Average Error: 0.0 → 0.0
Time: 1.4s
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 r701995 = x;
        double r701996 = y;
        double r701997 = r701996 * r701996;
        double r701998 = exp(r701997);
        double r701999 = r701995 * r701998;
        return r701999;
}
↓
double f(double x, double y) {
        double r702000 = x;
        double r702001 = y;
        double r702002 = r702001 * r702001;
        double r702003 = exp(r702002);
        double r702004 = r702000 * r702003;
        return r702004;
}

Error

Bits error versus x

Bits error versus y

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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 2020020 
(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))))