\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 r174484 = 1.0;
double r174485 = k;
double r174486 = sqrt(r174485);
double r174487 = r174484 / r174486;
double r174488 = 2.0;
double r174489 = atan2(1.0, 0.0);
double r174490 = r174488 * r174489;
double r174491 = n;
double r174492 = r174490 * r174491;
double r174493 = r174484 - r174485;
double r174494 = r174493 / r174488;
double r174495 = pow(r174492, r174494);
double r174496 = r174487 * r174495;
return r174496;
}
double f(double k, double n) {
double r174497 = 1.0;
double r174498 = k;
double r174499 = sqrt(r174498);
double r174500 = r174497 / r174499;
double r174501 = 2.0;
double r174502 = atan2(1.0, 0.0);
double r174503 = r174501 * r174502;
double r174504 = n;
double r174505 = r174503 * r174504;
double r174506 = r174497 - r174498;
double r174507 = r174506 / r174501;
double r174508 = 2.0;
double r174509 = r174507 / r174508;
double r174510 = pow(r174505, r174509);
double r174511 = r174510 * r174510;
double r174512 = r174500 * r174511;
return r174512;
}



Bits error versus k



Bits error versus n
Results
Initial program 0.4
rmApplied sqr-pow0.5
Final simplification0.5
herbie shell --seed 2019344 +o rules:numerics
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))