Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.4%

Time: 9.8s

Alternatives: 10

Speedup: 1.0×

• Could not converge on a ground truth

  1. k = 9.961399402265316e+248
  2. n = -1.2921489848371892e+60

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: 99.4% 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: 99.4% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 99.4%

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

2. Taylor expanded in k around 0

   \[\leadsto \color{blue}{\sqrt{\frac{1}{k}}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
3. Step-by-step derivation

   1. lower-sqrt.f64 N/A

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

   2. lower-/.f64 99.4

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

4. Applied rewrites 99.4%

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

Alternative 2: 99.4% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

function tmp = code(k, n)
	tmp = (1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ (0.5 + (-0.5 * 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[(0.5 + N[(-0.5 * k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.4%

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

2. Taylor expanded in k around 0

   \[\leadsto \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\left(\frac{1}{2} + \frac{-1}{2} \cdot k\right)}} \]
3. Step-by-step derivation

   1. lower-+.f64 N/A

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

   2. lower-*.f64 99.4

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

4. Applied rewrites 99.4%

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

Alternative 3: 98.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \left(2 \cdot \pi\right) \cdot n\\ \mathbf{if}\;k \leq 1:\\ \;\;\;\;\sqrt{\frac{1}{k}} \cdot {t\_0}^{\left(\frac{1}{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\sqrt{k}} \cdot {t\_0}^{\left(-0.5 \cdot k\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* (* 2.0 PI) n)))
   (if (<= k 1.0)
     (* (sqrt (/ 1.0 k)) (pow t_0 (/ 1.0 2.0)))
     (* (/ 1.0 (sqrt k)) (pow t_0 (* -0.5 k))))))

C

double code(double k, double n) {
	double t_0 = (2.0 * ((double) M_PI)) * n;
	double tmp;
	if (k <= 1.0) {
		tmp = sqrt((1.0 / k)) * pow(t_0, (1.0 / 2.0));
	} else {
		tmp = (1.0 / sqrt(k)) * pow(t_0, (-0.5 * k));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double t_0 = (2.0 * Math.PI) * n;
	double tmp;
	if (k <= 1.0) {
		tmp = Math.sqrt((1.0 / k)) * Math.pow(t_0, (1.0 / 2.0));
	} else {
		tmp = (1.0 / Math.sqrt(k)) * Math.pow(t_0, (-0.5 * k));
	}
	return tmp;
}

Python

def code(k, n):
	t_0 = (2.0 * math.pi) * n
	tmp = 0
	if k <= 1.0:
		tmp = math.sqrt((1.0 / k)) * math.pow(t_0, (1.0 / 2.0))
	else:
		tmp = (1.0 / math.sqrt(k)) * math.pow(t_0, (-0.5 * k))
	return tmp

Julia

function code(k, n)
	t_0 = Float64(Float64(2.0 * pi) * n)
	tmp = 0.0
	if (k <= 1.0)
		tmp = Float64(sqrt(Float64(1.0 / k)) * (t_0 ^ Float64(1.0 / 2.0)));
	else
		tmp = Float64(Float64(1.0 / sqrt(k)) * (t_0 ^ Float64(-0.5 * k)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	t_0 = (2.0 * pi) * n;
	tmp = 0.0;
	if (k <= 1.0)
		tmp = sqrt((1.0 / k)) * (t_0 ^ (1.0 / 2.0));
	else
		tmp = (1.0 / sqrt(k)) * (t_0 ^ (-0.5 * k));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := Block[{t$95$0 = N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision]}, If[LessEqual[k, 1.0], N[(N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision] * N[Power[t$95$0, N[(1.0 / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[t$95$0, N[(-0.5 * k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if k < 1

   1. Initial program 98.8%

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

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\sqrt{\frac{1}{k}}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   3. Step-by-step derivation

      1. lower-sqrt.f64 N/A

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

      2. lower-/.f64 98.8

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

   4. Applied rewrites 98.8%

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

   5. Taylor expanded in k around 0

      \[\leadsto \sqrt{\frac{1}{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{\color{blue}{1}}{2}\right)} \]
   6. Step-by-step derivation

