Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.4%
Time: 9.8s
Alternatives: 8
Speedup: 1.1×

Specification

?
\[\begin{array}{l} \\ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \end{array} \]
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
double code(double k, double n) {
	return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}
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));
}
def code(k, n):
	return (1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))
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 tmp = code(k, n)
	tmp = (1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0));
end
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]
\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}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 8 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 99.4% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \end{array} \]
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
double code(double k, double n) {
	return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}
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));
}
def code(k, n):
	return (1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))
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 tmp = code(k, n)
	tmp = (1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0));
end
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]
\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}

Alternative 1: 99.4% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := n \cdot \left(2 \cdot \pi\right)\\ \frac{\sqrt{t\_0}}{\sqrt{k \cdot {t\_0}^{k}}} \end{array} \end{array} \]
(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* n (* 2.0 PI)))) (/ (sqrt t_0) (sqrt (* k (pow t_0 k))))))
double code(double k, double n) {
	double t_0 = n * (2.0 * ((double) M_PI));
	return sqrt(t_0) / sqrt((k * pow(t_0, k)));
}
public static double code(double k, double n) {
	double t_0 = n * (2.0 * Math.PI);
	return Math.sqrt(t_0) / Math.sqrt((k * Math.pow(t_0, k)));
}
def code(k, n):
	t_0 = n * (2.0 * math.pi)
	return math.sqrt(t_0) / math.sqrt((k * math.pow(t_0, k)))
function code(k, n)
	t_0 = Float64(n * Float64(2.0 * pi))
	return Float64(sqrt(t_0) / sqrt(Float64(k * (t_0 ^ k))))
end
function tmp = code(k, n)
	t_0 = n * (2.0 * pi);
	tmp = sqrt(t_0) / sqrt((k * (t_0 ^ k)));
end
code[k_, n_] := Block[{t$95$0 = N[(n * N[(2.0 * Pi), $MachinePrecision]), $MachinePrecision]}, N[(N[Sqrt[t$95$0], $MachinePrecision] / N[Sqrt[N[(k * N[Power[t$95$0, k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := n \cdot \left(2 \cdot \pi\right)\\
\frac{\sqrt{t\_0}}{\sqrt{k \cdot {t\_0}^{k}}}
\end{array}
\end{array}
Derivation
  1. Initial program 99.1%

    \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
  2. Add Preprocessing
  3. Step-by-step derivation
    1. lift-*.f64N/A

      \[\leadsto \color{blue}{\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}} \]
    2. *-commutativeN/A

      \[\leadsto \color{blue}{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}}} \]
    3. lift-/.f64N/A

      \[\leadsto {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \color{blue}{\frac{1}{\sqrt{k}}} \]
    4. un-div-invN/A

      \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]
    5. lift-pow.f64N/A

      \[\leadsto \frac{\color{blue}{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}}{\sqrt{k}} \]
    6. lift-/.f64N/A

      \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\color{blue}{\left(\frac{1 - k}{2}\right)}}}{\sqrt{k}} \]
    7. lift--.f64N/A

      \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{\color{blue}{1 - k}}{2}\right)}}{\sqrt{k}} \]
    8. div-subN/A

      \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\color{blue}{\left(\frac{1}{2} - \frac{k}{2}\right)}}}{\sqrt{k}} \]
    9. metadata-evalN/A

      \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\color{blue}{\frac{1}{2}} - \frac{k}{2}\right)}}{\sqrt{k}} \]
    10. pow-subN/A

      \[\leadsto \frac{\color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\frac{1}{2}}}{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}}}{\sqrt{k}} \]
    11. associate-/l/N/A

      \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\frac{1}{2}}}{\sqrt{k} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}} \]
    12. lower-/.f64N/A

