Average Error: 30.0 → 0.1
Time: 12.2s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 8885.992756934142:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{3} \cdot \frac{1}{N}}{N \cdot N} - \left(\frac{\frac{1}{2}}{N \cdot N} - \frac{1}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 8885.992756934142:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r1358618 = N;
        double r1358619 = 1.0;
        double r1358620 = r1358618 + r1358619;
        double r1358621 = log(r1358620);
        double r1358622 = log(r1358618);
        double r1358623 = r1358621 - r1358622;
        return r1358623;
}
↓
double f(double N) {
        double r1358624 = N;
        double r1358625 = 8885.992756934142;
        bool r1358626 = r1358624 <= r1358625;
        double r1358627 = 1.0;
        double r1358628 = r1358627 + r1358624;
        double r1358629 = r1358628 / r1358624;
        double r1358630 = log(r1358629);
        double r1358631 = 0.3333333333333333;
        double r1358632 = r1358627 / r1358624;
        double r1358633 = r1358631 * r1358632;
        double r1358634 = r1358624 * r1358624;
        double r1358635 = r1358633 / r1358634;
        double r1358636 = 0.5;
        double r1358637 = r1358636 / r1358634;
        double r1358638 = r1358637 - r1358632;
        double r1358639 = r1358635 - r1358638;
        double r1358640 = r1358626 ? r1358630 : r1358639;
        return r1358640;
}

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 < 8885.992756934142

    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 8885.992756934142 < N

    1. Initial program 59.4

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

    3. Applied diff-log59.2

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

    4. Taylor expanded around inf 0.0

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

    5. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 8885.992756934142:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{3} \cdot \frac{1}{N}}{N \cdot N} - \left(\frac{\frac{1}{2}}{N \cdot N} - \frac{1}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019129 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))