\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 r88249 = 1.0;
double r88250 = k;
double r88251 = sqrt(r88250);
double r88252 = r88249 / r88251;
double r88253 = 2.0;
double r88254 = atan2(1.0, 0.0);
double r88255 = r88253 * r88254;
double r88256 = n;
double r88257 = r88255 * r88256;
double r88258 = r88249 - r88250;
double r88259 = r88258 / r88253;
double r88260 = pow(r88257, r88259);
double r88261 = r88252 * r88260;
return r88261;
}
double f(double k, double n) {
double r88262 = 1.0;
double r88263 = k;
double r88264 = sqrt(r88263);
double r88265 = r88262 / r88264;
double r88266 = 2.0;
double r88267 = r88262 - r88263;
double r88268 = r88267 / r88266;
double r88269 = pow(r88266, r88268);
double r88270 = atan2(1.0, 0.0);
double r88271 = pow(r88270, r88268);
double r88272 = r88269 * r88271;
double r88273 = r88265 * r88272;
double r88274 = n;
double r88275 = pow(r88274, r88268);
double r88276 = r88273 * r88275;
return r88276;
}



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 2019208 +o rules:numerics
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))