Average Error: 29.7 → 0.0
Time: 14.7s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 8023.454188180856363032944500446319580078:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{N \cdot \left(N \cdot N\right)} + \left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 8023.454188180856363032944500446319580078:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{0.3333333333333333148296162562473909929395}{N \cdot \left(N \cdot N\right)} + \left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right)\\

\end{array}
double f(double N) {
        double r2460894 = N;
        double r2460895 = 1.0;
        double r2460896 = r2460894 + r2460895;
        double r2460897 = log(r2460896);
        double r2460898 = log(r2460894);
        double r2460899 = r2460897 - r2460898;
        return r2460899;
}
↓
double f(double N) {
        double r2460900 = N;
        double r2460901 = 8023.454188180856;
        bool r2460902 = r2460900 <= r2460901;
        double r2460903 = 1.0;
        double r2460904 = r2460903 + r2460900;
        double r2460905 = r2460904 / r2460900;
        double r2460906 = log(r2460905);
        double r2460907 = 0.3333333333333333;
        double r2460908 = r2460900 * r2460900;
        double r2460909 = r2460900 * r2460908;
        double r2460910 = r2460907 / r2460909;
        double r2460911 = r2460903 / r2460900;
        double r2460912 = 0.5;
        double r2460913 = r2460912 / r2460908;
        double r2460914 = r2460911 - r2460913;
        double r2460915 = r2460910 + r2460914;
        double r2460916 = r2460902 ? r2460906 : r2460915;
        return r2460916;
}

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 < 8023.454188180856

    1. Initial program 0.1

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

    3. Applied diff-log0.1

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)}\]

    if 8023.454188180856 < N

    1. Initial program 59.6

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

    2. Taylor expanded around inf 0.0

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

    3. Simplified0.0

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{1}{N}, 1, \frac{\frac{1}{N}}{N} \cdot \left(\frac{0.3333333333333333148296162562473909929395}{N} - 0.5\right)\right)}\]

    4. Taylor expanded around 0 0.0

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

    5. Simplified0.0

      \[\leadsto \color{blue}{\frac{0.3333333333333333148296162562473909929395}{\left(N \cdot N\right) \cdot N} + \left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 8023.454188180856363032944500446319580078:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{N \cdot \left(N \cdot N\right)} + \left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019192 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1.0)) (log N)))