      \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\frac{1}{2}}}{\sqrt{k} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}} \]
  4. Applied rewrites99.3%

    \[\leadsto \color{blue}{\frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{\sqrt{k} \cdot {\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}}} \]
  5. Step-by-step derivation
    1. lift-*.f64N/A

      \[\leadsto \frac{\sqrt{\color{blue}{2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)}}}{\sqrt{k} \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{\left(k \cdot \frac{1}{2}\right)}} \]
    2. lift-*.f64N/A

      \[\leadsto \frac{\sqrt{2 \cdot \color{blue}{\left(\mathsf{PI}\left(\right) \cdot n\right)}}}{\sqrt{k} \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{\left(k \cdot \frac{1}{2}\right)}} \]
    3. associate-*r*N/A

      \[\leadsto \frac{\sqrt{\color{blue}{\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n}}}{\sqrt{k} \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{\left(k \cdot \frac{1}{2}\right)}} \]
    4. lift-*.f64N/A

      \[\leadsto \frac{\sqrt{\color{blue}{\left(2 \cdot \mathsf{PI}\left(\right)\right)} \cdot n}}{\sqrt{k} \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{\left(k \cdot \frac{1}{2}\right)}} \]
    5. *-commutativeN/A

      \[\leadsto \frac{\sqrt{\color{blue}{n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)}}}{\sqrt{k} \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{\left(k \cdot \frac{1}{2}\right)}} \]
    6. lower-*.f6499.3

      \[\leadsto \frac{\sqrt{\color{blue}{n \cdot \left(2 \cdot \pi\right)}}}{\sqrt{k} \cdot {\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}} \]
    7. lift-*.f64N/A

      \[\leadsto \frac{\sqrt{n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)}}{\color{blue}{\sqrt{k} \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{\left(k \cdot \frac{1}{2}\right)}}} \]
    8. lift-sqrt.f64N/A

      \[\leadsto \frac{\sqrt{n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)}}{\color{blue}{\sqrt{k}} \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{\left(k \cdot \frac{1}{2}\right)}} \]
    9. lift-pow.f64N/A

      \[\leadsto \frac{\sqrt{n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)}}{\sqrt{k} \cdot \color{blue}{{\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{\left(k \cdot \frac{1}{2}\right)}}} \]
    10. lift-*.f64N/A

      \[\leadsto \frac{\sqrt{n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)}}{\sqrt{k} \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{\color{blue}{\left(k \cdot \frac{1}{2}\right)}}} \]
    11. pow-unpowN/A

      \[\leadsto \frac{\sqrt{n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)}}{\sqrt{k} \cdot \color{blue}{{\left({\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{k}\right)}^{\frac{1}{2}}}} \]
    12. unpow1/2N/A

      \[\leadsto \frac{\sqrt{n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)}}{\sqrt{k} \cdot \color{blue}{\sqrt{{\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{k}}}} \]
    13. sqrt-unprodN/A

      \[\leadsto \frac{\sqrt{n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)}}{\color{blue}{\sqrt{k \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{k}}}} \]
    14. lower-sqrt.f64N/A

      \[\leadsto \frac{\sqrt{n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)}}{\color{blue}{\sqrt{k \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{k}}}} \]
    15. lower-*.f64N/A

      \[\leadsto \frac{\sqrt{n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)}}{\sqrt{\color{blue}{k \cdot {\left(2 \cdot \left(\mathsf{PI}\left(\right) \cdot n\right)\right)}^{k}}}} \]
    16. lower-pow.f6499.3

      \[\leadsto \frac{\sqrt{n \cdot \left(2 \cdot \pi\right)}}{\sqrt{k \cdot \color{blue}{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{k}}}} \]
  6. Applied rewrites99.3%

