Average Error: 29.5 → 0.1
Time: 26.0s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9307.91986860579:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) + \frac{\frac{\frac{1}{3}}{N \cdot N}}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9307.91986860579:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r2587773 = N;
        double r2587774 = 1.0;
        double r2587775 = r2587773 + r2587774;
        double r2587776 = log(r2587775);
        double r2587777 = log(r2587773);
        double r2587778 = r2587776 - r2587777;
        return r2587778;
}
↓
double f(double N) {
        double r2587779 = N;
        double r2587780 = 9307.91986860579;
        bool r2587781 = r2587779 <= r2587780;
        double r2587782 = 1.0;
        double r2587783 = r2587782 + r2587779;
        double r2587784 = r2587783 / r2587779;
        double r2587785 = log(r2587784);
        double r2587786 = r2587782 / r2587779;
        double r2587787 = -0.5;
        double r2587788 = r2587779 * r2587779;
        double r2587789 = r2587787 / r2587788;
        double r2587790 = r2587786 + r2587789;
        double r2587791 = 0.3333333333333333;
        double r2587792 = r2587791 / r2587788;
        double r2587793 = r2587792 / r2587779;
        double r2587794 = r2587790 + r2587793;
        double r2587795 = r2587781 ? r2587785 : r2587794;
        return r2587795;
}

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 < 9307.91986860579

    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 9307.91986860579 < 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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9307.91986860579:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) + \frac{\frac{\frac{1}{3}}{N \cdot N}}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019121 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))