Average Error: 29.3 → 0.1
Time: 10.1s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 11278.66194821723365748766809701919555664:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(\frac{1}{N} - \frac{0.5}{{N}^{2}}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 11278.66194821723365748766809701919555664:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(\frac{1}{N} - \frac{0.5}{{N}^{2}}\right)\\

\end{array}
double f(double N) {
        double r43555 = N;
        double r43556 = 1.0;
        double r43557 = r43555 + r43556;
        double r43558 = log(r43557);
        double r43559 = log(r43555);
        double r43560 = r43558 - r43559;
        return r43560;
}
↓
double f(double N) {
        double r43561 = N;
        double r43562 = 11278.661948217234;
        bool r43563 = r43561 <= r43562;
        double r43564 = 1.0;
        double r43565 = r43561 + r43564;
        double r43566 = r43565 / r43561;
        double r43567 = log(r43566);
        double r43568 = 0.3333333333333333;
        double r43569 = 3.0;
        double r43570 = pow(r43561, r43569);
        double r43571 = r43568 / r43570;
        double r43572 = r43564 / r43561;
        double r43573 = 0.5;
        double r43574 = 2.0;
        double r43575 = pow(r43561, r43574);
        double r43576 = r43573 / r43575;
        double r43577 = r43572 - r43576;
        double r43578 = r43571 + r43577;
        double r43579 = r43563 ? r43567 : r43578;
        return r43579;
}

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 < 11278.661948217234

    1. Initial program 0.1

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

    3. Applied diff-log0.1

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

    if 11278.661948217234 < N

    1. Initial program 59.6

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

    3. Applied diff-log59.3

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

    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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 11278.66194821723365748766809701919555664:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(\frac{1}{N} - \frac{0.5}{{N}^{2}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019208 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))