Average Error: 29.7 → 0.1
Time: 15.4s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 15002.08922444248310057446360588073730469:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{0.3333333333333333148296162562473909929395}{N} - 0.5\right) \cdot \frac{1}{N \cdot N} + \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 15002.08922444248310057446360588073730469:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r40341 = N;
        double r40342 = 1.0;
        double r40343 = r40341 + r40342;
        double r40344 = log(r40343);
        double r40345 = log(r40341);
        double r40346 = r40344 - r40345;
        return r40346;
}
↓
double f(double N) {
        double r40347 = N;
        double r40348 = 15002.089224442483;
        bool r40349 = r40347 <= r40348;
        double r40350 = 1.0;
        double r40351 = r40350 + r40347;
        double r40352 = r40351 / r40347;
        double r40353 = log(r40352);
        double r40354 = 0.3333333333333333;
        double r40355 = r40354 / r40347;
        double r40356 = 0.5;
        double r40357 = r40355 - r40356;
        double r40358 = 1.0;
        double r40359 = r40347 * r40347;
        double r40360 = r40358 / r40359;
        double r40361 = r40357 * r40360;
        double r40362 = r40350 / r40347;
        double r40363 = r40361 + r40362;
        double r40364 = r40349 ? r40353 : r40363;
        return r40364;
}

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 < 15002.089224442483

    1. Initial program 0.1

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

    2. Simplified0.1

      \[\leadsto \color{blue}{\log \left(1 + N\right) - \log N}\]
    3. Using strategy rm

    4. Applied diff-log0.1

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

    5. Simplified0.1

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

    if 15002.089224442483 < N

    1. Initial program 59.4

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

    2. Simplified59.4

      \[\leadsto \color{blue}{\log \left(1 + N\right) - \log N}\]
    3. Using strategy rm

    4. Applied diff-log59.1

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

    5. Simplified59.1

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

    6. Taylor expanded around inf 0.0

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

    7. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 15002.08922444248310057446360588073730469:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{0.3333333333333333148296162562473909929395}{N} - 0.5\right) \cdot \frac{1}{N \cdot N} + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019174 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1.0)) (log N)))