Average Error: 0.5 → 0.5
Time: 17.9s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\]
↓
\[\frac{1}{\frac{\sqrt{k}}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1}{2}\right)}}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}}}\]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\frac{1}{\frac{\sqrt{k}}{\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)}}}}
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
 :precision binary64
 (/
  1.0
  (/
   (sqrt k)
   (/ (pow (* (* 2.0 PI) n) (/ 1.0 2.0)) (pow (* (* 2.0 PI) n) (/ k 2.0))))))
double code(double k, double n) {
	return ((double) ((1.0 / ((double) sqrt(k))) * ((double) pow(((double) (((double) (2.0 * ((double) M_PI))) * n)), (((double) (1.0 - k)) / 2.0)))));
}
↓
double code(double k, double n) {
	return (1.0 / (((double) sqrt(k)) / (((double) pow(((double) (((double) (2.0 * ((double) M_PI))) * n)), (1.0 / 2.0))) / ((double) pow(((double) (((double) (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. Using strategy rm

  3. Applied div-sub_binary640.5

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

  4. Applied pow-sub_binary640.4

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

  5. Applied frac-times_binary640.4

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

  7. Applied associate-/l*_binary640.5

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

  8. Simplified0.5

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

  9. Final simplification0.5

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

Reproduce

herbie shell --seed 2020210 
(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))))