Migdal et al, Equation (51)

Percentage Accurate: 99.5% → 99.5%

Time: 15.2s

Alternatives: 12

Speedup: 1.0×

• could not determine a ground truth

  1. k = 2.00000000000000007e154
  2. n = 0.0

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, 0.6× speedup

Math

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

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* 2.0 (* PI n))))
   (/ (* (pow t_0 (* k -0.5)) (sqrt t_0)) (sqrt k))))

C

double code(double k, double n) {
	double t_0 = 2.0 * (((double) M_PI) * n);
	return (pow(t_0, (k * -0.5)) * sqrt(t_0)) / sqrt(k);
}

Java

public static double code(double k, double n) {
	double t_0 = 2.0 * (Math.PI * n);
	return (Math.pow(t_0, (k * -0.5)) * Math.sqrt(t_0)) / Math.sqrt(k);
}

Python

def code(k, n):
	t_0 = 2.0 * (math.pi * n)
	return (math.pow(t_0, (k * -0.5)) * math.sqrt(t_0)) / math.sqrt(k)

Julia

function code(k, n)
	t_0 = Float64(2.0 * Float64(pi * n))
	return Float64(Float64((t_0 ^ Float64(k * -0.5)) * sqrt(t_0)) / sqrt(k))
end

MATLAB

function tmp = code(k, n)
	t_0 = 2.0 * (pi * n);
	tmp = ((t_0 ^ (k * -0.5)) * sqrt(t_0)) / sqrt(k);
end

Wolfram

code[k_, n_] := Block[{t$95$0 = N[(2.0 * N[(Pi * n), $MachinePrecision]), $MachinePrecision]}, N[(N[(N[Power[t$95$0, N[(k * -0.5), $MachinePrecision]], $MachinePrecision] * N[Sqrt[t$95$0], $MachinePrecision]), $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]]

Derivation

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Alternative 2: 99.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \left(2 \cdot n\right)\\ \mathbf{if}\;k \leq 2.8 \cdot 10^{-52}:\\ \;\;\;\;\frac{\sqrt{t_0}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\sqrt{\frac{k}{{t_0}^{\left(1 - k\right)}}}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* PI (* 2.0 n))))
   (if (<= k 2.8e-52)
     (/ (sqrt t_0) (sqrt k))
     (/ 1.0 (sqrt (/ k (pow t_0 (- 1.0 k))))))))

C

double code(double k, double n) {
	double t_0 = ((double) M_PI) * (2.0 * n);
	double tmp;
	if (k <= 2.8e-52) {
		tmp = sqrt(t_0) / sqrt(k);
	} else {
		tmp = 1.0 / sqrt((k / pow(t_0, (1.0 - k))));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double t_0 = Math.PI * (2.0 * n);
	double tmp;
	if (k <= 2.8e-52) {
		tmp = Math.sqrt(t_0) / Math.sqrt(k);
	} else {
		tmp = 1.0 / Math.sqrt((k / Math.pow(t_0, (1.0 - k))));
	}
	return tmp;
}

Python

def code(k, n):
	t_0 = math.pi * (2.0 * n)
	tmp = 0
	if k <= 2.8e-52:
		tmp = math.sqrt(t_0) / math.sqrt(k)
	else:
		tmp = 1.0 / math.sqrt((k / math.pow(t_0, (1.0 - k))))
	return tmp

Julia

function code(k, n)
	t_0 = Float64(pi * Float64(2.0 * n))
	tmp = 0.0
	if (k <= 2.8e-52)
		tmp = Float64(sqrt(t_0) / sqrt(k));
	else
		tmp = Float64(1.0 / sqrt(Float64(k / (t_0 ^ Float64(1.0 - k)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	t_0 = pi * (2.0 * n);
	tmp = 0.0;
	if (k <= 2.8e-52)
		tmp = sqrt(t_0) / sqrt(k);
	else
		tmp = 1.0 / sqrt((k / (t_0 ^ (1.0 - k))));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := Block[{t$95$0 = N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[k, 2.8e-52], N[(N[Sqrt[t$95$0], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision], N[(1.0 / N[Sqrt[N[(k / N[Power[t$95$0, N[(1.0 - k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]

