Average Error: 29.7 → 0.1
Time: 10.7s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 3632.9677611080442:\\ \;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log N\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 3632.9677611080442:\\
\;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log 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 r52037 = N;
        double r52038 = 1.0;
        double r52039 = r52037 + r52038;
        double r52040 = log(r52039);
        double r52041 = log(r52037);
        double r52042 = r52040 - r52041;
        return r52042;
}
↓
double f(double N) {
        double r52043 = N;
        double r52044 = 3632.967761108044;
        bool r52045 = r52043 <= r52044;
        double r52046 = 1.0;
        double r52047 = r52043 + r52046;
        double r52048 = cbrt(r52047);
        double r52049 = r52048 * r52048;
        double r52050 = log(r52049);
        double r52051 = log(r52048);
        double r52052 = log(r52043);
        double r52053 = r52051 - r52052;
        double r52054 = r52050 + r52053;
        double r52055 = r52046 / r52043;
        double r52056 = 0.3333333333333333;
        double r52057 = 3.0;
        double r52058 = pow(r52043, r52057);
        double r52059 = r52056 / r52058;
        double r52060 = 0.5;
        double r52061 = r52043 * r52043;
        double r52062 = r52060 / r52061;
        double r52063 = r52059 - r52062;
        double r52064 = r52055 + r52063;
        double r52065 = r52045 ? r52054 : r52064;
        return r52065;
}

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 < 3632.967761108044

    1. Initial program 0.1

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

    3. Applied add-cube-cbrt0.1

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

    4. Applied log-prod0.1

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

    5. Applied associate--l+0.1

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

    if 3632.967761108044 < 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

      \[\leadsto \color{blue}{\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}}\]
    4. Using strategy rm

    5. Applied associate--l+0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 3632.9677611080442:\\ \;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log N\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020043 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))