Average Error: 2.2 → 0.1
Time: 59.9s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1.0 + 10.0 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 3307771.335745201:\\ \;\;\;\;\frac{a}{\frac{k \cdot \left(10.0 + k\right) + 1.0}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{a}{k} \cdot \frac{e^{m \cdot \log k}}{k} + \left(\frac{a}{k \cdot k} \cdot \frac{e^{m \cdot \log k}}{k \cdot k}\right) \cdot 99.0\right) - \frac{10.0 \cdot e^{m \cdot \log k}}{\frac{k \cdot \left(k \cdot k\right)}{a}}\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1.0 + 10.0 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 3307771.335745201:\\
\;\;\;\;\frac{a}{\frac{k \cdot \left(10.0 + k\right) + 1.0}{{k}^{m}}}\\

\mathbf{else}:\\
\;\;\;\;\left(\frac{a}{k} \cdot \frac{e^{m \cdot \log k}}{k} + \left(\frac{a}{k \cdot k} \cdot \frac{e^{m \cdot \log k}}{k \cdot k}\right) \cdot 99.0\right) - \frac{10.0 \cdot e^{m \cdot \log k}}{\frac{k \cdot \left(k \cdot k\right)}{a}}\\

\end{array}
double f(double a, double k, double m) {
        double r7199463 = a;
        double r7199464 = k;
        double r7199465 = m;
        double r7199466 = pow(r7199464, r7199465);
        double r7199467 = r7199463 * r7199466;
        double r7199468 = 1.0;
        double r7199469 = 10.0;
        double r7199470 = r7199469 * r7199464;
        double r7199471 = r7199468 + r7199470;
        double r7199472 = r7199464 * r7199464;
        double r7199473 = r7199471 + r7199472;
        double r7199474 = r7199467 / r7199473;
        return r7199474;
}
↓
double f(double a, double k, double m) {
        double r7199475 = k;
        double r7199476 = 3307771.335745201;
        bool r7199477 = r7199475 <= r7199476;
        double r7199478 = a;
        double r7199479 = 10.0;
        double r7199480 = r7199479 + r7199475;
        double r7199481 = r7199475 * r7199480;
        double r7199482 = 1.0;
        double r7199483 = r7199481 + r7199482;
        double r7199484 = m;
        double r7199485 = pow(r7199475, r7199484);
        double r7199486 = r7199483 / r7199485;
        double r7199487 = r7199478 / r7199486;
        double r7199488 = r7199478 / r7199475;
        double r7199489 = log(r7199475);
        double r7199490 = r7199484 * r7199489;
        double r7199491 = exp(r7199490);
        double r7199492 = r7199491 / r7199475;
        double r7199493 = r7199488 * r7199492;
        double r7199494 = r7199475 * r7199475;
        double r7199495 = r7199478 / r7199494;
        double r7199496 = r7199491 / r7199494;
        double r7199497 = r7199495 * r7199496;
        double r7199498 = 99.0;
        double r7199499 = r7199497 * r7199498;
        double r7199500 = r7199493 + r7199499;
        double r7199501 = r7199479 * r7199491;
        double r7199502 = r7199475 * r7199494;
        double r7199503 = r7199502 / r7199478;
        double r7199504 = r7199501 / r7199503;
        double r7199505 = r7199500 - r7199504;
        double r7199506 = r7199477 ? r7199487 : r7199505;
        return r7199506;
}

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 < 3307771.335745201

    1. Initial program 0.1

      \[\frac{a \cdot {k}^{m}}{\left(1.0 + 10.0 \cdot k\right) + k \cdot k}\]
    2. Using strategy rm

    3. Applied associate-/l*0.1

      \[\leadsto \color{blue}{\frac{a}{\frac{\left(1.0 + 10.0 \cdot k\right) + k \cdot k}{{k}^{m}}}}\]

    4. Simplified0.0

      \[\leadsto \frac{a}{\color{blue}{\frac{1.0 + k \cdot \left(k + 10.0\right)}{{k}^{m}}}}\]

    if 3307771.335745201 < k

    1. Initial program 5.6

      \[\frac{a \cdot {k}^{m}}{\left(1.0 + 10.0 \cdot k\right) + k \cdot k}\]
    2. Using strategy rm

    3. Applied associate-/l*5.6

      \[\leadsto \color{blue}{\frac{a}{\frac{\left(1.0 + 10.0 \cdot k\right) + k \cdot k}{{k}^{m}}}}\]

    4. Simplified5.6

      \[\leadsto \frac{a}{\color{blue}{\frac{1.0 + k \cdot \left(k + 10.0\right)}{{k}^{m}}}}\]
    5. Using strategy rm

    6. Applied clear-num5.8

      \[\leadsto \color{blue}{\frac{1}{\frac{\frac{1.0 + k \cdot \left(k + 10.0\right)}{{k}^{m}}}{a}}}\]

    7. Taylor expanded around inf 5.6

      \[\leadsto \color{blue}{\left(99.0 \cdot \frac{e^{-1 \cdot \left(\log \left(\frac{1}{k}\right) \cdot m\right)} \cdot a}{{k}^{4}} + \frac{e^{-1 \cdot \left(\log \left(\frac{1}{k}\right) \cdot m\right)} \cdot a}{{k}^{2}}\right) - 10.0 \cdot \frac{e^{-1 \cdot \left(\log \left(\frac{1}{k}\right) \cdot m\right)} \cdot a}{{k}^{3}}}\]

    8. Simplified0.2

      \[\leadsto \color{blue}{\left(99.0 \cdot \left(\frac{a}{k \cdot k} \cdot \frac{e^{-\log k \cdot \left(-m\right)}}{k \cdot k}\right) + \frac{a}{k} \cdot \frac{e^{-\log k \cdot \left(-m\right)}}{k}\right) - \frac{10.0 \cdot e^{-\log k \cdot \left(-m\right)}}{\frac{\left(k \cdot k\right) \cdot k}{a}}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 3307771.335745201:\\ \;\;\;\;\frac{a}{\frac{k \cdot \left(10.0 + k\right) + 1.0}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{a}{k} \cdot \frac{e^{m \cdot \log k}}{k} + \left(\frac{a}{k \cdot k} \cdot \frac{e^{m \cdot \log k}}{k \cdot k}\right) \cdot 99.0\right) - \frac{10.0 \cdot e^{m \cdot \log k}}{\frac{k \cdot \left(k \cdot k\right)}{a}}\\ \end{array}\]

Reproduce

herbie shell --seed 2019165 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1.0 (* 10.0 k)) (* k k))))