Average Error: 29.6 → 0.1
Time: 21.2s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9516.15297930401:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{\frac{1}{N}}{N}, \frac{-1}{2}, \mathsf{fma}\left(\frac{1}{N} \cdot \frac{\frac{1}{N}}{N}, \frac{1}{3}, \frac{1}{N}\right)\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9516.15297930401:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r700866 = N;
        double r700867 = 1.0;
        double r700868 = r700866 + r700867;
        double r700869 = log(r700868);
        double r700870 = log(r700866);
        double r700871 = r700869 - r700870;
        return r700871;
}
↓
double f(double N) {
        double r700872 = N;
        double r700873 = 9516.15297930401;
        bool r700874 = r700872 <= r700873;
        double r700875 = 1.0;
        double r700876 = r700875 + r700872;
        double r700877 = r700876 / r700872;
        double r700878 = log(r700877);
        double r700879 = r700875 / r700872;
        double r700880 = r700879 / r700872;
        double r700881 = -0.5;
        double r700882 = r700879 * r700880;
        double r700883 = 0.3333333333333333;
        double r700884 = fma(r700882, r700883, r700879);
        double r700885 = fma(r700880, r700881, r700884);
        double r700886 = r700874 ? r700878 : r700885;
        return r700886;
}

Error

Bits error versus N

Derivation

  1. Split input into 2 regimes

  2. if N < 9516.15297930401

    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 9516.15297930401 < N

    1. Initial program 59.4

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

    2. Simplified59.4

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

    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

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{\frac{1}{N}}{N}, \frac{-1}{2}, \mathsf{fma}\left(\frac{\frac{1}{N}}{N \cdot N}, \frac{1}{3}, \frac{1}{N}\right)\right)}\]
    5. Using strategy rm

    6. Applied add-log-exp0.1

      \[\leadsto \mathsf{fma}\left(\frac{\frac{1}{N}}{N}, \frac{-1}{2}, \mathsf{fma}\left(\color{blue}{\log \left(e^{\frac{\frac{1}{N}}{N \cdot N}}\right)}, \frac{1}{3}, \frac{1}{N}\right)\right)\]

    7. Simplified0.1

      \[\leadsto \mathsf{fma}\left(\frac{\frac{1}{N}}{N}, \frac{-1}{2}, \mathsf{fma}\left(\log \color{blue}{\left(e^{\frac{\frac{1}{N}}{N} \cdot \frac{1}{N}}\right)}, \frac{1}{3}, \frac{1}{N}\right)\right)\]
    8. Using strategy rm

    9. Applied rem-log-exp0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9516.15297930401:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{\frac{1}{N}}{N}, \frac{-1}{2}, \mathsf{fma}\left(\frac{1}{N} \cdot \frac{\frac{1}{N}}{N}, \frac{1}{3}, \frac{1}{N}\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019141 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))