Average Error: 0.5 → 0.6
Time: 9.6s
Precision: binary64
Cost: 19972
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
\[\begin{array}{l} \mathbf{if}\;k \leq 1.2237637377882354 \cdot 10^{-35}:\\ \;\;\;\;\frac{1}{\frac{\sqrt{k}}{\sqrt{n \cdot \left(2 \cdot \pi\right)}}}\\ \mathbf{else}:\\ \;\;\;\;{\left(\frac{k}{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}\right)}^{-0.5}\\ \end{array} \]
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
(FPCore (k n)
 :precision binary64
 (if (<= k 1.2237637377882354e-35)
   (/ 1.0 (/ (sqrt k) (sqrt (* n (* 2.0 PI)))))
   (pow (/ k (pow (* PI (* 2.0 n)) (- 1.0 k))) -0.5)))
double code(double k, double n) {
	return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}
double code(double k, double n) {
	double tmp;
	if (k <= 1.2237637377882354e-35) {
		tmp = 1.0 / (sqrt(k) / sqrt((n * (2.0 * ((double) M_PI)))));
	} else {
		tmp = pow((k / pow((((double) M_PI) * (2.0 * n)), (1.0 - k))), -0.5);
	}
	return tmp;
}
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));
}
public static double code(double k, double n) {
	double tmp;
	if (k <= 1.2237637377882354e-35) {
		tmp = 1.0 / (Math.sqrt(k) / Math.sqrt((n * (2.0 * Math.PI))));
	} else {
		tmp = Math.pow((k / Math.pow((Math.PI * (2.0 * n)), (1.0 - k))), -0.5);
	}
	return tmp;
}
def code(k, n):
	return (1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))
def code(k, n):
	tmp = 0
	if k <= 1.2237637377882354e-35:
		tmp = 1.0 / (math.sqrt(k) / math.sqrt((n * (2.0 * math.pi))))
	else:
		tmp = math.pow((k / math.pow((math.pi * (2.0 * n)), (1.0 - k))), -0.5)
	return tmp
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 code(k, n)
	tmp = 0.0
	if (k <= 1.2237637377882354e-35)
		tmp = Float64(1.0 / Float64(sqrt(k) / sqrt(Float64(n * Float64(2.0 * pi)))));
	else
		tmp = Float64(k / (Float64(pi * Float64(2.0 * n)) ^ Float64(1.0 - k))) ^ -0.5;
	end
	return tmp
end
function tmp = code(k, n)
	tmp = (1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0));
end
function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 1.2237637377882354e-35)
		tmp = 1.0 / (sqrt(k) / sqrt((n * (2.0 * pi))));
	else
		tmp = (k / ((pi * (2.0 * n)) ^ (1.0 - k))) ^ -0.5;
	end
	tmp_2 = tmp;
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]
code[k_, n_] := If[LessEqual[k, 1.2237637377882354e-35], N[(1.0 / N[(N[Sqrt[k], $MachinePrecision] / N[Sqrt[N[(n * N[(2.0 * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[Power[N[(k / N[Power[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], -0.5], $MachinePrecision]]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
\begin{array}{l}
\mathbf{if}\;k \leq 1.2237637377882354 \cdot 10^{-35}:\\
\;\;\;\;\frac{1}{\frac{\sqrt{k}}{\sqrt{n \cdot \left(2 \cdot \pi\right)}}}\\

\mathbf{else}:\\
\;\;\;\;{\left(\frac{k}{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}\right)}^{-0.5}\\


\end{array}

Error

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 2 regimes
  2. if k < 1.22376373778823541e-35

