92 lines
2.6 KiB
C
92 lines
2.6 KiB
C
#ifndef LEPTON_MSVC_ERFC_H_
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#define LEPTON_MSVC_ERFC_H_
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/*
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* Up to version 11 (VC++ 2012), Microsoft does not support the
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* standard C99 erf() and erfc() functions so we have to fake them here.
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* These were added in version 12 (VC++ 2013), which sets _MSC_VER=1800
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* (VC11 has _MSC_VER=1700).
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*/
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#if defined(_MSC_VER) || defined(__MINGW32__)
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#if !defined(M_PI)
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#define M_PI 3.14159265358979323846264338327950288
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#endif
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#endif
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#if defined(_MSC_VER)
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#if _MSC_VER <= 1700 // 1700 is VC11, 1800 is VC12
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/***************************
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* erf.cpp
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* author: Steve Strand
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* written: 29-Jan-04
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***************************/
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#include <cmath>
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static const double rel_error= 1E-12; //calculate 12 significant figures
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//you can adjust rel_error to trade off between accuracy and speed
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//but don't ask for > 15 figures (assuming usual 52 bit mantissa in a double)
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static double erfc(double x);
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static double erf(double x)
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//erf(x) = 2/sqrt(pi)*integral(exp(-t^2),t,0,x)
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// = 2/sqrt(pi)*[x - x^3/3 + x^5/5*2! - x^7/7*3! + ...]
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// = 1-erfc(x)
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{
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static const double two_sqrtpi= 1.128379167095512574; // 2/sqrt(pi)
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if (fabs(x) > 2.2) {
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return 1.0 - erfc(x); //use continued fraction when fabs(x) > 2.2
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}
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double sum= x, term= x, xsqr= x*x;
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int j= 1;
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do {
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term*= xsqr/j;
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sum-= term/(2*j+1);
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++j;
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term*= xsqr/j;
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sum+= term/(2*j+1);
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++j;
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} while (fabs(term)/sum > rel_error);
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return two_sqrtpi*sum;
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}
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static double erfc(double x)
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//erfc(x) = 2/sqrt(pi)*integral(exp(-t^2),t,x,inf)
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// = exp(-x^2)/sqrt(pi) * [1/x+ (1/2)/x+ (2/2)/x+ (3/2)/x+ (4/2)/x+ ...]
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// = 1-erf(x)
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//expression inside [] is a continued fraction so '+' means add to denominator only
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{
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static const double one_sqrtpi= 0.564189583547756287; // 1/sqrt(pi)
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if (fabs(x) < 2.2) {
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return 1.0 - erf(x); //use series when fabs(x) < 2.2
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}
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// Don't look for x==0 here!
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if (x < 0) { //continued fraction only valid for x>0
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return 2.0 - erfc(-x);
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}
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double a=1, b=x; //last two convergent numerators
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double c=x, d=x*x+0.5; //last two convergent denominators
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double q1, q2= b/d; //last two convergents (a/c and b/d)
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double n= 1.0, t;
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do {
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t= a*n+b*x;
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a= b;
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b= t;
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t= c*n+d*x;
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c= d;
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d= t;
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n+= 0.5;
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q1= q2;
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q2= b/d;
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} while (fabs(q1-q2)/q2 > rel_error);
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return one_sqrtpi*exp(-x*x)*q2;
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}
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#endif // _MSC_VER <= 1700
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#endif // _MSC_VER
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#endif // LEPTON_MSVC_ERFC_H_
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