Average Error: 0.0 → 0.0
Time: 2.5s
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 r811041 = x;
        double r811042 = y;
        double r811043 = r811042 * r811042;
        double r811044 = exp(r811043);
        double r811045 = r811041 * r811044;
        return r811045;
}
↓
double f(double x, double y) {
        double r811046 = x;
        double r811047 = y;
        double r811048 = exp(r811047);
        double r811049 = pow(r811048, r811047);
        double r811050 = r811046 * r811049;
        return r811050;
}

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 2020065 +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))))