Average Error: 0.5 → 0.4
Time: 5.2s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
↓
\[{2}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)} \cdot \left(\left({\left(\pi \cdot n\right)}^{\left(k \cdot -0.5\right)} \cdot \sqrt{\pi \cdot n}\right) \cdot {k}^{-0.5}\right) \]
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
 :precision binary64
 (*
  (pow 2.0 (fma k -0.5 0.5))
  (* (* (pow (* PI n) (* k -0.5)) (sqrt (* PI n))) (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 pow(2.0, fma(k, -0.5, 0.5)) * ((pow((((double) M_PI) * n), (k * -0.5)) * sqrt((((double) M_PI) * n))) * 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((2.0 ^ fma(k, -0.5, 0.5)) * Float64(Float64((Float64(pi * n) ^ Float64(k * -0.5)) * sqrt(Float64(pi * n))) * (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[Power[2.0, N[(k * -0.5 + 0.5), $MachinePrecision]], $MachinePrecision] * N[(N[(N[Power[N[(Pi * n), $MachinePrecision], N[(k * -0.5), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(Pi * n), $MachinePrecision]], $MachinePrecision]), $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)}
↓
{2}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)} \cdot \left(\left({\left(\pi \cdot n\right)}^{\left(k \cdot -0.5\right)} \cdot \sqrt{\pi \cdot n}\right) \cdot {k}^{-0.5}\right)

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. Simplified0.5

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

  3. Applied egg-rr0.5

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

  4. Applied egg-rr0.4

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

  5. Final simplification0.4

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

Reproduce

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))))