Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.5%

Time: 25.0s

Precision: binary64

Cost: 19968

Math

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

↓

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

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 k -0.5) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))

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(k, -0.5) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}

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(k, -0.5) * Math.pow(((2.0 * Math.PI) * n), ((1.0 - k) / 2.0));
}

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(k, -0.5) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))

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((k ^ -0.5) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0)))
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 = (k ^ -0.5) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0));
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[k, -0.5], $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.6%

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

2. Applied egg-rr 73.6%

   \[\leadsto \color{blue}{\left(e^{\mathsf{log1p}\left({k}^{-0.5}\right)} - 1\right)} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   Step-by-step derivation

   [Start] 99.6%	\[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   expm1-log1p-u [=>] 96.2%	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{1}{\sqrt{k}}\right)\right)} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   expm1-udef [=>] 73.6%	\[ \color{blue}{\left(e^{\mathsf{log1p}\left(\frac{1}{\sqrt{k}}\right)} - 1\right)} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   pow1/2 [=>] 73.6%	\[ \left(e^{\mathsf{log1p}\left(\frac{1}{\color{blue}{{k}^{0.5}}}\right)} - 1\right) \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   pow-flip [=>] 73.6%	\[ \left(e^{\mathsf{log1p}\left(\color{blue}{{k}^{\left(-0.5\right)}}\right)} - 1\right) \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   metadata-eval [=>] 73.6%	\[ \left(e^{\mathsf{log1p}\left({k}^{\color{blue}{-0.5}}\right)} - 1\right) \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

3. Simplified 99.7%

   \[\leadsto \color{blue}{{k}^{-0.5}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   Step-by-step derivation

   [Start] 73.6%	\[ \left(e^{\mathsf{log1p}\left({k}^{-0.5}\right)} - 1\right) \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   expm1-def [=>] 96.2%	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left({k}^{-0.5}\right)\right)} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   expm1-log1p [=>] 99.7%	\[ \color{blue}{{k}^{-0.5}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

4. Final simplification 99.7%

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

Alternatives

Alternative 1
Accuracy	99.5%
Cost	19968

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

Alternative 2
Accuracy	99.3%
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 9.2 \cdot 10^{-38}:\\ \;\;\;\;{k}^{-0.5} \cdot \sqrt{\left(2 \cdot \pi\right) \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 3
Accuracy	99.5%
Cost	19904

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

Alternative 4
Accuracy	49.4%
Cost	19648

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

Alternative 5
Accuracy	37.9%
Cost	13184

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

Alternative 6
Accuracy	37.9%
Cost	13184

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

Reproduce

herbie shell --seed 2023272 
(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))))