\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\sqrt{\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}} \cdot \sqrt{\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}double f(double k, double n) {
double r116307 = 1.0;
double r116308 = k;
double r116309 = sqrt(r116308);
double r116310 = r116307 / r116309;
double r116311 = 2.0;
double r116312 = atan2(1.0, 0.0);
double r116313 = r116311 * r116312;
double r116314 = n;
double r116315 = r116313 * r116314;
double r116316 = r116307 - r116308;
double r116317 = r116316 / r116311;
double r116318 = pow(r116315, r116317);
double r116319 = r116310 * r116318;
return r116319;
}
double f(double k, double n) {
double r116320 = 1.0;
double r116321 = k;
double r116322 = sqrt(r116321);
double r116323 = r116320 / r116322;
double r116324 = 2.0;
double r116325 = atan2(1.0, 0.0);
double r116326 = r116324 * r116325;
double r116327 = n;
double r116328 = r116326 * r116327;
double r116329 = r116320 - r116321;
double r116330 = r116329 / r116324;
double r116331 = pow(r116328, r116330);
double r116332 = r116323 * r116331;
double r116333 = sqrt(r116332);
double r116334 = r116333 * r116333;
return r116334;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.4
rmApplied add-sqr-sqrt0.5
Final simplification0.5
herbie shell --seed 2020045 +o rules:numerics
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))