Average Error: 29.1 → 0.1
Time: 13.6s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 3775.896918214588367845863103866577148438:\\ \;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(1 - \frac{0.5}{N}\right) \cdot \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 3775.896918214588367845863103866577148438:\\
\;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\

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

\end{array}
double f(double N) {
        double r50080 = N;
        double r50081 = 1.0;
        double r50082 = r50080 + r50081;
        double r50083 = log(r50082);
        double r50084 = log(r50080);
        double r50085 = r50083 - r50084;
        return r50085;
}
↓
double f(double N) {
        double r50086 = N;
        double r50087 = 3775.8969182145884;
        bool r50088 = r50086 <= r50087;
        double r50089 = 1.0;
        double r50090 = r50086 + r50089;
        double r50091 = log(r50090);
        double r50092 = log(r50091);
        double r50093 = exp(r50092);
        double r50094 = log(r50086);
        double r50095 = r50093 - r50094;
        double r50096 = 0.3333333333333333;
        double r50097 = 3.0;
        double r50098 = pow(r50086, r50097);
        double r50099 = r50096 / r50098;
        double r50100 = 0.5;
        double r50101 = r50100 / r50086;
        double r50102 = r50089 - r50101;
        double r50103 = 1.0;
        double r50104 = r50103 / r50086;
        double r50105 = r50102 * r50104;
        double r50106 = r50099 + r50105;
        double r50107 = r50088 ? r50095 : r50106;
        return r50107;
}

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 < 3775.8969182145884

    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 3775.8969182145884 < N

    1. Initial program 59.5

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

    2. Taylor expanded around inf 0.0

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

    3. Simplified0.0

      \[\leadsto \color{blue}{\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}}\]
    4. Using strategy rm

    5. Applied div-inv0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 3775.896918214588367845863103866577148438:\\ \;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(1 - \frac{0.5}{N}\right) \cdot \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019323 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))