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 std::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.Parameters
arg | - | floating point value |
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.
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
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.
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("expm1(1) = %f\n", expm1(1)); 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("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, 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
References
- 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)
See also
(C99)(C99) | computes e raised to the given power (ex) (function) |
(C99)(C99)(C99) | computes 2 raised to the given power (2x) (function) |
(C99)(C99)(C99) | computes natural (base-e) logarithm of 1 plus the given number (ln(1+x)) (function) |
C++ documentation for expm1 |
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