ref: 6280c0f17bf3919cf6c2506fec6edfa0a7ec10fe
dir: /sys/src/ape/lib/ap/math/erf.c/
#include <math.h>
#include <errno.h>
/*
C program for floating point error function
erf(x) returns the error function of its argument
erfc(x) returns 1 - erf(x)
erf(x) is defined by
${2 over sqrt(pi)} int from 0 to x e sup {-t sup 2} dt$
the entry for erfc is provided because of the
extreme loss of relative accuracy if erf(x) is
called for large x and the result subtracted
from 1. (e.g. for x= 10, 12 places are lost).
There are no error returns.
Calls exp.
Coefficients for large x are #5667 from Hart & Cheney (18.72D).
*/
#define M 7
#define N 9
static double torp = 1.1283791670955125738961589031;
static double p1[] = {
0.804373630960840172832162e5,
0.740407142710151470082064e4,
0.301782788536507577809226e4,
0.380140318123903008244444e2,
0.143383842191748205576712e2,
-.288805137207594084924010e0,
0.007547728033418631287834e0,
};
static double q1[] = {
0.804373630960840172826266e5,
0.342165257924628539769006e5,
0.637960017324428279487120e4,
0.658070155459240506326937e3,
0.380190713951939403753468e2,
0.100000000000000000000000e1,
0.0,
};
static double p2[] = {
0.18263348842295112592168999e4,
0.28980293292167655611275846e4,
0.2320439590251635247384768711e4,
0.1143262070703886173606073338e4,
0.3685196154710010637133875746e3,
0.7708161730368428609781633646e2,
0.9675807882987265400604202961e1,
0.5641877825507397413087057563e0,
0.0,
};
static double q2[] = {
0.18263348842295112595576438e4,
0.495882756472114071495438422e4,
0.60895424232724435504633068e4,
0.4429612803883682726711528526e4,
0.2094384367789539593790281779e4,
0.6617361207107653469211984771e3,
0.1371255960500622202878443578e3,
0.1714980943627607849376131193e2,
1.0,
};
double erfc(double);
double
erf(double arg)
{
int sign;
double argsq;
double d, n;
int i;
errno = 0;
sign = 1;
if(arg < 0) {
arg = -arg;
sign = -1;
}
if(arg < 0.5) {
argsq = arg*arg;
for(n=0,d=0,i=M-1; i>=0; i--) {
n = n*argsq + p1[i];
d = d*argsq + q1[i];
}
return sign*torp*arg*n/d;
}
if(arg >= 10)
return sign;
return sign*(1 - erfc(arg));
}
double
erfc(double arg)
{
double n, d;
int i;
errno = 0;
if(arg < 0)
return 2 - erfc(-arg);
if(arg < 0.5)
return 1 - erf(arg);
if(arg >= 10)
return 0;
for(n=0,d=0,i=N-1; i>=0; i--) {
n = n*arg + p2[i];
d = d*arg + q2[i];
}
return exp(-arg*arg)*n/d;
}