\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(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\frac{1 - k}{2}\right)}double f(double k, double n) {
double r3874996 = 1.0;
double r3874997 = k;
double r3874998 = sqrt(r3874997);
double r3874999 = r3874996 / r3874998;
double r3875000 = 2.0;
double r3875001 = atan2(1.0, 0.0);
double r3875002 = r3875000 * r3875001;
double r3875003 = n;
double r3875004 = r3875002 * r3875003;
double r3875005 = r3874996 - r3874997;
double r3875006 = r3875005 / r3875000;
double r3875007 = pow(r3875004, r3875006);
double r3875008 = r3874999 * r3875007;
return r3875008;
}
double f(double k, double n) {
double r3875009 = 1.0;
double r3875010 = k;
double r3875011 = sqrt(r3875010);
double r3875012 = r3875009 / r3875011;
double r3875013 = n;
double r3875014 = 2.0;
double r3875015 = atan2(1.0, 0.0);
double r3875016 = r3875014 * r3875015;
double r3875017 = r3875013 * r3875016;
double r3875018 = r3875009 - r3875010;
double r3875019 = r3875018 / r3875014;
double r3875020 = pow(r3875017, r3875019);
double r3875021 = r3875012 * r3875020;
return r3875021;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.4
Final simplification0.4
herbie shell --seed 2019130 +o rules:numerics
(FPCore (k n)
:name "Migdal et al, Equation (51)"
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))