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

Alternative 2: 98.2% accurate, 1.1× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 1:\\ \;\;\;\;\frac{\sqrt{n \cdot 2}}{\sqrt{\frac{k}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(k \cdot -0.5\right)}}{\sqrt{k}}\\ \end{array} \end{array} \]
(FPCore (k n)
 :precision binary64
 (if (<= k 1.0)
   (/ (sqrt (* n 2.0)) (sqrt (/ k PI)))
   (/ (pow (* n (* 2.0 PI)) (* k -0.5)) (sqrt k))))
double code(double k, double n) {
	double tmp;
	if (k <= 1.0) {
		tmp = sqrt((n * 2.0)) / sqrt((k / ((double) M_PI)));
	} else {
		tmp = pow((n * (2.0 * ((double) M_PI))), (k * -0.5)) / sqrt(k);
	}
	return tmp;
}
public static double code(double k, double n) {
	double tmp;
	if (k <= 1.0) {
		tmp = Math.sqrt((n * 2.0)) / Math.sqrt((k / Math.PI));
	} else {
		tmp = Math.pow((n * (2.0 * Math.PI)), (k * -0.5)) / Math.sqrt(k);
	}
	return tmp;
}
def code(k, n):
	tmp = 0
	if k <= 1.0:
		tmp = math.sqrt((n * 2.0)) / math.sqrt((k / math.pi))
	else:
		tmp = math.pow((n * (2.0 * math.pi)), (k * -0.5)) / math.sqrt(k)
	return tmp
function code(k, n)
	tmp = 0.0
	if (k <= 1.0)
		tmp = Float64(sqrt(Float64(n * 2.0)) / sqrt(Float64(k / pi)));
	else
		tmp = Float64((Float64(n * Float64(2.0 * pi)) ^ Float64(k * -0.5)) / sqrt(k));
	end
	return tmp
end
function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 1.0)
		tmp = sqrt((n * 2.0)) / sqrt((k / pi));
	else
		tmp = ((n * (2.0 * pi)) ^ (k * -0.5)) / sqrt(k);
	end
	tmp_2 = tmp;
end
code[k_, n_] := If[LessEqual[k, 1.0], N[(N[Sqrt[N[(n * 2.0), $MachinePrecision]], $MachinePrecision] / N[Sqrt[N[(k / Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Power[N[(n * N[(2.0 * Pi), $MachinePrecision]), $MachinePrecision], N[(k * -0.5), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
\mathbf{if}\;k \leq 1:\\
\;\;\;\;\frac{\sqrt{n \cdot 2}}{\sqrt{\frac{k}{\pi}}}\\

\mathbf{else}:\\
\;\;\;\;\frac{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(k \cdot -0.5\right)}}{\sqrt{k}}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if k < 1

    1. Initial program 98.3%

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
    2. Add Preprocessing
    3. Taylor expanded in k around 0

      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
    4. Step-by-step derivation
      1. lower-*.f64N/A

        \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
      2. lower-sqrt.f64N/A

        \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
      3. lower-/.f64N/A

        \[\leadsto \sqrt{\color{blue}{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
      4. lower-*.f64N/A

        \[\leadsto \sqrt{\frac{\color{blue}{n \cdot \mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
      5. lower-PI.f64N/A

        \[\leadsto \sqrt{\frac{n \cdot \color{blue}{\mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
      6. lower-sqrt.f6474.6

        \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
    5. Applied rewrites74.6%

      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. Applied rewrites74.6%

        \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \left(2 \cdot \pi\right)}{k}}} \]
      2. Step-by-step derivation
        1. Applied rewrites97.3%

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

        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. Add Preprocessing
        3. Taylor expanded in k around inf

          \[\leadsto \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\color{blue}{\left(\frac{-1}{2} \cdot k\right)}} \]
        4. Step-by-step derivation
          1. lower-*.f64100.0

            \[\leadsto \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\left(-0.5 \cdot k\right)}} \]
        5. Applied rewrites100.0%

          \[\leadsto \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\left(-0.5 \cdot k\right)}} \]
        6. Step-by-step derivation
          1. lift-*.f64N/A

            \[\leadsto \color{blue}{\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{-1}{2} \cdot k\right)}} \]
          2. *-commutativeN/A

            \[\leadsto \color{blue}{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{-1}{2} \cdot k\right)} \cdot \frac{1}{\sqrt{k}}} \]
          3. lift-/.f64N/A

            \[\leadsto {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{-1}{2} \cdot k\right)} \cdot \color{blue}{\frac{1}{\sqrt{k}}} \]
          4. un-div-invN/A

