Average Error: 0.0 → 0.0
Time: 6.7s
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 r61610 = x;
        double r61611 = r61610 * r61610;
        double r61612 = r61610 * r61611;
        double r61613 = r61612 + r61611;
        return r61613;
}
↓
double f(double x) {
        double r61614 = x;
        double r61615 = r61614 * r61614;
        double r61616 = r61615 + r61614;
        double r61617 = r61614 * r61616;
        return r61617;
}

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 2019291 
(FPCore (x)
  :name "Expression 3, p15"
  :precision binary64
  :pre (<= 0.0 x 2)

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

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