Average Error: 29.2 → 0.1
Time: 13.0s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 8915.293301236255501862615346908569335938:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} - \left(\frac{0.5}{N \cdot N} - \frac{\frac{0.3333333333333333148296162562473909929395}{N}}{N \cdot N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 8915.293301236255501862615346908569335938:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r3356711 = N;
        double r3356712 = 1.0;
        double r3356713 = r3356711 + r3356712;
        double r3356714 = log(r3356713);
        double r3356715 = log(r3356711);
        double r3356716 = r3356714 - r3356715;
        return r3356716;
}
↓
double f(double N) {
        double r3356717 = N;
        double r3356718 = 8915.293301236256;
        bool r3356719 = r3356717 <= r3356718;
        double r3356720 = 1.0;
        double r3356721 = r3356720 + r3356717;
        double r3356722 = r3356721 / r3356717;
        double r3356723 = log(r3356722);
        double r3356724 = r3356720 / r3356717;
        double r3356725 = 0.5;
        double r3356726 = r3356717 * r3356717;
        double r3356727 = r3356725 / r3356726;
        double r3356728 = 0.3333333333333333;
        double r3356729 = r3356728 / r3356717;
        double r3356730 = r3356729 / r3356726;
        double r3356731 = r3356727 - r3356730;
        double r3356732 = r3356724 - r3356731;
        double r3356733 = r3356719 ? r3356723 : r3356732;
        return r3356733;
}

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 < 8915.293301236256

    1. Initial program 0.1

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

    3. Applied add-log-exp0.1

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

    4. Simplified0.1

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

    if 8915.293301236256 < N

    1. Initial program 59.6

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

    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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 8915.293301236255501862615346908569335938:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} - \left(\frac{0.5}{N \cdot N} - \frac{\frac{0.3333333333333333148296162562473909929395}{N}}{N \cdot N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019172 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1.0)) (log N)))