Average Error: 0.5 → 0.5
Time: 6.1s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\]
↓
\[\sqrt{n} \cdot \frac{\frac{\sqrt{2} \cdot \sqrt{\pi}}{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\frac{k}{2}\right)}}}{\sqrt{k}}\]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\sqrt{n} \cdot \frac{\frac{\sqrt{2} \cdot \sqrt{\pi}}{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\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
 (*
  (sqrt n)
  (/ (/ (* (sqrt 2.0) (sqrt PI)) (pow (* n (* 2.0 PI)) (/ 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 sqrt(n) * (((sqrt(2.0) * sqrt((double) M_PI)) / pow((n * (2.0 * ((double) M_PI))), (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 div-sub_binary640.5

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

  5. Applied pow-sub_binary640.4

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

  6. Simplified0.4

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

  7. Simplified0.4

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

  9. Applied *-un-lft-identity_binary640.4

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

  10. Applied sqrt-prod_binary640.4

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

  11. Applied *-un-lft-identity_binary640.4

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

  12. Applied sqrt-prod_binary640.6

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

  13. Applied times-frac_binary640.6

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

  14. Applied times-frac_binary640.6

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

  15. Simplified0.6

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

  17. Applied sqrt-prod_binary640.5

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

  18. Final simplification0.5

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

Reproduce

herbie shell --seed 2020219 
(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))))