\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\frac{1}{\sqrt{k}} \cdot {\left(\left(\pi \cdot 2\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}double f(double k, double n) {
double r17219721 = 1.0;
double r17219722 = k;
double r17219723 = sqrt(r17219722);
double r17219724 = r17219721 / r17219723;
double r17219725 = 2.0;
double r17219726 = atan2(1.0, 0.0);
double r17219727 = r17219725 * r17219726;
double r17219728 = n;
double r17219729 = r17219727 * r17219728;
double r17219730 = r17219721 - r17219722;
double r17219731 = r17219730 / r17219725;
double r17219732 = pow(r17219729, r17219731);
double r17219733 = r17219724 * r17219732;
return r17219733;
}
double f(double k, double n) {
double r17219734 = 1.0;
double r17219735 = k;
double r17219736 = sqrt(r17219735);
double r17219737 = r17219734 / r17219736;
double r17219738 = atan2(1.0, 0.0);
double r17219739 = 2.0;
double r17219740 = r17219738 * r17219739;
double r17219741 = n;
double r17219742 = r17219740 * r17219741;
double r17219743 = r17219734 - r17219735;
double r17219744 = r17219743 / r17219739;
double r17219745 = pow(r17219742, r17219744);
double r17219746 = r17219737 * r17219745;
return r17219746;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.4
Simplified0.3
rmApplied div-inv0.4
Final simplification0.4
herbie shell --seed 2019104
(FPCore (k n)
:name "Migdal et al, Equation (51)"
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))