\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({2}^{\left(\frac{1 - k}{2}\right)} \cdot \left({\pi}^{\left(\frac{1 - k}{2}\right)} \cdot {n}^{\left(\frac{1 - k}{2}\right)}\right)\right)double f(double k, double n) {
double r150269 = 1.0;
double r150270 = k;
double r150271 = sqrt(r150270);
double r150272 = r150269 / r150271;
double r150273 = 2.0;
double r150274 = atan2(1.0, 0.0);
double r150275 = r150273 * r150274;
double r150276 = n;
double r150277 = r150275 * r150276;
double r150278 = r150269 - r150270;
double r150279 = r150278 / r150273;
double r150280 = pow(r150277, r150279);
double r150281 = r150272 * r150280;
return r150281;
}
double f(double k, double n) {
double r150282 = 1.0;
double r150283 = k;
double r150284 = sqrt(r150283);
double r150285 = r150282 / r150284;
double r150286 = 2.0;
double r150287 = r150282 - r150283;
double r150288 = r150287 / r150286;
double r150289 = pow(r150286, r150288);
double r150290 = atan2(1.0, 0.0);
double r150291 = pow(r150290, r150288);
double r150292 = n;
double r150293 = pow(r150292, r150288);
double r150294 = r150291 * r150293;
double r150295 = r150289 * r150294;
double r150296 = r150285 * r150295;
return r150296;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.5
rmApplied unpow-prod-down0.7
rmApplied unpow-prod-down0.6
Applied associate-*l*0.6
Final simplification0.6
herbie shell --seed 2020065
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))