Average Error: 1.9 → 0.1
Time: 1.1m
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 2.143710643040193 \cdot 10^{+69}:\\ \;\;\;\;\frac{1}{1 + \left(k + 10\right) \cdot k} \cdot \left({k}^{m} \cdot a\right)\\ \mathbf{else}:\\ \;\;\;\;99 \cdot \frac{a \cdot e^{m \cdot \log k}}{\left(k \cdot k\right) \cdot \left(k \cdot k\right)} + \left(e^{m \cdot \log k} \cdot \frac{\frac{a}{k}}{k} - \left(\frac{10}{k} \cdot \frac{\frac{a}{k}}{k}\right) \cdot e^{m \cdot \log k}\right)\\ \end{array}\]

Error

Bits error versus a

Bits error versus k

Bits error versus m

Derivation

  1. Split input into 2 regimes

  2. if k < 2.143710643040193e+69

    1. Initial program 0.1

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

    2. Simplified0.0

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

    4. Applied div-inv0.0

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

    if 2.143710643040193e+69 < k

    1. Initial program 6.2

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

    2. Simplified6.2

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

    4. Applied div-inv6.2

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

    5. Taylor expanded around -inf 63.0

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

    6. Simplified0.1

      \[\leadsto \color{blue}{99 \cdot \frac{e^{m \cdot \left(0 + \log k\right)} \cdot a}{\left(k \cdot k\right) \cdot \left(k \cdot k\right)} + \left(\frac{\frac{a}{k}}{k} \cdot e^{m \cdot \left(0 + \log k\right)} - \left(\frac{10}{k} \cdot \frac{\frac{a}{k}}{k}\right) \cdot e^{m \cdot \left(0 + \log k\right)}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 2.143710643040193 \cdot 10^{+69}:\\ \;\;\;\;\frac{1}{1 + \left(k + 10\right) \cdot k} \cdot \left({k}^{m} \cdot a\right)\\ \mathbf{else}:\\ \;\;\;\;99 \cdot \frac{a \cdot e^{m \cdot \log k}}{\left(k \cdot k\right) \cdot \left(k \cdot k\right)} + \left(e^{m \cdot \log k} \cdot \frac{\frac{a}{k}}{k} - \left(\frac{10}{k} \cdot \frac{\frac{a}{k}}{k}\right) \cdot e^{m \cdot \log k}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019100 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))