Average Error: 29.7 → 0.1
Time: 14.4s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 15002.08922444248310057446360588073730469:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} - \frac{0.5}{N \cdot N}\right) + \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 15002.08922444248310057446360588073730469:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} - \frac{0.5}{N \cdot N}\right) + \frac{1}{N}\\

\end{array}
double f(double N) {
        double r50910 = N;
        double r50911 = 1.0;
        double r50912 = r50910 + r50911;
        double r50913 = log(r50912);
        double r50914 = log(r50910);
        double r50915 = r50913 - r50914;
        return r50915;
}
↓
double f(double N) {
        double r50916 = N;
        double r50917 = 15002.089224442483;
        bool r50918 = r50916 <= r50917;
        double r50919 = 1.0;
        double r50920 = r50919 + r50916;
        double r50921 = r50920 / r50916;
        double r50922 = log(r50921);
        double r50923 = 0.3333333333333333;
        double r50924 = 3.0;
        double r50925 = pow(r50916, r50924);
        double r50926 = r50923 / r50925;
        double r50927 = 0.5;
        double r50928 = r50916 * r50916;
        double r50929 = r50927 / r50928;
        double r50930 = r50926 - r50929;
        double r50931 = r50919 / r50916;
        double r50932 = r50930 + r50931;
        double r50933 = r50918 ? r50922 : r50932;
        return r50933;
}

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 < 15002.089224442483

    1. Initial program 0.1

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

    2. Simplified0.1

      \[\leadsto \color{blue}{\log \left(1 + N\right) - \log N}\]
    3. Using strategy rm

    4. Applied diff-log0.1

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

    5. Simplified0.1

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

    if 15002.089224442483 < N

    1. Initial program 59.4

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

    2. Simplified59.4

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

    3. Taylor expanded around inf 0.0

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

    4. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 15002.08922444248310057446360588073730469:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} - \frac{0.5}{N \cdot N}\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019174 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1.0)) (log N)))