Average Error: 0.0 → 0.0
Time: 1.3s
Precision: 64
\[0.0 \le x \le 2\]
\[x \cdot \left(x \cdot x\right) + x \cdot x\]
↓
\[x \cdot \left(x \cdot x\right) + x \cdot x\]
x \cdot \left(x \cdot x\right) + x \cdot x
↓
x \cdot \left(x \cdot x\right) + x \cdot x
double f(double x) {
        double r71334 = x;
        double r71335 = r71334 * r71334;
        double r71336 = r71334 * r71335;
        double r71337 = r71336 + r71335;
        return r71337;
}
↓
double f(double x) {
        double r71338 = x;
        double r71339 = r71338 * r71338;
        double r71340 = r71338 * r71339;
        double r71341 = r71340 + r71339;
        return r71341;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original0.0
Target0.0
Herbie0.0
\[\left(\left(1 + x\right) \cdot x\right) \cdot x\]

Derivation

  1. Initial program 0.0

    \[x \cdot \left(x \cdot x\right) + x \cdot x\]

  2. Final simplification0.0

    \[\leadsto x \cdot \left(x \cdot x\right) + x \cdot x\]

Reproduce

herbie shell --seed 2020100 
(FPCore (x)
  :name "Expression 3, p15"
  :precision binary64
  :pre (<= 0.0 x 2)

  :herbie-target
  (* (* (+ 1 x) x) x)

  (+ (* x (* x x)) (* x x)))