Average Error: 0.0 → 0.0
Time: 1.5s
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 r90874 = x;
        double r90875 = r90874 * r90874;
        double r90876 = r90874 * r90875;
        double r90877 = r90876 + r90875;
        return r90877;
}
↓
double f(double x) {
        double r90878 = x;
        double r90879 = r90878 * r90878;
        double r90880 = r90879 + r90878;
        double r90881 = r90878 * r90880;
        return r90881;
}

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. Simplified0.0

    \[\leadsto \color{blue}{x \cdot \left(x \cdot x + x\right)}\]

  3. Final simplification0.0

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

Reproduce

herbie shell --seed 2020020 
(FPCore (x)
  :name "Expression 3, p15"
  :precision binary64
  :pre (<= 0.0 x 2)

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

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