Average Error: 2.2 → 0.1
Time: 5.0s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 1484471.41912973742:\\ \;\;\;\;\frac{{k}^{m}}{\frac{\mathsf{fma}\left(k, k, \mathsf{fma}\left(k, 10, 1\right)\right)}{a}}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 1484471.41912973742:\\
\;\;\;\;\frac{{k}^{m}}{\frac{\mathsf{fma}\left(k, k, \mathsf{fma}\left(k, 10, 1\right)\right)}{a}}\\

\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 r289284 = a;
        double r289285 = k;
        double r289286 = m;
        double r289287 = pow(r289285, r289286);
        double r289288 = r289284 * r289287;
        double r289289 = 1.0;
        double r289290 = 10.0;
        double r289291 = r289290 * r289285;
        double r289292 = r289289 + r289291;
        double r289293 = r289285 * r289285;
        double r289294 = r289292 + r289293;
        double r289295 = r289288 / r289294;
        return r289295;
}
↓
double f(double a, double k, double m) {
        double r289296 = k;
        double r289297 = 1484471.4191297374;
        bool r289298 = r289296 <= r289297;
        double r289299 = m;
        double r289300 = pow(r289296, r289299);
        double r289301 = 10.0;
        double r289302 = 1.0;
        double r289303 = fma(r289296, r289301, r289302);
        double r289304 = fma(r289296, r289296, r289303);
        double r289305 = a;
        double r289306 = r289304 / r289305;
        double r289307 = r289300 / r289306;
        double r289308 = -1.0;
        double r289309 = 1.0;
        double r289310 = r289309 / r289296;
        double r289311 = log(r289310);
        double r289312 = r289299 * r289311;
        double r289313 = r289308 * r289312;
        double r289314 = exp(r289313);
        double r289315 = r289314 / r289296;
        double r289316 = r289305 / r289296;
        double r289317 = 99.0;
        double r289318 = r289305 * r289314;
        double r289319 = 4.0;
        double r289320 = pow(r289296, r289319);
        double r289321 = r289318 / r289320;
        double r289322 = r289317 * r289321;
        double r289323 = 3.0;
        double r289324 = pow(r289296, r289323);
        double r289325 = r289318 / r289324;
        double r289326 = r289301 * r289325;
        double r289327 = r289322 - r289326;
        double r289328 = fma(r289315, r289316, r289327);
        double r289329 = r289298 ? r289307 : r289328;
        return r289329;
}

Error

Bits error versus a

Bits error versus k

Bits error versus m

Derivation

  1. Split input into 2 regimes

  2. if k < 1484471.4191297374

    1. Initial program 0.0

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

    2. Simplified0.1

      \[\leadsto \color{blue}{\frac{{k}^{m}}{\frac{\mathsf{fma}\left(k, k, \mathsf{fma}\left(k, 10, 1\right)\right)}{a}}}\]

    if 1484471.4191297374 < k

    1. Initial program 5.6

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

    2. Simplified5.7

      \[\leadsto \color{blue}{\frac{{k}^{m}}{\frac{\mathsf{fma}\left(k, k, \mathsf{fma}\left(k, 10, 1\right)\right)}{a}}}\]

    3. Taylor expanded around inf 5.6

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

    4. Simplified0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 1484471.41912973742:\\ \;\;\;\;\frac{{k}^{m}}{\frac{\mathsf{fma}\left(k, k, \mathsf{fma}\left(k, 10, 1\right)\right)}{a}}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020018 +o rules:numerics
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))