Average Error: 29.1 → 0.1
Time: 3.3s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 10132.834563235498:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;1 \cdot \left(\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\right)\\ \end{array}\]
\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}:\\
\;\;\;\;1 \cdot \left(\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\right)\\

\end{array}
double f(double N) {
        double r30257 = N;
        double r30258 = 1.0;
        double r30259 = r30257 + r30258;
        double r30260 = log(r30259);
        double r30261 = log(r30257);
        double r30262 = r30260 - r30261;
        return r30262;
}
↓
double f(double N) {
        double r30263 = N;
        double r30264 = 10132.834563235498;
        bool r30265 = r30263 <= r30264;
        double r30266 = 1.0;
        double r30267 = r30263 + r30266;
        double r30268 = r30267 / r30263;
        double r30269 = log(r30268);
        double r30270 = 1.0;
        double r30271 = 2.0;
        double r30272 = pow(r30263, r30271);
        double r30273 = r30270 / r30272;
        double r30274 = 0.3333333333333333;
        double r30275 = r30274 / r30263;
        double r30276 = 0.5;
        double r30277 = r30275 - r30276;
        double r30278 = r30273 * r30277;
        double r30279 = r30266 / r30263;
        double r30280 = r30278 + r30279;
        double r30281 = r30270 * r30280;
        double r30282 = r30265 ? r30269 : r30281;
        return r30282;
}

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

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

    5. Applied *-un-lft-identity0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 10132.834563235498:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;1 \cdot \left(\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\right)\\ \end{array}\]

Reproduce

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