Average Error: 0.0 → 0.0
Time: 1.4s
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 r100537 = x;
        double r100538 = r100537 * r100537;
        double r100539 = r100537 * r100538;
        double r100540 = r100539 + r100538;
        return r100540;
}

double f(double x) {
        double r100541 = x;
        double r100542 = r100541 * r100541;
        double r100543 = r100542 + r100541;
        double r100544 = r100541 * r100543;
        return r100544;
}

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 2020065 
(FPCore (x)
  :name "Expression 3, p15"
  :precision binary64
  :pre (<= 0.0 x 2)

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

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