Migdal et al, Equation (51)

Percentage Accurate: 99.5% → 99.5%

Time: 9.7s

Alternatives: 9

Speedup: 1.1×

• Could not converge on a ground truth

  1. k = 7.966076285234342e+258
  2. n = -3.1117481511385455e+47

Specification

Math

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

FPCore

(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (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));
}

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));
}

Python

def code(k, n):
	return (1.0 / math.sqrt(k)) * 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

MATLAB

function tmp = code(k, n)
	tmp = (1.0 / sqrt(k)) * (((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]

Initial Program: 99.5% accurate, 1.0× speedup

Math

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

FPCore

(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (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));
}

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));
}

Python

def code(k, n):
	return (1.0 / math.sqrt(k)) * 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

MATLAB

function tmp = code(k, n)
	tmp = (1.0 / sqrt(k)) * (((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]

Alternative 1: 99.5% accurate, 0.8× speedup

Math

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

FPCore

(FPCore (k n)
 :precision binary64
 (*
  (sqrt (/ 1.0 k))
  (/ (sqrt (* 2.0 (* PI n))) (pow (* (+ PI PI) n) (* 0.5 k)))))

C

double code(double k, double n) {
	return sqrt((1.0 / k)) * (sqrt((2.0 * (((double) M_PI) * n))) / pow(((((double) M_PI) + ((double) M_PI)) * n), (0.5 * k)));
}

Java

public static double code(double k, double n) {
	return Math.sqrt((1.0 / k)) * (Math.sqrt((2.0 * (Math.PI * n))) / Math.pow(((Math.PI + Math.PI) * n), (0.5 * k)));
}

Python

def code(k, n):
	return math.sqrt((1.0 / k)) * (math.sqrt((2.0 * (math.pi * n))) / math.pow(((math.pi + math.pi) * n), (0.5 * k)))

Julia

function code(k, n)
	return Float64(sqrt(Float64(1.0 / k)) * Float64(sqrt(Float64(2.0 * Float64(pi * n))) / (Float64(Float64(pi + pi) * n) ^ Float64(0.5 * k))))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt((1.0 / k)) * (sqrt((2.0 * (pi * n))) / (((pi + pi) * n) ^ (0.5 * k)));
end

Wolfram

code[k_, n_] := N[(N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[N[(2.0 * N[(Pi * n), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[Power[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision], N[(0.5 * k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.5%

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

3. Applied rewrites 99.5%

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

4. Taylor expanded in k around 0

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

6. Applied rewrites 99.5%

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

7. Taylor expanded in n around 0

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

9. Applied rewrites 99.5%

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

10. Taylor expanded in k around 0

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

12. Applied rewrites 99.5%

    \[\leadsto \sqrt{\frac{1}{k}} \cdot \frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\color{blue}{\left(0.5 \cdot k\right)}}} \]
13. Add Preprocessing

Alternative 2: 99.5% accurate, 1.1× speedup

Math

\[\begin{array}{l} \\ \frac{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\left(\mathsf{fma}\left(-0.5, k, 0.5\right)\right)}}{\sqrt{k}} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (/ (pow (* (+ PI PI) n) (fma -0.5 k 0.5)) (sqrt k)))

C

double code(double k, double n) {
	return pow(((((double) M_PI) + ((double) M_PI)) * n), fma(-0.5, k, 0.5)) / sqrt(k);
}

Julia

function code(k, n)
	return Float64((Float64(Float64(pi + pi) * n) ^ fma(-0.5, k, 0.5)) / sqrt(k))
end

Wolfram

code[k_, n_] := N[(N[Power[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision], N[(-0.5 * k + 0.5), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.5%

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

3. Applied rewrites 99.5%

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

4. Taylor expanded in k around 0

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

6. Applied rewrites 99.5%

   \[\leadsto \frac{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\color{blue}{\left(\mathsf{fma}\left(-0.5, k, 0.5\right)\right)}}}{\sqrt{k}} \]
7. Add Preprocessing

Alternative 3: 98.1% accurate, 1.1× speedup

Math

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

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 1.0)
   (* (sqrt (/ 1.0 k)) (* (sqrt (* PI n)) (sqrt 2.0)))
   (/ (pow (* (+ PI PI) n) (* -0.5 k)) (sqrt k))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 1.0) {
		tmp = sqrt((1.0 / k)) * (sqrt((((double) M_PI) * n)) * sqrt(2.0));
	} else {
		tmp = pow(((((double) M_PI) + ((double) M_PI)) * n), (-0.5 * k)) / sqrt(k);
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 1.0) {
		tmp = Math.sqrt((1.0 / k)) * (Math.sqrt((Math.PI * n)) * Math.sqrt(2.0));
	} else {
		tmp = Math.pow(((Math.PI + Math.PI) * n), (-0.5 * k)) / Math.sqrt(k);
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if k <= 1.0:
		tmp = math.sqrt((1.0 / k)) * (math.sqrt((math.pi * n)) * math.sqrt(2.0))
	else:
		tmp = math.pow(((math.pi + math.pi) * n), (-0.5 * k)) / math.sqrt(k)
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 1.0)
		tmp = Float64(sqrt(Float64(1.0 / k)) * Float64(sqrt(Float64(pi * n)) * sqrt(2.0)));
	else
		tmp = Float64((Float64(Float64(pi + pi) * n) ^ Float64(-0.5 * k)) / sqrt(k));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 1.0)
		tmp = sqrt((1.0 / k)) * (sqrt((pi * n)) * sqrt(2.0));
	else
		tmp = (((pi + pi) * n) ^ (-0.5 * k)) / sqrt(k);
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 1.0], N[(N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[N[(Pi * n), $MachinePrecision]], $MachinePrecision] * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Power[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision], N[(-0.5 * k), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 1

