\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}1 \cdot \frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1}{2}\right)}}{\sqrt{k} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}double f(double k, double n) {
double r155300 = 1.0;
double r155301 = k;
double r155302 = sqrt(r155301);
double r155303 = r155300 / r155302;
double r155304 = 2.0;
double r155305 = atan2(1.0, 0.0);
double r155306 = r155304 * r155305;
double r155307 = n;
double r155308 = r155306 * r155307;
double r155309 = r155300 - r155301;
double r155310 = r155309 / r155304;
double r155311 = pow(r155308, r155310);
double r155312 = r155303 * r155311;
return r155312;
}
double f(double k, double n) {
double r155313 = 1.0;
double r155314 = 2.0;
double r155315 = atan2(1.0, 0.0);
double r155316 = r155314 * r155315;
double r155317 = n;
double r155318 = r155316 * r155317;
double r155319 = r155313 / r155314;
double r155320 = pow(r155318, r155319);
double r155321 = k;
double r155322 = sqrt(r155321);
double r155323 = r155321 / r155314;
double r155324 = pow(r155318, r155323);
double r155325 = r155322 * r155324;
double r155326 = r155320 / r155325;
double r155327 = r155313 * r155326;
return r155327;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.5
rmApplied div-inv0.5
Applied associate-*l*0.5
Simplified0.5
rmApplied div-sub0.5
Applied pow-sub0.4
Applied associate-/l/0.4
Final simplification0.4
herbie shell --seed 2020059 +o rules:numerics
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))