Average Error: 29.1 → 0.1
Time: 12.4s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le \frac{671533156162795}{68719476736}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{\frac{6004799503160661}{18014398509481984}}{N} - \frac{1}{2}\right) + 1 \cdot \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le \frac{671533156162795}{68719476736}:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{\frac{6004799503160661}{18014398509481984}}{N} - \frac{1}{2}\right) + 1 \cdot \frac{1}{N}\\

\end{array}
double f(double N) {
        double r35495 = N;
        double r35496 = 1.0;
        double r35497 = r35495 + r35496;
        double r35498 = log(r35497);
        double r35499 = log(r35495);
        double r35500 = r35498 - r35499;
        return r35500;
}
↓
double f(double N) {
        double r35501 = N;
        double r35502 = 671533156162795.0;
        double r35503 = 68719476736.0;
        double r35504 = r35502 / r35503;
        bool r35505 = r35501 <= r35504;
        double r35506 = 1.0;
        double r35507 = r35501 + r35506;
        double r35508 = r35507 / r35501;
        double r35509 = log(r35508);
        double r35510 = 1.0;
        double r35511 = 2.0;
        double r35512 = pow(r35501, r35511);
        double r35513 = r35510 / r35512;
        double r35514 = 6004799503160661.0;
        double r35515 = 18014398509481984.0;
        double r35516 = r35514 / r35515;
        double r35517 = r35516 / r35501;
        double r35518 = 2.0;
        double r35519 = r35506 / r35518;
        double r35520 = r35517 - r35519;
        double r35521 = r35513 * r35520;
        double r35522 = r35510 / r35501;
        double r35523 = r35506 * r35522;
        double r35524 = r35521 + r35523;
        double r35525 = r35505 ? r35509 : r35524;
        return r35525;
}

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 < 9772.093561518632

    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 9772.093561518632 < N

    1. Initial program 59.6

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

    2. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\left(0.3333333333333333148296162562473909929395 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}}\]

    3. Simplified0.0

      \[\leadsto \color{blue}{\frac{1}{{N}^{2}} \cdot \left(\frac{\frac{6004799503160661}{18014398509481984}}{N} - \frac{1}{2}\right) + 1 \cdot \frac{1}{N}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le \frac{671533156162795}{68719476736}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{\frac{6004799503160661}{18014398509481984}}{N} - \frac{1}{2}\right) + 1 \cdot \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019303 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))