Average Error: 1.9 → 1.9
Time: 6.8s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\frac{a}{\frac{\left(1 + 10 \cdot k\right) + k \cdot k}{{k}^{m}}}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\frac{a}{\frac{\left(1 + 10 \cdot k\right) + k \cdot k}{{k}^{m}}}
double f(double a, double k, double m) {
        double r350801 = a;
        double r350802 = k;
        double r350803 = m;
        double r350804 = pow(r350802, r350803);
        double r350805 = r350801 * r350804;
        double r350806 = 1.0;
        double r350807 = 10.0;
        double r350808 = r350807 * r350802;
        double r350809 = r350806 + r350808;
        double r350810 = r350802 * r350802;
        double r350811 = r350809 + r350810;
        double r350812 = r350805 / r350811;
        return r350812;
}
↓
double f(double a, double k, double m) {
        double r350813 = a;
        double r350814 = 1.0;
        double r350815 = 10.0;
        double r350816 = k;
        double r350817 = r350815 * r350816;
        double r350818 = r350814 + r350817;
        double r350819 = r350816 * r350816;
        double r350820 = r350818 + r350819;
        double r350821 = m;
        double r350822 = pow(r350816, r350821);
        double r350823 = r350820 / r350822;
        double r350824 = r350813 / r350823;
        return r350824;
}

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 1.9

    \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
  2. Using strategy rm

  3. Applied associate-/l*1.9

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

  4. Final simplification1.9

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

Reproduce

herbie shell --seed 2019362 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))