Average Error: 0.0 → 0.0
Time: 896.0ms
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 r12287 = x;
        double r12288 = 1.0;
        double r12289 = 0.5;
        double r12290 = r12287 * r12289;
        double r12291 = r12288 - r12290;
        double r12292 = r12287 * r12291;
        return r12292;
}
↓
double f(double x) {
        double r12293 = x;
        double r12294 = 1.0;
        double r12295 = 0.5;
        double r12296 = r12293 * r12295;
        double r12297 = r12294 - r12296;
        double r12298 = r12293 * r12297;
        return r12298;
}

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 2019310 +o rules:numerics
(FPCore (x)
  :name "Numeric.SpecFunctions:log1p from math-functions-0.1.5.2, B"
  :precision binary64
  (* x (- 1 (* x 0.5))))