Average Error: 2.3 → 0.1
Time: 9.1s
Precision: 64
\[\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\]
↓
\[\begin{array}{l} \mathbf{if}\;k \le 3.98682903313430769394453442188420879001 \cdot 10^{138}:\\ \;\;\;\;\frac{\left(a \cdot {\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}\right) \cdot {\left(\sqrt[3]{k}\right)}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{99 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{4}} + \left(\frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a}{k} \cdot \frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}}{k} - \frac{10 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{3}}\right)\\ \end{array}\]
\frac{a \cdot {k}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}
↓
\begin{array}{l}
\mathbf{if}\;k \le 3.98682903313430769394453442188420879001 \cdot 10^{138}:\\
\;\;\;\;\frac{\left(a \cdot {\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}\right) \cdot {\left(\sqrt[3]{k}\right)}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\

\mathbf{else}:\\
\;\;\;\;\frac{99 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{4}} + \left(\frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a}{k} \cdot \frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}}{k} - \frac{10 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{3}}\right)\\

\end{array}
double f(double a, double k, double m) {
        double r188101 = a;
        double r188102 = k;
        double r188103 = m;
        double r188104 = pow(r188102, r188103);
        double r188105 = r188101 * r188104;
        double r188106 = 1.0;
        double r188107 = 10.0;
        double r188108 = r188107 * r188102;
        double r188109 = r188106 + r188108;
        double r188110 = r188102 * r188102;
        double r188111 = r188109 + r188110;
        double r188112 = r188105 / r188111;
        return r188112;
}
↓
double f(double a, double k, double m) {
        double r188113 = k;
        double r188114 = 3.986829033134308e+138;
        bool r188115 = r188113 <= r188114;
        double r188116 = a;
        double r188117 = cbrt(r188113);
        double r188118 = r188117 * r188117;
        double r188119 = m;
        double r188120 = pow(r188118, r188119);
        double r188121 = r188116 * r188120;
        double r188122 = pow(r188117, r188119);
        double r188123 = r188121 * r188122;
        double r188124 = 1.0;
        double r188125 = 10.0;
        double r188126 = r188125 * r188113;
        double r188127 = r188124 + r188126;
        double r188128 = r188113 * r188113;
        double r188129 = r188127 + r188128;
        double r188130 = r188123 / r188129;
        double r188131 = 99.0;
        double r188132 = 1.0;
        double r188133 = r188132 / r188113;
        double r188134 = -0.6666666666666666;
        double r188135 = pow(r188133, r188134);
        double r188136 = pow(r188135, r188119);
        double r188137 = r188136 * r188116;
        double r188138 = -0.3333333333333333;
        double r188139 = pow(r188133, r188138);
        double r188140 = pow(r188139, r188119);
        double r188141 = r188137 * r188140;
        double r188142 = r188131 * r188141;
        double r188143 = 4.0;
        double r188144 = pow(r188113, r188143);
        double r188145 = r188142 / r188144;
        double r188146 = r188137 / r188113;
        double r188147 = r188140 / r188113;
        double r188148 = r188146 * r188147;
        double r188149 = r188125 * r188141;
        double r188150 = 3.0;
        double r188151 = pow(r188113, r188150);
        double r188152 = r188149 / r188151;
        double r188153 = r188148 - r188152;
        double r188154 = r188145 + r188153;
        double r188155 = r188115 ? r188130 : r188154;
        return r188155;
}

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 < 3.986829033134308e+138

    1. Initial program 0.1

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

    3. Applied add-cube-cbrt0.1

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

    4. Applied unpow-prod-down0.1

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

    5. Applied associate-*r*0.1

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

    if 3.986829033134308e+138 < k

    1. Initial program 10.6

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

    3. Applied add-cube-cbrt10.6

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

    4. Applied unpow-prod-down10.6

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

    5. Applied associate-*r*10.6

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

    6. Taylor expanded around inf 10.6

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

    7. Simplified0.2

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;k \le 3.98682903313430769394453442188420879001 \cdot 10^{138}:\\ \;\;\;\;\frac{\left(a \cdot {\left(\sqrt[3]{k} \cdot \sqrt[3]{k}\right)}^{m}\right) \cdot {\left(\sqrt[3]{k}\right)}^{m}}{\left(1 + 10 \cdot k\right) + k \cdot k}\\ \mathbf{else}:\\ \;\;\;\;\frac{99 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{4}} + \left(\frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a}{k} \cdot \frac{{\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}}{k} - \frac{10 \cdot \left(\left({\left({\left(\frac{1}{k}\right)}^{\frac{-2}{3}}\right)}^{m} \cdot a\right) \cdot {\left({\left(\frac{1}{k}\right)}^{\frac{-1}{3}}\right)}^{m}\right)}{{k}^{3}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019304 
(FPCore (a k m)
  :name "Falkner and Boettcher, Appendix A"
  :precision binary64
  (/ (* a (pow k m)) (+ (+ 1 (* 10 k)) (* k k))))