Average Error: 29.9 → 0.0
Time: 15.3s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 7536.677708381748743704520165920257568359:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} - \left(\frac{0.5}{N \cdot N} - \frac{\frac{0.3333333333333333148296162562473909929395}{N}}{N \cdot N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 7536.677708381748743704520165920257568359:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r4841317 = N;
        double r4841318 = 1.0;
        double r4841319 = r4841317 + r4841318;
        double r4841320 = log(r4841319);
        double r4841321 = log(r4841317);
        double r4841322 = r4841320 - r4841321;
        return r4841322;
}
↓
double f(double N) {
        double r4841323 = N;
        double r4841324 = 7536.677708381749;
        bool r4841325 = r4841323 <= r4841324;
        double r4841326 = 1.0;
        double r4841327 = r4841326 + r4841323;
        double r4841328 = r4841327 / r4841323;
        double r4841329 = log(r4841328);
        double r4841330 = r4841326 / r4841323;
        double r4841331 = 0.5;
        double r4841332 = r4841323 * r4841323;
        double r4841333 = r4841331 / r4841332;
        double r4841334 = 0.3333333333333333;
        double r4841335 = r4841334 / r4841323;
        double r4841336 = r4841335 / r4841332;
        double r4841337 = r4841333 - r4841336;
        double r4841338 = r4841330 - r4841337;
        double r4841339 = r4841325 ? r4841329 : r4841338;
        return r4841339;
}

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 < 7536.677708381749

    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 7536.677708381749 < 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

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

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 7536.677708381748743704520165920257568359:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} - \left(\frac{0.5}{N \cdot N} - \frac{\frac{0.3333333333333333148296162562473909929395}{N}}{N \cdot N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019171 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1.0)) (log N)))