\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 r128892 = 1.0;
double r128893 = k;
double r128894 = sqrt(r128893);
double r128895 = r128892 / r128894;
double r128896 = 2.0;
double r128897 = atan2(1.0, 0.0);
double r128898 = r128896 * r128897;
double r128899 = n;
double r128900 = r128898 * r128899;
double r128901 = r128892 - r128893;
double r128902 = r128901 / r128896;
double r128903 = pow(r128900, r128902);
double r128904 = r128895 * r128903;
return r128904;
}
double f(double k, double n) {
double r128905 = 1.0;
double r128906 = k;
double r128907 = sqrt(r128906);
double r128908 = r128905 / r128907;
double r128909 = 2.0;
double r128910 = atan2(1.0, 0.0);
double r128911 = r128909 * r128910;
double r128912 = n;
double r128913 = r128911 * r128912;
double r128914 = r128905 - r128906;
double r128915 = r128914 / r128909;
double r128916 = 2.0;
double r128917 = r128915 / r128916;
double r128918 = pow(r128913, r128917);
double r128919 = r128918 * r128918;
double r128920 = r128908 * r128919;
return r128920;
}



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))))