std::expm1,std::expm1f,std::expm1l(3)

NAME
std::expm1,std::expm1f,std::expm1l - std::expm1,std::expm1f,std::expm1l

Synopsis
   Defined in header <cmath>
   float expm1 ( float num );
   double expm1 ( double num ); (until C++23)
   long double expm1 ( long double num );
   /* floating-point-type */ (since C++23)
               expm1 ( /* floating-point-type */ num ); (constexpr since C++26)
   float expm1f( float num ); (1) (2) (since C++11)
                        (constexpr since C++26)
   long double expm1l( long double num ); (3) (since C++11)
                        (constexpr since C++26)
   Additional overloads (since C++11)
   Defined in header <cmath>
   template< class Integer > (A) (constexpr since C++26)
   double expm1 ( Integer num );
   1-3) Computes the e (Euler's number, 2.7182818...) raised to the given power num,
   minus 1.0. This function is more accurate than the expression std::exp(num) - 1.0 if
   num is close to zero.
   The library provides overloads of std::expm1 for all cv-unqualified floating-point
   types as the type of the parameter.
   (since C++23)
   A) Additional overloads are provided for all integer types, which are (since C++11)
   treated as double.

Parameters
   num - floating-point or integer value

Return value
   If no errors occur enum
   -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 std::expm1 and std::log1p are useful for financial calculations, for
   example, when calculating small daily interest rates: (1+x)n
   -1 can be expressed as std::expm1(n * std::log1p(x)). These functions also simplify
   writing accurate inverse hyperbolic functions.
   For IEEE-compatible type double, overflow is guaranteed if 709.8 < num.
   The additional overloads are not required to be provided exactly as (A). They only
   need to be sufficient to ensure that for their argument num of integer type,
   std::expm1(num) has the same effect as std::expm1(static_cast<double>(num)).

Example
// Run this code
 #include <cerrno>
 #include <cfenv>
 #include <cmath>
 #include <cstring>
 #include <iostream>
 // #pragma STDC FENV_ACCESS ON
 int main()
 {
     std::cout << "expm1(1) = " << std::expm1(1) << '\n'
               << "Interest earned in 2 days on $100, compounded daily at 1%\n"
               << " on a 30/360 calendar = "
               << 100 * std::expm1(2 * std::log1p(0.01 / 360)) << '\n'
               << "exp(1e-16)-1 = " << std::exp(1e-16) - 1
               << ", but expm1(1e-16) = " << std::expm1(1e-16) << '\n';
     // special values
     std::cout << "expm1(-0) = " << std::expm1(-0.0) << '\n'
               << "expm1(-Inf) = " << std::expm1(-INFINITY) << '\n';
     // error handling
     errno = 0;
     std::feclearexcept(FE_ALL_EXCEPT);
     std::cout << "expm1(710) = " << std::expm1(710) << '\n';
     if (errno == ERANGE)
         std::cout << " errno == ERANGE: " << std::strerror(errno) << '\n';
     if (std::fetestexcept(FE_OVERFLOW))
         std::cout << " FE_OVERFLOW raised\n";
 }

Possible output:
 expm1(1) = 1.71828
 Interest earned in 2 days on $100, compounded daily at 1%
     on a 30/360 calendar = 0.00555563
 exp(1e-16)-1 = 0, but expm1(1e-16) = 1e-16
 expm1(-0) = -0
 expm1(-Inf) = -1
 expm1(710) = inf
     errno == ERANGE: Result too large
     FE_OVERFLOW raised

See also
   exp
   expf returns e raised to the given power (\({\small e^x}\)e^x)
   expl (function)
   (C++11)
   (C++11)
   exp2
   exp2f
   exp2l returns 2 raised to the given power (\({\small 2^x}\)2^x)
   (C++11) (function)
   (C++11)
   (C++11)
   log1p
   log1pf natural logarithm (to base e) of 1 plus the given number
   log1pl (\({\small\ln{(1+x)}}\)ln(1+x))
   (C++11) (function)
   (C++11)
   (C++11)
   C documentation for
   expm1