Migdal et al, Equation (51)

Average Error: 0.5 → 0.5

Time: 25.6s

Precision: binary64

Cost: 33216

Math

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

↓

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

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 (+ n n) (* -0.5 k))
  (/ (pow (+ n n) 0.5) (/ (sqrt k) (pow PI (* (+ k -1.0) -0.5))))))

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

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

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

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

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

3. Applied egg-rr 0.6

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

4. Simplified 0.6

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

   [Start] 0.6	\[ {\left({\left(n + n\right)}^{0.5}\right)}^{\left(1 - k\right)} \cdot \frac{{\pi}^{\left(\left(1 - k\right) \cdot 0.5\right)}}{\sqrt{k}} \]
   exponential.json-simplify-32 [=>] 0.6	\[ \color{blue}{{\left(n + n\right)}^{\left(0.5 \cdot \left(1 - k\right)\right)}} \cdot \frac{{\pi}^{\left(\left(1 - k\right) \cdot 0.5\right)}}{\sqrt{k}} \]
   rational.json-simplify-2 [=>] 0.6	\[ {\left(n + n\right)}^{\left(0.5 \cdot \left(1 - k\right)\right)} \cdot \frac{{\pi}^{\color{blue}{\left(0.5 \cdot \left(1 - k\right)\right)}}}{\sqrt{k}} \]

5. Applied egg-rr 0.5

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

6. Applied egg-rr 0.5

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

7. Final simplification 0.5

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

Alternatives

Alternative 1
Error	0.5
Cost	26752

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

Alternative 2
Error	0.5
Cost	26496

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

Alternative 3
Error	0.5
Cost	26240

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

Alternative 4
Error	0.5
Cost	19904

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

Alternative 5
Error	21.7
Cost	19840

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

Alternative 6
Error	21.7
Cost	19584

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

Reproduce

herbie shell --seed 2023073 
(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))))