/*	@(#)erf.c	4.1	12/25/82	*//*	C program for floating point error function	erf(x) returns the error function of its argument	erfc(x) returns 1.0-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 9int errno;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,};doubleerf(arg) double arg;{	double erfc();	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*1.);	return(sign*(1. - erfc(arg)));}doubleerfc(arg) double arg;{	double erf();	double exp();	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);}