Average Error: 29.2 → 0.1
Time: 31.6s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9550.567671573803:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\frac{1}{3}}{N \cdot \left(N \cdot N\right)} - \frac{\frac{1}{2}}{N \cdot N}\right) + \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9550.567671573803:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r5290460 = N;
        double r5290461 = 1.0;
        double r5290462 = r5290460 + r5290461;
        double r5290463 = log(r5290462);
        double r5290464 = log(r5290460);
        double r5290465 = r5290463 - r5290464;
        return r5290465;
}
↓
double f(double N) {
        double r5290466 = N;
        double r5290467 = 9550.567671573803;
        bool r5290468 = r5290466 <= r5290467;
        double r5290469 = 1.0;
        double r5290470 = r5290469 + r5290466;
        double r5290471 = r5290470 / r5290466;
        double r5290472 = log(r5290471);
        double r5290473 = 0.3333333333333333;
        double r5290474 = r5290466 * r5290466;
        double r5290475 = r5290466 * r5290474;
        double r5290476 = r5290473 / r5290475;
        double r5290477 = 0.5;
        double r5290478 = r5290477 / r5290474;
        double r5290479 = r5290476 - r5290478;
        double r5290480 = r5290469 / r5290466;
        double r5290481 = r5290479 + r5290480;
        double r5290482 = r5290468 ? r5290472 : r5290481;
        return r5290482;
}

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 < 9550.567671573803

    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 9550.567671573803 < N

    1. Initial program 59.6

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

    2. Simplified59.6

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

    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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9550.567671573803:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\frac{1}{3}}{N \cdot \left(N \cdot N\right)} - \frac{\frac{1}{2}}{N \cdot N}\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019128 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))