Average Error: 29.9 → 0.0
Time: 16.1s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9148.879937239351420430466532707214355469:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.3333333333333333148296162562473909929395}{N \cdot \left(N \cdot N\right)}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9148.879937239351420430466532707214355469:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r3549822 = N;
        double r3549823 = 1.0;
        double r3549824 = r3549822 + r3549823;
        double r3549825 = log(r3549824);
        double r3549826 = log(r3549822);
        double r3549827 = r3549825 - r3549826;
        return r3549827;
}
↓
double f(double N) {
        double r3549828 = N;
        double r3549829 = 9148.879937239351;
        bool r3549830 = r3549828 <= r3549829;
        double r3549831 = 1.0;
        double r3549832 = r3549831 + r3549828;
        double r3549833 = r3549832 / r3549828;
        double r3549834 = log(r3549833);
        double r3549835 = r3549831 / r3549828;
        double r3549836 = 0.3333333333333333;
        double r3549837 = r3549828 * r3549828;
        double r3549838 = r3549828 * r3549837;
        double r3549839 = r3549836 / r3549838;
        double r3549840 = r3549835 + r3549839;
        double r3549841 = 0.5;
        double r3549842 = r3549841 / r3549837;
        double r3549843 = r3549840 - r3549842;
        double r3549844 = r3549830 ? r3549834 : r3549843;
        return r3549844;
}

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 < 9148.879937239351

    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 9148.879937239351 < N

    1. Initial program 59.7

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

    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

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

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9148.879937239351420430466532707214355469:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.3333333333333333148296162562473909929395}{N \cdot \left(N \cdot N\right)}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019170 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1.0)) (log N)))