\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({\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{\frac{1 - k}{2}}{2}\right)} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{\frac{1 - k}{2}}{2}\right)}\right)double f(double k, double n) {
double r128498 = 1.0;
double r128499 = k;
double r128500 = sqrt(r128499);
double r128501 = r128498 / r128500;
double r128502 = 2.0;
double r128503 = atan2(1.0, 0.0);
double r128504 = r128502 * r128503;
double r128505 = n;
double r128506 = r128504 * r128505;
double r128507 = r128498 - r128499;
double r128508 = r128507 / r128502;
double r128509 = pow(r128506, r128508);
double r128510 = r128501 * r128509;
return r128510;
}
double f(double k, double n) {
double r128511 = 1.0;
double r128512 = k;
double r128513 = sqrt(r128512);
double r128514 = r128511 / r128513;
double r128515 = 2.0;
double r128516 = atan2(1.0, 0.0);
double r128517 = r128515 * r128516;
double r128518 = n;
double r128519 = r128517 * r128518;
double r128520 = r128511 - r128512;
double r128521 = r128520 / r128515;
double r128522 = 2.0;
double r128523 = r128521 / r128522;
double r128524 = pow(r128519, r128523);
double r128525 = r128524 * r128524;
double r128526 = r128514 * r128525;
return r128526;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.4
rmApplied sqr-pow0.5
Final simplification0.5
herbie shell --seed 2020056
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))