Average Error: 0.0 → 0.0
Time: 4.4s
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 r48128 = x;
        double r48129 = 1.0;
        double r48130 = 0.5;
        double r48131 = r48128 * r48130;
        double r48132 = r48129 - r48131;
        double r48133 = r48128 * r48132;
        return r48133;
}
↓
double f(double x) {
        double r48134 = x;
        double r48135 = 1.0;
        double r48136 = 0.5;
        double r48137 = r48134 * r48136;
        double r48138 = r48135 - r48137;
        double r48139 = r48134 * r48138;
        return r48139;
}

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 2019298 
(FPCore (x)
  :name "Numeric.SpecFunctions:log1p from math-functions-0.1.5.2, B"
  :precision binary64
  (* x (- 1 (* x 0.5))))