Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.6%
Time: 11.2s
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: 99.6% accurate, 0.9× speedup?

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

\\
\begin{array}{l}
t_0 := 2 \cdot \left(\pi \cdot n\right)\\
\frac{\sqrt{t\_0}}{\sqrt{k} \cdot {t\_0}^{\left(k \cdot 0.5\right)}}
\end{array}
\end{array}
Derivation
  1. Initial program 99.5%

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

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

Alternative 2: 61.4% accurate, 0.6× speedup?

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

\\
\begin{array}{l}
\mathbf{if}\;\frac{1}{\sqrt{k}} \cdot {\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\frac{1 - k}{2}\right)} \leq 0:\\
\;\;\;\;\left(\sqrt{\sqrt{\frac{1}{k} \cdot \frac{1}{k}}} \cdot \sqrt{\pi \cdot n}\right) \cdot \sqrt{2}\\

\mathbf{else}:\\
\;\;\;\;\frac{\sqrt{2 \cdot n}}{\sqrt{\frac{k}{\pi}}}\\


\end{array}
\end{array}
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)))) < 0.0

    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 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.f643.2

        \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
    5. Applied rewrites3.2%

      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
    6. Step-by-step derivation
      1. Applied rewrites3.2%

        \[\leadsto \frac{1}{\sqrt{\frac{k}{\pi \cdot n}}} \cdot \sqrt{\color{blue}{2}} \]
      2. Step-by-step derivation
        1. Applied rewrites3.2%

          \[\leadsto \left(\sqrt{\frac{1}{k}} \cdot \sqrt{\pi \cdot n}\right) \cdot \sqrt{\color{blue}{2}} \]
        2. Step-by-step derivation
          1. Applied rewrites52.8%

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

          if 0.0 < (*.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 99.3%

            \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
          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.f6451.7

              \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
          5. Applied rewrites51.7%

            \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
          6. Step-by-step derivation
            1. Applied rewrites51.9%

              \[\leadsto \color{blue}{\sqrt{\frac{2 \cdot \left(\pi \cdot n\right)}{k}}} \]
            2. Step-by-step derivation
              1. Applied rewrites51.9%

                \[\leadsto \sqrt{\left(2 \cdot n\right) \cdot \frac{\pi}{k}} \]
              2. Step-by-step derivation
                1. Applied rewrites70.6%

                  \[\leadsto \frac{\sqrt{2 \cdot n}}{\color{blue}{\sqrt{\frac{k}{\pi}}}} \]
              3. Recombined 2 regimes into one program.
              4. Final simplification65.6%

                \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{\sqrt{k}} \cdot {\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\frac{1 - k}{2}\right)} \leq 0:\\ \;\;\;\;\left(\sqrt{\sqrt{\frac{1}{k} \cdot \frac{1}{k}}} \cdot \sqrt{\pi \cdot n}\right) \cdot \sqrt{2}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot n}}{\sqrt{\frac{k}{\pi}}}\\ \end{array} \]
              5. Add Preprocessing

              Alternative 3: 99.5% accurate, 1.1× speedup?

              \[\begin{array}{l} \\ \frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\sqrt{k}} \end{array} \]
              (FPCore (k n)
               :precision binary64
               (/ (pow (* 2.0 (* PI n)) (fma k -0.5 0.5)) (sqrt k)))
              double code(double k, double n) {
              	return pow((2.0 * (((double) M_PI) * n)), fma(k, -0.5, 0.5)) / sqrt(k);
              }
              
              function code(k, n)
              	return Float64((Float64(2.0 * Float64(pi * n)) ^ fma(k, -0.5, 0.5)) / sqrt(k))
              end
              
              code[k_, n_] := N[(N[Power[N[(2.0 * N[(Pi * n), $MachinePrecision]), $MachinePrecision], N[(k * -0.5 + 0.5), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]
              
              \begin{array}{l}
              
              \\
              \frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\sqrt{k}}
              \end{array}
              
