Migdal et al, Equation (51)

Average Error: 0.77% → 0.82%

Time: 9.8s

Precision: binary64

Cost: 26624

Math

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

↓

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

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

Derivation

1. Initial program 0.77

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

2. Applied egg-rr 0.81

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

3. Simplified 0.82

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

   [Start] 0.81	\[ \frac{{2}^{\left(0.5 - k \cdot 0.5\right)}}{\frac{\sqrt{k}}{{\left(\pi \cdot n\right)}^{\left(0.5 - k \cdot 0.5\right)}}} \]
   associate-/r/ [=>] 0.82	\[ \color{blue}{\frac{{2}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}} \cdot {\left(\pi \cdot n\right)}^{\left(0.5 - k \cdot 0.5\right)}} \]
   *-commutative [=>] 0.82	\[ \color{blue}{{\left(\pi \cdot n\right)}^{\left(0.5 - k \cdot 0.5\right)} \cdot \frac{{2}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}}} \]
   *-commutative [=>] 0.82	\[ {\color{blue}{\left(n \cdot \pi\right)}}^{\left(0.5 - k \cdot 0.5\right)} \cdot \frac{{2}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}} \]
   cancel-sign-sub-inv [=>] 0.82	\[ {\left(n \cdot \pi\right)}^{\color{blue}{\left(0.5 + \left(-k\right) \cdot 0.5\right)}} \cdot \frac{{2}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}} \]
   metadata-eval [<=] 0.82	\[ {\left(n \cdot \pi\right)}^{\left(\color{blue}{1 \cdot 0.5} + \left(-k\right) \cdot 0.5\right)} \cdot \frac{{2}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}} \]
   mul-1-neg [<=] 0.82	\[ {\left(n \cdot \pi\right)}^{\left(1 \cdot 0.5 + \color{blue}{\left(-1 \cdot k\right)} \cdot 0.5\right)} \cdot \frac{{2}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}} \]
   distribute-rgt-in [<=] 0.82	\[ {\left(n \cdot \pi\right)}^{\color{blue}{\left(0.5 \cdot \left(1 + -1 \cdot k\right)\right)}} \cdot \frac{{2}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}} \]
   mul-1-neg [=>] 0.82	\[ {\left(n \cdot \pi\right)}^{\left(0.5 \cdot \left(1 + \color{blue}{\left(-k\right)}\right)\right)} \cdot \frac{{2}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}} \]
   sub-neg [<=] 0.82	\[ {\left(n \cdot \pi\right)}^{\left(0.5 \cdot \color{blue}{\left(1 - k\right)}\right)} \cdot \frac{{2}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}} \]
   cancel-sign-sub-inv [=>] 0.82	\[ {\left(n \cdot \pi\right)}^{\left(0.5 \cdot \left(1 - k\right)\right)} \cdot \frac{{2}^{\color{blue}{\left(0.5 + \left(-k\right) \cdot 0.5\right)}}}{\sqrt{k}} \]
   metadata-eval [<=] 0.82	\[ {\left(n \cdot \pi\right)}^{\left(0.5 \cdot \left(1 - k\right)\right)} \cdot \frac{{2}^{\left(\color{blue}{1 \cdot 0.5} + \left(-k\right) \cdot 0.5\right)}}{\sqrt{k}} \]
   mul-1-neg [<=] 0.82	\[ {\left(n \cdot \pi\right)}^{\left(0.5 \cdot \left(1 - k\right)\right)} \cdot \frac{{2}^{\left(1 \cdot 0.5 + \color{blue}{\left(-1 \cdot k\right)} \cdot 0.5\right)}}{\sqrt{k}} \]
   distribute-rgt-in [<=] 0.82	\[ {\left(n \cdot \pi\right)}^{\left(0.5 \cdot \left(1 - k\right)\right)} \cdot \frac{{2}^{\color{blue}{\left(0.5 \cdot \left(1 + -1 \cdot k\right)\right)}}}{\sqrt{k}} \]
   mul-1-neg [=>] 0.82	\[ {\left(n \cdot \pi\right)}^{\left(0.5 \cdot \left(1 - k\right)\right)} \cdot \frac{{2}^{\left(0.5 \cdot \left(1 + \color{blue}{\left(-k\right)}\right)\right)}}{\sqrt{k}} \]
   sub-neg [<=] 0.82	\[ {\left(n \cdot \pi\right)}^{\left(0.5 \cdot \left(1 - k\right)\right)} \cdot \frac{{2}^{\left(0.5 \cdot \color{blue}{\left(1 - k\right)}\right)}}{\sqrt{k}} \]

4. Final simplification 0.82

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

Alternatives

Alternative 1
Error	0.95%
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 3.2 \cdot 10^{-11}:\\ \;\;\;\;\sqrt{\frac{2}{k}} \cdot \sqrt{n \cdot \pi}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(n \cdot \left(\pi \cdot 2\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 2
Error	0.7%
Cost	19904

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

Alternative 3
Error	34.5%
Cost	19584

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

Alternative 4
Error	34.47%
Cost	19584

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

Alternative 5
Error	50.12%
Cost	13312

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

Alternative 6
Error	50.13%
Cost	13312

\[\frac{1}{\sqrt{\frac{\frac{\frac{k}{n}}{2}}{\pi}}} \]

Alternative 7
Error	50.97%
Cost	13184

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

Alternative 8
Error	50.97%
Cost	13184

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

Alternative 9
Error	50.97%
Cost	13184

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

Alternative 10
Error	51%
Cost	13184

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

Reproduce

herbie shell --seed 2023121 
(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))))