Average Error: 0.0 → 0.0
Time: 2.5s
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 r765694 = x;
        double r765695 = y;
        double r765696 = r765695 * r765695;
        double r765697 = exp(r765696);
        double r765698 = r765694 * r765697;
        return r765698;
}
↓
double f(double x, double y) {
        double r765699 = x;
        double r765700 = y;
        double r765701 = r765700 * r765700;
        double r765702 = exp(r765701);
        double r765703 = r765699 * r765702;
        return r765703;
}

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 2019353 +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))))