Average Error: 29.6 → 0.1
Time: 14.3s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 2.270642590573856978153344243764877319336 \cdot 10^{-5}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;\log \left(N + 1\right) - \log N \le 2.270642590573856978153344243764877319336 \cdot 10^{-5}:\\
\;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right)\\

\mathbf{else}:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\

\end{array}
double f(double N) {
        double r60890 = N;
        double r60891 = 1.0;
        double r60892 = r60890 + r60891;
        double r60893 = log(r60892);
        double r60894 = log(r60890);
        double r60895 = r60893 - r60894;
        return r60895;
}
↓
double f(double N) {
        double r60896 = N;
        double r60897 = 1.0;
        double r60898 = r60896 + r60897;
        double r60899 = log(r60898);
        double r60900 = log(r60896);
        double r60901 = r60899 - r60900;
        double r60902 = 2.270642590573857e-05;
        bool r60903 = r60901 <= r60902;
        double r60904 = 0.3333333333333333;
        double r60905 = 3.0;
        double r60906 = pow(r60896, r60905);
        double r60907 = r60904 / r60906;
        double r60908 = r60897 / r60896;
        double r60909 = 0.5;
        double r60910 = r60896 * r60896;
        double r60911 = r60909 / r60910;
        double r60912 = r60908 - r60911;
        double r60913 = r60907 + r60912;
        double r60914 = r60898 / r60896;
        double r60915 = log(r60914);
        double r60916 = r60903 ? r60913 : r60915;
        return r60916;
}

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 (- (log (+ N 1.0)) (log N)) < 2.270642590573857e-05

    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

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

    if 2.270642590573857e-05 < (- (log (+ N 1.0)) (log N))

    1. Initial program 0.1

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

    3. Applied diff-log0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 2.270642590573856978153344243764877319336 \cdot 10^{-5}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019235 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))