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

↓

\[\begin{array}{l} \mathbf{if}\;k \le 4.3016757667578286 \cdot 10^{+85}:\\ \;\;\;\;\frac{{k}^{m} \cdot a}{1 + \left(10 + k\right) \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{a}{k}}{k} \cdot \left({k}^{m} - \frac{{k}^{m}}{\frac{k}{10}}\right) + \frac{{k}^{m} \cdot 99}{\frac{{k}^{4}}{a}}\\ \end{array}\]

Derivation

Split input into 2 regimes
if k < 4.3016757667578286e+85
1. Initial program 0.1
  \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
2. Applied simplify0.0
  \[\leadsto \color{blue}{\frac{{k}^{m} \cdot a}{1 + \left(10 + k\right) \cdot k}}\]
if 4.3016757667578286e+85 < k
1. Initial program 7.0
  \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
2. Applied simplify7.0
  \[\leadsto \color{blue}{\frac{{k}^{m} \cdot a}{1 + \left(10 + k\right) \cdot k}}\]
3. Taylor expanded around inf 7.0
  \[\leadsto \color{blue}{\left(\frac{e^{-1 \cdot \left(\log \left(\frac{1}{k}\right) \cdot m\right)} \cdot a}{{k}^{2}} + 99 \cdot \frac{e^{-1 \cdot \left(\log \left(\frac{1}{k}\right) \cdot m\right)} \cdot a}{{k}^{4}}\right) - 10 \cdot \frac{e^{-1 \cdot \left(\log \left(\frac{1}{k}\right) \cdot m\right)} \cdot a}{{k}^{3}}}\]
4. Applied simplify0.1
  \[\leadsto \color{blue}{\frac{\frac{a}{k}}{k} \cdot \left({k}^{m} - \frac{{k}^{m}}{\frac{k}{10}}\right) + \frac{{k}^{m} \cdot 99}{\frac{{k}^{4}}{a}}}\]
Recombined 2 regimes into one program.

Runtime

herbie shell --seed '#(1064397287 3527694221 3797617954 1138343853 2854031332 1153838279)' 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))

Error

Derivation

`if k < 4.3016757667578286e+85`

`if 4.3016757667578286e+85 < k`

Runtime