Migdal et al, Equation (51)

Average Error: 0.5 → 0.5

Time: 10.5s

Precision: binary64

Cost: 19904

Math

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

↓

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

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
 (/ (pow (* PI (* 2.0 n)) (- 0.5 (/ k 2.0))) (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) {
	return pow((((double) M_PI) * (2.0 * n)), (0.5 - (k / 2.0))) / 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) {
	return Math.pow((Math.PI * (2.0 * n)), (0.5 - (k / 2.0))) / 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):
	return math.pow((math.pi * (2.0 * n)), (0.5 - (k / 2.0))) / 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)
	return Float64((Float64(pi * Float64(2.0 * n)) ^ Float64(0.5 - Float64(k / 2.0))) / 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)
	tmp = ((pi * (2.0 * n)) ^ (0.5 - (k / 2.0))) / 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_] := N[(N[Power[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision], N[(0.5 - N[(k / 2.0), $MachinePrecision]), $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 0.5

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

   [Start] 0.5	\[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   associate-*l/ [=>] 0.5	\[ \color{blue}{\frac{1 \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]
   *-lft-identity [=>] 0.5	\[ \frac{\color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}}{\sqrt{k}} \]
   sqr-pow [=>] 0.6	\[ \frac{\color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{\frac{1 - k}{2}}{2}\right)} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{\frac{1 - k}{2}}{2}\right)}}}{\sqrt{k}} \]
   sqr-pow [<=] 0.5	\[ \frac{\color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}}{\sqrt{k}} \]
   *-commutative [=>] 0.5	\[ \frac{{\left(\color{blue}{\left(\pi \cdot 2\right)} \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}} \]
   associate-*l* [=>] 0.5	\[ \frac{{\color{blue}{\left(\pi \cdot \left(2 \cdot n\right)\right)}}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}} \]
   div-sub [=>] 0.5	\[ \frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\color{blue}{\left(\frac{1}{2} - \frac{k}{2}\right)}}}{\sqrt{k}} \]
   metadata-eval [=>] 0.5	\[ \frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\color{blue}{0.5} - \frac{k}{2}\right)}}{\sqrt{k}} \]

3. Final simplification 0.5

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

Alternatives

Alternative 1
Error	0.7
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 7.5 \cdot 10^{-32}:\\ \;\;\;\;\sqrt{\frac{\pi}{k}} \cdot \sqrt{2 \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(n \cdot \left(\pi \cdot 2\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 2
Error	22.7
Cost	19584

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

Alternative 3
Error	22.7
Cost	19584

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

Alternative 4
Error	32.4
Cost	13248

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

Alternative 5
Error	32.4
Cost	13248

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

Alternative 6
Error	33.1
Cost	13184

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

Reproduce

herbie shell --seed 2023018 
(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))))