Average Error: 0.0 → 0.0
Time: 4.0s
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 r46199 = x;
        double r46200 = 1.0;
        double r46201 = 0.5;
        double r46202 = r46199 * r46201;
        double r46203 = r46200 - r46202;
        double r46204 = r46199 * r46203;
        return r46204;
}
↓
double f(double x) {
        double r46205 = x;
        double r46206 = 1.0;
        double r46207 = 0.5;
        double r46208 = r46205 * r46207;
        double r46209 = r46206 - r46208;
        double r46210 = r46205 * r46209;
        return r46210;
}

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 2020046 
(FPCore (x)
  :name "Numeric.SpecFunctions:log1p from math-functions-0.1.5.2, B"
  :precision binary64
  (* x (- 1 (* x 0.5))))