Average Error: 30.1 → 0.1
Time: 3.7s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 3926.44294337209203:\\ \;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{0.333333333333333315}{N} \cdot \frac{1}{{N}^{2}} + \left(\frac{1}{N} - \frac{\frac{0.5}{N}}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 3926.44294337209203:\\
\;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\

\mathbf{else}:\\
\;\;\;\;\frac{0.333333333333333315}{N} \cdot \frac{1}{{N}^{2}} + \left(\frac{1}{N} - \frac{\frac{0.5}{N}}{N}\right)\\

\end{array}
double f(double N) {
        double r38299 = N;
        double r38300 = 1.0;
        double r38301 = r38299 + r38300;
        double r38302 = log(r38301);
        double r38303 = log(r38299);
        double r38304 = r38302 - r38303;
        return r38304;
}
↓
double f(double N) {
        double r38305 = N;
        double r38306 = 3926.442943372092;
        bool r38307 = r38305 <= r38306;
        double r38308 = 1.0;
        double r38309 = r38305 + r38308;
        double r38310 = log(r38309);
        double r38311 = log(r38310);
        double r38312 = exp(r38311);
        double r38313 = log(r38305);
        double r38314 = r38312 - r38313;
        double r38315 = 0.3333333333333333;
        double r38316 = r38315 / r38305;
        double r38317 = 1.0;
        double r38318 = 2.0;
        double r38319 = pow(r38305, r38318);
        double r38320 = r38317 / r38319;
        double r38321 = r38316 * r38320;
        double r38322 = r38308 / r38305;
        double r38323 = 0.5;
        double r38324 = r38323 / r38305;
        double r38325 = r38324 / r38305;
        double r38326 = r38322 - r38325;
        double r38327 = r38321 + r38326;
        double r38328 = r38307 ? r38314 : r38327;
        return r38328;
}

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 < 3926.442943372092

    1. Initial program 0.1

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

    3. Applied add-exp-log0.1

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

    if 3926.442943372092 < N

    1. Initial program 59.4

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

    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

      \[\leadsto \color{blue}{\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}}\]
    4. Using strategy rm

    5. Applied sub-neg0.0

      \[\leadsto \frac{1}{{N}^{2}} \cdot \color{blue}{\left(\frac{0.333333333333333315}{N} + \left(-0.5\right)\right)} + \frac{1}{N}\]

    6. Applied distribute-rgt-in0.0

      \[\leadsto \color{blue}{\left(\frac{0.333333333333333315}{N} \cdot \frac{1}{{N}^{2}} + \left(-0.5\right) \cdot \frac{1}{{N}^{2}}\right)} + \frac{1}{N}\]

    7. Applied associate-+l+0.0

      \[\leadsto \color{blue}{\frac{0.333333333333333315}{N} \cdot \frac{1}{{N}^{2}} + \left(\left(-0.5\right) \cdot \frac{1}{{N}^{2}} + \frac{1}{N}\right)}\]

    8. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 3926.44294337209203:\\ \;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{0.333333333333333315}{N} \cdot \frac{1}{{N}^{2}} + \left(\frac{1}{N} - \frac{\frac{0.5}{N}}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020060 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))