(FPCore (k n) :precision binary64 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
(FPCore (k n) :precision binary64 (* (sqrt (* n (/ (* 2.0 PI) (pow (* 2.0 (* n PI)) k)))) (pow k -0.5)))
double code(double k, double n) {
return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}
double code(double k, double n) {
return sqrt((n * ((2.0 * ((double) M_PI)) / pow((2.0 * (n * ((double) M_PI))), k)))) * pow(k, -0.5);
}
public static double code(double k, double n) {
return (1.0 / Math.sqrt(k)) * Math.pow(((2.0 * Math.PI) * n), ((1.0 - k) / 2.0));
}
public static double code(double k, double n) {
return Math.sqrt((n * ((2.0 * Math.PI) / Math.pow((2.0 * (n * Math.PI)), k)))) * Math.pow(k, -0.5);
}
def code(k, n): return (1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))
def code(k, n): return math.sqrt((n * ((2.0 * math.pi) / math.pow((2.0 * (n * math.pi)), k)))) * math.pow(k, -0.5)
function code(k, n) return Float64(Float64(1.0 / sqrt(k)) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0))) end
function code(k, n) return Float64(sqrt(Float64(n * Float64(Float64(2.0 * pi) / (Float64(2.0 * Float64(n * pi)) ^ k)))) * (k ^ -0.5)) end
function tmp = code(k, n) tmp = (1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0)); end
function tmp = code(k, n) tmp = sqrt((n * ((2.0 * pi) / ((2.0 * (n * pi)) ^ k)))) * (k ^ -0.5); end
code[k_, n_] := N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
code[k_, n_] := N[(N[Sqrt[N[(n * N[(N[(2.0 * Pi), $MachinePrecision] / N[Power[N[(2.0 * N[(n * Pi), $MachinePrecision]), $MachinePrecision], k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[Power[k, -0.5], $MachinePrecision]), $MachinePrecision]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
\sqrt{n \cdot \frac{2 \cdot \pi}{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{k}}} \cdot {k}^{-0.5}



Bits error versus k



Bits error versus n
Results
Initial program 0.5
Simplified0.5
Taylor expanded in n around 0 3.5
Simplified0.5
Applied egg-rr0.4
Applied egg-rr0.4
Final simplification0.4
herbie shell --seed 2022155
(FPCore (k n)
:name "Migdal et al, Equation (51)"
:precision binary64
(* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))