Average Error: 2.0 → 0.1
Time: 33.3s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 1.304764198775255 \cdot 10^{+145}:\\ \;\;\;\;\frac{{k}^{m} \cdot a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{a}{k \cdot k} \cdot 99\right) \cdot \frac{e^{m \cdot \log k}}{k \cdot k} + \left(\frac{a}{k} \cdot \frac{e^{m \cdot \log k}}{k} - \frac{a}{k} \cdot \frac{e^{m \cdot \log k} \cdot 10}{k \cdot k}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 1.304764198775255 \cdot 10^{+145}:\\
\;\;\;\;\frac{{k}^{m} \cdot a}{1 + k \cdot \left(k + 10\right)}\\

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

\end{array}
double f(double a, double k, double m) {
        double r10159489 = a;
        double r10159490 = k;
        double r10159491 = m;
        double r10159492 = pow(r10159490, r10159491);
        double r10159493 = r10159489 * r10159492;
        double r10159494 = 1.0;
        double r10159495 = 10.0;
        double r10159496 = r10159495 * r10159490;
        double r10159497 = r10159494 + r10159496;
        double r10159498 = r10159490 * r10159490;
        double r10159499 = r10159497 + r10159498;
        double r10159500 = r10159493 / r10159499;
        return r10159500;
}
↓
double f(double a, double k, double m) {
        double r10159501 = k;
        double r10159502 = 1.304764198775255e+145;
        bool r10159503 = r10159501 <= r10159502;
        double r10159504 = m;
        double r10159505 = pow(r10159501, r10159504);
        double r10159506 = a;
        double r10159507 = r10159505 * r10159506;
        double r10159508 = 1.0;
        double r10159509 = 10.0;
        double r10159510 = r10159501 + r10159509;
        double r10159511 = r10159501 * r10159510;
        double r10159512 = r10159508 + r10159511;
        double r10159513 = r10159507 / r10159512;
        double r10159514 = r10159501 * r10159501;
        double r10159515 = r10159506 / r10159514;
        double r10159516 = 99.0;
        double r10159517 = r10159515 * r10159516;
        double r10159518 = log(r10159501);
        double r10159519 = r10159504 * r10159518;
        double r10159520 = exp(r10159519);
        double r10159521 = r10159520 / r10159514;
        double r10159522 = r10159517 * r10159521;
        double r10159523 = r10159506 / r10159501;
        double r10159524 = r10159520 / r10159501;
        double r10159525 = r10159523 * r10159524;
        double r10159526 = r10159520 * r10159509;
        double r10159527 = r10159526 / r10159514;
        double r10159528 = r10159523 * r10159527;
        double r10159529 = r10159525 - r10159528;
        double r10159530 = r10159522 + r10159529;
        double r10159531 = r10159503 ? r10159513 : r10159530;
        return r10159531;
}

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 < 1.304764198775255e+145

    1. Initial program 0.1

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

    2. Simplified0.1

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

    4. Applied *-un-lft-identity0.1

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

    5. Applied associate-*r*0.1

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

    6. Simplified0.1

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

    if 1.304764198775255e+145 < k

    1. Initial program 10.1

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

    2. Simplified10.1

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

    3. Taylor expanded around inf 10.1

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

    4. Simplified0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 1.304764198775255 \cdot 10^{+145}:\\ \;\;\;\;\frac{{k}^{m} \cdot a}{1 + k \cdot \left(k + 10\right)}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{a}{k \cdot k} \cdot 99\right) \cdot \frac{e^{m \cdot \log k}}{k \cdot k} + \left(\frac{a}{k} \cdot \frac{e^{m \cdot \log k}}{k} - \frac{a}{k} \cdot \frac{e^{m \cdot \log k} \cdot 10}{k \cdot k}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019149 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))