Average Error: 0.5 → 0.4
Time: 16.8s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\]
↓
\[\frac{\sqrt{\frac{1}{k}} \cdot \sqrt{2 \cdot \left(n \cdot \pi\right)}}{{\left(\sqrt{2 \cdot \left(n \cdot \pi\right)}\right)}^{k}}\]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\frac{\sqrt{\frac{1}{k}} \cdot \sqrt{2 \cdot \left(n \cdot \pi\right)}}{{\left(\sqrt{2 \cdot \left(n \cdot \pi\right)}\right)}^{k}}
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
 :precision binary64
 (/
  (* (sqrt (/ 1.0 k)) (sqrt (* 2.0 (* n PI))))
  (pow (sqrt (* 2.0 (* n PI))) k)))
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) {
	return (sqrt(1.0 / k) * sqrt(2.0 * (n * ((double) M_PI)))) / pow(sqrt(2.0 * (n * ((double) M_PI))), k);
}

Error

Bits error versus k

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  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. Simplified0.4

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

  3. Taylor expanded around 0 3.4

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

  4. Simplified0.5

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

  6. Applied pow-sub_binary64_8360.4

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

  7. Applied associate-*r/_binary64_7020.4

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

  8. Final simplification0.4

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

Reproduce

herbie shell --seed 2021098 
(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))))