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expm1, expm1f, expm1l
Defined in header <math.h> |
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(1) | (since C99) |
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(2) | (since C99) |
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(3) | (since C99) |
Defined in header <tgmath.h> |
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(4) | (since C99) |
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.
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
- 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)
See also
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(C99)(C99)
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computes e raised to the given power (\({\small e^x}\)ex) (function) |
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(C99)(C99)(C99)
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computes 2 raised to the given power (\({\small 2^x}\)2x) (function) |
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(C99)(C99)(C99)
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computes natural (base-e) logarithm of 1 plus the given number (\({\small \ln{(1+x)} }\)ln(1+x)) (function) |
C++ documentation for expm1 |
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