Average Error: 29.5 → 0.1
Time: 39.0s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9307.91986860579:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) - \frac{\frac{-1}{3}}{N \cdot \left(N \cdot N\right)}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9307.91986860579:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) - \frac{\frac{-1}{3}}{N \cdot \left(N \cdot N\right)}\\

\end{array}
double f(double N) {
        double r2093893 = N;
        double r2093894 = 1.0;
        double r2093895 = r2093893 + r2093894;
        double r2093896 = log(r2093895);
        double r2093897 = log(r2093893);
        double r2093898 = r2093896 - r2093897;
        return r2093898;
}
↓
double f(double N) {
        double r2093899 = N;
        double r2093900 = 9307.91986860579;
        bool r2093901 = r2093899 <= r2093900;
        double r2093902 = 1.0;
        double r2093903 = r2093902 + r2093899;
        double r2093904 = r2093903 / r2093899;
        double r2093905 = log(r2093904);
        double r2093906 = r2093902 / r2093899;
        double r2093907 = -0.5;
        double r2093908 = r2093899 * r2093899;
        double r2093909 = r2093907 / r2093908;
        double r2093910 = r2093906 + r2093909;
        double r2093911 = -0.3333333333333333;
        double r2093912 = r2093899 * r2093908;
        double r2093913 = r2093911 / r2093912;
        double r2093914 = r2093910 - r2093913;
        double r2093915 = r2093901 ? r2093905 : r2093914;
        return r2093915;
}

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 < 9307.91986860579

    1. Initial program 0.1

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

    2. Simplified0.1

      \[\leadsto \color{blue}{\mathsf{log1p}\left(N\right) - \log N}\]
    3. Using strategy rm

    4. Applied log1p-udef0.1

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

    5. Applied diff-log0.1

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

    if 9307.91986860579 < N

    1. Initial program 59.5

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

    2. Simplified59.5

      \[\leadsto \color{blue}{\mathsf{log1p}\left(N\right) - \log N}\]

    3. Taylor expanded around -inf 0.0

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

    4. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9307.91986860579:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) - \frac{\frac{-1}{3}}{N \cdot \left(N \cdot N\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2019121 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))