\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\left(\frac{1}{\sqrt{k}} \cdot \left({2}^{\left(\frac{1 - k}{2}\right)} \cdot {\pi}^{\left(\frac{1 - k}{2}\right)}\right)\right) \cdot {n}^{\left(\frac{1 - k}{2}\right)}double f(double k, double n) {
double r112040 = 1.0;
double r112041 = k;
double r112042 = sqrt(r112041);
double r112043 = r112040 / r112042;
double r112044 = 2.0;
double r112045 = atan2(1.0, 0.0);
double r112046 = r112044 * r112045;
double r112047 = n;
double r112048 = r112046 * r112047;
double r112049 = r112040 - r112041;
double r112050 = r112049 / r112044;
double r112051 = pow(r112048, r112050);
double r112052 = r112043 * r112051;
return r112052;
}
double f(double k, double n) {
double r112053 = 1.0;
double r112054 = k;
double r112055 = sqrt(r112054);
double r112056 = r112053 / r112055;
double r112057 = 2.0;
double r112058 = r112053 - r112054;
double r112059 = r112058 / r112057;
double r112060 = pow(r112057, r112059);
double r112061 = atan2(1.0, 0.0);
double r112062 = pow(r112061, r112059);
double r112063 = r112060 * r112062;
double r112064 = r112056 * r112063;
double r112065 = n;
double r112066 = pow(r112065, r112059);
double r112067 = r112064 * r112066;
return r112067;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.4
rmApplied unpow-prod-down0.6
Applied associate-*r*0.6
rmApplied unpow-prod-down0.5
Final simplification0.5
herbie shell --seed 2019305 +o rules:numerics
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))