Average Error: 0.0 → 0.0
Time: 9.1s
Precision: binary64
\[x \cdot e^{y \cdot y}\]
\[x \cdot {\left(e^{y}\right)}^{y}\]
x \cdot e^{y \cdot y}
x \cdot {\left(e^{y}\right)}^{y}
double code(double x, double y) {
	return ((double) (x * ((double) exp(((double) (y * y))))));
}
double code(double x, double y) {
	return ((double) (x * ((double) pow(((double) exp(y)), y))));
}

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. Using strategy rm
  3. Applied add-log-exp0.0

    \[\leadsto x \cdot e^{\color{blue}{\log \left(e^{y}\right)} \cdot y}\]
  4. Applied exp-to-pow0.0

    \[\leadsto x \cdot \color{blue}{{\left(e^{y}\right)}^{y}}\]
  5. Final simplification0.0

    \[\leadsto x \cdot {\left(e^{y}\right)}^{y}\]

Reproduce

herbie shell --seed 2020182 
(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))))