Average Error: 0.0 → 0.0
Time: 1.1s
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 r50644 = x;
        double r50645 = 1.0;
        double r50646 = 0.5;
        double r50647 = r50644 * r50646;
        double r50648 = r50645 - r50647;
        double r50649 = r50644 * r50648;
        return r50649;
}
↓
double f(double x) {
        double r50650 = x;
        double r50651 = 1.0;
        double r50652 = 0.5;
        double r50653 = r50650 * r50652;
        double r50654 = r50651 - r50653;
        double r50655 = r50650 * r50654;
        return r50655;
}

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 2020036 
(FPCore (x)
  :name "Numeric.SpecFunctions:log1p from math-functions-0.1.5.2, B"
  :precision binary64
  (* x (- 1 (* x 0.5))))