Average Error: 2.2 → 0.1
Time: 50.0s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 3.158753162179787 \cdot 10^{+109}:\\ \;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(a \cdot 99\right) \cdot e^{\log k \cdot m}}{\left(k \cdot k\right) \cdot \left(k \cdot k\right)} + \left(\frac{\frac{a \cdot e^{\log k \cdot m}}{k}}{k} + \frac{\left(a \cdot e^{\log k \cdot m}\right) \cdot -10}{\left(k \cdot k\right) \cdot k}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 3.158753162179787 \cdot 10^{+109}:\\
\;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\

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

\end{array}
double f(double a, double k, double m) {
        double r9595401 = a;
        double r9595402 = k;
        double r9595403 = m;
        double r9595404 = pow(r9595402, r9595403);
        double r9595405 = r9595401 * r9595404;
        double r9595406 = 1.0;
        double r9595407 = 10.0;
        double r9595408 = r9595407 * r9595402;
        double r9595409 = r9595406 + r9595408;
        double r9595410 = r9595402 * r9595402;
        double r9595411 = r9595409 + r9595410;
        double r9595412 = r9595405 / r9595411;
        return r9595412;
}
↓
double f(double a, double k, double m) {
        double r9595413 = k;
        double r9595414 = 3.158753162179787e+109;
        bool r9595415 = r9595413 <= r9595414;
        double r9595416 = a;
        double r9595417 = 1.0;
        double r9595418 = 10.0;
        double r9595419 = r9595413 + r9595418;
        double r9595420 = r9595419 * r9595413;
        double r9595421 = r9595417 + r9595420;
        double r9595422 = m;
        double r9595423 = pow(r9595413, r9595422);
        double r9595424 = r9595421 / r9595423;
        double r9595425 = r9595416 / r9595424;
        double r9595426 = 99.0;
        double r9595427 = r9595416 * r9595426;
        double r9595428 = log(r9595413);
        double r9595429 = r9595428 * r9595422;
        double r9595430 = exp(r9595429);
        double r9595431 = r9595427 * r9595430;
        double r9595432 = r9595413 * r9595413;
        double r9595433 = r9595432 * r9595432;
        double r9595434 = r9595431 / r9595433;
        double r9595435 = r9595416 * r9595430;
        double r9595436 = r9595435 / r9595413;
        double r9595437 = r9595436 / r9595413;
        double r9595438 = -10.0;
        double r9595439 = r9595435 * r9595438;
        double r9595440 = r9595432 * r9595413;
        double r9595441 = r9595439 / r9595440;
        double r9595442 = r9595437 + r9595441;
        double r9595443 = r9595434 + r9595442;
        double r9595444 = r9595415 ? r9595425 : r9595443;
        return r9595444;
}

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 < 3.158753162179787e+109

    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 3.158753162179787e+109 < k

    1. Initial program 8.8

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

    2. Simplified8.8

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

    4. Applied clear-num9.0

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

    5. Taylor expanded around 0 9.0

      \[\leadsto \frac{1}{\frac{\frac{\color{blue}{\left({k}^{2} + 10 \cdot k\right)} + 1}{{k}^{m}}}{a}}\]

    6. Simplified9.0

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

    7. Taylor expanded around inf 8.8

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

    8. Simplified0.3

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 3.158753162179787 \cdot 10^{+109}:\\ \;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(a \cdot 99\right) \cdot e^{\log k \cdot m}}{\left(k \cdot k\right) \cdot \left(k \cdot k\right)} + \left(\frac{\frac{a \cdot e^{\log k \cdot m}}{k}}{k} + \frac{\left(a \cdot e^{\log k \cdot m}\right) \cdot -10}{\left(k \cdot k\right) \cdot k}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019163 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))