   7. Applied rewrites 96.3%

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

   if 1 < k

   1. Initial program 100.0%

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

   2. Taylor expanded in k around inf

      \[\leadsto \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\left(\frac{-1}{2} \cdot k\right)}} \]
   3. Step-by-step derivation

      1. lower-*.f64 99.7

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

   4. Applied rewrites 99.7%

      \[\leadsto \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\left(-0.5 \cdot k\right)}} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Alternative 4: 98.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{1}{\sqrt{k}}\\ t_1 := \left(2 \cdot \pi\right) \cdot n\\ \mathbf{if}\;k \leq 1:\\ \;\;\;\;t\_0 \cdot {t\_1}^{0.5}\\ \mathbf{else}:\\ \;\;\;\;t\_0 \cdot {t\_1}^{\left(-0.5 \cdot k\right)}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (let* ((t_0 (/ 1.0 (sqrt k))) (t_1 (* (* 2.0 PI) n)))
   (if (<= k 1.0) (* t_0 (pow t_1 0.5)) (* t_0 (pow t_1 (* -0.5 k))))))

C

double code(double k, double n) {
	double t_0 = 1.0 / sqrt(k);
	double t_1 = (2.0 * ((double) M_PI)) * n;
	double tmp;
	if (k <= 1.0) {
		tmp = t_0 * pow(t_1, 0.5);
	} else {
		tmp = t_0 * pow(t_1, (-0.5 * k));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double t_0 = 1.0 / Math.sqrt(k);
	double t_1 = (2.0 * Math.PI) * n;
	double tmp;
	if (k <= 1.0) {
		tmp = t_0 * Math.pow(t_1, 0.5);
	} else {
		tmp = t_0 * Math.pow(t_1, (-0.5 * k));
	}
	return tmp;
}

Python

def code(k, n):
	t_0 = 1.0 / math.sqrt(k)
	t_1 = (2.0 * math.pi) * n
	tmp = 0
	if k <= 1.0:
		tmp = t_0 * math.pow(t_1, 0.5)
	else:
		tmp = t_0 * math.pow(t_1, (-0.5 * k))
	return tmp

Julia

function code(k, n)
	t_0 = Float64(1.0 / sqrt(k))
	t_1 = Float64(Float64(2.0 * pi) * n)
	tmp = 0.0
	if (k <= 1.0)
		tmp = Float64(t_0 * (t_1 ^ 0.5));
	else
		tmp = Float64(t_0 * (t_1 ^ Float64(-0.5 * k)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	t_0 = 1.0 / sqrt(k);
	t_1 = (2.0 * pi) * n;
	tmp = 0.0;
	if (k <= 1.0)
		tmp = t_0 * (t_1 ^ 0.5);
	else
		tmp = t_0 * (t_1 ^ (-0.5 * k));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := Block[{t$95$0 = N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision]}, If[LessEqual[k, 1.0], N[(t$95$0 * N[Power[t$95$1, 0.5], $MachinePrecision]), $MachinePrecision], N[(t$95$0 * N[Power[t$95$1, N[(-0.5 * k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if k < 1

   1. Initial program 98.8%

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

   2. Taylor expanded in k around 0

      \[\leadsto \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\frac{1}{2}}} \]
   3. Step-by-step derivation

   4. Applied rewrites 96.3%

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

   if 1 < k

   1. Initial program 100.0%

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

   2. Taylor expanded in k around inf

      \[\leadsto \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\left(\frac{-1}{2} \cdot k\right)}} \]
   3. Step-by-step derivation

      1. lower-*.f64 99.7

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

   4. Applied rewrites 99.7%

      \[\leadsto \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\left(-0.5 \cdot k\right)}} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Alternative 5: 60.8% accurate, 1.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 4.2 \cdot 10^{-13}:\\ \;\;\;\;\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{0.5}\\ \mathbf{else}:\\ \;\;\;\;\frac{n \cdot \left(\sqrt{\frac{k \cdot \pi}{n}} \cdot \sqrt{2}\right)}{k}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= k 4.2e-13)
   (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) 0.5))
   (/ (* n (* (sqrt (/ (* k PI) n)) (sqrt 2.0))) k)))

