Average Error: 0.0 → 0.0
Time: 1.0s
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 r94575 = x;
        double r94576 = r94575 * r94575;
        double r94577 = r94575 * r94576;
        double r94578 = r94577 + r94576;
        return r94578;
}
↓
double f(double x) {
        double r94579 = x;
        double r94580 = r94579 * r94579;
        double r94581 = r94579 * r94580;
        double r94582 = r94581 + r94580;
        return r94582;
}

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 2019354 
(FPCore (x)
  :name "Expression 3, p15"
  :precision binary64
  :pre (<= 0.0 x 2)

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

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