Average Error: 0.5 → 0.5
Time: 8.8s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\]
↓
\[{\left(2 \cdot \pi\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{\frac{\sqrt{n}}{{n}^{\left(\frac{k}{2}\right)}}}{\sqrt{k}}\]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
{\left(2 \cdot \pi\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{\frac{\sqrt{n}}{{n}^{\left(\frac{k}{2}\right)}}}{\sqrt{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
 (*
  (pow (* 2.0 PI) (/ (- 1.0 k) 2.0))
  (/ (/ (sqrt n) (pow n (/ k 2.0))) (sqrt 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 pow((2.0 * ((double) M_PI)), ((1.0 - k) / 2.0)) * ((sqrt(n) / pow(n, (k / 2.0))) / sqrt(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. Simplified0.5

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

  4. Applied *-un-lft-identity_binary64_7600.5

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

  5. Applied sqrt-prod_binary64_7760.5

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

  6. Applied unpow-prod-down_binary64_8390.6

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

  7. Applied times-frac_binary64_7660.6

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

  8. Simplified0.6

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

  10. Applied div-sub_binary64_7650.6

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

  11. Applied pow-sub_binary64_8360.5

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

  12. Simplified0.5

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

  13. Final simplification0.5

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

Reproduce

herbie shell --seed 2020354 
(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))))