Average Error: 29.2 → 0.1
Time: 12.9s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 3050.16836541700968155055306851863861084:\\ \;\;\;\;\left(\sqrt[3]{\log \left(N + 1\right)} \cdot \sqrt[3]{\log \left(N + 1\right)}\right) \cdot \sqrt[3]{\log \left(N + 1\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 3050.16836541700968155055306851863861084:\\
\;\;\;\;\left(\sqrt[3]{\log \left(N + 1\right)} \cdot \sqrt[3]{\log \left(N + 1\right)}\right) \cdot \sqrt[3]{\log \left(N + 1\right)} - \log N\\

\mathbf{else}:\\
\;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\

\end{array}
double f(double N) {
        double r31132 = N;
        double r31133 = 1.0;
        double r31134 = r31132 + r31133;
        double r31135 = log(r31134);
        double r31136 = log(r31132);
        double r31137 = r31135 - r31136;
        return r31137;
}
↓
double f(double N) {
        double r31138 = N;
        double r31139 = 3050.1683654170097;
        bool r31140 = r31138 <= r31139;
        double r31141 = 1.0;
        double r31142 = r31138 + r31141;
        double r31143 = log(r31142);
        double r31144 = cbrt(r31143);
        double r31145 = r31144 * r31144;
        double r31146 = r31145 * r31144;
        double r31147 = log(r31138);
        double r31148 = r31146 - r31147;
        double r31149 = 0.3333333333333333;
        double r31150 = 3.0;
        double r31151 = pow(r31138, r31150);
        double r31152 = r31149 / r31151;
        double r31153 = 0.5;
        double r31154 = r31153 / r31138;
        double r31155 = r31141 - r31154;
        double r31156 = r31155 / r31138;
        double r31157 = r31152 + r31156;
        double r31158 = r31140 ? r31148 : r31157;
        return r31158;
}

Error

Bits error versus N

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 2 regimes

  2. if N < 3050.1683654170097

    1. Initial program 0.1

      \[\log \left(N + 1\right) - \log N\]
    2. Using strategy rm

    3. Applied add-cube-cbrt0.1

      \[\leadsto \color{blue}{\left(\sqrt[3]{\log \left(N + 1\right)} \cdot \sqrt[3]{\log \left(N + 1\right)}\right) \cdot \sqrt[3]{\log \left(N + 1\right)}} - \log N\]

    if 3050.1683654170097 < N

    1. Initial program 59.4

      \[\log \left(N + 1\right) - \log N\]

    2. Taylor expanded around inf 0.1

      \[\leadsto \color{blue}{\left(0.3333333333333333148296162562473909929395 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}}\]

    3. Simplified0.1

      \[\leadsto \color{blue}{\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 3050.16836541700968155055306851863861084:\\ \;\;\;\;\left(\sqrt[3]{\log \left(N + 1\right)} \cdot \sqrt[3]{\log \left(N + 1\right)}\right) \cdot \sqrt[3]{\log \left(N + 1\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019325 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))