Average Error: 2.1 → 0.1
Time: 6.3s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 3.945215226703958 \cdot 10^{153}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 3.945215226703958 \cdot 10^{153}:\\
\;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\

\end{array}
double f(double a, double k, double m) {
        double r164069 = a;
        double r164070 = k;
        double r164071 = m;
        double r164072 = pow(r164070, r164071);
        double r164073 = r164069 * r164072;
        double r164074 = 1.0;
        double r164075 = 10.0;
        double r164076 = r164075 * r164070;
        double r164077 = r164074 + r164076;
        double r164078 = r164070 * r164070;
        double r164079 = r164077 + r164078;
        double r164080 = r164073 / r164079;
        return r164080;
}
↓
double f(double a, double k, double m) {
        double r164081 = k;
        double r164082 = 3.945215226703958e+153;
        bool r164083 = r164081 <= r164082;
        double r164084 = a;
        double r164085 = m;
        double r164086 = pow(r164081, r164085);
        double r164087 = r164084 * r164086;
        double r164088 = 1.0;
        double r164089 = 10.0;
        double r164090 = r164089 * r164081;
        double r164091 = r164088 + r164090;
        double r164092 = r164081 * r164081;
        double r164093 = r164091 + r164092;
        double r164094 = r164087 / r164093;
        double r164095 = -1.0;
        double r164096 = 1.0;
        double r164097 = r164096 / r164081;
        double r164098 = log(r164097);
        double r164099 = r164085 * r164098;
        double r164100 = r164095 * r164099;
        double r164101 = exp(r164100);
        double r164102 = r164101 / r164081;
        double r164103 = r164084 / r164081;
        double r164104 = 99.0;
        double r164105 = r164084 * r164101;
        double r164106 = 4.0;
        double r164107 = pow(r164081, r164106);
        double r164108 = r164105 / r164107;
        double r164109 = r164104 * r164108;
        double r164110 = 3.0;
        double r164111 = pow(r164081, r164110);
        double r164112 = r164105 / r164111;
        double r164113 = r164089 * r164112;
        double r164114 = r164109 - r164113;
        double r164115 = fma(r164102, r164103, r164114);
        double r164116 = r164083 ? r164094 : r164115;
        return r164116;
}

Error

Bits error versus a

Bits error versus k

Bits error versus m

Derivation

  1. Split input into 2 regimes

  2. if k < 3.945215226703958e+153

    1. Initial program 0.0

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

    if 3.945215226703958e+153 < k

    1. Initial program 10.8

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

    2. Taylor expanded around inf 10.8

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

    3. Simplified0.1

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 3.945215226703958 \cdot 10^{153}:\\ \;\;\;\;\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{k}, \frac{a}{k}, 99 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{4}} - 10 \cdot \frac{a \cdot e^{-1 \cdot \left(m \cdot \log \left(\frac{1}{k}\right)\right)}}{{k}^{3}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020034 +o rules:numerics
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))