Average Error: 30.1 → 0.1
Time: 3.8s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 3926.44294337209203:\\ \;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \frac{0.333333333333333315}{N} + \left(\frac{1}{N} - \frac{\frac{0.5}{N}}{N}\right)\\ \end{array}\]
\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{1}{{N}^{2}} \cdot \frac{0.333333333333333315}{N} + \left(\frac{1}{N} - \frac{\frac{0.5}{N}}{N}\right)\\

\end{array}
double f(double N) {
        double r30611 = N;
        double r30612 = 1.0;
        double r30613 = r30611 + r30612;
        double r30614 = log(r30613);
        double r30615 = log(r30611);
        double r30616 = r30614 - r30615;
        return r30616;
}
↓
double f(double N) {
        double r30617 = N;
        double r30618 = 3926.442943372092;
        bool r30619 = r30617 <= r30618;
        double r30620 = 1.0;
        double r30621 = r30617 + r30620;
        double r30622 = log(r30621);
        double r30623 = log(r30622);
        double r30624 = exp(r30623);
        double r30625 = log(r30617);
        double r30626 = r30624 - r30625;
        double r30627 = 1.0;
        double r30628 = 2.0;
        double r30629 = pow(r30617, r30628);
        double r30630 = r30627 / r30629;
        double r30631 = 0.3333333333333333;
        double r30632 = r30631 / r30617;
        double r30633 = r30630 * r30632;
        double r30634 = r30620 / r30617;
        double r30635 = 0.5;
        double r30636 = r30635 / r30617;
        double r30637 = r30636 / r30617;
        double r30638 = r30634 - r30637;
        double r30639 = r30633 + r30638;
        double r30640 = r30619 ? r30626 : r30639;
        return r30640;
}

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-lft-in0.0

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

    7. Applied associate-+l+0.0

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

    8. Simplified0.0

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

  4. Final simplification0.1

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

Reproduce

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