Average Error: 0.0 → 0.0
Time: 1.3s
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 r68356 = x;
        double r68357 = 1.0;
        double r68358 = 0.5;
        double r68359 = r68356 * r68358;
        double r68360 = r68357 - r68359;
        double r68361 = r68356 * r68360;
        return r68361;
}
↓
double f(double x) {
        double r68362 = x;
        double r68363 = 1.0;
        double r68364 = 0.5;
        double r68365 = r68362 * r68364;
        double r68366 = r68363 - r68365;
        double r68367 = r68362 * r68366;
        return r68367;
}

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 2019322 
(FPCore (x)
  :name "Numeric.SpecFunctions:log1p from math-functions-0.1.5.2, B"
  :precision binary64
  (* x (- 1 (* x 0.5))))