   1. Initial program 98.9%

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

   3. Applied rewrites 99.2%

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

   4. Taylor expanded in k around 0

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

   6. Applied rewrites 99.2%

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

   7. Taylor expanded in k around 0

      \[\leadsto \sqrt{\frac{1}{k}} \cdot \color{blue}{\left(\sqrt{n \cdot \mathsf{PI}\left(\right)} \cdot \sqrt{2}\right)} \]

   9. Applied rewrites 96.7%

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

   11. Applied rewrites 96.6%

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

   if 1 < k

   1. Initial program 100.0%

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

   3. Applied rewrites 100.0%

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

   4. Taylor expanded in k around inf

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

   6. Applied rewrites 99.7%

      \[\leadsto \frac{{\left(\left(\pi + \pi\right) \cdot n\right)}^{\color{blue}{\left(-0.5 \cdot k\right)}}}{\sqrt{k}} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Alternative 4: 49.2% accurate, 2.0× speedup

Math

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

FPCore

(FPCore (k n)
 :precision binary64
 (* (sqrt (/ 1.0 k)) (* (sqrt (* PI n)) (sqrt 2.0))))

C

double code(double k, double n) {
	return sqrt((1.0 / k)) * (sqrt((((double) M_PI) * n)) * sqrt(2.0));
}

Java

public static double code(double k, double n) {
	return Math.sqrt((1.0 / k)) * (Math.sqrt((Math.PI * n)) * Math.sqrt(2.0));
}

Python

def code(k, n):
	return math.sqrt((1.0 / k)) * (math.sqrt((math.pi * n)) * math.sqrt(2.0))

Julia

function code(k, n)
	return Float64(sqrt(Float64(1.0 / k)) * Float64(sqrt(Float64(pi * n)) * sqrt(2.0)))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt((1.0 / k)) * (sqrt((pi * n)) * sqrt(2.0));
end

Wolfram

code[k_, n_] := N[(N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[N[(Pi * n), $MachinePrecision]], $MachinePrecision] * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.5%

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

3. Applied rewrites 99.5%

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

4. Taylor expanded in k around 0

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

6. Applied rewrites 99.5%

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

7. Taylor expanded in k around 0

   \[\leadsto \sqrt{\frac{1}{k}} \cdot \color{blue}{\left(\sqrt{n \cdot \mathsf{PI}\left(\right)} \cdot \sqrt{2}\right)} \]

9. Applied rewrites 49.1%

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

11. Applied rewrites 49.1%

    \[\leadsto \sqrt{\frac{1}{k}} \cdot \left(\sqrt{\pi \cdot n} \cdot \color{blue}{\sqrt{2}}\right) \]
12. Add Preprocessing

Alternative 5: 49.1% accurate, 2.0× speedup

Math

\[\begin{array}{l} \\ \sqrt{\frac{1}{k}} \cdot \left(\sqrt{\pi + \pi} \cdot \sqrt{n}\right) \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (* (sqrt (/ 1.0 k)) (* (sqrt (+ PI PI)) (sqrt n))))

C

double code(double k, double n) {
	return sqrt((1.0 / k)) * (sqrt((((double) M_PI) + ((double) M_PI))) * sqrt(n));
}

Java

public static double code(double k, double n) {
	return Math.sqrt((1.0 / k)) * (Math.sqrt((Math.PI + Math.PI)) * Math.sqrt(n));
}

Python

def code(k, n):
	return math.sqrt((1.0 / k)) * (math.sqrt((math.pi + math.pi)) * math.sqrt(n))

Julia

function code(k, n)
	return Float64(sqrt(Float64(1.0 / k)) * Float64(sqrt(Float64(pi + pi)) * sqrt(n)))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt((1.0 / k)) * (sqrt((pi + pi)) * sqrt(n));
end

Wolfram

code[k_, n_] := N[(N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[N[(Pi + Pi), $MachinePrecision]], $MachinePrecision] * N[Sqrt[n], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.5%

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

3. Applied rewrites 99.5%

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

4. Taylor expanded in k around 0

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

6. Applied rewrites 99.5%

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

7. Taylor expanded in k around 0

   \[\leadsto \sqrt{\frac{1}{k}} \cdot \color{blue}{\left(\sqrt{n \cdot \mathsf{PI}\left(\right)} \cdot \sqrt{2}\right)} \]

