Average Error: 0.5 → 0.5
Time: 7.1s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\]
↓
\[\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{0.5}}{\sqrt{k} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}\]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{0.5}}{\sqrt{k} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
 :precision binary64
 (/ (pow (* (* 2.0 PI) n) 0.5) (* (sqrt k) (pow (* (* 2.0 PI) n) (/ 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));
}
↓
double code(double k, double n) {
	return pow(((2.0 * ((double) M_PI)) * n), 0.5) / (sqrt(k) * pow(((2.0 * ((double) M_PI)) * n), (k / 2.0)));
}

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. Simplified0.5

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

  4. Applied div-sub_binary64_7680.5

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

  5. Applied pow-sub_binary64_8390.5

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

  6. Applied associate-/l/_binary64_7100.5

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

  7. Final simplification0.5

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

Reproduce

herbie shell --seed 2020289 
(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))))