Average Error: 2.0 → 0.1
Time: 50.1s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 3.899886463060430542828981650483975689114 \cdot 10^{109}:\\ \;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\log k \cdot m} \cdot a}{k \cdot k} \cdot \frac{99}{k \cdot k} + \left(\frac{e^{\log k \cdot m}}{k} \cdot \frac{a}{k} - \frac{10 \cdot e^{\log k \cdot m}}{\frac{k \cdot \left(k \cdot k\right)}{a}}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 3.899886463060430542828981650483975689114 \cdot 10^{109}:\\
\;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\

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

\end{array}
double f(double a, double k, double m) {
        double r11677887 = a;
        double r11677888 = k;
        double r11677889 = m;
        double r11677890 = pow(r11677888, r11677889);
        double r11677891 = r11677887 * r11677890;
        double r11677892 = 1.0;
        double r11677893 = 10.0;
        double r11677894 = r11677893 * r11677888;
        double r11677895 = r11677892 + r11677894;
        double r11677896 = r11677888 * r11677888;
        double r11677897 = r11677895 + r11677896;
        double r11677898 = r11677891 / r11677897;
        return r11677898;
}
↓
double f(double a, double k, double m) {
        double r11677899 = k;
        double r11677900 = 3.8998864630604305e+109;
        bool r11677901 = r11677899 <= r11677900;
        double r11677902 = a;
        double r11677903 = 1.0;
        double r11677904 = 10.0;
        double r11677905 = r11677899 + r11677904;
        double r11677906 = r11677905 * r11677899;
        double r11677907 = r11677903 + r11677906;
        double r11677908 = m;
        double r11677909 = pow(r11677899, r11677908);
        double r11677910 = r11677907 / r11677909;
        double r11677911 = r11677902 / r11677910;
        double r11677912 = log(r11677899);
        double r11677913 = r11677912 * r11677908;
        double r11677914 = exp(r11677913);
        double r11677915 = r11677914 * r11677902;
        double r11677916 = r11677899 * r11677899;
        double r11677917 = r11677915 / r11677916;
        double r11677918 = 99.0;
        double r11677919 = r11677918 / r11677916;
        double r11677920 = r11677917 * r11677919;
        double r11677921 = r11677914 / r11677899;
        double r11677922 = r11677902 / r11677899;
        double r11677923 = r11677921 * r11677922;
        double r11677924 = r11677904 * r11677914;
        double r11677925 = r11677899 * r11677916;
        double r11677926 = r11677925 / r11677902;
        double r11677927 = r11677924 / r11677926;
        double r11677928 = r11677923 - r11677927;
        double r11677929 = r11677920 + r11677928;
        double r11677930 = r11677901 ? r11677911 : r11677929;
        return r11677930;
}

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 < 3.8998864630604305e+109

    1. Initial program 0.1

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

    2. Simplified0.1

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

    4. Applied *-commutative0.1

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

    if 3.8998864630604305e+109 < k

    1. Initial program 7.9

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

    2. Simplified7.9

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

    3. Taylor expanded around inf 7.9

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

    4. Simplified0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 3.899886463060430542828981650483975689114 \cdot 10^{109}:\\ \;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{e^{\log k \cdot m} \cdot a}{k \cdot k} \cdot \frac{99}{k \cdot k} + \left(\frac{e^{\log k \cdot m}}{k} \cdot \frac{a}{k} - \frac{10 \cdot e^{\log k \cdot m}}{\frac{k \cdot \left(k \cdot k\right)}{a}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019174 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))