Migdal et al, Equation (51)

Average Error: 0.5 → 0.5

Time: 9.4s

Precision: binary64

Cost: 32896

Math

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

↓

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

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 (* n (* 2.0 PI))))
   (/ (* (pow t_0 (* k -0.5)) (sqrt t_0)) (sqrt k))))

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 = n * (2.0 * ((double) M_PI));
	return (pow(t_0, (k * -0.5)) * sqrt(t_0)) / sqrt(k);
}

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 = n * (2.0 * Math.PI);
	return (Math.pow(t_0, (k * -0.5)) * Math.sqrt(t_0)) / Math.sqrt(k);
}

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 = n * (2.0 * math.pi)
	return (math.pow(t_0, (k * -0.5)) * math.sqrt(t_0)) / math.sqrt(k)

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(n * Float64(2.0 * pi))
	return Float64(Float64((t_0 ^ Float64(k * -0.5)) * sqrt(t_0)) / sqrt(k))
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 = n * (2.0 * pi);
	tmp = ((t_0 ^ (k * -0.5)) * sqrt(t_0)) / sqrt(k);
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[(n * N[(2.0 * Pi), $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]]

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. Simplified 0.4

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

3. Applied egg-rr 0.5

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

4. Final simplification 0.5

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

Alternatives

Alternative 1
Error	0.5
Cost	26176

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

Alternative 2
Error	0.6
Cost	19972

\[\begin{array}{l} \mathbf{if}\;k \leq 1.1011707219099613 \cdot 10^{-48}:\\ \;\;\;\;\frac{1}{\sqrt{k} \cdot {\left(2 \cdot \left(n \cdot \pi\right)\right)}^{-0.5}}\\ \mathbf{else}:\\ \;\;\;\;{\left(\frac{k}{{\left(\pi \cdot \left(n + n\right)\right)}^{\left(1 - k\right)}}\right)}^{-0.5}\\ \end{array} \]

Alternative 3
Error	0.5
Cost	19968

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

Alternative 4
Error	0.7
Cost	19908

\[\begin{array}{l} t_0 := 2 \cdot \left(n \cdot \pi\right)\\ \mathbf{if}\;k \leq 1.1907048836177118 \cdot 10^{-48}:\\ \;\;\;\;\frac{1}{\sqrt{k} \cdot {t_0}^{-0.5}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{t_0}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 5
Error	17.7
Cost	19844

\[\begin{array}{l} \mathbf{if}\;k \leq 2.7 \cdot 10^{+131}:\\ \;\;\;\;\frac{\sqrt{n}}{\sqrt{0.5 \cdot \frac{k}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;{\left({\left(\frac{\pi \cdot \left(n \cdot 2\right)}{k}\right)}^{3}\right)}^{0.16666666666666666}\\ \end{array} \]

Alternative 6
Error	20.9
Cost	19780

\[\begin{array}{l} \mathbf{if}\;k \leq 2.4 \cdot 10^{+91}:\\ \;\;\;\;\frac{\sqrt{n}}{\sqrt{0.5 \cdot \frac{k}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(\left(n + n\right) \cdot \frac{\pi}{k}\right)}^{1.5}}\\ \end{array} \]

Alternative 7
Error	22.1
Cost	19584

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

Alternative 8
Error	32.3
Cost	13248

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

Alternative 9
Error	33.0
Cost	13184

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

Alternative 10
Error	32.9
Cost	13184

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

Reproduce

herbie shell --seed 2022221 
(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))))