Average Error: 29.1 → 0.1
Time: 11.0s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 10132.834563235498:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 10132.834563235498:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\

\end{array}
double f(double N) {
        double r32970 = N;
        double r32971 = 1.0;
        double r32972 = r32970 + r32971;
        double r32973 = log(r32972);
        double r32974 = log(r32970);
        double r32975 = r32973 - r32974;
        return r32975;
}
↓
double f(double N) {
        double r32976 = N;
        double r32977 = 10132.834563235498;
        bool r32978 = r32976 <= r32977;
        double r32979 = 1.0;
        double r32980 = r32976 + r32979;
        double r32981 = r32980 / r32976;
        double r32982 = log(r32981);
        double r32983 = r32979 / r32976;
        double r32984 = 0.3333333333333333;
        double r32985 = 3.0;
        double r32986 = pow(r32976, r32985);
        double r32987 = r32984 / r32986;
        double r32988 = r32983 + r32987;
        double r32989 = 0.5;
        double r32990 = r32976 * r32976;
        double r32991 = r32989 / r32990;
        double r32992 = r32988 - r32991;
        double r32993 = r32978 ? r32982 : r32992;
        return r32993;
}

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 < 10132.834563235498

    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 10132.834563235498 < N

    1. Initial program 59.5

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

    2. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\left(0.333333333333333315 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}}\]

    3. Simplified0.0

      \[\leadsto \color{blue}{\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 10132.834563235498:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]

Reproduce

herbie shell --seed 2020046 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))