Average Error: 29.6 → 0.1
Time: 27.1s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 8386.707428118045:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{3}}{N \cdot \left(N \cdot N\right)} + \left(\frac{\frac{-1}{2}}{N \cdot N} + \frac{1}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 8386.707428118045:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r1541301 = N;
        double r1541302 = 1.0;
        double r1541303 = r1541301 + r1541302;
        double r1541304 = log(r1541303);
        double r1541305 = log(r1541301);
        double r1541306 = r1541304 - r1541305;
        return r1541306;
}
↓
double f(double N) {
        double r1541307 = N;
        double r1541308 = 8386.707428118045;
        bool r1541309 = r1541307 <= r1541308;
        double r1541310 = 1.0;
        double r1541311 = r1541310 + r1541307;
        double r1541312 = r1541311 / r1541307;
        double r1541313 = log(r1541312);
        double r1541314 = 0.3333333333333333;
        double r1541315 = r1541307 * r1541307;
        double r1541316 = r1541307 * r1541315;
        double r1541317 = r1541314 / r1541316;
        double r1541318 = -0.5;
        double r1541319 = r1541318 / r1541315;
        double r1541320 = r1541310 / r1541307;
        double r1541321 = r1541319 + r1541320;
        double r1541322 = r1541317 + r1541321;
        double r1541323 = r1541309 ? r1541313 : r1541322;
        return r1541323;
}

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 < 8386.707428118045

    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 8386.707428118045 < N

    1. Initial program 59.4

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

    2. Taylor expanded around inf 0.0

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

    3. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 8386.707428118045:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{3}}{N \cdot \left(N \cdot N\right)} + \left(\frac{\frac{-1}{2}}{N \cdot N} + \frac{1}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019152 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))