Average Error: 0.5 → 0.4
Time: 9.1s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\]
↓
\[1 \cdot \frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(\frac{1}{2}\right)}}{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{k}{2}\right)} \cdot \sqrt{k}}\]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
1 \cdot \frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(\frac{1}{2}\right)}}{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{k}{2}\right)} \cdot \sqrt{k}}
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 ((double) (1.0 * (((double) pow(((double) (2.0 * ((double) (((double) M_PI) * n)))), (1.0 / 2.0))) / ((double) (((double) pow(((double) (((double) M_PI) * ((double) (2.0 * n)))), (k / 2.0))) * ((double) 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

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

  4. Applied div-sub0.5

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

  5. Applied pow-sub0.4

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

  6. Applied associate-/l/0.4

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

  7. Simplified0.4

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

  8. Final simplification0.4

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

Reproduce

herbie shell --seed 2020196 
(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))))