Average Error: 0.5 → 0.4
Time: 7.3s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
↓
\[\frac{\sqrt{\left(n \cdot \pi\right) \cdot \left(2 \cdot {\left({\left(\left(n \cdot \pi\right) \cdot 2\right)}^{k}\right)}^{-1}\right)}}{\sqrt{k}} \]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\frac{\sqrt{\left(n \cdot \pi\right) \cdot \left(2 \cdot {\left({\left(\left(n \cdot \pi\right) \cdot 2\right)}^{k}\right)}^{-1}\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 PI) (* 2.0 (pow (pow (* (* n PI) 2.0) k) -1.0)))) (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 * ((double) M_PI)) * (2.0 * pow(pow(((n * ((double) M_PI)) * 2.0), k), -1.0)))) / 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. Simplified0.4

    \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\sqrt{k}}} \]

  3. Applied egg-rr0.4

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

  4. Taylor expanded in k around inf 0.5

    \[\leadsto \frac{\color{blue}{\sqrt{n \cdot \pi} \cdot \left(\sqrt{2} \cdot e^{-0.5 \cdot \left(k \cdot \log \left(2 \cdot \left(n \cdot \pi\right)\right)\right)}\right)}}{\sqrt{k}} \]

  5. Simplified0.4

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

  6. Applied egg-rr0.4

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

  7. Final simplification0.4

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

Reproduce

herbie shell --seed 2022125 
(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))))