Average Error: 29.1 → 0.1
Time: 1.0m
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 3.790391140512383572058752179145812988281 \cdot 10^{-6}:\\ \;\;\;\;\left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right) + \frac{\frac{0.3333333333333333148296162562473909929395}{N}}{N \cdot N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;\log \left(N + 1\right) - \log N \le 3.790391140512383572058752179145812988281 \cdot 10^{-6}:\\
\;\;\;\;\left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right) + \frac{\frac{0.3333333333333333148296162562473909929395}{N}}{N \cdot N}\\

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

\end{array}
double f(double N) {
        double r3399337 = N;
        double r3399338 = 1.0;
        double r3399339 = r3399337 + r3399338;
        double r3399340 = log(r3399339);
        double r3399341 = log(r3399337);
        double r3399342 = r3399340 - r3399341;
        return r3399342;
}
↓
double f(double N) {
        double r3399343 = N;
        double r3399344 = 1.0;
        double r3399345 = r3399343 + r3399344;
        double r3399346 = log(r3399345);
        double r3399347 = log(r3399343);
        double r3399348 = r3399346 - r3399347;
        double r3399349 = 3.7903911405123836e-06;
        bool r3399350 = r3399348 <= r3399349;
        double r3399351 = r3399344 / r3399343;
        double r3399352 = 0.5;
        double r3399353 = r3399343 * r3399343;
        double r3399354 = r3399352 / r3399353;
        double r3399355 = r3399351 - r3399354;
        double r3399356 = 0.3333333333333333;
        double r3399357 = r3399356 / r3399343;
        double r3399358 = r3399357 / r3399353;
        double r3399359 = r3399355 + r3399358;
        double r3399360 = r3399345 / r3399343;
        double r3399361 = log(r3399360);
        double r3399362 = r3399350 ? r3399359 : r3399361;
        return r3399362;
}

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)) < 3.7903911405123836e-06

    1. Initial program 59.8

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

    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

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

    if 3.7903911405123836e-06 < (- (log (+ N 1.0)) (log N))

    1. Initial program 0.2

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

    \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 3.790391140512383572058752179145812988281 \cdot 10^{-6}:\\ \;\;\;\;\left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right) + \frac{\frac{0.3333333333333333148296162562473909929395}{N}}{N \cdot N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019200 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1.0)) (log N)))