Average Error: 0.5 → 0.4
Time: 5.9s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
↓
\[\begin{array}{l} t_0 := \pi \cdot \left(n \cdot 2\right)\\ \frac{{t_0}^{\left(k \cdot -0.5\right)} \cdot \sqrt{t_0}}{\sqrt{k}} \end{array} \]
(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 (* PI (* n 2.0))))
   (/ (* (pow t_0 (* k -0.5)) (sqrt t_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 = ((double) M_PI) * (n * 2.0);
	return (pow(t_0, (k * -0.5)) * sqrt(t_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 = Math.PI * (n * 2.0);
	return (Math.pow(t_0, (k * -0.5)) * Math.sqrt(t_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 = math.pi * (n * 2.0)
	return (math.pow(t_0, (k * -0.5)) * math.sqrt(t_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(pi * Float64(n * 2.0))
	return Float64(Float64((t_0 ^ Float64(k * -0.5)) * sqrt(t_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 = pi * (n * 2.0);
	tmp = ((t_0 ^ (k * -0.5)) * sqrt(t_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[(Pi * N[(n * 2.0), $MachinePrecision]), $MachinePrecision]}, N[(N[(N[Power[t$95$0, N[(k * -0.5), $MachinePrecision]], $MachinePrecision] * N[Sqrt[t$95$0], $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 := \pi \cdot \left(n \cdot 2\right)\\
\frac{{t_0}^{\left(k \cdot -0.5\right)} \cdot \sqrt{t_0}}{\sqrt{k}}
\end{array}

Error

Bits error versus k

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

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-rr0.4

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

  4. Final simplification0.4

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

Reproduce

herbie shell --seed 2022156 
(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))))