Average Error: 0.0 → 0.0
Time: 5.4s
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 r38953 = x;
        double r38954 = 1.0;
        double r38955 = 0.5;
        double r38956 = r38953 * r38955;
        double r38957 = r38954 - r38956;
        double r38958 = r38953 * r38957;
        return r38958;
}
↓
double f(double x) {
        double r38959 = x;
        double r38960 = 1.0;
        double r38961 = 0.5;
        double r38962 = r38959 * r38961;
        double r38963 = r38960 - r38962;
        double r38964 = r38959 * r38963;
        return r38964;
}

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 2020047 
(FPCore (x)
  :name "Numeric.SpecFunctions:log1p from math-functions-0.1.5.2, B"
  :precision binary64
  (* x (- 1 (* x 0.5))))