            \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{-1}{2} \cdot k\right)}}{\sqrt{k}}} \]
          5. lower-/.f64100.0

            \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(-0.5 \cdot k\right)}}{\sqrt{k}}} \]
          6. lift-*.f64N/A

            \[\leadsto \frac{{\color{blue}{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}}^{\left(\frac{-1}{2} \cdot k\right)}}{\sqrt{k}} \]
          7. *-commutativeN/A

            \[\leadsto \frac{{\color{blue}{\left(n \cdot \left(2 \cdot \mathsf{PI}\left(\right)\right)\right)}}^{\left(\frac{-1}{2} \cdot k\right)}}{\sqrt{k}} \]
          8. lower-*.f64100.0

            \[\leadsto \frac{{\color{blue}{\left(n \cdot \left(2 \cdot \pi\right)\right)}}^{\left(-0.5 \cdot k\right)}}{\sqrt{k}} \]
        7. Applied rewrites100.0%

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

      Alternative 3: 99.5% accurate, 1.1× speedup?

      \[\begin{array}{l} \\ \frac{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\sqrt{k}} \end{array} \]
      (FPCore (k n)
       :precision binary64
       (/ (pow (* n (* 2.0 PI)) (fma k -0.5 0.5)) (sqrt k)))
      double code(double k, double n) {
      	return pow((n * (2.0 * ((double) M_PI))), fma(k, -0.5, 0.5)) / sqrt(k);
      }
      
      function code(k, n)
      	return Float64((Float64(n * Float64(2.0 * pi)) ^ fma(k, -0.5, 0.5)) / sqrt(k))
      end
      
      code[k_, n_] := N[(N[Power[N[(n * N[(2.0 * Pi), $MachinePrecision]), $MachinePrecision], N[(k * -0.5 + 0.5), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]
      
      \begin{array}{l}
      
      \\
      \frac{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\sqrt{k}}
      \end{array}
      
      Derivation
      1. Initial program 99.1%

        \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
      2. Add Preprocessing
      3. Step-by-step derivation
        1. lift-*.f64N/A

          \[\leadsto \color{blue}{\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}} \]
        2. *-commutativeN/A

          \[\leadsto \color{blue}{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}}} \]
        3. lift-/.f64N/A

          \[\leadsto {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \color{blue}{\frac{1}{\sqrt{k}}} \]
        4. un-div-invN/A

          \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]
        5. lift-pow.f64N/A

          \[\leadsto \frac{\color{blue}{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}}{\sqrt{k}} \]
        6. lift-/.f64N/A

          \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\color{blue}{\left(\frac{1 - k}{2}\right)}}}{\sqrt{k}} \]
        7. lift--.f64N/A

          \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{\color{blue}{1 - k}}{2}\right)}}{\sqrt{k}} \]
        8. div-subN/A

          \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\color{blue}{\left(\frac{1}{2} - \frac{k}{2}\right)}}}{\sqrt{k}} \]
        9. metadata-evalN/A

          \[\leadsto \frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\color{blue}{\frac{1}{2}} - \frac{k}{2}\right)}}{\sqrt{k}} \]
        10. pow-subN/A

          \[\leadsto \frac{\color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\frac{1}{2}}}{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}}}{\sqrt{k}} \]
        11. associate-/l/N/A

          \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\frac{1}{2}}}{\sqrt{k} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}} \]
        12. lower-/.f64N/A

          \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\frac{1}{2}}}{\sqrt{k} \cdot {\left(\left(2 \cdot \mathsf{PI}\left(\right)\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}} \]
      4. Applied rewrites99.3%

        \[\leadsto \color{blue}{\frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{\sqrt{k} \cdot {\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}}} \]
      5. Applied rewrites99.2%

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

      Alternative 4: 49.8% accurate, 3.2× speedup?

