#include #include #include static const double rel_error= 1E-12; //calculate 12 significant figures //you can adjust rel_error to trade off between accuracy and speed //but don't ask for > 15 figures (assuming usual 52 bit mantissa in a double) double erf(double x) //erf(x) = 2/sqrt(pi)*integral(exp(-t^2),t,0,x) // = 2/sqrt(pi)*[x - x^3/3 + x^5/5*2! - x^7/7*3! + ...] // = 1-erfc(x) { static const double two_sqrtpi= 1.128379167095512574; // 2/sqrt(pi) if (fabs(x) > 2.2) { return 1.0 - erfc(x); //use continued fraction when fabs(x) > 2.2 } double sum= x, term= x, xsqr= x*x; int j= 1; do { term*= xsqr/j; sum-= term/(2*j+1); ++j; term*= xsqr/j; sum+= term/(2*j+1); ++j; } while (fabs(term/sum) > rel_error); return two_sqrtpi*sum; } double erfc(double x) //erfc(x) = 2/sqrt(pi)*integral(exp(-t^2),t,x,inf) // = exp(-x^2)/sqrt(pi) * [1/x+ (1/2)/x+ (2/2)/x+ (3/2)/x+ (4/2)/x+ ...] // = 1-erf(x) //expression inside [] is a continued fraction so '+' means add to denominator only { static const double one_sqrtpi= 0.564189583547756287; // 1/sqrt(pi) if (fabs(x) < 2.2) { return 1.0 - erf(x); //use series when fabs(x) < 2.2 } if (std::signbit(x)) { //continued fraction only valid for x>0 return 2.0 - erfc(-x); } double a=1, b=x; //last two convergent numerators double c=x, d=x*x+0.5; //last two convergent denominators double q1, q2= b/d; //last two convergents (a/c and b/d) double n= 1.0, t; do { t= a*n+b*x; a= b; b= t; t= c*n+d*x; c= d; d= t; n+= 0.5; q1= q2; q2= b/d; } while (fabs(q1-q2)/q2 > rel_error); return one_sqrtpi*exp(-x*x)*q2; }