Average Error: 29.7 → 0.1
Time: 14.9s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 7951.823313362932822201400995254516601562:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.3333333333333333148296162562473909929395}{N \cdot \left(N \cdot N\right)}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 7951.823313362932822201400995254516601562:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

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

\end{array}
double f(double N) {
        double r3153217 = N;
        double r3153218 = 1.0;
        double r3153219 = r3153217 + r3153218;
        double r3153220 = log(r3153219);
        double r3153221 = log(r3153217);
        double r3153222 = r3153220 - r3153221;
        return r3153222;
}
↓
double f(double N) {
        double r3153223 = N;
        double r3153224 = 7951.823313362933;
        bool r3153225 = r3153223 <= r3153224;
        double r3153226 = 1.0;
        double r3153227 = r3153226 + r3153223;
        double r3153228 = r3153227 / r3153223;
        double r3153229 = log(r3153228);
        double r3153230 = r3153226 / r3153223;
        double r3153231 = 0.3333333333333333;
        double r3153232 = r3153223 * r3153223;
        double r3153233 = r3153223 * r3153232;
        double r3153234 = r3153231 / r3153233;
        double r3153235 = r3153230 + r3153234;
        double r3153236 = 0.5;
        double r3153237 = r3153236 / r3153232;
        double r3153238 = r3153235 - r3153237;
        double r3153239 = r3153225 ? r3153229 : r3153238;
        return r3153239;
}

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 < 7951.823313362933

    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 7951.823313362933 < 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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 7951.823313362932822201400995254516601562:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.3333333333333333148296162562473909929395}{N \cdot \left(N \cdot N\right)}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019169 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1.0)) (log N)))