erfc, erfcf, erfcl
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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.Contents
[edit] Parameters
arg
-
floating-point value
[edit] Return value
If no errors occur, value of the complementary error function of arg, that is \(\frac{2}{\sqrt{\pi} }\int_{arg}^{\infty}{e^{-{t^2} }\mathsf{d}t}\) 2
√π
∫∞arge-t2
dt or \({\small 1-\operatorname{erf}(arg)}\)1-erf(arg), is returned.
If a range error occurs due to underflow, the correct result (after rounding) is returned.
[edit] 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.
[edit] Notes
For the IEEE-compatible type double, underflow is guaranteed if arg > 26.55.
[edit] 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
[edit] 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)
[edit] See also
C++ documentation for erfc
[edit] External links
Weisstein, Eric W. "Erfc." From MathWorld — A Wolfram Web Resource.