9. Applied rewrites 49.1%

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

11. Applied rewrites 49.0%

    \[\leadsto \sqrt{\frac{1}{k}} \cdot \left(\sqrt{\pi + \pi} \cdot \color{blue}{\sqrt{n}}\right) \]
12. Add Preprocessing

Alternative 6: 49.0% accurate, 2.7× speedup

Math

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

FPCore

(FPCore (k n) :precision binary64 (/ (sqrt (* (+ PI PI) n)) (sqrt k)))

C

double code(double k, double n) {
	return sqrt(((((double) M_PI) + ((double) M_PI)) * n)) / sqrt(k);
}

Java

public static double code(double k, double n) {
	return Math.sqrt(((Math.PI + Math.PI) * n)) / Math.sqrt(k);
}

Python

def code(k, n):
	return math.sqrt(((math.pi + math.pi) * n)) / math.sqrt(k)

Julia

function code(k, n)
	return Float64(sqrt(Float64(Float64(pi + pi) * n)) / sqrt(k))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt(((pi + pi) * n)) / sqrt(k);
end

Wolfram

code[k_, n_] := N[(N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.5%

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

2. Taylor expanded in k around 0

   \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]

4. Applied rewrites 37.9%

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

6. Applied rewrites 49.2%

   \[\leadsto \frac{\sqrt{\left(\pi + \pi\right) \cdot n}}{\color{blue}{\sqrt{k}}} \]
7. Add Preprocessing

Alternative 7: 37.9% accurate, 3.1× speedup

Math

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

FPCore

(FPCore (k n) :precision binary64 (sqrt (* (* n (/ PI k)) 2.0)))

C

double code(double k, double n) {
	return sqrt(((n * (((double) M_PI) / k)) * 2.0));
}

Java

public static double code(double k, double n) {
	return Math.sqrt(((n * (Math.PI / k)) * 2.0));
}

Python

def code(k, n):
	return math.sqrt(((n * (math.pi / k)) * 2.0))

Julia

function code(k, n)
	return sqrt(Float64(Float64(n * Float64(pi / k)) * 2.0))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt(((n * (pi / k)) * 2.0));
end

Wolfram

code[k_, n_] := N[Sqrt[N[(N[(n * N[(Pi / k), $MachinePrecision]), $MachinePrecision] * 2.0), $MachinePrecision]], $MachinePrecision]

Derivation

1. Initial program 99.5%

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

2. Taylor expanded in k around 0

   \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]

4. Applied rewrites 37.9%

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

6. Applied rewrites 37.9%

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

8. Applied rewrites 37.9%

   \[\leadsto \sqrt{\left(n \cdot \frac{\pi}{k}\right) \cdot 2} \]
9. Add Preprocessing

Alternative 8: 37.9% accurate, 3.1× speedup

Math

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

FPCore

(FPCore (k n) :precision binary64 (sqrt (/ (* (+ PI PI) n) k)))

C

double code(double k, double n) {
	return sqrt((((((double) M_PI) + ((double) M_PI)) * n) / k));
}

Java

public static double code(double k, double n) {
	return Math.sqrt((((Math.PI + Math.PI) * n) / k));
}

Python

def code(k, n):
	return math.sqrt((((math.pi + math.pi) * n) / k))

Julia

function code(k, n)
	return sqrt(Float64(Float64(Float64(pi + pi) * n) / k))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt((((pi + pi) * n) / k));
end

Wolfram

code[k_, n_] := N[Sqrt[N[(N[(N[(Pi + Pi), $MachinePrecision] * n), $MachinePrecision] / k), $MachinePrecision]], $MachinePrecision]

Derivation

1. Initial program 99.5%

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

2. Taylor expanded in k around 0

   \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]

4. Applied rewrites 37.9%

   \[\leadsto \color{blue}{\sqrt{\frac{\left(\pi + \pi\right) \cdot n}{k}}} \]
5. Add Preprocessing

Alternative 9: 37.9% accurate, 3.1× speedup

Math

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

FPCore

(FPCore (k n) :precision binary64 (sqrt (* (+ PI PI) (/ n k))))

C

double code(double k, double n) {
	return sqrt(((((double) M_PI) + ((double) M_PI)) * (n / k)));
}

Java

public static double code(double k, double n) {
	return Math.sqrt(((Math.PI + Math.PI) * (n / k)));
}

Python

def code(k, n):
	return math.sqrt(((math.pi + math.pi) * (n / k)))

Julia

function code(k, n)
	return sqrt(Float64(Float64(pi + pi) * Float64(n / k)))
end

MATLAB

function tmp = code(k, n)
	tmp = sqrt(((pi + pi) * (n / k)));
end

Wolfram

code[k_, n_] := N[Sqrt[N[(N[(Pi + Pi), $MachinePrecision] * N[(n / k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]

Derivation

1. Initial program 99.5%

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

2. Taylor expanded in k around 0

   \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]

4. Applied rewrites 37.9%

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

6. Applied rewrites 37.9%

   \[\leadsto \sqrt{\left(\pi + \pi\right) \cdot \frac{n}{k}} \]
7. Add Preprocessing

Reproduce

herbie shell --seed 2025121 
(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))))