              Derivation
              1. Initial program 99.5%

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

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

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

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

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

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

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

              Alternative 4: 49.8% accurate, 3.2× speedup?

              \[\begin{array}{l} \\ \frac{\sqrt{2 \cdot n}}{\sqrt{\frac{k}{\pi}}} \end{array} \]
              (FPCore (k n) :precision binary64 (/ (sqrt (* 2.0 n)) (sqrt (/ k PI))))
              double code(double k, double n) {
              	return sqrt((2.0 * n)) / sqrt((k / ((double) M_PI)));
              }
              
              public static double code(double k, double n) {
              	return Math.sqrt((2.0 * n)) / Math.sqrt((k / Math.PI));
              }
              
              def code(k, n):
              	return math.sqrt((2.0 * n)) / math.sqrt((k / math.pi))
              
              function code(k, n)
              	return Float64(sqrt(Float64(2.0 * n)) / sqrt(Float64(k / pi)))
              end
              
              function tmp = code(k, n)
              	tmp = sqrt((2.0 * n)) / sqrt((k / pi));
              end
              
              code[k_, n_] := N[(N[Sqrt[N[(2.0 * n), $MachinePrecision]], $MachinePrecision] / N[Sqrt[N[(k / Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
              
              \begin{array}{l}
              
              \\
              \frac{\sqrt{2 \cdot n}}{\sqrt{\frac{k}{\pi}}}
              \end{array}
              
              Derivation
              1. Initial program 99.5%

                \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
              2. 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.f6438.1

                  \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
              5. Applied rewrites38.1%

                \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
              6. Step-by-step derivation
                1. Applied rewrites38.2%

                  \[\leadsto \color{blue}{\sqrt{\frac{2 \cdot \left(\pi \cdot n\right)}{k}}} \]
                2. Step-by-step derivation
                  1. Applied rewrites38.2%

                    \[\leadsto \sqrt{\left(2 \cdot n\right) \cdot \frac{\pi}{k}} \]
                  2. Step-by-step derivation
                    1. Applied rewrites51.7%

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

                    Alternative 5: 49.8% accurate, 3.6× speedup?

                    \[\begin{array}{l} \\ \sqrt{\pi \cdot n} \cdot \sqrt{\frac{2}{k}} \end{array} \]
                    (FPCore (k n) :precision binary64 (* (sqrt (* PI n)) (sqrt (/ 2.0 k))))
                    double code(double k, double n) {
                    	return sqrt((((double) M_PI) * n)) * sqrt((2.0 / k));
                    }
                    
                    public static double code(double k, double n) {
                    	return Math.sqrt((Math.PI * n)) * Math.sqrt((2.0 / k));
                    }
                    
                    def code(k, n):
                    	return math.sqrt((math.pi * n)) * math.sqrt((2.0 / k))
                    
                    function code(k, n)
                    	return Float64(sqrt(Float64(pi * n)) * sqrt(Float64(2.0 / k)))
                    end
                    
                    function tmp = code(k, n)
                    	tmp = sqrt((pi * n)) * sqrt((2.0 / k));
                    end
                    
                    code[k_, n_] := N[(N[Sqrt[N[(Pi * n), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(2.0 / k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
                    
                    \begin{array}{l}
                    
                    \\
                    \sqrt{\pi \cdot n} \cdot \sqrt{\frac{2}{k}}
                    \end{array}
                    
                    Derivation
                    1. Initial program 99.5%

                      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
                    2. 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.f6438.1

                        \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
                    5. Applied rewrites38.1%

                      \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
                    6. Step-by-step derivation
                      1. Applied rewrites38.2%

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

                          \[\leadsto \sqrt{\frac{2}{k}} \cdot \color{blue}{\sqrt{\pi \cdot n}} \]
                        2. Final simplification51.6%

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

                        Alternative 6: 49.8% 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 99.5%

                          \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
                        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.f6438.1

                            \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
                        5. Applied rewrites38.1%

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

                            \[\leadsto \sqrt{n} \cdot \color{blue}{\sqrt{\frac{\pi}{k} \cdot 2}} \]
                          2. Final simplification51.6%

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

                          Alternative 7: 38.5% accurate, 4.8× speedup?

