Average Error: 0.1 → 0
Time: 1.8s
Precision: 64
\[\left(d1 \cdot \left(\left(\left(\left(\left(d1 \cdot \left(d1 \cdot d1\right)\right) \cdot d1\right) \cdot d1\right) \cdot \left(d1 \cdot d1\right)\right) \cdot d1\right)\right) \cdot d1\]
↓
\[{d1}^{10}\]
\left(d1 \cdot \left(\left(\left(\left(\left(d1 \cdot \left(d1 \cdot d1\right)\right) \cdot d1\right) \cdot d1\right) \cdot \left(d1 \cdot d1\right)\right) \cdot d1\right)\right) \cdot d1
↓
{d1}^{10}
double f(double d1) {
        double r208028 = d1;
        double r208029 = r208028 * r208028;
        double r208030 = r208028 * r208029;
        double r208031 = r208030 * r208028;
        double r208032 = r208031 * r208028;
        double r208033 = r208032 * r208029;
        double r208034 = r208033 * r208028;
        double r208035 = r208028 * r208034;
        double r208036 = r208035 * r208028;
        return r208036;
}
↓
double f(double d1) {
        double r208037 = d1;
        double r208038 = 10.0;
        double r208039 = pow(r208037, r208038);
        return r208039;
}

Error

Bits error versus d1

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original0.1
Target0
Herbie0
\[{d1}^{10}\]

Derivation

  1. Initial program 0.1

    \[\left(d1 \cdot \left(\left(\left(\left(\left(d1 \cdot \left(d1 \cdot d1\right)\right) \cdot d1\right) \cdot d1\right) \cdot \left(d1 \cdot d1\right)\right) \cdot d1\right)\right) \cdot d1\]

  2. Simplified0

    \[\leadsto \color{blue}{{d1}^{10}}\]

  3. Final simplification0

    \[\leadsto {d1}^{10}\]

Reproduce

herbie shell --seed 2020047 +o rules:numerics
(FPCore (d1)
  :name "FastMath test5"
  :precision binary64

  :herbie-target
  (pow d1 10)

  (* (* d1 (* (* (* (* (* d1 (* d1 d1)) d1) d1) (* d1 d1)) d1)) d1))