C

double code(double k, double n) {
	double tmp;
	if (k <= 4.2e-13) {
		tmp = (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), 0.5);
	} else {
		tmp = (n * (sqrt(((k * ((double) M_PI)) / n)) * sqrt(2.0))) / k;
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (k <= 4.2e-13) {
		tmp = (1.0 / Math.sqrt(k)) * Math.pow(((2.0 * Math.PI) * n), 0.5);
	} else {
		tmp = (n * (Math.sqrt(((k * Math.PI) / n)) * Math.sqrt(2.0))) / k;
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if k <= 4.2e-13:
		tmp = (1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), 0.5)
	else:
		tmp = (n * (math.sqrt(((k * math.pi) / n)) * math.sqrt(2.0))) / k
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (k <= 4.2e-13)
		tmp = Float64(Float64(1.0 / sqrt(k)) * (Float64(Float64(2.0 * pi) * n) ^ 0.5));
	else
		tmp = Float64(Float64(n * Float64(sqrt(Float64(Float64(k * pi) / n)) * sqrt(2.0))) / k);
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 4.2e-13)
		tmp = (1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ 0.5);
	else
		tmp = (n * (sqrt(((k * pi) / n)) * sqrt(2.0))) / k;
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[k, 4.2e-13], N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], 0.5], $MachinePrecision]), $MachinePrecision], N[(N[(n * N[(N[Sqrt[N[(N[(k * Pi), $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / k), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 4.19999999999999977e-13

   1. Initial program 99.0%

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

   2. Taylor expanded in k around 0

      \[\leadsto \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\frac{1}{2}}} \]
   3. Step-by-step derivation

   4. Applied rewrites 98.8%

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

   if 4.19999999999999977e-13 < k

   1. Initial program 99.7%

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

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{\frac{-1}{2} \cdot \left(\sqrt{{k}^{3} \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \left(\log \left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right) \cdot \sqrt{2}\right)\right) + \sqrt{k \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \sqrt{2}}{k}} \]
   3. Step-by-step derivation

      1. lower-/.f64 N/A

         \[\leadsto \frac{\frac{-1}{2} \cdot \left(\sqrt{{k}^{3} \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \left(\log \left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right) \cdot \sqrt{2}\right)\right) + \sqrt{k \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \sqrt{2}}{\color{blue}{k}} \]

   4. Applied rewrites 34.9%

      \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(-0.5, \sqrt{{k}^{3} \cdot \left(n \cdot \pi\right)} \cdot \left(\log \left(2 \cdot \left(n \cdot \pi\right)\right) \cdot \sqrt{2}\right), \sqrt{k \cdot \left(n \cdot \pi\right)} \cdot \sqrt{2}\right)}{k}} \]

   5. Taylor expanded in n around inf

      \[\leadsto \frac{n \cdot \left(\frac{-1}{2} \cdot \left(\sqrt{\frac{{k}^{3} \cdot \mathsf{PI}\left(\right)}{n}} \cdot \left(\sqrt{2} \cdot \left(\log \left(2 \cdot \mathsf{PI}\left(\right)\right) + -1 \cdot \log \left(\frac{1}{n}\right)\right)\right)\right) + \sqrt{\frac{k \cdot \mathsf{PI}\left(\right)}{n}} \cdot \sqrt{2}\right)}{k} \]
   6. Step-by-step derivation

      1. lower-*.f64 N/A

         \[\leadsto \frac{n \cdot \left(\frac{-1}{2} \cdot \left(\sqrt{\frac{{k}^{3} \cdot \mathsf{PI}\left(\right)}{n}} \cdot \left(\sqrt{2} \cdot \left(\log \left(2 \cdot \mathsf{PI}\left(\right)\right) + -1 \cdot \log \left(\frac{1}{n}\right)\right)\right)\right) + \sqrt{\frac{k \cdot \mathsf{PI}\left(\right)}{n}} \cdot \sqrt{2}\right)}{k} \]