Derivation

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Alternative 3: 99.5% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 4: 99.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 1.75 \cdot 10^{-53}:\\ \;\;\;\;\sqrt{\frac{\pi}{k}} \cdot \sqrt{2 \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 1.75e-53)
   (* (sqrt (/ PI k)) (sqrt (* 2.0 n)))
   (sqrt (/ (pow (* PI (* 2.0 n)) (- 1.0 k)) k))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 1.75e-53) {
		tmp = sqrt((((double) M_PI) / k)) * sqrt((2.0 * n));
	} else {
		tmp = sqrt((pow((((double) M_PI) * (2.0 * n)), (1.0 - k)) / k));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 1.75e-53) {
		tmp = Math.sqrt((Math.PI / k)) * Math.sqrt((2.0 * n));
	} else {
		tmp = Math.sqrt((Math.pow((Math.PI * (2.0 * n)), (1.0 - k)) / k));
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if k <= 1.75e-53:
		tmp = math.sqrt((math.pi / k)) * math.sqrt((2.0 * n))
	else:
		tmp = math.sqrt((math.pow((math.pi * (2.0 * n)), (1.0 - k)) / k))
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 1.75e-53)
		tmp = Float64(sqrt(Float64(pi / k)) * sqrt(Float64(2.0 * n)));
	else
		tmp = sqrt(Float64((Float64(pi * Float64(2.0 * n)) ^ Float64(1.0 - k)) / k));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 1.75e-53)
		tmp = sqrt((pi / k)) * sqrt((2.0 * n));
	else
		tmp = sqrt((((pi * (2.0 * n)) ^ (1.0 - k)) / k));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 1.75e-53], N[(N[Sqrt[N[(Pi / k), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(2.0 * n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sqrt[N[(N[Power[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision] / k), $MachinePrecision]], $MachinePrecision]]

Derivation

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Alternative 5: 99.5% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 6: 50.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 1.2 \cdot 10^{+229}:\\ \;\;\;\;\frac{1}{\frac{\sqrt{k}}{\sqrt{\pi \cdot \left(2 \cdot n\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(n \cdot \left(\pi \cdot \frac{2}{k}\right)\right)}^{1.5}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 1.2e+229)
   (/ 1.0 (/ (sqrt k) (sqrt (* PI (* 2.0 n)))))
   (cbrt (pow (* n (* PI (/ 2.0 k))) 1.5))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 1.2e+229) {
		tmp = 1.0 / (sqrt(k) / sqrt((((double) M_PI) * (2.0 * n))));
	} else {
		tmp = cbrt(pow((n * (((double) M_PI) * (2.0 / k))), 1.5));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 1.2e+229) {
		tmp = 1.0 / (Math.sqrt(k) / Math.sqrt((Math.PI * (2.0 * n))));
	} else {
		tmp = Math.cbrt(Math.pow((n * (Math.PI * (2.0 / k))), 1.5));
	}
	return tmp;
}

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 1.2e+229)
		tmp = Float64(1.0 / Float64(sqrt(k) / sqrt(Float64(pi * Float64(2.0 * n)))));
	else
		tmp = cbrt((Float64(n * Float64(pi * Float64(2.0 / k))) ^ 1.5));
	end
	return tmp
end

Wolfram

code[k_, n_] := If[LessEqual[k, 1.2e+229], N[(1.0 / N[(N[Sqrt[k], $MachinePrecision] / N[Sqrt[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Power[N[Power[N[(n * N[(Pi * N[(2.0 / k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 1.5], $MachinePrecision], 1/3], $MachinePrecision]]

