Average Error: 0.5 → 0.4
Time: 8.0s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
↓
\[\begin{array}{l} t_0 := \sqrt{\pi \cdot \left(n \cdot 2\right)}\\ \mathsf{fma}\left(\frac{0.5}{{t_0}^{k}}, \frac{t_0 + t_0}{\sqrt{k}}, 0\right) \end{array} \]
(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 (sqrt (* PI (* n 2.0)))))
   (fma (/ 0.5 (pow t_0 k)) (/ (+ t_0 t_0) (sqrt k)) 0.0)))
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 = sqrt((((double) M_PI) * (n * 2.0)));
	return fma((0.5 / pow(t_0, k)), ((t_0 + t_0) / sqrt(k)), 0.0);
}

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 = sqrt(Float64(pi * Float64(n * 2.0)))
	return fma(Float64(0.5 / (t_0 ^ k)), Float64(Float64(t_0 + t_0) / sqrt(k)), 0.0)
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[Sqrt[N[(Pi * N[(n * 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, N[(N[(0.5 / N[Power[t$95$0, k], $MachinePrecision]), $MachinePrecision] * N[(N[(t$95$0 + t$95$0), $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] + 0.0), $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 := \sqrt{\pi \cdot \left(n \cdot 2\right)}\\
\mathsf{fma}\left(\frac{0.5}{{t_0}^{k}}, \frac{t_0 + t_0}{\sqrt{k}}, 0\right)
\end{array}

Error

Bits error versus k

Bits error versus n

Derivation

  1. Initial program 0.5

    \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

  2. Simplified0.4

    \[\leadsto \color{blue}{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}}} \]

  3. Applied egg-rr0.4

    \[\leadsto \frac{\color{blue}{\left(\sqrt{\pi \cdot \left(n \cdot 2\right)} + \sqrt{\pi \cdot \left(n \cdot 2\right)}\right) \cdot \frac{0.5}{{\left(\sqrt{\pi \cdot \left(n \cdot 2\right)}\right)}^{k}}}}{\sqrt{k}} \]

  4. Applied egg-rr0.4

    \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{0.5}{{\left(\sqrt{\pi \cdot \left(n \cdot 2\right)}\right)}^{k}}, \frac{\sqrt{\pi \cdot \left(n \cdot 2\right)} + \sqrt{\pi \cdot \left(n \cdot 2\right)}}{\sqrt{k}}, 0\right)} \]

  5. Final simplification0.4

    \[\leadsto \mathsf{fma}\left(\frac{0.5}{{\left(\sqrt{\pi \cdot \left(n \cdot 2\right)}\right)}^{k}}, \frac{\sqrt{\pi \cdot \left(n \cdot 2\right)} + \sqrt{\pi \cdot \left(n \cdot 2\right)}}{\sqrt{k}}, 0\right) \]

Reproduce

herbie shell --seed 2022194 
(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))))