      2. lower-fma.f64 N/A

         \[\leadsto \frac{n \cdot \mathsf{fma}\left(\frac{-1}{2}, \sqrt{\frac{{k}^{3} \cdot \mathsf{PI}\left(\right)}{n}} \cdot \left(\sqrt{2} \cdot \left(\log \left(2 \cdot \mathsf{PI}\left(\right)\right) + -1 \cdot \log \left(\frac{1}{n}\right)\right)\right), \sqrt{\frac{k \cdot \mathsf{PI}\left(\right)}{n}} \cdot \sqrt{2}\right)}{k} \]

   7. Applied rewrites 43.8%

      \[\leadsto \frac{n \cdot \mathsf{fma}\left(-0.5, \sqrt{\frac{{k}^{3} \cdot \pi}{n}} \cdot \left(\sqrt{2} \cdot \left(\log \left(2 \cdot \pi\right) + -1 \cdot \log \left(\frac{1}{n}\right)\right)\right), \sqrt{\frac{k \cdot \pi}{n}} \cdot \sqrt{2}\right)}{k} \]

   8. Taylor expanded in k around 0

      \[\leadsto \frac{n \cdot \left(\sqrt{\frac{k \cdot \mathsf{PI}\left(\right)}{n}} \cdot \sqrt{2}\right)}{k} \]
   9. Step-by-step derivation

      1. lift-*.f64 N/A

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

      2. lift-PI.f64 N/A

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

      3. lift-/.f64 N/A

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

      4. lift-sqrt.f64 N/A

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

      5. lift-sqrt.f64 N/A

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

      6. lift-*.f64 27.7

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

   10. Applied rewrites 27.7%

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

Alternative 6: 60.7% accurate, 1.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;n \leq 10^{-18}:\\ \;\;\;\;\frac{n \cdot \left(\sqrt{\frac{k \cdot \pi}{n}} \cdot \sqrt{2}\right)}{k}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\sqrt{k}} \cdot \left(\sqrt{n \cdot \pi} \cdot \sqrt{2}\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= n 1e-18)
   (/ (* n (* (sqrt (/ (* k PI) n)) (sqrt 2.0))) k)
   (* (/ 1.0 (sqrt k)) (* (sqrt (* n PI)) (sqrt 2.0)))))

C

double code(double k, double n) {
	double tmp;
	if (n <= 1e-18) {
		tmp = (n * (sqrt(((k * ((double) M_PI)) / n)) * sqrt(2.0))) / k;
	} else {
		tmp = (1.0 / sqrt(k)) * (sqrt((n * ((double) M_PI))) * sqrt(2.0));
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (n <= 1e-18) {
		tmp = (n * (Math.sqrt(((k * Math.PI) / n)) * Math.sqrt(2.0))) / k;
	} else {
		tmp = (1.0 / Math.sqrt(k)) * (Math.sqrt((n * Math.PI)) * Math.sqrt(2.0));
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if n <= 1e-18:
		tmp = (n * (math.sqrt(((k * math.pi) / n)) * math.sqrt(2.0))) / k
	else:
		tmp = (1.0 / math.sqrt(k)) * (math.sqrt((n * math.pi)) * math.sqrt(2.0))
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (n <= 1e-18)
		tmp = Float64(Float64(n * Float64(sqrt(Float64(Float64(k * pi) / n)) * sqrt(2.0))) / k);
	else
		tmp = Float64(Float64(1.0 / sqrt(k)) * Float64(sqrt(Float64(n * pi)) * sqrt(2.0)));
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (n <= 1e-18)
		tmp = (n * (sqrt(((k * pi) / n)) * sqrt(2.0))) / k;
	else
		tmp = (1.0 / sqrt(k)) * (sqrt((n * pi)) * sqrt(2.0));
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[n, 1e-18], N[(N[(n * N[(N[Sqrt[N[(N[(k * Pi), $MachinePrecision] / n), $MachinePrecision]], $MachinePrecision] * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / k), $MachinePrecision], N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[(N[Sqrt[N[(n * Pi), $MachinePrecision]], $MachinePrecision] * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if n < 1.0000000000000001e-18