      \[\begin{array}{l} \\ \frac{\sqrt{n \cdot 2}}{\sqrt{\frac{k}{\pi}}} \end{array} \]
      (FPCore (k n) :precision binary64 (/ (sqrt (* n 2.0)) (sqrt (/ k PI))))
      double code(double k, double n) {
      	return sqrt((n * 2.0)) / sqrt((k / ((double) M_PI)));
      }
      
      public static double code(double k, double n) {
      	return Math.sqrt((n * 2.0)) / Math.sqrt((k / Math.PI));
      }
      
      def code(k, n):
      	return math.sqrt((n * 2.0)) / math.sqrt((k / math.pi))
      
      function code(k, n)
      	return Float64(sqrt(Float64(n * 2.0)) / sqrt(Float64(k / pi)))
      end
      
      function tmp = code(k, n)
      	tmp = sqrt((n * 2.0)) / sqrt((k / pi));
      end
      
      code[k_, n_] := N[(N[Sqrt[N[(n * 2.0), $MachinePrecision]], $MachinePrecision] / N[Sqrt[N[(k / Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
      
      \begin{array}{l}
      
      \\
      \frac{\sqrt{n \cdot 2}}{\sqrt{\frac{k}{\pi}}}
      \end{array}
      
      Derivation
      1. Initial program 99.1%

        \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
      2. Add Preprocessing
      3. Taylor expanded in k around 0

        \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
      4. Step-by-step derivation
        1. lower-*.f64N/A

          \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
        2. lower-sqrt.f64N/A

          \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
        3. lower-/.f64N/A

          \[\leadsto \sqrt{\color{blue}{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
        4. lower-*.f64N/A

          \[\leadsto \sqrt{\frac{\color{blue}{n \cdot \mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
        5. lower-PI.f64N/A

          \[\leadsto \sqrt{\frac{n \cdot \color{blue}{\mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
        6. lower-sqrt.f6439.4

          \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
      5. Applied rewrites39.4%

        \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
      6. Step-by-step derivation
        1. Applied rewrites39.4%

          \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \left(2 \cdot \pi\right)}{k}}} \]
        2. Step-by-step derivation
          1. Applied rewrites51.1%

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

          Alternative 5: 49.7% accurate, 3.6× speedup?

          \[\begin{array}{l} \\ \sqrt{\frac{2}{k}} \cdot \sqrt{n \cdot \pi} \end{array} \]
          (FPCore (k n) :precision binary64 (* (sqrt (/ 2.0 k)) (sqrt (* n PI))))
          double code(double k, double n) {
          	return sqrt((2.0 / k)) * sqrt((n * ((double) M_PI)));
          }
          
          public static double code(double k, double n) {
          	return Math.sqrt((2.0 / k)) * Math.sqrt((n * Math.PI));
          }
          
          def code(k, n):
          	return math.sqrt((2.0 / k)) * math.sqrt((n * math.pi))
          
          function code(k, n)
          	return Float64(sqrt(Float64(2.0 / k)) * sqrt(Float64(n * pi)))
          end
          
          function tmp = code(k, n)
          	tmp = sqrt((2.0 / k)) * sqrt((n * pi));
          end
          
          code[k_, n_] := N[(N[Sqrt[N[(2.0 / k), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(n * Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
          
          \begin{array}{l}
          
          \\
          \sqrt{\frac{2}{k}} \cdot \sqrt{n \cdot \pi}
          \end{array}
          
          Derivation
          1. Initial program 99.1%

            \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
          2. Add Preprocessing
          3. Taylor expanded in k around 0

            \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
          4. Step-by-step derivation
            1. lower-*.f64N/A

              \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
            2. lower-sqrt.f64N/A

              \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
            3. lower-/.f64N/A

              \[\leadsto \sqrt{\color{blue}{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
            4. lower-*.f64N/A

              \[\leadsto \sqrt{\frac{\color{blue}{n \cdot \mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
            5. lower-PI.f64N/A

              \[\leadsto \sqrt{\frac{n \cdot \color{blue}{\mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
            6. lower-sqrt.f6439.4

