Average Error: 0.3 → 0.4
Time: 7.7s
Precision: 64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\]
↓
\[\frac{1}{\sqrt{k}} \cdot \left(\sqrt{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}} \cdot \sqrt{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}\right)\]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\frac{1}{\sqrt{k}} \cdot \left(\sqrt{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}} \cdot \sqrt{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}\right)
double f(double k, double n) {
        double r130206 = 1.0;
        double r130207 = k;
        double r130208 = sqrt(r130207);
        double r130209 = r130206 / r130208;
        double r130210 = 2.0;
        double r130211 = atan2(1.0, 0.0);
        double r130212 = r130210 * r130211;
        double r130213 = n;
        double r130214 = r130212 * r130213;
        double r130215 = r130206 - r130207;
        double r130216 = r130215 / r130210;
        double r130217 = pow(r130214, r130216);
        double r130218 = r130209 * r130217;
        return r130218;
}
↓
double f(double k, double n) {
        double r130219 = 1.0;
        double r130220 = k;
        double r130221 = sqrt(r130220);
        double r130222 = r130219 / r130221;
        double r130223 = 2.0;
        double r130224 = atan2(1.0, 0.0);
        double r130225 = r130223 * r130224;
        double r130226 = n;
        double r130227 = r130225 * r130226;
        double r130228 = r130219 - r130220;
        double r130229 = r130228 / r130223;
        double r130230 = pow(r130227, r130229);
        double r130231 = sqrt(r130230);
        double r130232 = r130231 * r130231;
        double r130233 = r130222 * r130232;
        return r130233;
}

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 0.3

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

  3. Applied add-sqr-sqrt0.4

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

  4. Final simplification0.4

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

Reproduce

herbie shell --seed 2020001 
(FPCore (k n)
  :name "Migdal et al, Equation (51)"
  :precision binary64
  (* (/ 1 (sqrt k)) (pow (* (* 2 PI) n) (/ (- 1 k) 2))))