Average Error: 0.0 → 0.0
Time: 24.7s
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 r595681 = x;
        double r595682 = y;
        double r595683 = r595682 * r595682;
        double r595684 = exp(r595683);
        double r595685 = r595681 * r595684;
        return r595685;
}

double f(double x, double y) {
        double r595686 = x;
        double r595687 = y;
        double r595688 = exp(r595687);
        double r595689 = pow(r595688, r595687);
        double r595690 = r595686 * r595689;
        return r595690;
}

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 2019303 +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))))