Average Error: 28.9 → 0.1
Time: 15.0s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 4795.848553905726475932169705629348754883:\\ \;\;\;\;\sqrt[3]{{\left(\log \left(N + 1\right)\right)}^{3}} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 4795.848553905726475932169705629348754883:\\
\;\;\;\;\sqrt[3]{{\left(\log \left(N + 1\right)\right)}^{3}} - \log N\\

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

\end{array}
double f(double N) {
        double r36707 = N;
        double r36708 = 1.0;
        double r36709 = r36707 + r36708;
        double r36710 = log(r36709);
        double r36711 = log(r36707);
        double r36712 = r36710 - r36711;
        return r36712;
}
↓
double f(double N) {
        double r36713 = N;
        double r36714 = 4795.8485539057265;
        bool r36715 = r36713 <= r36714;
        double r36716 = 1.0;
        double r36717 = r36713 + r36716;
        double r36718 = log(r36717);
        double r36719 = 3.0;
        double r36720 = pow(r36718, r36719);
        double r36721 = cbrt(r36720);
        double r36722 = log(r36713);
        double r36723 = r36721 - r36722;
        double r36724 = 0.3333333333333333;
        double r36725 = pow(r36713, r36719);
        double r36726 = r36724 / r36725;
        double r36727 = 0.5;
        double r36728 = r36727 / r36713;
        double r36729 = r36716 - r36728;
        double r36730 = r36729 / r36713;
        double r36731 = r36726 + r36730;
        double r36732 = r36715 ? r36723 : r36731;
        return r36732;
}

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 < 4795.8485539057265

    1. Initial program 0.1

      \[\log \left(N + 1\right) - \log N\]
    2. Using strategy rm

    3. Applied add-cbrt-cube0.1

      \[\leadsto \color{blue}{\sqrt[3]{\left(\log \left(N + 1\right) \cdot \log \left(N + 1\right)\right) \cdot \log \left(N + 1\right)}} - \log N\]

    4. Simplified0.1

      \[\leadsto \sqrt[3]{\color{blue}{{\left(\log \left(N + 1\right)\right)}^{3}}} - \log N\]

    if 4795.8485539057265 < N

    1. Initial program 59.4

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

    2. Taylor expanded around inf 0.1

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

    3. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 4795.848553905726475932169705629348754883:\\ \;\;\;\;\sqrt[3]{{\left(\log \left(N + 1\right)\right)}^{3}} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019326 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))