Average Error: 2.1 → 0.1
Time: 4.4m
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 11365813.86397301:\\ \;\;\;\;\frac{a}{\frac{\left(10 \cdot k + 1\right) + k \cdot k}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;99 \cdot \frac{a \cdot e^{m \cdot \log k}}{\left(k \cdot k\right) \cdot \left(k \cdot k\right)} + \left(e^{m \cdot \log k} \cdot \frac{\frac{a}{k}}{k} - \left(\frac{10}{k} \cdot \frac{\frac{a}{k}}{k}\right) \cdot e^{m \cdot \log k}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 11365813.86397301:\\
\;\;\;\;\frac{a}{\frac{\left(10 \cdot k + 1\right) + k \cdot k}{{k}^{m}}}\\

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

\end{array}
double f(double a, double k, double m) {
        double r85116879 = a;
        double r85116880 = k;
        double r85116881 = m;
        double r85116882 = pow(r85116880, r85116881);
        double r85116883 = r85116879 * r85116882;
        double r85116884 = 1.0;
        double r85116885 = 10.0;
        double r85116886 = r85116885 * r85116880;
        double r85116887 = r85116884 + r85116886;
        double r85116888 = r85116880 * r85116880;
        double r85116889 = r85116887 + r85116888;
        double r85116890 = r85116883 / r85116889;
        return r85116890;
}
↓
double f(double a, double k, double m) {
        double r85116891 = k;
        double r85116892 = 11365813.86397301;
        bool r85116893 = r85116891 <= r85116892;
        double r85116894 = a;
        double r85116895 = 10.0;
        double r85116896 = r85116895 * r85116891;
        double r85116897 = 1.0;
        double r85116898 = r85116896 + r85116897;
        double r85116899 = r85116891 * r85116891;
        double r85116900 = r85116898 + r85116899;
        double r85116901 = m;
        double r85116902 = pow(r85116891, r85116901);
        double r85116903 = r85116900 / r85116902;
        double r85116904 = r85116894 / r85116903;
        double r85116905 = 99.0;
        double r85116906 = log(r85116891);
        double r85116907 = r85116901 * r85116906;
        double r85116908 = exp(r85116907);
        double r85116909 = r85116894 * r85116908;
        double r85116910 = r85116899 * r85116899;
        double r85116911 = r85116909 / r85116910;
        double r85116912 = r85116905 * r85116911;
        double r85116913 = r85116894 / r85116891;
        double r85116914 = r85116913 / r85116891;
        double r85116915 = r85116908 * r85116914;
        double r85116916 = r85116895 / r85116891;
        double r85116917 = r85116916 * r85116914;
        double r85116918 = r85116917 * r85116908;
        double r85116919 = r85116915 - r85116918;
        double r85116920 = r85116912 + r85116919;
        double r85116921 = r85116893 ? r85116904 : r85116920;
        return r85116921;
}

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 < 11365813.86397301

    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 associate-/l*0.1

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

    if 11365813.86397301 < k

    1. Initial program 5.6

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

    2. Taylor expanded around -inf 63.0

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

    3. Simplified0.2

      \[\leadsto \color{blue}{99 \cdot \frac{e^{m \cdot \left(0 + \log k\right)} \cdot a}{\left(k \cdot k\right) \cdot \left(k \cdot k\right)} + \left(\frac{\frac{a}{k}}{k} \cdot e^{m \cdot \left(0 + \log k\right)} - \left(\frac{10}{k} \cdot \frac{\frac{a}{k}}{k}\right) \cdot e^{m \cdot \left(0 + \log k\right)}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 11365813.86397301:\\ \;\;\;\;\frac{a}{\frac{\left(10 \cdot k + 1\right) + k \cdot k}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;99 \cdot \frac{a \cdot e^{m \cdot \log k}}{\left(k \cdot k\right) \cdot \left(k \cdot k\right)} + \left(e^{m \cdot \log k} \cdot \frac{\frac{a}{k}}{k} - \left(\frac{10}{k} \cdot \frac{\frac{a}{k}}{k}\right) \cdot e^{m \cdot \log k}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019112 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))