Average Error: 0.0 → 0.0
Time: 5.9s
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 r107051 = x;
        double r107052 = r107051 * r107051;
        double r107053 = r107051 * r107052;
        double r107054 = r107053 + r107052;
        return r107054;
}
↓
double f(double x) {
        double r107055 = x;
        double r107056 = r107055 * r107055;
        double r107057 = r107055 * r107056;
        double r107058 = r107057 + r107056;
        return r107058;
}

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 2020046 
(FPCore (x)
  :name "Expression 3, p15"
  :precision binary64
  :pre (<= 0.0 x 2)

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

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