Average Error: 29.8 → 0.1
Time: 7.3s
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 r32766 = N;
        double r32767 = 1.0;
        double r32768 = r32766 + r32767;
        double r32769 = log(r32768);
        double r32770 = log(r32766);
        double r32771 = r32769 - r32770;
        return r32771;
}
↓
double f(double N) {
        double r32772 = N;
        double r32773 = 9562.64518051556;
        bool r32774 = r32772 <= r32773;
        double r32775 = 1.0;
        double r32776 = r32772 + r32775;
        double r32777 = r32776 / r32772;
        double r32778 = log(r32777);
        double r32779 = r32775 / r32772;
        double r32780 = 0.3333333333333333;
        double r32781 = 3.0;
        double r32782 = pow(r32772, r32781);
        double r32783 = r32780 / r32782;
        double r32784 = 0.5;
        double r32785 = r32772 * r32772;
        double r32786 = r32784 / r32785;
        double r32787 = r32783 - r32786;
        double r32788 = r32779 + r32787;
        double r32789 = r32774 ? r32778 : r32788;
        return r32789;
}

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 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))