Average Error: 0.0 → 0.0
Time: 1.0s
Precision: 64
\[x \cdot \left(1 - x \cdot 0.5\right)\]
↓
\[x \cdot \left(1 - x \cdot 0.5\right)\]
x \cdot \left(1 - x \cdot 0.5\right)
↓
x \cdot \left(1 - x \cdot 0.5\right)
double f(double x) {
        double r37689 = x;
        double r37690 = 1.0;
        double r37691 = 0.5;
        double r37692 = r37689 * r37691;
        double r37693 = r37690 - r37692;
        double r37694 = r37689 * r37693;
        return r37694;
}
↓
double f(double x) {
        double r37695 = x;
        double r37696 = 1.0;
        double r37697 = 0.5;
        double r37698 = r37695 * r37697;
        double r37699 = r37696 - r37698;
        double r37700 = r37695 * r37699;
        return r37700;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Initial program 0.0

    \[x \cdot \left(1 - x \cdot 0.5\right)\]

  2. Final simplification0.0

    \[\leadsto x \cdot \left(1 - x \cdot 0.5\right)\]

Reproduce

herbie shell --seed 2020064 
(FPCore (x)
  :name "Numeric.SpecFunctions:log1p from math-functions-0.1.5.2, B"
  :precision binary64
  (* x (- 1 (* x 0.5))))