Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.5%
Time: 9.0s
Alternatives: 7
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 7 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: 99.5% accurate, 0.9× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \left(2 \cdot \pi\right) \cdot n\\ \frac{\sqrt{t\_0}}{{t\_0}^{\left(0.5 \cdot k\right)} \cdot \sqrt{k}} \end{array} \end{array} \]
(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* (* 2.0 PI) n)))
   (/ (sqrt t_0) (* (pow t_0 (* 0.5 k)) (sqrt k)))))
double code(double k, double n) {
	double t_0 = (2.0 * ((double) M_PI)) * n;
	return sqrt(t_0) / (pow(t_0, (0.5 * k)) * sqrt(k));
}
public static double code(double k, double n) {
	double t_0 = (2.0 * Math.PI) * n;
	return Math.sqrt(t_0) / (Math.pow(t_0, (0.5 * k)) * Math.sqrt(k));
}
def code(k, n):
	t_0 = (2.0 * math.pi) * n
	return math.sqrt(t_0) / (math.pow(t_0, (0.5 * k)) * math.sqrt(k))
function code(k, n)
	t_0 = Float64(Float64(2.0 * pi) * n)
	return Float64(sqrt(t_0) / Float64((t_0 ^ Float64(0.5 * k)) * sqrt(k)))
end
function tmp = code(k, n)
	t_0 = (2.0 * pi) * n;
	tmp = sqrt(t_0) / ((t_0 ^ (0.5 * k)) * sqrt(k));
end
code[k_, n_] := Block[{t$95$0 = N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision]}, N[(N[Sqrt[t$95$0], $MachinePrecision] / N[(N[Power[t$95$0, N[(0.5 * k), $MachinePrecision]], $MachinePrecision] * N[Sqrt[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \left(2 \cdot \pi\right) \cdot n\\
\frac{\sqrt{t\_0}}{{t\_0}^{\left(0.5 \cdot k\right)} \cdot \sqrt{k}}
\end{array}
\end{array}
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. 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 rewrites99.7%

    \[\leadsto \color{blue}{\frac{\sqrt{n \cdot \left(\pi \cdot 2\right)}}{\sqrt{k} \cdot {\left(n \cdot \left(\pi \cdot 2\right)\right)}^{\left(0.5 \cdot k\right)}}} \]
  5. Final simplification99.7%

    \[\leadsto \frac{\sqrt{\left(2 \cdot \pi\right) \cdot n}}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(0.5 \cdot k\right)} \cdot \sqrt{k}} \]
  6. Add Preprocessing

Alternative 2: 99.4% accurate, 1.1× speedup?

\[\begin{array}{l} \\ \sqrt{\frac{1}{k}} \cdot {\left(\left(2 \cdot n\right) \cdot \pi\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)} \end{array} \]
(FPCore (k n)
 :precision binary64
 (* (sqrt (/ 1.0 k)) (pow (* (* 2.0 n) PI) (fma k -0.5 0.5))))
double code(double k, double n) {
	return sqrt((1.0 / k)) * pow(((2.0 * n) * ((double) M_PI)), fma(k, -0.5, 0.5));
}
function code(k, n)
	return Float64(sqrt(Float64(1.0 / k)) * (Float64(Float64(2.0 * n) * pi) ^ fma(k, -0.5, 0.5)))
end
code[k_, n_] := N[(N[Sqrt[N[(1.0 / k), $MachinePrecision]], $MachinePrecision] * N[Power[N[(N[(2.0 * n), $MachinePrecision] * Pi), $MachinePrecision], N[(k * -0.5 + 0.5), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
\sqrt{\frac{1}{k}} \cdot {\left(\left(2 \cdot n\right) \cdot \pi\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}
\end{array}
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. Add Preprocessing
  3. Taylor expanded in k around inf

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

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

      \[\leadsto \color{blue}{e^{\frac{1}{2} \cdot \left(\log \left(2 \cdot \left(n \cdot \mathsf{PI}\left(\right)\right)\right) \cdot \left(1 - k\right)\right)} \cdot \sqrt{\frac{1}{k}}} \]
  5. Applied rewrites99.5%

    \[\leadsto \color{blue}{{\left(\left(2 \cdot n\right) \cdot \pi\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)} \cdot \sqrt{\frac{1}{k}}} \]
  6. Final simplification99.5%

    \[\leadsto \sqrt{\frac{1}{k}} \cdot {\left(\left(2 \cdot n\right) \cdot \pi\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)} \]
  7. Add Preprocessing

Alternative 3: 99.5% accurate, 1.1× speedup?

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

\\
\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\mathsf{fma}\left(-0.5, k, 0.5\right)\right)}}{\sqrt{k}}
\end{array}
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. 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 rewrites99.7%

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

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

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

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

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

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

Alternative 4: 49.9% accurate, 3.2× speedup?

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

\\
\frac{\sqrt{n}}{\sqrt{\frac{k}{2 \cdot \pi}}}
\end{array}
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. 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. *-commutativeN/A

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

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

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

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

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

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

      \[\leadsto \sqrt{2} \cdot \sqrt{\frac{\color{blue}{\mathsf{PI}\left(\right) \cdot n}}{k}} \]
    8. lower-PI.f6439.1

      \[\leadsto \sqrt{2} \cdot \sqrt{\frac{\color{blue}{\pi} \cdot n}{k}} \]
  5. Applied rewrites39.1%

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

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

        \[\leadsto \sqrt{\left(\pi \cdot 2\right) \cdot \frac{n}{k}} \]
      2. Applied rewrites49.8%

        \[\leadsto \frac{\sqrt{n}}{\color{blue}{\sqrt{\frac{k}{\pi \cdot 2}}}} \]
      3. Final simplification49.8%

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

      Alternative 5: 49.9% accurate, 3.6× speedup?

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

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

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

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

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

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

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

          \[\leadsto \sqrt{2} \cdot \sqrt{\frac{\color{blue}{\mathsf{PI}\left(\right) \cdot n}}{k}} \]
        8. lower-PI.f6439.1

          \[\leadsto \sqrt{2} \cdot \sqrt{\frac{\color{blue}{\pi} \cdot n}{k}} \]
      5. Applied rewrites39.1%

        \[\leadsto \color{blue}{\sqrt{2} \cdot \sqrt{\frac{\pi \cdot n}{k}}} \]
      6. Step-by-step derivation
        1. Applied rewrites49.8%

          \[\leadsto \sqrt{n} \cdot \color{blue}{\sqrt{\frac{2 \cdot \pi}{k}}} \]
        2. Final simplification49.8%

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

        Alternative 6: 38.2% accurate, 4.0× speedup?

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

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

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

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

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

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

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

            \[\leadsto \sqrt{2} \cdot \sqrt{\frac{\color{blue}{\mathsf{PI}\left(\right) \cdot n}}{k}} \]
          8. lower-PI.f6439.1

            \[\leadsto \sqrt{2} \cdot \sqrt{\frac{\color{blue}{\pi} \cdot n}{k}} \]
        5. Applied rewrites39.1%

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

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

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

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

            Alternative 7: 38.2% accurate, 4.8× speedup?

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

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

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

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

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

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

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

                \[\leadsto \sqrt{2} \cdot \sqrt{\frac{\color{blue}{\mathsf{PI}\left(\right) \cdot n}}{k}} \]
              8. lower-PI.f6439.1

                \[\leadsto \sqrt{2} \cdot \sqrt{\frac{\color{blue}{\pi} \cdot n}{k}} \]
            5. Applied rewrites39.1%

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

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

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

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

                Reproduce

                ?
                herbie shell --seed 2024240 
                (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))))