Average Error: 2.0 → 0.1
Time: 20.9s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 7.485561233862986 \cdot 10^{+31}:\\ \;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{e^{\log k \cdot m}}{k} + \left(-10 \cdot \frac{e^{\log k \cdot m}}{\frac{\left(k \cdot k\right) \cdot k}{a}} + \frac{e^{\log k \cdot m} \cdot 99}{\frac{\left(k \cdot k\right) \cdot \left(k \cdot k\right)}{a}}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 7.485561233862986 \cdot 10^{+31}:\\
\;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\

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

\end{array}
double f(double a, double k, double m) {
        double r7069676 = a;
        double r7069677 = k;
        double r7069678 = m;
        double r7069679 = pow(r7069677, r7069678);
        double r7069680 = r7069676 * r7069679;
        double r7069681 = 1.0;
        double r7069682 = 10.0;
        double r7069683 = r7069682 * r7069677;
        double r7069684 = r7069681 + r7069683;
        double r7069685 = r7069677 * r7069677;
        double r7069686 = r7069684 + r7069685;
        double r7069687 = r7069680 / r7069686;
        return r7069687;
}
↓
double f(double a, double k, double m) {
        double r7069688 = k;
        double r7069689 = 7.485561233862986e+31;
        bool r7069690 = r7069688 <= r7069689;
        double r7069691 = a;
        double r7069692 = 1.0;
        double r7069693 = 10.0;
        double r7069694 = r7069688 + r7069693;
        double r7069695 = r7069694 * r7069688;
        double r7069696 = r7069692 + r7069695;
        double r7069697 = m;
        double r7069698 = pow(r7069688, r7069697);
        double r7069699 = r7069696 / r7069698;
        double r7069700 = r7069691 / r7069699;
        double r7069701 = r7069691 / r7069688;
        double r7069702 = log(r7069688);
        double r7069703 = r7069702 * r7069697;
        double r7069704 = exp(r7069703);
        double r7069705 = r7069704 / r7069688;
        double r7069706 = r7069701 * r7069705;
        double r7069707 = -10.0;
        double r7069708 = r7069688 * r7069688;
        double r7069709 = r7069708 * r7069688;
        double r7069710 = r7069709 / r7069691;
        double r7069711 = r7069704 / r7069710;
        double r7069712 = r7069707 * r7069711;
        double r7069713 = 99.0;
        double r7069714 = r7069704 * r7069713;
        double r7069715 = r7069708 * r7069708;
        double r7069716 = r7069715 / r7069691;
        double r7069717 = r7069714 / r7069716;
        double r7069718 = r7069712 + r7069717;
        double r7069719 = r7069706 + r7069718;
        double r7069720 = r7069690 ? r7069700 : r7069719;
        return r7069720;
}

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 < 7.485561233862986e+31

    1. Initial program 0.1

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

    2. Simplified0.0

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

    if 7.485561233862986e+31 < k

    1. Initial program 5.9

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

    2. Simplified5.9

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

    3. Taylor expanded around inf 5.9

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

    4. Simplified0.2

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 7.485561233862986 \cdot 10^{+31}:\\ \;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{a}{k} \cdot \frac{e^{\log k \cdot m}}{k} + \left(-10 \cdot \frac{e^{\log k \cdot m}}{\frac{\left(k \cdot k\right) \cdot k}{a}} + \frac{e^{\log k \cdot m} \cdot 99}{\frac{\left(k \cdot k\right) \cdot \left(k \cdot k\right)}{a}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019164 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))