Average Error: 2.2 → 0.1
Time: 29.6s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 4.500625807678366712530494498236264862465 \cdot 10^{133}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{1 + k \cdot \left(10 + k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{m \cdot \log k}}{k} \cdot \frac{a}{k} + \left(\frac{99}{k \cdot k} - \frac{10}{k}\right) \cdot \left(\frac{e^{m \cdot \log k}}{k} \cdot \frac{a}{k}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 4.500625807678366712530494498236264862465 \cdot 10^{133}:\\
\;\;\;\;a \cdot \frac{{k}^{m}}{1 + k \cdot \left(10 + k\right)}\\

\mathbf{else}:\\
\;\;\;\;\frac{e^{m \cdot \log k}}{k} \cdot \frac{a}{k} + \left(\frac{99}{k \cdot k} - \frac{10}{k}\right) \cdot \left(\frac{e^{m \cdot \log k}}{k} \cdot \frac{a}{k}\right)\\

\end{array}
double f(double a, double k, double m) {
        double r8947621 = a;
        double r8947622 = k;
        double r8947623 = m;
        double r8947624 = pow(r8947622, r8947623);
        double r8947625 = r8947621 * r8947624;
        double r8947626 = 1.0;
        double r8947627 = 10.0;
        double r8947628 = r8947627 * r8947622;
        double r8947629 = r8947626 + r8947628;
        double r8947630 = r8947622 * r8947622;
        double r8947631 = r8947629 + r8947630;
        double r8947632 = r8947625 / r8947631;
        return r8947632;
}
↓
double f(double a, double k, double m) {
        double r8947633 = k;
        double r8947634 = 4.5006258076783667e+133;
        bool r8947635 = r8947633 <= r8947634;
        double r8947636 = a;
        double r8947637 = m;
        double r8947638 = pow(r8947633, r8947637);
        double r8947639 = 1.0;
        double r8947640 = 10.0;
        double r8947641 = r8947640 + r8947633;
        double r8947642 = r8947633 * r8947641;
        double r8947643 = r8947639 + r8947642;
        double r8947644 = r8947638 / r8947643;
        double r8947645 = r8947636 * r8947644;
        double r8947646 = log(r8947633);
        double r8947647 = r8947637 * r8947646;
        double r8947648 = exp(r8947647);
        double r8947649 = r8947648 / r8947633;
        double r8947650 = r8947636 / r8947633;
        double r8947651 = r8947649 * r8947650;
        double r8947652 = 99.0;
        double r8947653 = r8947633 * r8947633;
        double r8947654 = r8947652 / r8947653;
        double r8947655 = r8947640 / r8947633;
        double r8947656 = r8947654 - r8947655;
        double r8947657 = r8947656 * r8947651;
        double r8947658 = r8947651 + r8947657;
        double r8947659 = r8947635 ? r8947645 : r8947658;
        return r8947659;
}

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. Split input into 2 regimes

  2. if k < 4.5006258076783667e+133

    1. Initial program 0.1

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

    3. Applied *-un-lft-identity0.1

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

    4. Applied times-frac0.1

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

    5. Simplified0.1

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

    6. Simplified0.1

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

    if 4.5006258076783667e+133 < k

    1. Initial program 9.9

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

    2. Taylor expanded around inf 9.9

      \[\leadsto \color{blue}{\left(\frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{2}} + 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}}\right) - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}}\]

    3. Simplified0.1

      \[\leadsto \color{blue}{\frac{a}{k} \cdot \frac{e^{\log k \cdot m}}{k} + \left(\frac{a}{k} \cdot \frac{e^{\log k \cdot m}}{k}\right) \cdot \left(\frac{99}{k \cdot k} - \frac{10}{k}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 4.500625807678366712530494498236264862465 \cdot 10^{133}:\\ \;\;\;\;a \cdot \frac{{k}^{m}}{1 + k \cdot \left(10 + k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{m \cdot \log k}}{k} \cdot \frac{a}{k} + \left(\frac{99}{k \cdot k} - \frac{10}{k}\right) \cdot \left(\frac{e^{m \cdot \log k}}{k} \cdot \frac{a}{k}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019200 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))