Average Error: 0.0 → 0.0
Time: 7.2s
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 r658435 = x;
        double r658436 = y;
        double r658437 = r658436 * r658436;
        double r658438 = exp(r658437);
        double r658439 = r658435 * r658438;
        return r658439;
}
↓
double f(double x, double y) {
        double r658440 = x;
        double r658441 = y;
        double r658442 = r658441 * r658441;
        double r658443 = exp(r658442);
        double r658444 = r658440 * r658443;
        return r658444;
}

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 2019208 
(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))))