Average Error: 29.1 → 0.1
Time: 11.2s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 10132.834563235498:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 10132.834563235498:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\

\end{array}
double f(double N) {
        double r29050 = N;
        double r29051 = 1.0;
        double r29052 = r29050 + r29051;
        double r29053 = log(r29052);
        double r29054 = log(r29050);
        double r29055 = r29053 - r29054;
        return r29055;
}
↓
double f(double N) {
        double r29056 = N;
        double r29057 = 10132.834563235498;
        bool r29058 = r29056 <= r29057;
        double r29059 = 1.0;
        double r29060 = r29056 + r29059;
        double r29061 = r29060 / r29056;
        double r29062 = log(r29061);
        double r29063 = r29059 / r29056;
        double r29064 = 0.3333333333333333;
        double r29065 = 3.0;
        double r29066 = pow(r29056, r29065);
        double r29067 = r29064 / r29066;
        double r29068 = r29063 + r29067;
        double r29069 = 0.5;
        double r29070 = r29056 * r29056;
        double r29071 = r29069 / r29070;
        double r29072 = r29068 - r29071;
        double r29073 = r29058 ? r29062 : r29072;
        return r29073;
}

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 < 10132.834563235498

    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 10132.834563235498 < N

    1. Initial program 59.5

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

    2. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\left(0.333333333333333315 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}}\]

    3. Simplified0.0

      \[\leadsto \color{blue}{\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 10132.834563235498:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]

Reproduce

herbie shell --seed 2020046 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))