Average Error: 2.2 → 0.1
Time: 52.7s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 3.158753162179787 \cdot 10^{+109}:\\ \;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(a \cdot 99\right) \cdot e^{\log k \cdot m}}{\left(k \cdot k\right) \cdot \left(k \cdot k\right)} + \left(\frac{\frac{a \cdot e^{\log k \cdot m}}{k}}{k} + \frac{\left(a \cdot e^{\log k \cdot m}\right) \cdot -10}{\left(k \cdot k\right) \cdot k}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 3.158753162179787 \cdot 10^{+109}:\\
\;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\

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

\end{array}
double f(double a, double k, double m) {
        double r10287200 = a;
        double r10287201 = k;
        double r10287202 = m;
        double r10287203 = pow(r10287201, r10287202);
        double r10287204 = r10287200 * r10287203;
        double r10287205 = 1.0;
        double r10287206 = 10.0;
        double r10287207 = r10287206 * r10287201;
        double r10287208 = r10287205 + r10287207;
        double r10287209 = r10287201 * r10287201;
        double r10287210 = r10287208 + r10287209;
        double r10287211 = r10287204 / r10287210;
        return r10287211;
}
↓
double f(double a, double k, double m) {
        double r10287212 = k;
        double r10287213 = 3.158753162179787e+109;
        bool r10287214 = r10287212 <= r10287213;
        double r10287215 = a;
        double r10287216 = 1.0;
        double r10287217 = 10.0;
        double r10287218 = r10287212 + r10287217;
        double r10287219 = r10287218 * r10287212;
        double r10287220 = r10287216 + r10287219;
        double r10287221 = m;
        double r10287222 = pow(r10287212, r10287221);
        double r10287223 = r10287220 / r10287222;
        double r10287224 = r10287215 / r10287223;
        double r10287225 = 99.0;
        double r10287226 = r10287215 * r10287225;
        double r10287227 = log(r10287212);
        double r10287228 = r10287227 * r10287221;
        double r10287229 = exp(r10287228);
        double r10287230 = r10287226 * r10287229;
        double r10287231 = r10287212 * r10287212;
        double r10287232 = r10287231 * r10287231;
        double r10287233 = r10287230 / r10287232;
        double r10287234 = r10287215 * r10287229;
        double r10287235 = r10287234 / r10287212;
        double r10287236 = r10287235 / r10287212;
        double r10287237 = -10.0;
        double r10287238 = r10287234 * r10287237;
        double r10287239 = r10287231 * r10287212;
        double r10287240 = r10287238 / r10287239;
        double r10287241 = r10287236 + r10287240;
        double r10287242 = r10287233 + r10287241;
        double r10287243 = r10287214 ? r10287224 : r10287242;
        return r10287243;
}

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 < 3.158753162179787e+109

    1. Initial program 0.1

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

    2. Simplified0.0

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

    if 3.158753162179787e+109 < k

    1. Initial program 8.8

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

    2. Simplified8.8

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

    4. Applied clear-num9.0

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

    5. Taylor expanded around 0 9.0

      \[\leadsto \frac{1}{\frac{\frac{\color{blue}{\left({k}^{2} + 10 \cdot k\right)} + 1}{{k}^{m}}}{a}}\]

    6. Simplified9.0

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

    7. Taylor expanded around inf 8.8

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

    8. Simplified0.3

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 3.158753162179787 \cdot 10^{+109}:\\ \;\;\;\;\frac{a}{\frac{1 + \left(k + 10\right) \cdot k}{{k}^{m}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\left(a \cdot 99\right) \cdot e^{\log k \cdot m}}{\left(k \cdot k\right) \cdot \left(k \cdot k\right)} + \left(\frac{\frac{a \cdot e^{\log k \cdot m}}{k}}{k} + \frac{\left(a \cdot e^{\log k \cdot m}\right) \cdot -10}{\left(k \cdot k\right) \cdot k}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019163 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))