    1. Initial program 0.5

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
    2. Applied egg-rr0.5

      \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{k}}{\sqrt{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}}}} \]
    3. Taylor expanded in k around 0 0.5

      \[\leadsto \frac{1}{\frac{\sqrt{k}}{\sqrt{\color{blue}{2 \cdot \left(n \cdot \pi\right)}}}} \]
    4. Simplified0.5

      \[\leadsto \frac{1}{\frac{\sqrt{k}}{\sqrt{\color{blue}{n \cdot \left(2 \cdot \pi\right)}}}} \]
      Proof
      (*.f64 n (*.f64 2 (PI.f64))): 0 points increase in error, 0 points decrease in error
      (*.f64 n (Rewrite=> *-commutative_binary64 (*.f64 (PI.f64) 2))): 0 points increase in error, 0 points decrease in error
      (Rewrite<= associate-*l*_binary64 (*.f64 (*.f64 n (PI.f64)) 2)): 0 points increase in error, 0 points decrease in error
      (Rewrite<= *-commutative_binary64 (*.f64 2 (*.f64 n (PI.f64)))): 0 points increase in error, 0 points decrease in error

    if 1.22376373778823541e-35 < k

    1. Initial program 0.5

      \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
    2. Applied egg-rr0.6

      \[\leadsto \color{blue}{\frac{1}{\frac{\sqrt{k}}{\sqrt{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}}}} \]
    3. Applied egg-rr0.7

      \[\leadsto \color{blue}{{\left(\frac{k}{{\left(\left(2 \cdot n\right) \cdot \pi\right)}^{\left(1 - k\right)}}\right)}^{-0.5}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1.2237637377882354 \cdot 10^{-35}:\\ \;\;\;\;\frac{1}{\frac{\sqrt{k}}{\sqrt{n \cdot \left(2 \cdot \pi\right)}}}\\ \mathbf{else}:\\ \;\;\;\;{\left(\frac{k}{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}\right)}^{-0.5}\\ \end{array} \]

Alternatives

Alternative 1
Error0.5
Cost26304
\[\frac{1}{\frac{\sqrt{k}}{\sqrt{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}}} \]
Alternative 2
Error0.5
Cost20032
\[\sqrt{\frac{1}{k}} \cdot {\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\frac{1 - k}{2}\right)} \]
Alternative 3
Error0.6
Cost19908
\[\begin{array}{l} \mathbf{if}\;k \leq 1.2237637377882354 \cdot 10^{-35}:\\ \;\;\;\;\frac{1}{\frac{\sqrt{k}}{\sqrt{n \cdot \left(2 \cdot \pi\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]
Alternative 4
Error0.5
Cost19904
\[\frac{{\left(\pi \cdot \left(n + n\right)\right)}^{\left(\left(1 - k\right) \cdot 0.5\right)}}{\sqrt{k}} \]
Alternative 5
Error2.8
Cost19844
\[\begin{array}{l} \mathbf{if}\;k \leq 1.2237637377882354 \cdot 10^{-35}:\\ \;\;\;\;\frac{1}{\frac{\sqrt{k}}{\sqrt{n \cdot \left(2 \cdot \pi\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{2 \cdot \left(n \cdot \left(\left(1 + \frac{\pi}{k}\right) + -1\right)\right)}\\ \end{array} \]
Alternative 6
Error2.7
Cost19716
\[\begin{array}{l} \mathbf{if}\;k \leq 1.2237637377882354 \cdot 10^{-35}:\\ \;\;\;\;\sqrt{2 \cdot n} \cdot \sqrt{\frac{\pi}{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{2 \cdot \left(n \cdot \left(\left(1 + \frac{\pi}{k}\right) + -1\right)\right)}\\ \end{array} \]
Alternative 7
Error13.0
Cost13440
\[\sqrt{2 \cdot \left(n \cdot \left(\left(1 + \frac{\pi}{k}\right) + -1\right)\right)} \]
Alternative 8
Error31.8
Cost13248
\[{\left(\frac{k}{2 \cdot \left(\pi \cdot n\right)}\right)}^{-0.5} \]
Alternative 9
Error32.5
Cost13184
\[\sqrt{2 \cdot \left(n \cdot \frac{\pi}{k}\right)} \]
Alternative 10
Error32.4
Cost13184
\[\sqrt{2 \cdot \left(\pi \cdot \frac{n}{k}\right)} \]

Error

Reproduce

herbie shell --seed 2022308 
(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))))