Average Error: 2.2 → 0.2
Time: 39.1s
Precision: 64
Internal Precision: 576
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
\[\begin{array}{l} \mathbf{if}\;k \le 2.070234980506773 \cdot 10^{+107}:\\ \;\;\;\;(e^{\log_* (1 + \frac{1}{(k \cdot \left(k + 10\right) + 1)_*})} - 1)^* \cdot \left({k}^{m} \cdot a\right)\\ \mathbf{else}:\\ \;\;\;\;(\left(\frac{{\left(e^{m}\right)}^{\left(\log k\right)}}{\frac{{k}^{4}}{a}}\right) \cdot 99 + \left((\left(\frac{-10}{k}\right) \cdot \left(\frac{{\left(e^{m}\right)}^{\left(\log k\right)}}{\frac{k}{\frac{a}{k}}}\right) + \left(\frac{{\left(e^{m}\right)}^{\left(\log k\right)}}{\frac{k}{\frac{a}{k}}}\right))_*\right))_*\\ \end{array}\]

Error

Bits error versus a

Bits error versus k

Bits error versus m

Derivation

  1. Split input into 2 regimes
  2. if k < 2.070234980506773e+107

    1. Initial program 0.1

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
    2. Initial simplification0.0

      \[\leadsto \frac{{k}^{m} \cdot a}{(k \cdot \left(k + 10\right) + 1)_*}\]
    3. Using strategy rm
    4. Applied div-inv0.0

      \[\leadsto \color{blue}{\left({k}^{m} \cdot a\right) \cdot \frac{1}{(k \cdot \left(k + 10\right) + 1)_*}}\]
    5. Using strategy rm
    6. Applied expm1-log1p-u0.1

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

    if 2.070234980506773e+107 < k

    1. Initial program 9.0

      \[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
    2. Initial simplification9.0

      \[\leadsto \frac{{k}^{m} \cdot a}{(k \cdot \left(k + 10\right) + 1)_*}\]
    3. Using strategy rm
    4. Applied div-inv9.0

      \[\leadsto \color{blue}{\left({k}^{m} \cdot a\right) \cdot \frac{1}{(k \cdot \left(k + 10\right) + 1)_*}}\]
    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. Simplified0.6

      \[\leadsto \color{blue}{(\left(\frac{{\left(e^{m}\right)}^{\left(\log k\right)}}{\frac{{k}^{4}}{a}}\right) \cdot 99 + \left((\left(\frac{-10}{k}\right) \cdot \left(\frac{{\left(e^{m}\right)}^{\left(\log k\right)}}{\frac{k}{\frac{a}{k}}}\right) + \left(\frac{{\left(e^{m}\right)}^{\left(\log k\right)}}{\frac{k}{\frac{a}{k}}}\right))_*\right))_*}\]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 2.070234980506773 \cdot 10^{+107}:\\ \;\;\;\;(e^{\log_* (1 + \frac{1}{(k \cdot \left(k + 10\right) + 1)_*})} - 1)^* \cdot \left({k}^{m} \cdot a\right)\\ \mathbf{else}:\\ \;\;\;\;(\left(\frac{{\left(e^{m}\right)}^{\left(\log k\right)}}{\frac{{k}^{4}}{a}}\right) \cdot 99 + \left((\left(\frac{-10}{k}\right) \cdot \left(\frac{{\left(e^{m}\right)}^{\left(\log k\right)}}{\frac{k}{\frac{a}{k}}}\right) + \left(\frac{{\left(e^{m}\right)}^{\left(\log k\right)}}{\frac{k}{\frac{a}{k}}}\right))_*\right))_*\\ \end{array}\]

Runtime

Time bar (total: 39.1s)Debug logProfile

BaselineHerbieOracleSpan%
Regimes2.20.20.12.196.9%
herbie shell --seed 2018263 +o rules:numerics
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))