Average Error: 29.1 → 0.1
Time: 16.3s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9772.093561518631759099662303924560546875:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1}{\frac{N}{1 - \frac{0.5}{N}}}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9772.093561518631759099662303924560546875:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1}{\frac{N}{1 - \frac{0.5}{N}}}\\

\end{array}
double f(double N) {
        double r49672 = N;
        double r49673 = 1.0;
        double r49674 = r49672 + r49673;
        double r49675 = log(r49674);
        double r49676 = log(r49672);
        double r49677 = r49675 - r49676;
        return r49677;
}
↓
double f(double N) {
        double r49678 = N;
        double r49679 = 9772.093561518632;
        bool r49680 = r49678 <= r49679;
        double r49681 = 1.0;
        double r49682 = r49678 + r49681;
        double r49683 = r49682 / r49678;
        double r49684 = log(r49683);
        double r49685 = 0.3333333333333333;
        double r49686 = 3.0;
        double r49687 = pow(r49678, r49686);
        double r49688 = r49685 / r49687;
        double r49689 = 1.0;
        double r49690 = 0.5;
        double r49691 = r49690 / r49678;
        double r49692 = r49681 - r49691;
        double r49693 = r49678 / r49692;
        double r49694 = r49689 / r49693;
        double r49695 = r49688 + r49694;
        double r49696 = r49680 ? r49684 : r49695;
        return r49696;
}

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 < 9772.093561518632

    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 9772.093561518632 < N

    1. Initial program 59.6

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

    2. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\left(0.3333333333333333148296162562473909929395 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}}\]

    3. Simplified0.0

      \[\leadsto \color{blue}{\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}}\]
    4. Using strategy rm

    5. Applied clear-num0.0

      \[\leadsto \frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \color{blue}{\frac{1}{\frac{N}{1 - \frac{0.5}{N}}}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9772.093561518631759099662303924560546875:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1}{\frac{N}{1 - \frac{0.5}{N}}}\\ \end{array}\]

Reproduce

herbie shell --seed 2019303 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))