Average Error: 0.0 → 0.0
Time: 2.1s
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 r102654 = x;
        double r102655 = r102654 * r102654;
        double r102656 = r102654 * r102655;
        double r102657 = r102656 + r102655;
        return r102657;
}

double f(double x) {
        double r102658 = x;
        double r102659 = r102658 * r102658;
        double r102660 = r102659 + r102658;
        double r102661 = r102658 * r102660;
        return r102661;
}

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 2020057 
(FPCore (x)
  :name "Expression 3, p15"
  :precision binary64
  :pre (<= 0.0 x 2)

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

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