Migdal et al, Equation (51)

Average Error: 0.5 → 0.5

Time: 8.3s

Precision: binary64

Math

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

↓

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

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
 (fabs (/ (pow (sqrt (/ (* PI -2.0) (/ -1.0 n))) (- 1.0 k)) (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 fabs((pow(sqrt(((((double) M_PI) * -2.0) / (-1.0 / n))), (1.0 - k)) / 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.abs((Math.pow(Math.sqrt(((Math.PI * -2.0) / (-1.0 / n))), (1.0 - k)) / 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.fabs((math.pow(math.sqrt(((math.pi * -2.0) / (-1.0 / n))), (1.0 - k)) / 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 abs(Float64((sqrt(Float64(Float64(pi * -2.0) / Float64(-1.0 / n))) ^ Float64(1.0 - k)) / 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 = abs(((sqrt(((pi * -2.0) / (-1.0 / n))) ^ (1.0 - k)) / 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[Abs[N[(N[Power[N[Sqrt[N[(N[(Pi * -2.0), $MachinePrecision] / N[(-1.0 / n), $MachinePrecision]), $MachinePrecision]], $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]

Error

Bits error versus k

Bits error versus n

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

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

3. Applied egg-rr 0.4

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

4. Taylor expanded in n around -inf 64.0

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

5. Simplified 0.5

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

6. Final simplification 0.5

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

Reproduce

herbie shell --seed 2022131 
(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))))