Average Error: 29.7 → 0.1
Time: 8.9s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 3995.2538179566035978496074676513671875:\\ \;\;\;\;\sqrt{\log \left(N + 1\right)} \cdot \sqrt{\log \left(N + 1\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} - \left(\frac{\frac{0.5}{N}}{N} - 0.3333333333333333148296162562473909929395 \cdot \frac{1}{{N}^{3}}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 3995.2538179566035978496074676513671875:\\
\;\;\;\;\sqrt{\log \left(N + 1\right)} \cdot \sqrt{\log \left(N + 1\right)} - \log N\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{N} - \left(\frac{\frac{0.5}{N}}{N} - 0.3333333333333333148296162562473909929395 \cdot \frac{1}{{N}^{3}}\right)\\

\end{array}
double f(double N) {
        double r20806 = N;
        double r20807 = 1.0;
        double r20808 = r20806 + r20807;
        double r20809 = log(r20808);
        double r20810 = log(r20806);
        double r20811 = r20809 - r20810;
        return r20811;
}
↓
double f(double N) {
        double r20812 = N;
        double r20813 = 3995.2538179566036;
        bool r20814 = r20812 <= r20813;
        double r20815 = 1.0;
        double r20816 = r20812 + r20815;
        double r20817 = log(r20816);
        double r20818 = sqrt(r20817);
        double r20819 = r20818 * r20818;
        double r20820 = log(r20812);
        double r20821 = r20819 - r20820;
        double r20822 = r20815 / r20812;
        double r20823 = 0.5;
        double r20824 = r20823 / r20812;
        double r20825 = r20824 / r20812;
        double r20826 = 0.3333333333333333;
        double r20827 = 1.0;
        double r20828 = 3.0;
        double r20829 = pow(r20812, r20828);
        double r20830 = r20827 / r20829;
        double r20831 = r20826 * r20830;
        double r20832 = r20825 - r20831;
        double r20833 = r20822 - r20832;
        double r20834 = r20814 ? r20821 : r20833;
        return r20834;
}

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 < 3995.2538179566036

    1. Initial program 0.1

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

    3. Applied add-sqr-sqrt0.1

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

    if 3995.2538179566036 < N

    1. Initial program 59.5

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

    3. Applied add-sqr-sqrt59.9

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

    4. Taylor expanded around inf 0.0

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

    5. Simplified0.0

      \[\leadsto \color{blue}{\frac{1}{N} - \left(\frac{\frac{0.5}{N}}{N} - 0.3333333333333333148296162562473909929395 \cdot \frac{1}{{N}^{3}}\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 3995.2538179566035978496074676513671875:\\ \;\;\;\;\sqrt{\log \left(N + 1\right)} \cdot \sqrt{\log \left(N + 1\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} - \left(\frac{\frac{0.5}{N}}{N} - 0.3333333333333333148296162562473909929395 \cdot \frac{1}{{N}^{3}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019298 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))