\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(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\frac{1 - k}{2}\right)}double f(double k, double n) {
double r3381907 = 1.0;
double r3381908 = k;
double r3381909 = sqrt(r3381908);
double r3381910 = r3381907 / r3381909;
double r3381911 = 2.0;
double r3381912 = atan2(1.0, 0.0);
double r3381913 = r3381911 * r3381912;
double r3381914 = n;
double r3381915 = r3381913 * r3381914;
double r3381916 = r3381907 - r3381908;
double r3381917 = r3381916 / r3381911;
double r3381918 = pow(r3381915, r3381917);
double r3381919 = r3381910 * r3381918;
return r3381919;
}
double f(double k, double n) {
double r3381920 = 1.0;
double r3381921 = k;
double r3381922 = sqrt(r3381921);
double r3381923 = r3381920 / r3381922;
double r3381924 = n;
double r3381925 = 2.0;
double r3381926 = atan2(1.0, 0.0);
double r3381927 = r3381925 * r3381926;
double r3381928 = r3381924 * r3381927;
double r3381929 = r3381920 - r3381921;
double r3381930 = r3381929 / r3381925;
double r3381931 = pow(r3381928, r3381930);
double r3381932 = r3381923 * r3381931;
return r3381932;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.4
Final simplification0.4
herbie shell --seed 2019163
(FPCore (k n)
:name "Migdal et al, Equation (51)"
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))