              \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
          5. Applied rewrites39.4%

            \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
          6. Step-by-step derivation
            1. Applied rewrites39.4%

              \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \left(2 \cdot \pi\right)}{k}}} \]
            2. Step-by-step derivation
              1. Applied rewrites39.5%

                \[\leadsto \sqrt{\left(n \cdot \pi\right) \cdot \frac{2}{k}} \]
              2. Step-by-step derivation
                1. Applied rewrites51.1%

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

                Alternative 6: 49.7% accurate, 3.6× speedup?

                \[\begin{array}{l} \\ \sqrt{n} \cdot \sqrt{2 \cdot \frac{\pi}{k}} \end{array} \]
                (FPCore (k n) :precision binary64 (* (sqrt n) (sqrt (* 2.0 (/ PI k)))))
                double code(double k, double n) {
                	return sqrt(n) * sqrt((2.0 * (((double) M_PI) / k)));
                }
                
                public static double code(double k, double n) {
                	return Math.sqrt(n) * Math.sqrt((2.0 * (Math.PI / k)));
                }
                
                def code(k, n):
                	return math.sqrt(n) * math.sqrt((2.0 * (math.pi / k)))
                
                function code(k, n)
                	return Float64(sqrt(n) * sqrt(Float64(2.0 * Float64(pi / k))))
                end
                
                function tmp = code(k, n)
                	tmp = sqrt(n) * sqrt((2.0 * (pi / k)));
                end
                
                code[k_, n_] := N[(N[Sqrt[n], $MachinePrecision] * N[Sqrt[N[(2.0 * N[(Pi / k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
                
                \begin{array}{l}
                
                \\
                \sqrt{n} \cdot \sqrt{2 \cdot \frac{\pi}{k}}
                \end{array}
                
                Derivation
                1. Initial program 99.1%

                  \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
                2. Add Preprocessing
                3. Taylor expanded in k around 0

                  \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
                4. Step-by-step derivation
                  1. lower-*.f64N/A

                    \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
                  2. lower-sqrt.f64N/A

                    \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
                  3. lower-/.f64N/A

                    \[\leadsto \sqrt{\color{blue}{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
                  4. lower-*.f64N/A

                    \[\leadsto \sqrt{\frac{\color{blue}{n \cdot \mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
                  5. lower-PI.f64N/A

                    \[\leadsto \sqrt{\frac{n \cdot \color{blue}{\mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
                  6. lower-sqrt.f6439.4

                    \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
                5. Applied rewrites39.4%

                  \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
                6. Step-by-step derivation
                  1. Applied rewrites51.0%

                    \[\leadsto \sqrt{n} \cdot \color{blue}{\sqrt{\frac{\pi}{k} \cdot 2}} \]
                  2. Final simplification51.0%

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

                  Alternative 7: 38.2% accurate, 4.8× speedup?

                  \[\begin{array}{l} \\ \sqrt{\pi \cdot \frac{n \cdot 2}{k}} \end{array} \]
                  (FPCore (k n) :precision binary64 (sqrt (* PI (/ (* n 2.0) k))))
                  double code(double k, double n) {
                  	return sqrt((((double) M_PI) * ((n * 2.0) / k)));
                  }
                  
                  public static double code(double k, double n) {
                  	return Math.sqrt((Math.PI * ((n * 2.0) / k)));
                  }
                  
                  def code(k, n):
                  	return math.sqrt((math.pi * ((n * 2.0) / k)))
                  
                  function code(k, n)
                  	return sqrt(Float64(pi * Float64(Float64(n * 2.0) / k)))
                  end
                  
                  function tmp = code(k, n)
                  	tmp = sqrt((pi * ((n * 2.0) / k)));
                  end
                  
                  code[k_, n_] := N[Sqrt[N[(Pi * N[(N[(n * 2.0), $MachinePrecision] / k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
                  
