Average Error: 2.2 → 0.1
Time: 28.8s
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 r9571071 = a;
        double r9571072 = k;
        double r9571073 = m;
        double r9571074 = pow(r9571072, r9571073);
        double r9571075 = r9571071 * r9571074;
        double r9571076 = 1.0;
        double r9571077 = 10.0;
        double r9571078 = r9571077 * r9571072;
        double r9571079 = r9571076 + r9571078;
        double r9571080 = r9571072 * r9571072;
        double r9571081 = r9571079 + r9571080;
        double r9571082 = r9571075 / r9571081;
        return r9571082;
}
↓
double f(double a, double k, double m) {
        double r9571083 = k;
        double r9571084 = 4.5006258076783667e+133;
        bool r9571085 = r9571083 <= r9571084;
        double r9571086 = a;
        double r9571087 = m;
        double r9571088 = pow(r9571083, r9571087);
        double r9571089 = 1.0;
        double r9571090 = 10.0;
        double r9571091 = r9571090 + r9571083;
        double r9571092 = r9571083 * r9571091;
        double r9571093 = r9571089 + r9571092;
        double r9571094 = r9571088 / r9571093;
        double r9571095 = r9571086 * r9571094;
        double r9571096 = log(r9571083);
        double r9571097 = r9571087 * r9571096;
        double r9571098 = exp(r9571097);
        double r9571099 = r9571098 / r9571083;
        double r9571100 = r9571086 / r9571083;
        double r9571101 = r9571099 * r9571100;
        double r9571102 = 99.0;
        double r9571103 = r9571083 * r9571083;
        double r9571104 = r9571102 / r9571103;
        double r9571105 = r9571090 / r9571083;
        double r9571106 = r9571104 - r9571105;
        double r9571107 = r9571106 * r9571101;
        double r9571108 = r9571101 + r9571107;
        double r9571109 = r9571085 ? r9571095 : r9571108;
        return r9571109;
}

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))))