Average Error: 0.0 → 0.0
Time: 598.0ms
Precision: 64
\[0.0 \le x \le 2\]
\[x \cdot \left(x \cdot x\right) + x \cdot x\]
↓
\[x \cdot \left(x \cdot x\right) + x \cdot x\]
x \cdot \left(x \cdot x\right) + x \cdot x
↓
x \cdot \left(x \cdot x\right) + x \cdot x
double f(double x) {
        double r103516 = x;
        double r103517 = r103516 * r103516;
        double r103518 = r103516 * r103517;
        double r103519 = r103518 + r103517;
        return r103519;
}
↓
double f(double x) {
        double r103520 = x;
        double r103521 = r103520 * r103520;
        double r103522 = r103520 * r103521;
        double r103523 = r103522 + r103521;
        return r103523;
}

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

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

Reproduce

herbie shell --seed 2020027 
(FPCore (x)
  :name "Expression 3, p15"
  :precision binary64
  :pre (<= 0.0 x 2)

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

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