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