Average Error: 0.0 → 0.0
Time: 15.0s
Precision: 64
\[x \cdot e^{y \cdot y}\]
↓
\[{\left(e^{y}\right)}^{y} \cdot x\]
x \cdot e^{y \cdot y}
↓
{\left(e^{y}\right)}^{y} \cdot x
double f(double x, double y) {
        double r28614516 = x;
        double r28614517 = y;
        double r28614518 = r28614517 * r28614517;
        double r28614519 = exp(r28614518);
        double r28614520 = r28614516 * r28614519;
        return r28614520;
}
↓
double f(double x, double y) {
        double r28614521 = y;
        double r28614522 = exp(r28614521);
        double r28614523 = pow(r28614522, r28614521);
        double r28614524 = x;
        double r28614525 = r28614523 * r28614524;
        return r28614525;
}

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

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

Reproduce

herbie shell --seed 2019163 +o rules:numerics
(FPCore (x y)
  :name "Data.Number.Erf:$dmerfcx from erf-2.0.0.0"

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

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