Average Error: 0.5 → 0.5
Time: 9.4s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
\[\begin{array}{l} t_0 := 2 \cdot \left(n \cdot \pi\right)\\ \sqrt{\frac{1}{k}} \cdot \frac{\sqrt{t_0}}{\sqrt{{t_0}^{k}}} \end{array} \]
\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)\\
\sqrt{\frac{1}{k}} \cdot \frac{\sqrt{t_0}}{\sqrt{{t_0}^{k}}}
\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))))
   (* (sqrt (/ 1.0 k)) (/ (sqrt t_0) (sqrt (pow t_0 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) {
	double t_0 = 2.0 * (n * ((double) M_PI));
	return sqrt((1.0 / k)) * (sqrt(t_0) / sqrt(pow(t_0, 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. Taylor expanded in n around 0 3.5

    \[\leadsto \color{blue}{\sqrt{\frac{1}{k}} \cdot e^{0.5 \cdot \left(\left(1 - k\right) \cdot \left(\log n + \log \left(2 \cdot \pi\right)\right)\right)}} \]
  4. Simplified0.5

    \[\leadsto \color{blue}{\sqrt{\frac{1}{k}} \cdot \sqrt{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(1 - k\right)}}} \]
  5. Applied egg-rr0.5

    \[\leadsto \sqrt{\frac{1}{k}} \cdot \color{blue}{\frac{\sqrt{2 \cdot \left(n \cdot \pi\right)}}{\sqrt{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{k}}}} \]
  6. Final simplification0.5

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

Reproduce

herbie shell --seed 2022127 
(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))))