Average Error: 29.2 → 0.1
Time: 13.7s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 10223.09734470723924459889531135559082031:\\ \;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log N\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 10223.09734470723924459889531135559082031:\\
\;\;\;\;\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{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\

\end{array}
double f(double N) {
        double r56226 = N;
        double r56227 = 1.0;
        double r56228 = r56226 + r56227;
        double r56229 = log(r56228);
        double r56230 = log(r56226);
        double r56231 = r56229 - r56230;
        return r56231;
}
↓
double f(double N) {
        double r56232 = N;
        double r56233 = 10223.09734470724;
        bool r56234 = r56232 <= r56233;
        double r56235 = 1.0;
        double r56236 = r56232 + r56235;
        double r56237 = cbrt(r56236);
        double r56238 = r56237 * r56237;
        double r56239 = log(r56238);
        double r56240 = log(r56237);
        double r56241 = log(r56232);
        double r56242 = r56240 - r56241;
        double r56243 = r56239 + r56242;
        double r56244 = 0.3333333333333333;
        double r56245 = 3.0;
        double r56246 = pow(r56232, r56245);
        double r56247 = r56244 / r56246;
        double r56248 = 0.5;
        double r56249 = r56248 / r56232;
        double r56250 = r56235 - r56249;
        double r56251 = r56250 / r56232;
        double r56252 = r56247 + r56251;
        double r56253 = r56234 ? r56243 : r56252;
        return r56253;
}

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 < 10223.09734470724

    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 10223.09734470724 < N

    1. Initial program 59.5

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

    2. Taylor expanded around inf 0.0

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

    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

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 10223.09734470723924459889531135559082031:\\ \;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log N\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019325 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))