Average Error: 28.8 → 0.1
Time: 4.5s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 2.35842836585 \cdot 10^{-5}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;\log \left(N + 1\right) - \log N \le 2.35842836585 \cdot 10^{-5}:\\
\;\;\;\;\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 r52178 = N;
        double r52179 = 1.0;
        double r52180 = r52178 + r52179;
        double r52181 = log(r52180);
        double r52182 = log(r52178);
        double r52183 = r52181 - r52182;
        return r52183;
}
↓
double f(double N) {
        double r52184 = N;
        double r52185 = 1.0;
        double r52186 = r52184 + r52185;
        double r52187 = log(r52186);
        double r52188 = log(r52184);
        double r52189 = r52187 - r52188;
        double r52190 = 2.3584283658451e-05;
        bool r52191 = r52189 <= r52190;
        double r52192 = 1.0;
        double r52193 = 2.0;
        double r52194 = pow(r52184, r52193);
        double r52195 = r52192 / r52194;
        double r52196 = 0.3333333333333333;
        double r52197 = r52196 / r52184;
        double r52198 = 0.5;
        double r52199 = r52197 - r52198;
        double r52200 = r52195 * r52199;
        double r52201 = r52185 / r52184;
        double r52202 = r52200 + r52201;
        double r52203 = r52186 / r52184;
        double r52204 = log(r52203);
        double r52205 = r52191 ? r52202 : r52204;
        return r52205;
}

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)) < 2.3584283658451e-05

    1. Initial program 59.6

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

    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 2.3584283658451e-05 < (- (log (+ N 1.0)) (log N))

    1. Initial program 0.1

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

    3. Applied diff-log0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 2.35842836585 \cdot 10^{-5}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020035 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))