The function that numerically approximates the inverse error (erf) function $[\large \frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt]^{-1}$ is captured in the C function below - it's a fragment of the source code of my (now free) iOS App SciStatCalc. Note the one iteration of Halley's method to refine the result.

Feel free to use it, if you wish to do so.

Feel free to use it, if you wish to do so.

double inverf(double x)

{

double w,p;

w = -log((1.00-x)*(1.00+x));

if ( w < 5.000000 ) {

w = w - 2.500000;

p= 2.81022636e-08;

p= 3.43273939e-07 + p*w;

p= -3.5233877e-06 + p*w;

p = -4.39150654e-06 + p*w;

p = 0.00021858087 + p*w;

p = -0.00125372503 + p*w;

p = -0.00417768164 + p*w;

p = 0.246640727 + p*w;

p = 1.50140941 + p*w;

} else

{

w = sqrtf(w) - 3.000000;

p = -0.000200214257;

p = 0.000100950558 + p*w;

p = 0.00134934322 + p*w;

p = -0.00367342844 + p*w;

p = 0.00573950773 + p*w;

p = -0.0076224613 + p*w;

p = 0.00943887047 + p*w;

p = 1.00167406 + p*w;

p = 2.83297682 + p*w;

}

double res_ra = p*x; // assign to rational estimate variable

// Halley's method to refine estimate of inverse erf

double res_hm = 0;

double fx = (erf(res_ra) - x);

double df = (2.0/sqrt(M_PI))*exp(-(res_ra * res_ra));

double d2f = -2 * res_ra * df;

res_hm = res_ra - (2*fx*df)/((2*df*df) - (fx*d2f));

return res_hm;

}

{

double res;

res = (inverf((2*p) - 1) *sig*sqrt(2.0)) + mu;

return res;

}

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