exp (arg)-1.0 if arg is close to zero.4) Type-generic macro: If
arg has type long double, expm1l is called. Otherwise, if arg has integer type or the type double, expm1 is called. Otherwise, expm1f is called.[edit] Parameters
arg
-
floating-point value
[edit] Return value
If no errors occur earg
-1 is returned.
If a range error due to overflow 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.
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
- 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.
For IEEE-compatible type double, overflow is guaranteed if 709.8 < arg.
[edit] Example
Run this code
#include <errno.h> #include <fenv.h> #include <float.h> #include <math.h> #include <stdio.h> // #pragma STDC FENV_ACCESS ON int main(void) { printf ("expm1(1) = %f\n", expm1(1)); printf ("Interest earned in 2 days on 100,ドル compounded daily at 1%%\n" " on a 30/360 calendar = %f\n", 100*expm1(2*log1p (0.01/360))); printf ("exp(1e-16)-1 = %g, but expm1(1e-16) = %g\n", exp (1e-16)-1, expm1(1e-16)); // special values printf ("expm1(-0) = %f\n", expm1(-0.0)); printf ("expm1(-Inf) = %f\n", expm1(-INFINITY)); //error handling errno = 0; feclearexcept (FE_ALL_EXCEPT ); printf ("expm1(710) = %f\n", expm1(710)); if (errno == ERANGE ) perror (" errno == ERANGE"); if (fetestexcept (FE_OVERFLOW )) puts (" FE_OVERFLOW raised"); }
Possible output:
expm1(1) = 1.718282 Interest earned in 2 days on 100,ドル compounded daily at 1% on a 30/360 calendar = 0.005556 exp(1e-16)-1 = 0, but expm1(1e-16) = 1e-16 expm1(-0) = -0.000000 expm1(-Inf) = -1.000000 expm1(710) = inf errno == ERANGE: Result too large FE_OVERFLOW raised
[edit] References
- C23 standard (ISO/IEC 9899:2024):
- 7.12.6.3 The expm1 functions (p: TBD)
- 7.25 Type-generic math <tgmath.h> (p: TBD)
- F.10.3.3 The expm1 functions (p: TBD)
- C17 standard (ISO/IEC 9899:2018):
- 7.12.6.3 The expm1 functions (p: 177)
- 7.25 Type-generic math <tgmath.h> (p: 272-273)
- F.10.3.3 The expm1 functions (p: 379)
- C11 standard (ISO/IEC 9899:2011):
- 7.12.6.3 The expm1 functions (p: 243)
- 7.25 Type-generic math <tgmath.h> (p: 373-375)
- F.10.3.3 The expm1 functions (p: 521)
- C99 standard (ISO/IEC 9899:1999):
- 7.12.6.3 The expm1 functions (p: 223-224)
- 7.22 Type-generic math <tgmath.h> (p: 335-337)
- F.9.3.3 The expm1 functions (p: 458)
[edit] See also
(C99)(C99)(C99)
(function) [edit]
C++ documentation for expm1