Average Error: 29.1 → 0.1
Time: 18.5s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 3775.896918214588367845863103866577148438:\\ \;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 3775.896918214588367845863103866577148438:\\
\;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\

\mathbf{else}:\\
\;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right)\\

\end{array}
double f(double N) {
        double r35124 = N;
        double r35125 = 1.0;
        double r35126 = r35124 + r35125;
        double r35127 = log(r35126);
        double r35128 = log(r35124);
        double r35129 = r35127 - r35128;
        return r35129;
}
↓
double f(double N) {
        double r35130 = N;
        double r35131 = 3775.8969182145884;
        bool r35132 = r35130 <= r35131;
        double r35133 = 1.0;
        double r35134 = r35130 + r35133;
        double r35135 = log(r35134);
        double r35136 = log(r35135);
        double r35137 = exp(r35136);
        double r35138 = log(r35130);
        double r35139 = r35137 - r35138;
        double r35140 = 0.3333333333333333;
        double r35141 = 3.0;
        double r35142 = pow(r35130, r35141);
        double r35143 = r35140 / r35142;
        double r35144 = r35133 / r35130;
        double r35145 = 0.5;
        double r35146 = r35130 * r35130;
        double r35147 = r35145 / r35146;
        double r35148 = r35144 - r35147;
        double r35149 = r35143 + r35148;
        double r35150 = r35132 ? r35139 : r35149;
        return r35150;
}

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 < 3775.8969182145884

    1. Initial program 0.1

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

    3. Applied add-exp-log0.1

      \[\leadsto \color{blue}{e^{\log \left(\log \left(N + 1\right)\right)}} - \log N\]

    if 3775.8969182145884 < 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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 3775.896918214588367845863103866577148438:\\ \;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \left(\frac{1}{N} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019323 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))