Derivation

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Alternative 7: 54.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 5.3 \cdot 10^{+147}:\\ \;\;\;\;\frac{1}{\frac{\sqrt{k}}{\sqrt{\pi \cdot \left(2 \cdot n\right)}}}\\ \mathbf{else}:\\ \;\;\;\;{\left({\left(n \cdot \left(\pi \cdot \frac{2}{k}\right)\right)}^{3}\right)}^{0.16666666666666666}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 5.3e+147)
   (/ 1.0 (/ (sqrt k) (sqrt (* PI (* 2.0 n)))))
   (pow (pow (* n (* PI (/ 2.0 k))) 3.0) 0.16666666666666666)))

C

double code(double k, double n) {
	double tmp;
	if (k <= 5.3e+147) {
		tmp = 1.0 / (sqrt(k) / sqrt((((double) M_PI) * (2.0 * n))));
	} else {
		tmp = pow(pow((n * (((double) M_PI) * (2.0 / k))), 3.0), 0.16666666666666666);
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 5.3e+147) {
		tmp = 1.0 / (Math.sqrt(k) / Math.sqrt((Math.PI * (2.0 * n))));
	} else {
		tmp = Math.pow(Math.pow((n * (Math.PI * (2.0 / k))), 3.0), 0.16666666666666666);
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if k <= 5.3e+147:
		tmp = 1.0 / (math.sqrt(k) / math.sqrt((math.pi * (2.0 * n))))
	else:
		tmp = math.pow(math.pow((n * (math.pi * (2.0 / k))), 3.0), 0.16666666666666666)
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 5.3e+147)
		tmp = Float64(1.0 / Float64(sqrt(k) / sqrt(Float64(pi * Float64(2.0 * n)))));
	else
		tmp = (Float64(n * Float64(pi * Float64(2.0 / k))) ^ 3.0) ^ 0.16666666666666666;
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 5.3e+147)
		tmp = 1.0 / (sqrt(k) / sqrt((pi * (2.0 * n))));
	else
		tmp = ((n * (pi * (2.0 / k))) ^ 3.0) ^ 0.16666666666666666;
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 5.3e+147], N[(1.0 / N[(N[Sqrt[k], $MachinePrecision] / N[Sqrt[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Power[N[Power[N[(n * N[(Pi * N[(2.0 / k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision], 0.16666666666666666], $MachinePrecision]]

Derivation

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Alternative 8: 50.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 1.35 \cdot 10^{+229}:\\ \;\;\;\;\sqrt{\frac{\pi}{k}} \cdot \sqrt{2 \cdot n}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(n \cdot \left(\pi \cdot \frac{2}{k}\right)\right)}^{1.5}}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 1.35e+229)
   (* (sqrt (/ PI k)) (sqrt (* 2.0 n)))
   (cbrt (pow (* n (* PI (/ 2.0 k))) 1.5))))

C

double code(double k, double n) {
	double tmp;
	if (k <= 1.35e+229) {
		tmp = sqrt((((double) M_PI) / k)) * sqrt((2.0 * n));
	} else {
		tmp = cbrt(pow((n * (((double) M_PI) * (2.0 / k))), 1.5));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 1.35e+229) {
		tmp = Math.sqrt((Math.PI / k)) * Math.sqrt((2.0 * n));
	} else {
		tmp = Math.cbrt(Math.pow((n * (Math.PI * (2.0 / k))), 1.5));
	}
	return tmp;
}

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 1.35e+229)
		tmp = Float64(sqrt(Float64(pi / k)) * sqrt(Float64(2.0 * n)));
	else
		tmp = cbrt((Float64(n * Float64(pi * Float64(2.0 / k))) ^ 1.5));
	end
	return tmp
end

Wolfram

code[k_, n_] := If[LessEqual[k, 1.35e+229], N[(N[Sqrt[N[(Pi / k), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(2.0 * n), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Power[N[Power[N[(n * N[(Pi * N[(2.0 / k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 1.5], $MachinePrecision], 1/3], $MachinePrecision]]

Derivation

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Alternative 9: 49.6% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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Alternative 10: 49.6% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

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1. Add Preprocessing

Alternative 11: 37.8% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Alternative 12: 37.9% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

&prev;&pcontext;&pcontext2;&ctx;
1. Add Preprocessing

Reproduce

herbie shell --seed 2023364 
(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))))