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



Bits error versus k



Bits error versus n
Results
Initial program 0.5
Simplified0.5
Applied egg-rr0.4
Applied egg-rr0.4
Final simplification0.4
herbie shell --seed 2022139
(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))))