(FPCore (k n) :precision binary64 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
(FPCore (k n) :precision binary64 (let* ((t_0 (* n (* PI 2.0)))) (/ (sqrt (* t_0 (pow (pow (pow t_0 k) -0.5) 2.0))) (sqrt k))))
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) {
double t_0 = n * (((double) M_PI) * 2.0);
return sqrt((t_0 * pow(pow(pow(t_0, k), -0.5), 2.0))) / sqrt(k);
}
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) {
double t_0 = n * (Math.PI * 2.0);
return Math.sqrt((t_0 * Math.pow(Math.pow(Math.pow(t_0, k), -0.5), 2.0))) / Math.sqrt(k);
}
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): t_0 = n * (math.pi * 2.0) return math.sqrt((t_0 * math.pow(math.pow(math.pow(t_0, k), -0.5), 2.0))) / math.sqrt(k)
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) t_0 = Float64(n * Float64(pi * 2.0)) return Float64(sqrt(Float64(t_0 * (((t_0 ^ k) ^ -0.5) ^ 2.0))) / sqrt(k)) 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) t_0 = n * (pi * 2.0); tmp = sqrt((t_0 * (((t_0 ^ k) ^ -0.5) ^ 2.0))) / sqrt(k); 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_] := Block[{t$95$0 = N[(n * N[(Pi * 2.0), $MachinePrecision]), $MachinePrecision]}, N[(N[Sqrt[N[(t$95$0 * N[Power[N[Power[N[Power[t$95$0, k], $MachinePrecision], -0.5], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
\begin{array}{l}
t_0 := n \cdot \left(\pi \cdot 2\right)\\
\frac{\sqrt{t_0 \cdot {\left({\left({t_0}^{k}\right)}^{-0.5}\right)}^{2}}}{\sqrt{k}}
\end{array}



Bits error versus k



Bits error versus n
Results
Initial program 0.5
Simplified0.4
Applied egg-rr0.3
Taylor expanded in k around inf 0.4
Applied egg-rr0.4
Final simplification0.4
herbie shell --seed 2022162
(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))))