Migdal et al, Equation (51)

Average Error: 0.5 → 0.7

Time: 11.2s

Precision: binary64

Cost: 32896

Math

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

↓

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

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 (* 2.0 PI) (+ 0.5 (* k -0.5))) (sqrt k)) (sqrt (pow n (- 1.0 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)), (0.5 + (k * -0.5))) / sqrt(k)) * sqrt(pow(n, (1.0 - 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), (0.5 + (k * -0.5))) / Math.sqrt(k)) * Math.sqrt(Math.pow(n, (1.0 - 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), (0.5 + (k * -0.5))) / math.sqrt(k)) * math.sqrt(math.pow(n, (1.0 - 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) ^ Float64(0.5 + Float64(k * -0.5))) / sqrt(k)) * sqrt((n ^ Float64(1.0 - 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) ^ (0.5 + (k * -0.5))) / sqrt(k)) * sqrt((n ^ (1.0 - 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[(N[Power[N[(2.0 * Pi), $MachinePrecision], N[(0.5 + N[(k * -0.5), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[Power[n, N[(1.0 - k), $MachinePrecision]], $MachinePrecision]], $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. Applied egg-rr 0.7

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

3. Simplified 0.7

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

   [Start] 0.7	\[ \frac{{\left(2 \cdot \pi\right)}^{\left(0.5 - k \cdot 0.5\right)}}{\frac{\sqrt{k}}{{\left({n}^{\left(1 - k\right)}\right)}^{0.5}}} \]
   associate-/r/ [=>] 0.7	\[ \color{blue}{\frac{{\left(2 \cdot \pi\right)}^{\left(0.5 - k \cdot 0.5\right)}}{\sqrt{k}} \cdot {\left({n}^{\left(1 - k\right)}\right)}^{0.5}} \]
   *-commutative [=>] 0.7	\[ \frac{{\left(2 \cdot \pi\right)}^{\left(0.5 - \color{blue}{0.5 \cdot k}\right)}}{\sqrt{k}} \cdot {\left({n}^{\left(1 - k\right)}\right)}^{0.5} \]
   unpow1/2 [=>] 0.7	\[ \frac{{\left(2 \cdot \pi\right)}^{\left(0.5 - 0.5 \cdot k\right)}}{\sqrt{k}} \cdot \color{blue}{\sqrt{{n}^{\left(1 - k\right)}}} \]

4. Final simplification 0.7

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

Alternatives

Alternative 1
Error	0.6
Cost	26624

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

Alternative 2
Error	0.5
Cost	19968

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

Alternative 3
Error	0.5
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 6.6 \cdot 10^{-23}:\\ \;\;\;\;\sqrt{\frac{\pi}{k}} \cdot \sqrt{2 \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 4
Error	0.5
Cost	19904

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

Alternative 5
Error	22.7
Cost	19584

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

Alternative 6
Error	33.0
Cost	13312

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

Alternative 7
Error	33.6
Cost	13184

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

Alternative 8
Error	33.6
Cost	13184

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

Reproduce

herbie shell --seed 2023083 
(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))))