Average Error: 0.0 → 0.0
Time: 5.8s
Precision: 64
\[x \cdot e^{y \cdot y}\]
↓
\[x \cdot e^{y \cdot y}\]
x \cdot e^{y \cdot y}
↓
x \cdot e^{y \cdot y}
double f(double x, double y) {
        double r438656 = x;
        double r438657 = y;
        double r438658 = r438657 * r438657;
        double r438659 = exp(r438658);
        double r438660 = r438656 * r438659;
        return r438660;
}
↓
double f(double x, double y) {
        double r438661 = x;
        double r438662 = y;
        double r438663 = r438662 * r438662;
        double r438664 = exp(r438663);
        double r438665 = r438661 * r438664;
        return r438665;
}

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. Final simplification0.0

    \[\leadsto x \cdot e^{y \cdot y}\]

Reproduce

herbie shell --seed 2019235 +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))))