Migdal et al, Equation (51)

Average Accuracy: 99.2% → 99.3%

Time: 13.0s

Precision: binary64

Cost: 32960

Math

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

↓

\[\begin{array}{l} t_0 := 2 \cdot \left(\pi \cdot n\right)\\ \frac{\sqrt{t_0} \cdot {k}^{-0.5}}{{t_0}^{\left(k \cdot 0.5\right)}} \end{array} \]

FPCore

(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))))
   (/ (* (sqrt t_0) (pow k -0.5)) (pow t_0 (* k 0.5)))))

C

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 (sqrt(t_0) * pow(k, -0.5)) / pow(t_0, (k * 0.5));
}

Java

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.sqrt(t_0) * Math.pow(k, -0.5)) / Math.pow(t_0, (k * 0.5));
}

Python

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.sqrt(t_0) * math.pow(k, -0.5)) / math.pow(t_0, (k * 0.5))

Julia

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(2.0 * Float64(pi * n))
	return Float64(Float64(sqrt(t_0) * (k ^ -0.5)) / (t_0 ^ Float64(k * 0.5)))
end

MATLAB

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 = (sqrt(t_0) * (k ^ -0.5)) / (t_0 ^ (k * 0.5));
end

Wolfram

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[(2.0 * N[(Pi * n), $MachinePrecision]), $MachinePrecision]}, N[(N[(N[Sqrt[t$95$0], $MachinePrecision] * N[Power[k, -0.5], $MachinePrecision]), $MachinePrecision] / N[Power[t$95$0, N[(k * 0.5), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 99.2%

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

2. Applied egg-rr 99.3%

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

3. Final simplification 99.3%

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

Alternatives

Alternative 1
Accuracy	99.3%
Cost	32896

\[\begin{array}{l} t_0 := \pi \cdot \left(2 \cdot n\right)\\ \frac{\sqrt{t_0} \cdot {t_0}^{\left(k \cdot -0.5\right)}}{\sqrt{k}} \end{array} \]

Alternative 2
Accuracy	99.3%
Cost	19968

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

Alternative 3
Accuracy	99.3%
Cost	19968

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

Alternative 4
Accuracy	99.2%
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 1.15 \cdot 10^{-17}:\\ \;\;\;\;\sqrt{2 \cdot \left(\pi \cdot n\right)} \cdot {k}^{-0.5}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 5
Accuracy	99.3%
Cost	19904

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

Alternative 6
Accuracy	66.9%
Cost	19780

\[\begin{array}{l} \mathbf{if}\;k \leq 5 \cdot 10^{+216}:\\ \;\;\;\;\frac{\sqrt{n \cdot \left(2 \cdot \pi\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(n \cdot \left(\pi \cdot \frac{2}{k}\right)\right)}^{1.5}}\\ \end{array} \]

Alternative 7
Accuracy	66.9%
Cost	19780

\[\begin{array}{l} \mathbf{if}\;k \leq 5 \cdot 10^{+217}:\\ \;\;\;\;\sqrt{2 \cdot \left(\pi \cdot n\right)} \cdot {k}^{-0.5}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(n \cdot \left(\pi \cdot \frac{2}{k}\right)\right)}^{1.5}}\\ \end{array} \]

Alternative 8
Accuracy	65.7%
Cost	19584

\[\sqrt{2 \cdot n} \cdot \sqrt{\frac{\pi}{k}} \]

Alternative 9
Accuracy	65.8%
Cost	19584

\[\frac{\sqrt{2 \cdot n}}{\sqrt{\frac{k}{\pi}}} \]

Alternative 10
Accuracy	65.7%
Cost	19584

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

Alternative 11
Accuracy	49.8%
Cost	13312

\[\frac{1}{\sqrt{k \cdot \frac{0.5}{\pi \cdot n}}} \]

Alternative 12
Accuracy	49.8%
Cost	13312

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

Alternative 13
Accuracy	48.9%
Cost	13184

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

Reproduce

herbie shell --seed 2023122 
(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))))