Average Error: 29.8 → 0.1
Time: 10.6s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9562.6451805155593:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9562.6451805155593:\\
\;\;\;\;\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 r46597 = N;
        double r46598 = 1.0;
        double r46599 = r46597 + r46598;
        double r46600 = log(r46599);
        double r46601 = log(r46597);
        double r46602 = r46600 - r46601;
        return r46602;
}
↓
double f(double N) {
        double r46603 = N;
        double r46604 = 9562.64518051556;
        bool r46605 = r46603 <= r46604;
        double r46606 = 1.0;
        double r46607 = r46603 + r46606;
        double r46608 = r46607 / r46603;
        double r46609 = log(r46608);
        double r46610 = r46606 / r46603;
        double r46611 = 0.3333333333333333;
        double r46612 = 3.0;
        double r46613 = pow(r46603, r46612);
        double r46614 = r46611 / r46613;
        double r46615 = 0.5;
        double r46616 = r46603 * r46603;
        double r46617 = r46615 / r46616;
        double r46618 = r46614 - r46617;
        double r46619 = r46610 + r46618;
        double r46620 = r46605 ? r46609 : r46619;
        return r46620;
}

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 < 9562.64518051556

    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 9562.64518051556 < 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

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9562.6451805155593:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020047 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))