Average Error: 29.1 → 0.1
Time: 9.7s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 10132.834563235498:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]
\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}:\\
\;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\

\end{array}
double f(double N) {
        double r36127 = N;
        double r36128 = 1.0;
        double r36129 = r36127 + r36128;
        double r36130 = log(r36129);
        double r36131 = log(r36127);
        double r36132 = r36130 - r36131;
        return r36132;
}
↓
double f(double N) {
        double r36133 = N;
        double r36134 = 10132.834563235498;
        bool r36135 = r36133 <= r36134;
        double r36136 = 1.0;
        double r36137 = r36133 + r36136;
        double r36138 = r36137 / r36133;
        double r36139 = log(r36138);
        double r36140 = r36136 / r36133;
        double r36141 = 0.3333333333333333;
        double r36142 = 3.0;
        double r36143 = pow(r36133, r36142);
        double r36144 = r36141 / r36143;
        double r36145 = 0.5;
        double r36146 = r36133 * r36133;
        double r36147 = r36145 / r36146;
        double r36148 = r36144 - r36147;
        double r36149 = r36140 + r36148;
        double r36150 = r36135 ? r36139 : r36149;
        return r36150;
}

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

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

  4. Final simplification0.1

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

Reproduce

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