std::erfc,std::erfcf,std::erfcl(3)

NAME
std::erfc,std::erfcf,std::erfcl - std::erfc,std::erfcf,std::erfcl

Synopsis
   Defined in header <cmath>
   float erfc ( float num );
   double erfc ( double num ); (until C++23)
   long double erfc ( long double num );
   /* floating-point-type */ (since C++23)
               erfc ( /* floating-point-type */ num ); (constexpr since C++26)
   float erfcf( float num ); (1) (2) (since C++11)
                        (constexpr since C++26)
   long double erfcl( 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 erfc ( Integer num );
   1-3) Computes the complementary error function of num, that is 1.0 - std::erf(num),
   but without loss of precision for large num.
   The library provides overloads of std::erfc 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, value of the complementary error function of num, that is
   \(\frac{2}{\sqrt{\pi} }\int_{num}^{\infty}{e^{-{t^2} }\mathsf{d}t}\)
   2
   √
   π
   ∫∞
   nume^-t2
   dt or \({\small 1-\operatorname{erf}(num)}\)1-erf(num), 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 is returned.
     * If the argument is -∞, 2 is returned.
     * If the argument is NaN, NaN is returned.

Notes
   For the IEEE-compatible type double, underflow is guaranteed if num > 26.55.
   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::erfc(num) has the same effect as std::erfc(static_cast<double>(num)).

Example
// Run this code
 #include <cmath>
 #include <iomanip>
 #include <iostream>
 double normalCDF(double x) // Phi(-∞, x) aka N(x)
 {
     return std::erfc(-x / std::sqrt(2)) / 2;
 }
 int main()
 {
     std::cout << "normal cumulative distribution function:\n"
               << std::fixed << std::setprecision(2);
     for (double n = 0; n < 1; n += 0.1)
         std::cout << "normalCDF(" << n << ") = " << 100 * normalCDF(n) << "%\n";
     std::cout << "special values:\n"
               << "erfc(-Inf) = " << std::erfc(-INFINITY) << '\n'
               << "erfc(Inf) = " << std::erfc(INFINITY) << '\n';
 }

Output:
 normal cumulative distribution function:
 normalCDF(0.00) = 50.00%
 normalCDF(0.10) = 53.98%
 normalCDF(0.20) = 57.93%
 normalCDF(0.30) = 61.79%
 normalCDF(0.40) = 65.54%
 normalCDF(0.50) = 69.15%
 normalCDF(0.60) = 72.57%
 normalCDF(0.70) = 75.80%
 normalCDF(0.80) = 78.81%
 normalCDF(0.90) = 81.59%
 normalCDF(1.00) = 84.13%
 special values:
 erfc(-Inf) = 2.00
 erfc(Inf) = 0.00

See also
   erf
   erff
   erfl error function
   (C++11) (function)
   (C++11)
   (C++11)
   C documentation for
   erfc

External links
   Weisstein, Eric W. "Erfc." From MathWorld — A Wolfram Web Resource.