Average Error: 29.2 → 0.1
Time: 43.6s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9094.170836439409:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) + \frac{\frac{\frac{1}{3}}{N \cdot N}}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9094.170836439409:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) + \frac{\frac{\frac{1}{3}}{N \cdot N}}{N}\\

\end{array}
double f(double N) {
        double r2311294 = N;
        double r2311295 = 1.0;
        double r2311296 = r2311294 + r2311295;
        double r2311297 = log(r2311296);
        double r2311298 = log(r2311294);
        double r2311299 = r2311297 - r2311298;
        return r2311299;
}
↓
double f(double N) {
        double r2311300 = N;
        double r2311301 = 9094.170836439409;
        bool r2311302 = r2311300 <= r2311301;
        double r2311303 = 1.0;
        double r2311304 = r2311303 + r2311300;
        double r2311305 = r2311304 / r2311300;
        double r2311306 = log(r2311305);
        double r2311307 = r2311303 / r2311300;
        double r2311308 = -0.5;
        double r2311309 = r2311300 * r2311300;
        double r2311310 = r2311308 / r2311309;
        double r2311311 = r2311307 + r2311310;
        double r2311312 = 0.3333333333333333;
        double r2311313 = r2311312 / r2311309;
        double r2311314 = r2311313 / r2311300;
        double r2311315 = r2311311 + r2311314;
        double r2311316 = r2311302 ? r2311306 : r2311315;
        return r2311316;
}

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 < 9094.170836439409

    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 9094.170836439409 < N

    1. Initial program 59.5

      \[\log \left(N + 1\right) - \log N\]
    2. Using strategy rm

    3. Applied diff-log59.2

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

    4. Taylor expanded around inf 0.0

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

    5. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9094.170836439409:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) + \frac{\frac{\frac{1}{3}}{N \cdot N}}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019112 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))