                          \[\begin{array}{l} \\ \sqrt{\left(2 \cdot \pi\right) \cdot \frac{n}{k}} \end{array} \]
                          (FPCore (k n) :precision binary64 (sqrt (* (* 2.0 PI) (/ n k))))
                          double code(double k, double n) {
                          	return sqrt(((2.0 * ((double) M_PI)) * (n / k)));
                          }
                          
                          public static double code(double k, double n) {
                          	return Math.sqrt(((2.0 * Math.PI) * (n / k)));
                          }
                          
                          def code(k, n):
                          	return math.sqrt(((2.0 * math.pi) * (n / k)))
                          
                          function code(k, n)
                          	return sqrt(Float64(Float64(2.0 * pi) * Float64(n / k)))
                          end
                          
                          function tmp = code(k, n)
                          	tmp = sqrt(((2.0 * pi) * (n / k)));
                          end
                          
                          code[k_, n_] := N[Sqrt[N[(N[(2.0 * Pi), $MachinePrecision] * N[(n / k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
                          
                          \begin{array}{l}
                          
                          \\
                          \sqrt{\left(2 \cdot \pi\right) \cdot \frac{n}{k}}
                          \end{array}
                          
                          Derivation
                          1. Initial program 99.5%

                            \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
                          2. 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.f6438.1

                              \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
                          5. Applied rewrites38.1%

                            \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
                          6. Step-by-step derivation
                            1. Applied rewrites38.2%

                              \[\leadsto \color{blue}{\sqrt{\frac{2 \cdot \left(\pi \cdot n\right)}{k}}} \]
                            2. Step-by-step derivation
                              1. Applied rewrites38.2%

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

                              Alternative 8: 38.5% accurate, 4.8× speedup?

                              \[\begin{array}{l} \\ \sqrt{\left(2 \cdot n\right) \cdot \frac{\pi}{k}} \end{array} \]
                              (FPCore (k n) :precision binary64 (sqrt (* (* 2.0 n) (/ PI k))))
                              double code(double k, double n) {
                              	return sqrt(((2.0 * n) * (((double) M_PI) / k)));
                              }
                              
                              public static double code(double k, double n) {
                              	return Math.sqrt(((2.0 * n) * (Math.PI / k)));
                              }
                              
                              def code(k, n):
                              	return math.sqrt(((2.0 * n) * (math.pi / k)))
                              
                              function code(k, n)
                              	return sqrt(Float64(Float64(2.0 * n) * Float64(pi / k)))
                              end
                              
                              function tmp = code(k, n)
                              	tmp = sqrt(((2.0 * n) * (pi / k)));
                              end
                              
                              code[k_, n_] := N[Sqrt[N[(N[(2.0 * n), $MachinePrecision] * N[(Pi / k), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
                              
                              \begin{array}{l}
                              
                              \\
                              \sqrt{\left(2 \cdot n\right) \cdot \frac{\pi}{k}}
                              \end{array}
                              
                              Derivation
                              1. Initial program 99.5%

                                \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
                              2. 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.f6438.1

                                  \[\leadsto \sqrt{\frac{n \cdot \pi}{k}} \cdot \color{blue}{\sqrt{2}} \]
                              5. Applied rewrites38.1%

                                \[\leadsto \color{blue}{\sqrt{\frac{n \cdot \pi}{k}} \cdot \sqrt{2}} \]
                              6. Step-by-step derivation
                                1. Applied rewrites38.2%

                                  \[\leadsto \color{blue}{\sqrt{\frac{2 \cdot \left(\pi \cdot n\right)}{k}}} \]
                                2. Step-by-step derivation
                                  1. Applied rewrites38.2%

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

                                  Reproduce

                                  ?
                                  herbie shell --seed 2024221 
                                  (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))))