Average Error: 29.1 → 0.1
Time: 15.5s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9772.093561518631759099662303924560546875:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1}{\frac{N}{1 - \frac{0.5}{N}}}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9772.093561518631759099662303924560546875:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\

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

\end{array}
double f(double N) {
        double r44465 = N;
        double r44466 = 1.0;
        double r44467 = r44465 + r44466;
        double r44468 = log(r44467);
        double r44469 = log(r44465);
        double r44470 = r44468 - r44469;
        return r44470;
}
↓
double f(double N) {
        double r44471 = N;
        double r44472 = 9772.093561518632;
        bool r44473 = r44471 <= r44472;
        double r44474 = 1.0;
        double r44475 = r44471 + r44474;
        double r44476 = r44475 / r44471;
        double r44477 = log(r44476);
        double r44478 = 0.3333333333333333;
        double r44479 = 3.0;
        double r44480 = pow(r44471, r44479);
        double r44481 = r44478 / r44480;
        double r44482 = 1.0;
        double r44483 = 0.5;
        double r44484 = r44483 / r44471;
        double r44485 = r44474 - r44484;
        double r44486 = r44471 / r44485;
        double r44487 = r44482 / r44486;
        double r44488 = r44481 + r44487;
        double r44489 = r44473 ? r44477 : r44488;
        return r44489;
}

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

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

    5. Applied clear-num0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9772.093561518631759099662303924560546875:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1}{\frac{N}{1 - \frac{0.5}{N}}}\\ \end{array}\]

Reproduce

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