Average Error: 29.2 → 0.1
Time: 5.8s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 1.113298076 \cdot 10^{-7}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;\log \left(N + 1\right) - \log N \le 1.113298076 \cdot 10^{-7}:\\
\;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\

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

\end{array}
double f(double N) {
        double r50320 = N;
        double r50321 = 1.0;
        double r50322 = r50320 + r50321;
        double r50323 = log(r50322);
        double r50324 = log(r50320);
        double r50325 = r50323 - r50324;
        return r50325;
}
↓
double f(double N) {
        double r50326 = N;
        double r50327 = 1.0;
        double r50328 = r50326 + r50327;
        double r50329 = log(r50328);
        double r50330 = log(r50326);
        double r50331 = r50329 - r50330;
        double r50332 = 1.1132980759498423e-07;
        bool r50333 = r50331 <= r50332;
        double r50334 = 1.0;
        double r50335 = 2.0;
        double r50336 = pow(r50326, r50335);
        double r50337 = r50334 / r50336;
        double r50338 = 0.3333333333333333;
        double r50339 = r50338 / r50326;
        double r50340 = 0.5;
        double r50341 = r50339 - r50340;
        double r50342 = r50337 * r50341;
        double r50343 = r50327 / r50326;
        double r50344 = r50342 + r50343;
        double r50345 = r50328 / r50326;
        double r50346 = log(r50345);
        double r50347 = r50333 ? r50344 : r50346;
        return r50347;
}

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)) < 1.1132980759498423e-07

    1. Initial program 59.9

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

    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

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

    if 1.1132980759498423e-07 < (- (log (+ N 1.0)) (log N))

    1. Initial program 0.3

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

    3. Applied diff-log0.2

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 1.113298076 \cdot 10^{-7}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020036 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))