\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\frac{1}{\frac{\sqrt{k}}{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\frac{1}{2} - \frac{k}{2}\right)}}}double f(double k, double n) {
double r2639797 = 1.0;
double r2639798 = k;
double r2639799 = sqrt(r2639798);
double r2639800 = r2639797 / r2639799;
double r2639801 = 2.0;
double r2639802 = atan2(1.0, 0.0);
double r2639803 = r2639801 * r2639802;
double r2639804 = n;
double r2639805 = r2639803 * r2639804;
double r2639806 = r2639797 - r2639798;
double r2639807 = r2639806 / r2639801;
double r2639808 = pow(r2639805, r2639807);
double r2639809 = r2639800 * r2639808;
return r2639809;
}
double f(double k, double n) {
double r2639810 = 1.0;
double r2639811 = k;
double r2639812 = sqrt(r2639811);
double r2639813 = n;
double r2639814 = 2.0;
double r2639815 = atan2(1.0, 0.0);
double r2639816 = r2639814 * r2639815;
double r2639817 = r2639813 * r2639816;
double r2639818 = 0.5;
double r2639819 = r2639811 / r2639814;
double r2639820 = r2639818 - r2639819;
double r2639821 = pow(r2639817, r2639820);
double r2639822 = r2639812 / r2639821;
double r2639823 = r2639810 / r2639822;
return r2639823;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.4
Simplified0.3
rmApplied clear-num0.4
Final simplification0.4
herbie shell --seed 2019153 +o rules:numerics
(FPCore (k n)
:name "Migdal et al, Equation (51)"
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))