Average Error: 29.1 → 0.1
Time: 14.8s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9233.861277203294:\\ \;\;\;\;\log \left(\sqrt{\frac{1 + N}{N}}\right) + \log \left(\sqrt{\frac{1 + N}{N}}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \frac{\frac{-1}{2} + \frac{\frac{1}{3}}{N}}{N \cdot N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9233.861277203294:\\
\;\;\;\;\log \left(\sqrt{\frac{1 + N}{N}}\right) + \log \left(\sqrt{\frac{1 + N}{N}}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{N} + \frac{\frac{-1}{2} + \frac{\frac{1}{3}}{N}}{N \cdot N}\\

\end{array}
double f(double N) {
        double r2487785 = N;
        double r2487786 = 1.0;
        double r2487787 = r2487785 + r2487786;
        double r2487788 = log(r2487787);
        double r2487789 = log(r2487785);
        double r2487790 = r2487788 - r2487789;
        return r2487790;
}
↓
double f(double N) {
        double r2487791 = N;
        double r2487792 = 9233.861277203294;
        bool r2487793 = r2487791 <= r2487792;
        double r2487794 = 1.0;
        double r2487795 = r2487794 + r2487791;
        double r2487796 = r2487795 / r2487791;
        double r2487797 = sqrt(r2487796);
        double r2487798 = log(r2487797);
        double r2487799 = r2487798 + r2487798;
        double r2487800 = r2487794 / r2487791;
        double r2487801 = -0.5;
        double r2487802 = 0.3333333333333333;
        double r2487803 = r2487802 / r2487791;
        double r2487804 = r2487801 + r2487803;
        double r2487805 = r2487791 * r2487791;
        double r2487806 = r2487804 / r2487805;
        double r2487807 = r2487800 + r2487806;
        double r2487808 = r2487793 ? r2487799 : r2487807;
        return r2487808;
}

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 < 9233.861277203294

    1. Initial program 0.1

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

    3. Applied add-log-exp0.2

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

    4. Simplified0.1

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

    6. Applied add-sqr-sqrt0.2

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

    7. Applied log-prod0.1

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

    if 9233.861277203294 < N

    1. Initial program 59.5

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

    2. Taylor expanded around inf 0.0

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

    3. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9233.861277203294:\\ \;\;\;\;\log \left(\sqrt{\frac{1 + N}{N}}\right) + \log \left(\sqrt{\frac{1 + N}{N}}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \frac{\frac{-1}{2} + \frac{\frac{1}{3}}{N}}{N \cdot N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019162 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))