Average Error: 0.0 → 0.0
Time: 22.3s
Precision: 64
\[x \cdot e^{y \cdot y}\]
\[x \cdot {\left(e^{y}\right)}^{y}\]
x \cdot e^{y \cdot y}
x \cdot {\left(e^{y}\right)}^{y}
double f(double x, double y) {
        double r521312 = x;
        double r521313 = y;
        double r521314 = r521313 * r521313;
        double r521315 = exp(r521314);
        double r521316 = r521312 * r521315;
        return r521316;
}

double f(double x, double y) {
        double r521317 = x;
        double r521318 = y;
        double r521319 = exp(r521318);
        double r521320 = pow(r521319, r521318);
        double r521321 = r521317 * r521320;
        return r521321;
}

Error

Bits error versus x

Bits error versus y

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original0.0
Target0.0
Herbie0.0
\[x \cdot {\left(e^{y}\right)}^{y}\]

Derivation

  1. Initial program 0.0

    \[x \cdot e^{y \cdot y}\]
  2. Using strategy rm
  3. Applied add-log-exp0.0

    \[\leadsto x \cdot e^{\color{blue}{\log \left(e^{y}\right)} \cdot y}\]
  4. Applied exp-to-pow0.0

    \[\leadsto x \cdot \color{blue}{{\left(e^{y}\right)}^{y}}\]
  5. Final simplification0.0

    \[\leadsto x \cdot {\left(e^{y}\right)}^{y}\]

Reproduce

herbie shell --seed 2019322 +o rules:numerics
(FPCore (x y)
  :name "Data.Number.Erf:$dmerfcx from erf-2.0.0.0"
  :precision binary64

  :herbie-target
  (* x (pow (exp y) y))

  (* x (exp (* y y))))