Average Error: 2.2 → 0.1
Time: 10.1s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 1.8129058405319809 \cdot 10^{106}:\\ \;\;\;\;\frac{a \cdot {\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}}{\frac{\left(1 + 10 \cdot k\right) + k \cdot k}{{\left(\sqrt[3]{k}\right)}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 1.8129058405319809 \cdot 10^{106}:\\
\;\;\;\;\frac{a \cdot {\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}}{\frac{\left(1 + 10 \cdot k\right) + k \cdot k}{{\left(\sqrt[3]{k}\right)}^{m}}}\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\

\end{array}
double f(double a, double k, double m) {
        double r256414 = a;
        double r256415 = k;
        double r256416 = m;
        double r256417 = pow(r256415, r256416);
        double r256418 = r256414 * r256417;
        double r256419 = 1.0;
        double r256420 = 10.0;
        double r256421 = r256420 * r256415;
        double r256422 = r256419 + r256421;
        double r256423 = r256415 * r256415;
        double r256424 = r256422 + r256423;
        double r256425 = r256418 / r256424;
        return r256425;
}
↓
double f(double a, double k, double m) {
        double r256426 = k;
        double r256427 = 1.812905840531981e+106;
        bool r256428 = r256426 <= r256427;
        double r256429 = a;
        double r256430 = cbrt(r256426);
        double r256431 = r256430 * r256430;
        double r256432 = m;
        double r256433 = pow(r256431, r256432);
        double r256434 = r256429 * r256433;
        double r256435 = 1.0;
        double r256436 = 10.0;
        double r256437 = r256436 * r256426;
        double r256438 = r256435 + r256437;
        double r256439 = r256426 * r256426;
        double r256440 = r256438 + r256439;
        double r256441 = pow(r256430, r256432);
        double r256442 = r256440 / r256441;
        double r256443 = r256434 / r256442;
        double r256444 = -1.0;
        double r256445 = 1.0;
        double r256446 = r256445 / r256426;
        double r256447 = log(r256446);
        double r256448 = r256432 * r256447;
        double r256449 = r256444 * r256448;
        double r256450 = exp(r256449);
        double r256451 = r256450 / r256426;
        double r256452 = r256429 / r256426;
        double r256453 = 99.0;
        double r256454 = r256429 * r256450;
        double r256455 = 4.0;
        double r256456 = pow(r256426, r256455);
        double r256457 = r256454 / r256456;
        double r256458 = r256453 * r256457;
        double r256459 = 3.0;
        double r256460 = pow(r256426, r256459);
        double r256461 = r256454 / r256460;
        double r256462 = r256436 * r256461;
        double r256463 = r256458 - r256462;
        double r256464 = fma(r256451, r256452, r256463);
        double r256465 = r256428 ? r256443 : r256464;
        return r256465;
}

Error

Bits error versus a

Bits error versus k

Bits error versus m

Derivation

  1. Split input into 2 regimes

  2. if k < 1.812905840531981e+106

    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 add-cube-cbrt0.1

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

    4. Applied unpow-prod-down0.1

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

    5. Applied associate-*r*0.1

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

    7. Applied associate-/l*0.1

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

    if 1.812905840531981e+106 < k

    1. Initial program 8.6

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

    2. Taylor expanded around inf 8.6

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

    3. Simplified0.2

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 1.8129058405319809 \cdot 10^{106}:\\ \;\;\;\;\frac{a \cdot {\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}}{\frac{\left(1 + 10 \cdot k\right) + k \cdot k}{{\left(\sqrt[3]{k}\right)}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020056 +o rules:numerics
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))