   1. Initial program 99.4%

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

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{\frac{-1}{2} \cdot \left(\sqrt{{k}^{3} \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \left(\log \left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right) \cdot \sqrt{2}\right)\right) + \sqrt{k \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \sqrt{2}}{k}} \]
   3. Step-by-step derivation

      1. lower-/.f64 N/A

         \[\leadsto \frac{\frac{-1}{2} \cdot \left(\sqrt{{k}^{3} \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \left(\log \left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right) \cdot \sqrt{2}\right)\right) + \sqrt{k \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \sqrt{2}}{\color{blue}{k}} \]

   4. Applied rewrites 58.9%

      \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(-0.5, \sqrt{{k}^{3} \cdot \left(n \cdot \pi\right)} \cdot \left(\log \left(2 \cdot \left(n \cdot \pi\right)\right) \cdot \sqrt{2}\right), \sqrt{k \cdot \left(n \cdot \pi\right)} \cdot \sqrt{2}\right)}{k}} \]

   5. Taylor expanded in n around inf

      \[\leadsto \frac{n \cdot \left(\frac{-1}{2} \cdot \left(\sqrt{\frac{{k}^{3} \cdot \mathsf{PI}\left(\right)}{n}} \cdot \left(\sqrt{2} \cdot \left(\log \left(2 \cdot \mathsf{PI}\left(\right)\right) + -1 \cdot \log \left(\frac{1}{n}\right)\right)\right)\right) + \sqrt{\frac{k \cdot \mathsf{PI}\left(\right)}{n}} \cdot \sqrt{2}\right)}{k} \]
   6. Step-by-step derivation

      1. lower-*.f64 N/A

         \[\leadsto \frac{n \cdot \left(\frac{-1}{2} \cdot \left(\sqrt{\frac{{k}^{3} \cdot \mathsf{PI}\left(\right)}{n}} \cdot \left(\sqrt{2} \cdot \left(\log \left(2 \cdot \mathsf{PI}\left(\right)\right) + -1 \cdot \log \left(\frac{1}{n}\right)\right)\right)\right) + \sqrt{\frac{k \cdot \mathsf{PI}\left(\right)}{n}} \cdot \sqrt{2}\right)}{k} \]

      2. lower-fma.f64 N/A

         \[\leadsto \frac{n \cdot \mathsf{fma}\left(\frac{-1}{2}, \sqrt{\frac{{k}^{3} \cdot \mathsf{PI}\left(\right)}{n}} \cdot \left(\sqrt{2} \cdot \left(\log \left(2 \cdot \mathsf{PI}\left(\right)\right) + -1 \cdot \log \left(\frac{1}{n}\right)\right)\right), \sqrt{\frac{k \cdot \mathsf{PI}\left(\right)}{n}} \cdot \sqrt{2}\right)}{k} \]

   7. Applied rewrites 91.4%

      \[\leadsto \frac{n \cdot \mathsf{fma}\left(-0.5, \sqrt{\frac{{k}^{3} \cdot \pi}{n}} \cdot \left(\sqrt{2} \cdot \left(\log \left(2 \cdot \pi\right) + -1 \cdot \log \left(\frac{1}{n}\right)\right)\right), \sqrt{\frac{k \cdot \pi}{n}} \cdot \sqrt{2}\right)}{k} \]

   8. Taylor expanded in k around 0

      \[\leadsto \frac{n \cdot \left(\sqrt{\frac{k \cdot \mathsf{PI}\left(\right)}{n}} \cdot \sqrt{2}\right)}{k} \]
   9. Step-by-step derivation

