Average Error: 29.4 → 0.1
Time: 17.2s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 1.452548268332520819967612624168395996094 \cdot 10^{-7}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{\frac{N + 1}{N}} \cdot \sqrt{\frac{N + 1}{N}}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;\log \left(N + 1\right) - \log N \le 1.452548268332520819967612624168395996094 \cdot 10^{-7}:\\
\;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\

\mathbf{else}:\\
\;\;\;\;\log \left(\sqrt{\frac{N + 1}{N}} \cdot \sqrt{\frac{N + 1}{N}}\right)\\

\end{array}
double f(double N) {
        double r48849 = N;
        double r48850 = 1.0;
        double r48851 = r48849 + r48850;
        double r48852 = log(r48851);
        double r48853 = log(r48849);
        double r48854 = r48852 - r48853;
        return r48854;
}
↓
double f(double N) {
        double r48855 = N;
        double r48856 = 1.0;
        double r48857 = r48855 + r48856;
        double r48858 = log(r48857);
        double r48859 = log(r48855);
        double r48860 = r48858 - r48859;
        double r48861 = 1.4525482683325208e-07;
        bool r48862 = r48860 <= r48861;
        double r48863 = 0.3333333333333333;
        double r48864 = 3.0;
        double r48865 = pow(r48855, r48864);
        double r48866 = r48863 / r48865;
        double r48867 = 0.5;
        double r48868 = r48867 / r48855;
        double r48869 = r48856 - r48868;
        double r48870 = r48869 / r48855;
        double r48871 = r48866 + r48870;
        double r48872 = r48857 / r48855;
        double r48873 = sqrt(r48872);
        double r48874 = r48873 * r48873;
        double r48875 = log(r48874);
        double r48876 = r48862 ? r48871 : r48875;
        return r48876;
}

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 (- (log (+ N 1.0)) (log N)) < 1.4525482683325208e-07

    1. Initial program 60.0

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

    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

      \[\leadsto \color{blue}{\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}}\]

    if 1.4525482683325208e-07 < (- (log (+ N 1.0)) (log N))

    1. Initial program 0.3

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

    3. Applied diff-log0.2

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

    5. Applied add-sqr-sqrt0.2

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 1.452548268332520819967612624168395996094 \cdot 10^{-7}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\sqrt{\frac{N + 1}{N}} \cdot \sqrt{\frac{N + 1}{N}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019304 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))