Average Error: 0.5 → 0.6
Time: 7.6s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\]
↓
\[\left({\pi}^{\left(\frac{1 - k}{2}\right)} \cdot \left({2}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}}\right)\right) \cdot {n}^{\left(\frac{1 - k}{2}\right)}\]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\left({\pi}^{\left(\frac{1 - k}{2}\right)} \cdot \left({2}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}}\right)\right) \cdot {n}^{\left(\frac{1 - k}{2}\right)}
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
 :precision binary64
 (*
  (*
   (pow PI (/ (- 1.0 k) 2.0))
   (* (pow 2.0 (/ (- 1.0 k) 2.0)) (/ 1.0 (sqrt k))))
  (pow n (/ (- 1.0 k) 2.0))))
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) (((double) (((double) pow(((double) M_PI), (((double) (1.0 - k)) / 2.0))) * ((double) (((double) pow(2.0, (((double) (1.0 - k)) / 2.0))) * (1.0 / ((double) sqrt(k))))))) * ((double) pow(n, (((double) (1.0 - 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 Error: 0.5 bits

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

  3. Applied unpow-prod-downError: 0.7 bits

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

  5. Applied unpow-prod-downError: 0.6 bits

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

  7. Applied associate-*r*Error: 0.6 bits

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

  8. SimplifiedError: 0.6 bits

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

  9. Final simplificationError: 0.6 bits

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

Reproduce

herbie shell --seed 2020203 
(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))))