Average Error: 2.1 → 2.1
Time: 7.4s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\frac{{k}^{m}}{k \cdot \left(10 + k\right) + 1} \cdot a\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\frac{{k}^{m}}{k \cdot \left(10 + k\right) + 1} \cdot a
double f(double a, double k, double m) {
        double r612 = a;
        double r613 = k;
        double r614 = m;
        double r615 = pow(r613, r614);
        double r616 = r612 * r615;
        double r617 = 1.0;
        double r618 = 10.0;
        double r619 = r618 * r613;
        double r620 = r617 + r619;
        double r621 = r613 * r613;
        double r622 = r620 + r621;
        double r623 = r616 / r622;
        return r623;
}
↓
double f(double a, double k, double m) {
        double r624 = k;
        double r625 = m;
        double r626 = pow(r624, r625);
        double r627 = 10.0;
        double r628 = r627 + r624;
        double r629 = r624 * r628;
        double r630 = 1.0;
        double r631 = r629 + r630;
        double r632 = r626 / r631;
        double r633 = a;
        double r634 = r632 * r633;
        return r634;
}

Error

Bits error versus a

Bits error versus k

Bits error versus m

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Initial program 2.1

    \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]

  2. Simplified2.1

    \[\leadsto \color{blue}{\frac{{k}^{m}}{k \cdot \left(10 + k\right) + 1} \cdot a}\]

  3. Final simplification2.1

    \[\leadsto \frac{{k}^{m}}{k \cdot \left(10 + k\right) + 1} \cdot a\]

Reproduce

herbie shell --seed 2020025 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))