Average Error: 29.8 → 0.1
Time: 9.0s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9562.6451805155593:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9562.6451805155593:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\

\end{array}
double f(double N) {
        double r41843 = N;
        double r41844 = 1.0;
        double r41845 = r41843 + r41844;
        double r41846 = log(r41845);
        double r41847 = log(r41843);
        double r41848 = r41846 - r41847;
        return r41848;
}
↓
double f(double N) {
        double r41849 = N;
        double r41850 = 9562.64518051556;
        bool r41851 = r41849 <= r41850;
        double r41852 = 1.0;
        double r41853 = r41849 + r41852;
        double r41854 = r41853 / r41849;
        double r41855 = log(r41854);
        double r41856 = r41852 / r41849;
        double r41857 = 0.3333333333333333;
        double r41858 = 3.0;
        double r41859 = pow(r41849, r41858);
        double r41860 = r41857 / r41859;
        double r41861 = 0.5;
        double r41862 = r41849 * r41849;
        double r41863 = r41861 / r41862;
        double r41864 = r41860 - r41863;
        double r41865 = r41856 + r41864;
        double r41866 = r41851 ? r41855 : r41865;
        return r41866;
}

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 < 9562.64518051556

    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 9562.64518051556 < 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

      \[\leadsto \color{blue}{\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}}\]
    4. Using strategy rm

    5. Applied associate--l+0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9562.6451805155593:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020047 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))