Migdal et al, Equation (51)

Average Error: 0.5 → 0.4

Time: 39.0s

Precision: binary64

Cost: 19904

Math

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

↓

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

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 (* (* 2.0 PI) n) (/ (- 1.0 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(((2.0 * ((double) M_PI)) * n), ((1.0 - 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(((2.0 * Math.PI) * n), ((1.0 - 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(((2.0 * math.pi) * n), ((1.0 - 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(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - 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 = (((2.0 * pi) * n) ^ ((1.0 - 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[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $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

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

   [Start] 0.5	\[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   rational_best-simplify-1 [=>] 0.5	\[ \color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}}} \]
   rational_best-simplify-55 [=>] 0.4	\[ \color{blue}{1 \cdot \frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]
   rational_best-simplify-1 [=>] 0.4	\[ \color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}} \cdot 1} \]
   rational_best-simplify-7 [=>] 0.4	\[ \color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]

3. Final simplification 0.4

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

Alternatives

Alternative 1
Error	1.4
Cost	19904

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

Alternative 2
Error	22.5
Cost	19584

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

Alternative 3
Error	22.5
Cost	19584

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

Alternative 4
Error	33.5
Cost	13184

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

Alternative 5
Error	33.5
Cost	13184

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

Reproduce

herbie shell --seed 2023099 
(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))))