Average Error: 0.0 → 0.0
Time: 1.5s
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 r765126 = x;
        double r765127 = y;
        double r765128 = r765127 * r765127;
        double r765129 = exp(r765128);
        double r765130 = r765126 * r765129;
        return r765130;
}
↓
double f(double x, double y) {
        double r765131 = x;
        double r765132 = y;
        double r765133 = r765132 * r765132;
        double r765134 = exp(r765133);
        double r765135 = r765131 * r765134;
        return r765135;
}

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 2019356 
(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))))