Average Error: 0.0 → 0.0
Time: 6.6s
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 r49099 = x;
        double r49100 = 1.0;
        double r49101 = 0.5;
        double r49102 = r49099 * r49101;
        double r49103 = r49100 - r49102;
        double r49104 = r49099 * r49103;
        return r49104;
}
↓
double f(double x) {
        double r49105 = x;
        double r49106 = 1.0;
        double r49107 = 0.5;
        double r49108 = r49105 * r49107;
        double r49109 = r49106 - r49108;
        double r49110 = r49105 * r49109;
        return r49110;
}

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 2019303 
(FPCore (x)
  :name "Numeric.SpecFunctions:log1p from math-functions-0.1.5.2, B"
  :precision binary64
  (* x (- 1 (* x 0.5))))