                  \begin{array}{l}
                  
                  \\
                  \sqrt{\pi \cdot \frac{n \cdot 2}{k}}
                  \end{array}
                  
                  Derivation
                  1. Initial program 99.1%

                    \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
                  2. Add Preprocessing
                  3. Taylor expanded in k around 0

                    \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
                  4. Step-by-step derivation
                    1. lower-*.f64N/A

                      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
                    2. lower-sqrt.f64N/A

                      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
                    3. lower-/.f64N/A

                      \[\leadsto \sqrt{\color{blue}{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
                    4. lower-*.f64N/A

                      \[\leadsto \sqrt{\frac{\color{blue}{n \cdot \mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
                    5. lower-PI.f64N/A

                      \[\leadsto \sqrt{\frac{n \cdot \color{blue}{\mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
                    6. lower-sqrt.f6439.4

                      \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
                  5. Applied rewrites39.4%

                    \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
                  6. Step-by-step derivation
                    1. Applied rewrites39.4%

                      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \left(2 \cdot \pi\right)}{k}}} \]
                    2. Step-by-step derivation
                      1. Applied rewrites39.5%

                        \[\leadsto \sqrt{\frac{n \cdot 2}{k} \cdot \pi} \]
                      2. Final simplification39.5%

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

                      Alternative 8: 38.1% accurate, 4.8× speedup?

                      \[\begin{array}{l} \\ \sqrt{\frac{2}{k} \cdot \left(n \cdot \pi\right)} \end{array} \]
                      (FPCore (k n) :precision binary64 (sqrt (* (/ 2.0 k) (* n PI))))
                      double code(double k, double n) {
                      	return sqrt(((2.0 / k) * (n * ((double) M_PI))));
                      }
                      
                      public static double code(double k, double n) {
                      	return Math.sqrt(((2.0 / k) * (n * Math.PI)));
                      }
                      
                      def code(k, n):
                      	return math.sqrt(((2.0 / k) * (n * math.pi)))
                      
                      function code(k, n)
                      	return sqrt(Float64(Float64(2.0 / k) * Float64(n * pi)))
                      end
                      
                      function tmp = code(k, n)
                      	tmp = sqrt(((2.0 / k) * (n * pi)));
                      end
                      
                      code[k_, n_] := N[Sqrt[N[(N[(2.0 / k), $MachinePrecision] * N[(n * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
                      
                      \begin{array}{l}
                      
                      \\
                      \sqrt{\frac{2}{k} \cdot \left(n \cdot \pi\right)}
                      \end{array}
                      
                      Derivation
                      1. Initial program 99.1%

                        \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
                      2. Add Preprocessing
                      3. Taylor expanded in k around 0

                        \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
                      4. Step-by-step derivation
                        1. lower-*.f64N/A

                          \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}} \cdot \sqrt{2}} \]
                        2. lower-sqrt.f64N/A

                          \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
                        3. lower-/.f64N/A

                          \[\leadsto \sqrt{\color{blue}{\frac{n \cdot \mathsf{PI}\left(\right)}{k}}} \cdot \sqrt{2} \]
                        4. lower-*.f64N/A

                          \[\leadsto \sqrt{\frac{\color{blue}{n \cdot \mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
                        5. lower-PI.f64N/A

                          \[\leadsto \sqrt{\frac{n \cdot \color{blue}{\mathsf{PI}\left(\right)}}{k}} \cdot \sqrt{2} \]
                        6. lower-sqrt.f6439.4

                          \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
                      5. Applied rewrites39.4%

                        \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
                      6. Step-by-step derivation
                        1. Applied rewrites39.4%

                          \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \left(2 \cdot \pi\right)}{k}}} \]
                        2. Step-by-step derivation
                          1. Applied rewrites39.5%

                            \[\leadsto \sqrt{\left(n \cdot \pi\right) \cdot \frac{2}{k}} \]
                          2. Final simplification39.5%

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

                          Reproduce

                          ?
                          herbie shell --seed 2024223 
                          (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))))