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Showing content from https://en.cppreference.com/w/cpp/language/../algorithm/../../cpp/../c/numeric/math/expm1.html below:

expm1, expm1f, expm1l - cppreference.com

float       expm1f( float arg );

double      expm1( double arg );

long double expm1l( long double arg );

#define expm1( arg )

Computes the

(Euler's number,

) raised to the given power

, minus

. This function is more accurate than the expression

if

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.

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.

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.

#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, 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

• 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)

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