\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 r140888 = 1.0;
double r140889 = k;
double r140890 = sqrt(r140889);
double r140891 = r140888 / r140890;
double r140892 = 2.0;
double r140893 = atan2(1.0, 0.0);
double r140894 = r140892 * r140893;
double r140895 = n;
double r140896 = r140894 * r140895;
double r140897 = r140888 - r140889;
double r140898 = r140897 / r140892;
double r140899 = pow(r140896, r140898);
double r140900 = r140891 * r140899;
return r140900;
}
double f(double k, double n) {
double r140901 = 1.0;
double r140902 = k;
double r140903 = sqrt(r140902);
double r140904 = r140901 / r140903;
double r140905 = 2.0;
double r140906 = atan2(1.0, 0.0);
double r140907 = r140905 * r140906;
double r140908 = n;
double r140909 = r140907 * r140908;
double r140910 = r140901 - r140902;
double r140911 = r140910 / r140905;
double r140912 = 2.0;
double r140913 = r140911 / r140912;
double r140914 = pow(r140909, r140913);
double r140915 = r140904 * r140914;
double r140916 = r140915 * r140914;
return r140916;
}



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 2019346
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))