Average Error: 0.0 → 0.0
Time: 5.8s
Precision: 64
\[0.0 \le x \le 2\]
\[x \cdot \left(x \cdot x\right) + x \cdot x\]
↓
\[x \cdot \left(x \cdot x + x\right)\]
x \cdot \left(x \cdot x\right) + x \cdot x
↓
x \cdot \left(x \cdot x + x\right)
double f(double x) {
        double r89258 = x;
        double r89259 = r89258 * r89258;
        double r89260 = r89258 * r89259;
        double r89261 = r89260 + r89259;
        return r89261;
}
↓
double f(double x) {
        double r89262 = x;
        double r89263 = r89262 * r89262;
        double r89264 = r89263 + r89262;
        double r89265 = r89262 * r89264;
        return r89265;
}

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

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

Reproduce

herbie shell --seed 2019294 
(FPCore (x)
  :name "Expression 3, p15"
  :precision binary64
  :pre (<= 0.0 x 2)

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

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