Average Error: 29.5 → 0.1
Time: 15.4s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 4683.545058486523:\\ \;\;\;\;\frac{\log \left(1 + N\right) \cdot \log \left(1 + N\right) - \log N \cdot \log N}{\log \left(1 + N\right) + \log N}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{-1}{2}}{N \cdot N} + \left(\frac{\frac{1}{3}}{\left(N \cdot N\right) \cdot N} + \frac{1}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 4683.545058486523:\\
\;\;\;\;\frac{\log \left(1 + N\right) \cdot \log \left(1 + N\right) - \log N \cdot \log N}{\log \left(1 + N\right) + \log N}\\

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

\end{array}
double f(double N) {
        double r553118 = N;
        double r553119 = 1.0;
        double r553120 = r553118 + r553119;
        double r553121 = log(r553120);
        double r553122 = log(r553118);
        double r553123 = r553121 - r553122;
        return r553123;
}
↓
double f(double N) {
        double r553124 = N;
        double r553125 = 4683.545058486523;
        bool r553126 = r553124 <= r553125;
        double r553127 = 1.0;
        double r553128 = r553127 + r553124;
        double r553129 = log(r553128);
        double r553130 = r553129 * r553129;
        double r553131 = log(r553124);
        double r553132 = r553131 * r553131;
        double r553133 = r553130 - r553132;
        double r553134 = r553129 + r553131;
        double r553135 = r553133 / r553134;
        double r553136 = -0.5;
        double r553137 = r553124 * r553124;
        double r553138 = r553136 / r553137;
        double r553139 = 0.3333333333333333;
        double r553140 = r553137 * r553124;
        double r553141 = r553139 / r553140;
        double r553142 = r553127 / r553124;
        double r553143 = r553141 + r553142;
        double r553144 = r553138 + r553143;
        double r553145 = r553126 ? r553135 : r553144;
        return r553145;
}

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 < 4683.545058486523

    1. Initial program 0.1

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

    3. Applied flip--0.1

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

    if 4683.545058486523 < N

    1. Initial program 59.5

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

    3. Applied diff-log59.3

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

    4. Taylor expanded around -inf 0.1

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

    5. Simplified0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 4683.545058486523:\\ \;\;\;\;\frac{\log \left(1 + N\right) \cdot \log \left(1 + N\right) - \log N \cdot \log N}{\log \left(1 + N\right) + \log N}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{-1}{2}}{N \cdot N} + \left(\frac{\frac{1}{3}}{\left(N \cdot N\right) \cdot N} + \frac{1}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019151 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))