\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({2}^{\left(\frac{1 - k}{2}\right)} \cdot {\pi}^{\left(\frac{1 - k}{2}\right)}\right)\right) \cdot {n}^{\left(\frac{1 - k}{2}\right)}double f(double k, double n) {
double r82073 = 1.0;
double r82074 = k;
double r82075 = sqrt(r82074);
double r82076 = r82073 / r82075;
double r82077 = 2.0;
double r82078 = atan2(1.0, 0.0);
double r82079 = r82077 * r82078;
double r82080 = n;
double r82081 = r82079 * r82080;
double r82082 = r82073 - r82074;
double r82083 = r82082 / r82077;
double r82084 = pow(r82081, r82083);
double r82085 = r82076 * r82084;
return r82085;
}
double f(double k, double n) {
double r82086 = 1.0;
double r82087 = k;
double r82088 = sqrt(r82087);
double r82089 = r82086 / r82088;
double r82090 = 2.0;
double r82091 = r82086 - r82087;
double r82092 = r82091 / r82090;
double r82093 = pow(r82090, r82092);
double r82094 = atan2(1.0, 0.0);
double r82095 = pow(r82094, r82092);
double r82096 = r82093 * r82095;
double r82097 = r82089 * r82096;
double r82098 = n;
double r82099 = pow(r82098, r82092);
double r82100 = r82097 * r82099;
return r82100;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.4
rmApplied unpow-prod-down0.5
Applied associate-*r*0.5
rmApplied unpow-prod-down0.5
Final simplification0.5
herbie shell --seed 2019325
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))