Average Error: 2.0 → 0.2
Time: 47.4s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 105336881957004620240781312:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\left(a \cdot e^{\log \left({k}^{m}\right)}\right) \cdot 99}{{k}^{4}} - \frac{a \cdot e^{\log \left({k}^{m}\right)}}{{k}^{3}} \cdot 10\right) + \frac{a}{k} \cdot \frac{e^{\log \left({k}^{m}\right)}}{k}\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 105336881957004620240781312:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\

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

\end{array}
double f(double a, double k, double m) {
        double r252232 = a;
        double r252233 = k;
        double r252234 = m;
        double r252235 = pow(r252233, r252234);
        double r252236 = r252232 * r252235;
        double r252237 = 1.0;
        double r252238 = 10.0;
        double r252239 = r252238 * r252233;
        double r252240 = r252237 + r252239;
        double r252241 = r252233 * r252233;
        double r252242 = r252240 + r252241;
        double r252243 = r252236 / r252242;
        return r252243;
}
↓
double f(double a, double k, double m) {
        double r252244 = k;
        double r252245 = 1.0533688195700462e+26;
        bool r252246 = r252244 <= r252245;
        double r252247 = a;
        double r252248 = m;
        double r252249 = pow(r252244, r252248);
        double r252250 = r252247 * r252249;
        double r252251 = 1.0;
        double r252252 = 10.0;
        double r252253 = r252252 * r252244;
        double r252254 = r252251 + r252253;
        double r252255 = r252244 * r252244;
        double r252256 = r252254 + r252255;
        double r252257 = r252250 / r252256;
        double r252258 = log(r252249);
        double r252259 = exp(r252258);
        double r252260 = r252247 * r252259;
        double r252261 = 99.0;
        double r252262 = r252260 * r252261;
        double r252263 = 4.0;
        double r252264 = pow(r252244, r252263);
        double r252265 = r252262 / r252264;
        double r252266 = 3.0;
        double r252267 = pow(r252244, r252266);
        double r252268 = r252260 / r252267;
        double r252269 = r252268 * r252252;
        double r252270 = r252265 - r252269;
        double r252271 = r252247 / r252244;
        double r252272 = r252259 / r252244;
        double r252273 = r252271 * r252272;
        double r252274 = r252270 + r252273;
        double r252275 = r252246 ? r252257 : r252274;
        return r252275;
}

Error

Bits error versus a

Bits error versus k

Bits error versus m

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 2 regimes

  2. if k < 1.0533688195700462e+26

    1. Initial program 0.1

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

    if 1.0533688195700462e+26 < k

    1. Initial program 5.7

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

    3. Applied clear-num6.0

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

    4. Simplified5.9

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

    5. Taylor expanded around inf 5.7

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

    6. Simplified0.4

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 105336881957004620240781312:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\left(a \cdot e^{\log \left({k}^{m}\right)}\right) \cdot 99}{{k}^{4}} - \frac{a \cdot e^{\log \left({k}^{m}\right)}}{{k}^{3}} \cdot 10\right) + \frac{a}{k} \cdot \frac{e^{\log \left({k}^{m}\right)}}{k}\\ \end{array}\]

Reproduce

herbie shell --seed 2019323 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))