Average Error: 30.1 → 0.1
Time: 16.1s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9454.187458193411657703109085559844970703:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1}{N}\right) - \frac{\frac{0.5}{N}}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9454.187458193411657703109085559844970703:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1}{N}\right) - \frac{\frac{0.5}{N}}{N}\\

\end{array}
double f(double N) {
        double r50614 = N;
        double r50615 = 1.0;
        double r50616 = r50614 + r50615;
        double r50617 = log(r50616);
        double r50618 = log(r50614);
        double r50619 = r50617 - r50618;
        return r50619;
}
↓
double f(double N) {
        double r50620 = N;
        double r50621 = 9454.187458193412;
        bool r50622 = r50620 <= r50621;
        double r50623 = 1.0;
        double r50624 = r50620 + r50623;
        double r50625 = r50624 / r50620;
        double r50626 = log(r50625);
        double r50627 = 0.3333333333333333;
        double r50628 = 3.0;
        double r50629 = pow(r50620, r50628);
        double r50630 = r50627 / r50629;
        double r50631 = r50623 / r50620;
        double r50632 = r50630 + r50631;
        double r50633 = 0.5;
        double r50634 = r50633 / r50620;
        double r50635 = r50634 / r50620;
        double r50636 = r50632 - r50635;
        double r50637 = r50622 ? r50626 : r50636;
        return r50637;
}

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 < 9454.187458193412

    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 9454.187458193412 < N

    1. Initial program 59.4

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

    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

      \[\leadsto \color{blue}{\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}}\]
    4. Using strategy rm

    5. Applied div-sub0.0

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

    6. Applied associate-+r-0.0

      \[\leadsto \color{blue}{\left(\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1}{N}\right) - \frac{\frac{0.5}{N}}{N}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9454.187458193411657703109085559844970703:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1}{N}\right) - \frac{\frac{0.5}{N}}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019322 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))