Average Error: 29.2 → 0.1
Time: 14.8s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 8915.293301236255501862615346908569335938:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\frac{0.3333333333333333148296162562473909929395}{N}}{N \cdot N} - \frac{0.5}{N \cdot N}\right) + \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 8915.293301236255501862615346908569335938:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r3415465 = N;
        double r3415466 = 1.0;
        double r3415467 = r3415465 + r3415466;
        double r3415468 = log(r3415467);
        double r3415469 = log(r3415465);
        double r3415470 = r3415468 - r3415469;
        return r3415470;
}
↓
double f(double N) {
        double r3415471 = N;
        double r3415472 = 8915.293301236256;
        bool r3415473 = r3415471 <= r3415472;
        double r3415474 = 1.0;
        double r3415475 = r3415474 + r3415471;
        double r3415476 = r3415475 / r3415471;
        double r3415477 = log(r3415476);
        double r3415478 = 0.3333333333333333;
        double r3415479 = r3415478 / r3415471;
        double r3415480 = r3415471 * r3415471;
        double r3415481 = r3415479 / r3415480;
        double r3415482 = 0.5;
        double r3415483 = r3415482 / r3415480;
        double r3415484 = r3415481 - r3415483;
        double r3415485 = r3415474 / r3415471;
        double r3415486 = r3415484 + r3415485;
        double r3415487 = r3415473 ? r3415477 : r3415486;
        return r3415487;
}

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 < 8915.293301236256

    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 8915.293301236256 < 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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 8915.293301236255501862615346908569335938:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{\frac{0.3333333333333333148296162562473909929395}{N}}{N \cdot N} - \frac{0.5}{N \cdot N}\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019172 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1.0)) (log N)))