\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(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{\frac{1 - k}{2}}{2}\right)}\right) \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{\frac{1 - k}{2}}{2}\right)}double f(double k, double n) {
double r131392 = 1.0;
double r131393 = k;
double r131394 = sqrt(r131393);
double r131395 = r131392 / r131394;
double r131396 = 2.0;
double r131397 = atan2(1.0, 0.0);
double r131398 = r131396 * r131397;
double r131399 = n;
double r131400 = r131398 * r131399;
double r131401 = r131392 - r131393;
double r131402 = r131401 / r131396;
double r131403 = pow(r131400, r131402);
double r131404 = r131395 * r131403;
return r131404;
}
double f(double k, double n) {
double r131405 = 1.0;
double r131406 = k;
double r131407 = sqrt(r131406);
double r131408 = r131405 / r131407;
double r131409 = 2.0;
double r131410 = atan2(1.0, 0.0);
double r131411 = r131409 * r131410;
double r131412 = n;
double r131413 = r131411 * r131412;
double r131414 = r131405 - r131406;
double r131415 = r131414 / r131409;
double r131416 = 2.0;
double r131417 = r131415 / r131416;
double r131418 = pow(r131413, r131417);
double r131419 = r131408 * r131418;
double r131420 = r131419 * r131418;
return r131420;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.4
rmApplied sqr-pow0.5
Applied associate-*r*0.5
Final simplification0.5
herbie shell --seed 2019354
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))