Average Error: 2.1 → 0.1
Time: 33.3s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 3.641563002348689 \cdot 10^{+130}:\\ \;\;\;\;\frac{{k}^{m} \cdot a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{e^{\log k \cdot m}}{k} \cdot \frac{a}{k} + \left(\frac{e^{\log k \cdot m}}{k} \cdot \frac{a}{k \cdot k}\right) \cdot -10\right) + \frac{99 \cdot e^{\log k \cdot m}}{\frac{\left(k \cdot k\right) \cdot \left(k \cdot k\right)}{a}}\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 3.641563002348689 \cdot 10^{+130}:\\
\;\;\;\;\frac{{k}^{m} \cdot a}{1 + k \cdot \left(k + 10\right)}\\

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

\end{array}
double f(double a, double k, double m) {
        double r6375371 = a;
        double r6375372 = k;
        double r6375373 = m;
        double r6375374 = pow(r6375372, r6375373);
        double r6375375 = r6375371 * r6375374;
        double r6375376 = 1.0;
        double r6375377 = 10.0;
        double r6375378 = r6375377 * r6375372;
        double r6375379 = r6375376 + r6375378;
        double r6375380 = r6375372 * r6375372;
        double r6375381 = r6375379 + r6375380;
        double r6375382 = r6375375 / r6375381;
        return r6375382;
}
↓
double f(double a, double k, double m) {
        double r6375383 = k;
        double r6375384 = 3.641563002348689e+130;
        bool r6375385 = r6375383 <= r6375384;
        double r6375386 = m;
        double r6375387 = pow(r6375383, r6375386);
        double r6375388 = a;
        double r6375389 = r6375387 * r6375388;
        double r6375390 = 1.0;
        double r6375391 = 10.0;
        double r6375392 = r6375383 + r6375391;
        double r6375393 = r6375383 * r6375392;
        double r6375394 = r6375390 + r6375393;
        double r6375395 = r6375389 / r6375394;
        double r6375396 = log(r6375383);
        double r6375397 = r6375396 * r6375386;
        double r6375398 = exp(r6375397);
        double r6375399 = r6375398 / r6375383;
        double r6375400 = r6375388 / r6375383;
        double r6375401 = r6375399 * r6375400;
        double r6375402 = r6375383 * r6375383;
        double r6375403 = r6375388 / r6375402;
        double r6375404 = r6375399 * r6375403;
        double r6375405 = -10.0;
        double r6375406 = r6375404 * r6375405;
        double r6375407 = r6375401 + r6375406;
        double r6375408 = 99.0;
        double r6375409 = r6375408 * r6375398;
        double r6375410 = r6375402 * r6375402;
        double r6375411 = r6375410 / r6375388;
        double r6375412 = r6375409 / r6375411;
        double r6375413 = r6375407 + r6375412;
        double r6375414 = r6375385 ? r6375395 : r6375413;
        return r6375414;
}

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 < 3.641563002348689e+130

    1. Initial program 0.1

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

    2. Simplified0.0

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

    if 3.641563002348689e+130 < k

    1. Initial program 8.9

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

    2. Simplified8.9

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

    3. Taylor expanded around inf 8.9

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

    4. Simplified0.2

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 3.641563002348689 \cdot 10^{+130}:\\ \;\;\;\;\frac{{k}^{m} \cdot a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{e^{\log k \cdot m}}{k} \cdot \frac{a}{k} + \left(\frac{e^{\log k \cdot m}}{k} \cdot \frac{a}{k \cdot k}\right) \cdot -10\right) + \frac{99 \cdot e^{\log k \cdot m}}{\frac{\left(k \cdot k\right) \cdot \left(k \cdot k\right)}{a}}\\ \end{array}\]

Reproduce

herbie shell --seed 2019143 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))