erfc, erfcf, erfcl
From cppreference.com
| Defined in header <math.h>
|
||
float erfcf( float arg );
|
(1) | (since C99) |
double erfc( double arg );
|
(2) | (since C99) |
long double erfcl( long double arg );
|
(3) | (since C99) |
| Defined in header <tgmath.h>
|
||
#define erfc( arg )
|
(4) | (since C99) |
1-3) Computes the complementary error function of
arg, that is 1.0 - erf(arg), but without loss of precision for large arg.4) Type-generic macro: If
arg has type long double, erfcl is called. Otherwise, if arg has integer type or the type double, erfc is called. Otherwise, erfcf is called.Parameters
| arg | - | floating-point value |
Return value
If no errors occur, value of the complementary error function of arg, that is {{mathjax-or|\(\frac{2}{\sqrt{\pi} }\int_{arg}^{\infty}{e^{-{t^2} }\mathsf{d}t}\)|
| 2 |
| √π |
∫∞
arge-t2
dt}} or 1-erf(arg), 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 arg > 26.55.
Example
Run this code
#include <math.h>
#include <stdio.h>
double normalCDF(double x) // Phi(-∞, x) aka N(x)
{
return erfc(-x / sqrt(2)) / 2;
}
int main(void)
{
puts("normal cumulative distribution function:");
for (double n = 0; n < 1; n += 0.1)
printf("normalCDF(%.2f) %5.2f%%\n", n, 100 * normalCDF(n));
printf("special values:\n"
"erfc(-Inf) = %f\n"
"erfc(Inf) = %f\n",
erfc(-INFINITY),
erfc(INFINITY));
}
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) = 2.000000
erfc(Inf) = 0.000000
References
- C23 standard (ISO/IEC 9899:2024):
- 7.12.8.2 The erfc functions (p: 249-250)
- 7.25 Type-generic math <tgmath.h> (p: 373-375)
- F.10.5.2 The erfc functions (p: 525)
- C17 standard (ISO/IEC 9899:2018):
- 7.12.8.2 The erfc functions (p: 249-250)
- 7.25 Type-generic math <tgmath.h> (p: 373-375)
- F.10.5.2 The erfc functions (p: 525)
- C11 standard (ISO/IEC 9899:2011):
- 7.12.8.2 The erfc functions (p: 249-250)
- 7.25 Type-generic math <tgmath.h> (p: 373-375)
- F.10.5.2 The erfc functions (p: 525)
- C99 standard (ISO/IEC 9899:1999):
- 7.12.8.2 The erfc functions (p: 230)
- 7.22 Type-generic math <tgmath.h> (p: 335-337)
- F.9.5.2 The erfc functions (p: 462)
See also
(C99)(C99)(C99) |
computes error function (function) |
C++ documentation for erfc
| |
External links
| Weisstein, Eric W. "Erfc." From MathWorld — A Wolfram Web Resource. |