      1. lift-*.f64 N/A

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

      2. lift-PI.f64 N/A

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

      3. lift-/.f64 N/A

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

      4. lift-sqrt.f64 N/A

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

      5. lift-sqrt.f64 N/A

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

      6. lift-*.f64 74.4

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

   10. Applied rewrites 74.4%

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

   if 1.0000000000000001e-18 < n

   1. Initial program 99.4%

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

   2. Taylor expanded in k around 0

      \[\leadsto \frac{1}{\sqrt{k}} \cdot \color{blue}{\left(\sqrt{n \cdot \mathsf{PI}\left(\right)} \cdot \sqrt{2}\right)} \]
   3. Step-by-step derivation

      1. lower-*.f64 N/A

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

      2. lower-sqrt.f64 N/A

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

      3. lower-*.f64 N/A

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

      4. lift-PI.f64 N/A

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

      5. lower-sqrt.f64 48.6

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

   4. Applied rewrites 48.6%

      \[\leadsto \frac{1}{\sqrt{k}} \cdot \color{blue}{\left(\sqrt{n \cdot \pi} \cdot \sqrt{2}\right)} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Alternative 7: 48.2% accurate, 0.7× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \leq 10^{+142}:\\ \;\;\;\;\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{k \cdot \left(n \cdot \pi\right)} \cdot \sqrt{2}}{k}\\ \end{array} \end{array} \]

FPCore

(FPCore (k n)
 :precision binary64
 (if (<= (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))) 1e+142)
   (* (sqrt (/ (* n PI) k)) (sqrt 2.0))
   (/ (* (sqrt (* k (* n PI))) (sqrt 2.0)) k)))

C

double code(double k, double n) {
	double tmp;
	if (((1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0))) <= 1e+142) {
		tmp = sqrt(((n * ((double) M_PI)) / k)) * sqrt(2.0);
	} else {
		tmp = (sqrt((k * (n * ((double) M_PI)))) * sqrt(2.0)) / k;
	}
	return tmp;
}

Java

public static double code(double k, double n) {
	double tmp;
	if (((1.0 / Math.sqrt(k)) * Math.pow(((2.0 * Math.PI) * n), ((1.0 - k) / 2.0))) <= 1e+142) {
		tmp = Math.sqrt(((n * Math.PI) / k)) * Math.sqrt(2.0);
	} else {
		tmp = (Math.sqrt((k * (n * Math.PI))) * Math.sqrt(2.0)) / k;
	}
	return tmp;
}

Python

def code(k, n):
	tmp = 0
	if ((1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))) <= 1e+142:
		tmp = math.sqrt(((n * math.pi) / k)) * math.sqrt(2.0)
	else:
		tmp = (math.sqrt((k * (n * math.pi))) * math.sqrt(2.0)) / k
	return tmp

Julia

function code(k, n)
	tmp = 0.0
	if (Float64(Float64(1.0 / sqrt(k)) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0))) <= 1e+142)
		tmp = Float64(sqrt(Float64(Float64(n * pi) / k)) * sqrt(2.0));
	else
		tmp = Float64(Float64(sqrt(Float64(k * Float64(n * pi))) * sqrt(2.0)) / k);
	end
	return tmp
end

MATLAB

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (((1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0))) <= 1e+142)
		tmp = sqrt(((n * pi) / k)) * sqrt(2.0);
	else
		tmp = (sqrt((k * (n * pi))) * sqrt(2.0)) / k;
	end
	tmp_2 = tmp;
end

Wolfram

code[k_, n_] := If[LessEqual[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], 1e+142], N[(N[Sqrt[N[(N[(n * Pi), $MachinePrecision] / k), $MachinePrecision]], $MachinePrecision] * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[N[(k * N[(n * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision] / k), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 (/.f64 #s(literal 1 binary64) (sqrt.f64 k)) (pow.f64 (*.f64 (*.f64 #s(literal 2 binary64) (PI.f64)) n) (/.f64 (-.f64 #s(literal 1 binary64) k) #s(literal 2 binary64)))) < 1.00000000000000005e142

   1. Initial program 99.3%

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

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
   3. Step-by-step derivation

