expm1, expm1f, expm1l
From cppreference.com
| Defined in header <math.h>
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||
| 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
Run this code
#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 raisedReferences
- 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
| |