Average Error: 29.1 → 0.1
Time: 3.5s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 8017.65951815249264:\\ \;\;\;\;\log \left(\sqrt{\frac{N + 1}{N}} \cdot \sqrt{\frac{N + 1}{N}}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(0.333333333333333315 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 8017.65951815249264:\\
\;\;\;\;\log \left(\sqrt{\frac{N + 1}{N}} \cdot \sqrt{\frac{N + 1}{N}}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(0.333333333333333315 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}\\

\end{array}
double f(double N) {
        double r36466 = N;
        double r36467 = 1.0;
        double r36468 = r36466 + r36467;
        double r36469 = log(r36468);
        double r36470 = log(r36466);
        double r36471 = r36469 - r36470;
        return r36471;
}
↓
double f(double N) {
        double r36472 = N;
        double r36473 = 8017.659518152493;
        bool r36474 = r36472 <= r36473;
        double r36475 = 1.0;
        double r36476 = r36472 + r36475;
        double r36477 = r36476 / r36472;
        double r36478 = sqrt(r36477);
        double r36479 = r36478 * r36478;
        double r36480 = log(r36479);
        double r36481 = 0.3333333333333333;
        double r36482 = 1.0;
        double r36483 = 3.0;
        double r36484 = pow(r36472, r36483);
        double r36485 = r36482 / r36484;
        double r36486 = r36481 * r36485;
        double r36487 = r36482 / r36472;
        double r36488 = r36475 * r36487;
        double r36489 = r36486 + r36488;
        double r36490 = 0.5;
        double r36491 = 2.0;
        double r36492 = pow(r36472, r36491);
        double r36493 = r36482 / r36492;
        double r36494 = r36490 * r36493;
        double r36495 = r36489 - r36494;
        double r36496 = r36474 ? r36480 : r36495;
        return r36496;
}

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 < 8017.659518152493

    1. Initial program 0.1

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

    3. Applied diff-log0.1

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)}\]
    4. Using strategy rm

    5. Applied add-sqr-sqrt0.1

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

    if 8017.659518152493 < N

    1. Initial program 59.6

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

    3. Applied diff-log59.4

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)}\]
    4. Using strategy rm

    5. Applied add-sqr-sqrt59.5

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

    6. 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. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 8017.65951815249264:\\ \;\;\;\;\log \left(\sqrt{\frac{N + 1}{N}} \cdot \sqrt{\frac{N + 1}{N}}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(0.333333333333333315 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}\\ \end{array}\]

Reproduce

herbie shell --seed 2020039 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))