Average Error: 0.0 → 0.0
Time: 4.3s
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 r53157 = x;
        double r53158 = 1.0;
        double r53159 = 0.5;
        double r53160 = r53157 * r53159;
        double r53161 = r53158 - r53160;
        double r53162 = r53157 * r53161;
        return r53162;
}
↓
double f(double x) {
        double r53163 = x;
        double r53164 = 1.0;
        double r53165 = 0.5;
        double r53166 = r53163 * r53165;
        double r53167 = r53164 - r53166;
        double r53168 = r53163 * r53167;
        return r53168;
}

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 2019297 
(FPCore (x)
  :name "Numeric.SpecFunctions:log1p from math-functions-0.1.5.2, B"
  :precision binary64
  (* x (- 1 (* x 0.5))))