      1. lower-*.f64 N/A

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

      2. lower-sqrt.f64 N/A

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

      3. lower-/.f64 N/A

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

      4. lower-*.f64 N/A

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

      5. lift-PI.f64 N/A

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

      6. lower-sqrt.f64 55.9

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

   4. Applied rewrites 55.9%

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

   if 1.00000000000000005e142 < (*.f64 (/.f64 #s(literal 1 binary64) (sqrt.f64 k)) (pow.f64 (*.f64 (*.f64 #s(literal 2 binary64) (PI.f64)) n) (/.f64 (-.f64 #s(literal 1 binary64) k) #s(literal 2 binary64))))

   1. Initial program 99.6%

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

   2. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\frac{\frac{-1}{2} \cdot \left(\sqrt{{k}^{3} \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \left(\log \left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right) \cdot \sqrt{2}\right)\right) + \sqrt{k \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \sqrt{2}}{k}} \]
   3. Step-by-step derivation

      1. lower-/.f64 N/A

         \[\leadsto \frac{\frac{-1}{2} \cdot \left(\sqrt{{k}^{3} \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \left(\log \left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right) \cdot \sqrt{2}\right)\right) + \sqrt{k \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \sqrt{2}}{\color{blue}{k}} \]

   4. Applied rewrites 78.4%

      \[\leadsto \color{blue}{\frac{\mathsf{fma}\left(-0.5, \sqrt{{k}^{3} \cdot \left(n \cdot \pi\right)} \cdot \left(\log \left(2 \cdot \left(n \cdot \pi\right)\right) \cdot \sqrt{2}\right), \sqrt{k \cdot \left(n \cdot \pi\right)} \cdot \sqrt{2}\right)}{k}} \]

   5. Taylor expanded in k around 0

      \[\leadsto \frac{\sqrt{k \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)} \cdot \sqrt{2}}{k} \]
   6. Step-by-step derivation

      1. lift-*.f64 N/A

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

      2. lift-PI.f64 N/A

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

      3. lift-*.f64 N/A

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

      4. lift-sqrt.f64 N/A

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

      5. lift-sqrt.f64 N/A

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

      6. lift-*.f64 36.6

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

   7. Applied rewrites 36.6%

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

Alternative 8: 48.2% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 99.4%

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

2. Taylor expanded in k around 0

   \[\leadsto \frac{1}{\sqrt{k}} \cdot \color{blue}{\left(\sqrt{n \cdot \mathsf{PI}\left(\right)} \cdot \sqrt{2}\right)} \]
3. Step-by-step derivation

   1. lower-*.f64 N/A

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

   2. lower-sqrt.f64 N/A

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

   3. lower-*.f64 N/A

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

   4. lift-PI.f64 N/A

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

   5. lower-sqrt.f64 48.2

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

4. Applied rewrites 48.2%

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

Alternative 9: 48.2% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 99.4%

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

2. Taylor expanded in k around 0

   \[\leadsto \color{blue}{\sqrt{\frac{1}{k}}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
3. Step-by-step derivation

   1. lower-sqrt.f64 N/A

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

   2. lower-/.f64 99.4

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

4. Applied rewrites 99.4%

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

5. Taylor expanded in k around 0

   \[\leadsto \sqrt{\frac{1}{k}} \cdot \color{blue}{\left(\sqrt{n \cdot \mathsf{PI}\left(\right)} \cdot \sqrt{2}\right)} \]
6. Step-by-step derivation

   1. lower-*.f64 N/A

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

   2. lower-sqrt.f64 N/A

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

   3. lift-*.f64 N/A

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

   4. lift-PI.f64 N/A

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

   5. lift-sqrt.f64 48.2

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

7. Applied rewrites 48.2%

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

Alternative 10: 37.0% accurate, 2.7× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 99.4%

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

2. Taylor expanded in k around 0

   \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
3. Step-by-step derivation

   1. lower-*.f64 N/A

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

   2. lower-sqrt.f64 N/A

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

   3. lower-/.f64 N/A

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

   4. lower-*.f64 N/A

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

   5. lift-PI.f64 N/A

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

   6. lower-sqrt.f64 37.0

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

4. Applied rewrites 37.0%

   \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
5. Add Preprocessing

Reproduce

herbie shell --seed 2025124 
(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))))