Average Error: 29.5 → 0.0
Time: 15.2s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 11216.499008092209:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{3}}{N \cdot \left(N \cdot N\right)} + \left(\frac{1}{N} - \frac{\frac{1}{2}}{N \cdot N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 11216.499008092209:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r932254 = N;
        double r932255 = 1.0;
        double r932256 = r932254 + r932255;
        double r932257 = log(r932256);
        double r932258 = log(r932254);
        double r932259 = r932257 - r932258;
        return r932259;
}
↓
double f(double N) {
        double r932260 = N;
        double r932261 = 11216.499008092209;
        bool r932262 = r932260 <= r932261;
        double r932263 = 1.0;
        double r932264 = r932263 + r932260;
        double r932265 = r932264 / r932260;
        double r932266 = log(r932265);
        double r932267 = 0.3333333333333333;
        double r932268 = r932260 * r932260;
        double r932269 = r932260 * r932268;
        double r932270 = r932267 / r932269;
        double r932271 = r932263 / r932260;
        double r932272 = 0.5;
        double r932273 = r932272 / r932268;
        double r932274 = r932271 - r932273;
        double r932275 = r932270 + r932274;
        double r932276 = r932262 ? r932266 : r932275;
        return r932276;
}

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 < 11216.499008092209

    1. Initial program 0.1

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

    3. Applied add-log-exp0.1

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

    4. Simplified0.1

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

    if 11216.499008092209 < N

    1. Initial program 59.7

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

    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

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

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 11216.499008092209:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{3}}{N \cdot \left(N \cdot N\right)} + \left(\frac{1}{N} - \frac{\frac{1}{2}}{N \cdot N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019132 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))