Migdal et al, Equation (51)

Average Accuracy: 99.2% → 99.3%

Time: 12.0s

Precision: binary64

Cost: 33024

Math

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

↓

\[\begin{array}{l} t_0 := \pi \cdot \left(2 \cdot n\right)\\ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{t_0}}{{t_0}^{\left(k \cdot 0.5\right)}} \end{array} \]

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 (* PI (* 2.0 n))))
   (* (/ 1.0 (sqrt k)) (/ (sqrt t_0) (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 = ((double) M_PI) * (2.0 * n);
	return (1.0 / sqrt(k)) * (sqrt(t_0) / 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 = Math.PI * (2.0 * n);
	return (1.0 / Math.sqrt(k)) * (Math.sqrt(t_0) / 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 = math.pi * (2.0 * n)
	return (1.0 / math.sqrt(k)) * (math.sqrt(t_0) / 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(pi * Float64(2.0 * n))
	return Float64(Float64(1.0 / sqrt(k)) * Float64(sqrt(t_0) / (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 = pi * (2.0 * n);
	tmp = (1.0 / sqrt(k)) * (sqrt(t_0) / (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[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision]}, N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[(N[Sqrt[t$95$0], $MachinePrecision] / N[Power[t$95$0, N[(k * 0.5), $MachinePrecision]], $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%

   \[\leadsto \frac{1}{\sqrt{k}} \cdot \color{blue}{\frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}}} \]

3. Simplified 99.3%

   \[\leadsto \frac{1}{\sqrt{k}} \cdot \color{blue}{\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}}} \]
   Proof

   [Start] 99.3	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}} \]
   associate-*r* [=>] 99.3	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{\color{blue}{\left(2 \cdot \pi\right) \cdot n}}}{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}} \]
   *-commutative [=>] 99.3	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{\color{blue}{\left(\pi \cdot 2\right)} \cdot n}}{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}} \]
   associate-*l* [=>] 99.3	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{\color{blue}{\pi \cdot \left(2 \cdot n\right)}}}{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}} \]
   associate-*r* [=>] 99.3	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{{\color{blue}{\left(\left(2 \cdot \pi\right) \cdot n\right)}}^{\left(k \cdot 0.5\right)}} \]
   *-commutative [=>] 99.3	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{{\left(\color{blue}{\left(\pi \cdot 2\right)} \cdot n\right)}^{\left(k \cdot 0.5\right)}} \]
   associate-*l* [=>] 99.3	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{{\color{blue}{\left(\pi \cdot \left(2 \cdot n\right)\right)}}^{\left(k \cdot 0.5\right)}} \]

4. Final simplification 99.3%

   \[\leadsto \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}} \]

Alternatives

Alternative 1
Accuracy	99.4%
Cost	32896

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

Alternative 2
Accuracy	99.1%
Cost	20036

\[\begin{array}{l} \mathbf{if}\;k \leq 4.7 \cdot 10^{-17}:\\ \;\;\;\;{k}^{-0.5} \cdot \sqrt{n \cdot \left(\pi \cdot 2\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\frac{\pi \cdot -2}{\frac{-1}{n}}\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 3
Accuracy	99.3%
Cost	19968

\[{k}^{-0.5} \cdot {\left(n \cdot \left(\pi \cdot 2\right)\right)}^{\left(\frac{1 - k}{2}\right)} \]

Alternative 4
Accuracy	99.1%
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 9.6 \cdot 10^{-17}:\\ \;\;\;\;{k}^{-0.5} \cdot \sqrt{n \cdot \left(\pi \cdot 2\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot \left(n + n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 5
Accuracy	99.3%
Cost	19904

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

Alternative 6
Accuracy	70.6%
Cost	19844

\[\begin{array}{l} \mathbf{if}\;k \leq 4.5 \cdot 10^{+153}:\\ \;\;\;\;{k}^{-0.5} \cdot \sqrt{n \cdot \left(\pi \cdot 2\right)}\\ \mathbf{else}:\\ \;\;\;\;{\left({\left(\pi \cdot \frac{n}{k}\right)}^{3} \cdot 8\right)}^{0.16666666666666666}\\ \end{array} \]

Alternative 7
Accuracy	64.5%
Cost	19648

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

Alternative 8
Accuracy	64.5%
Cost	19584

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

Alternative 9
Accuracy	48.9%
Cost	13248

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

Alternative 10
Accuracy	48.9%
Cost	13248

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

Alternative 11
Accuracy	48.0%
Cost	13184

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

Alternative 12
Accuracy	48.0%
Cost	13184

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

Reproduce

herbie shell --seed 2023125 
(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))))