Average Error: 29.4 → 0.0
Time: 48.0s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 7453.229305032585:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) - \frac{\frac{-1}{3}}{N \cdot \left(N \cdot N\right)}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 7453.229305032585:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r2972984 = N;
        double r2972985 = 1.0;
        double r2972986 = r2972984 + r2972985;
        double r2972987 = log(r2972986);
        double r2972988 = log(r2972984);
        double r2972989 = r2972987 - r2972988;
        return r2972989;
}
↓
double f(double N) {
        double r2972990 = N;
        double r2972991 = 7453.229305032585;
        bool r2972992 = r2972990 <= r2972991;
        double r2972993 = 1.0;
        double r2972994 = r2972993 + r2972990;
        double r2972995 = r2972994 / r2972990;
        double r2972996 = log(r2972995);
        double r2972997 = r2972993 / r2972990;
        double r2972998 = -0.5;
        double r2972999 = r2972990 * r2972990;
        double r2973000 = r2972998 / r2972999;
        double r2973001 = r2972997 + r2973000;
        double r2973002 = -0.3333333333333333;
        double r2973003 = r2972990 * r2972999;
        double r2973004 = r2973002 / r2973003;
        double r2973005 = r2973001 - r2973004;
        double r2973006 = r2972992 ? r2972996 : r2973005;
        return r2973006;
}

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 < 7453.229305032585

    1. Initial program 0.1

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

    2. Simplified0.1

      \[\leadsto \color{blue}{\log_* (1 + N) - \log N}\]
    3. Using strategy rm

    4. Applied log1p-udef0.1

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

    5. Applied diff-log0.1

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

    if 7453.229305032585 < N

    1. Initial program 59.5

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

    2. Simplified59.5

      \[\leadsto \color{blue}{\log_* (1 + N) - \log N}\]
    3. Using strategy rm

    4. Applied log1p-udef59.5

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

    5. Applied diff-log59.3

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

    6. Taylor expanded around -inf 0.0

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

    7. Simplified0.0

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

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 7453.229305032585:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) - \frac{\frac{-1}{3}}{N \cdot \left(N \cdot N\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2019119 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))