/C

# expm1, expm1f, expm1l

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.

### 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

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
• 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)

 expexpfexpl (C99)(C99) computes e raised to the given power ($${\small e^x}$$ex) (function) exp2exp2fexp2l (C99)(C99)(C99) computes 2 raised to the given power ($${\small 2^x}$$2x) (function) log1plog1pflog1pl (C99)(C99)(C99) computes natural (base-e) logarithm of 1 plus the given number ($${\small \ln{(1+x)} }$$ln(1+x)) (function) C++ documentation for expm1