Average Error: 0.5 → 0.5
Time: 7.9s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
↓
\[\begin{array}{l} t_0 := 2 \cdot \left(n \cdot \pi\right)\\ {k}^{-0.5} \cdot \frac{\sqrt{t_0}}{{t_0}^{\left(\frac{k}{2}\right)}} \end{array} \]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\begin{array}{l}
t_0 := 2 \cdot \left(n \cdot \pi\right)\\
{k}^{-0.5} \cdot \frac{\sqrt{t_0}}{{t_0}^{\left(\frac{k}{2}\right)}}
\end{array}

(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* 2.0 (* n PI))))
   (* (pow k -0.5) (/ (sqrt t_0) (pow t_0 (/ k 2.0))))))
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) {
	double t_0 = 2.0 * (n * ((double) M_PI));
	return pow(k, -0.5) * (sqrt(t_0) / pow(t_0, (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_binary640.5

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

  5. Simplified0.5

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

  6. Simplified0.5

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

  8. Applied pow1/2_binary640.5

    \[\leadsto \frac{1}{\color{blue}{{k}^{0.5}}} \cdot \frac{\sqrt{2 \cdot \left(n \cdot \pi\right)}}{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(\frac{k}{2}\right)}} \]

  9. Applied pow-flip_binary640.5

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

  10. Simplified0.5

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

  11. Final simplification0.5

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

Reproduce

herbie shell --seed 2021204 
(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))))