Average Error: 29.6 → 0.1
Time: 3.8s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 10602.0020683454586:\\ \;\;\;\;\log \left(\left(N + 1\right) \cdot \frac{1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 10602.0020683454586:\\
\;\;\;\;\log \left(\left(N + 1\right) \cdot \frac{1}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\

\end{array}
double f(double N) {
        double r50052 = N;
        double r50053 = 1.0;
        double r50054 = r50052 + r50053;
        double r50055 = log(r50054);
        double r50056 = log(r50052);
        double r50057 = r50055 - r50056;
        return r50057;
}
↓
double f(double N) {
        double r50058 = N;
        double r50059 = 10602.002068345459;
        bool r50060 = r50058 <= r50059;
        double r50061 = 1.0;
        double r50062 = r50058 + r50061;
        double r50063 = 1.0;
        double r50064 = r50063 / r50058;
        double r50065 = r50062 * r50064;
        double r50066 = log(r50065);
        double r50067 = 2.0;
        double r50068 = pow(r50058, r50067);
        double r50069 = r50063 / r50068;
        double r50070 = 0.3333333333333333;
        double r50071 = r50070 / r50058;
        double r50072 = 0.5;
        double r50073 = r50071 - r50072;
        double r50074 = r50069 * r50073;
        double r50075 = r50061 / r50058;
        double r50076 = r50074 + r50075;
        double r50077 = r50060 ? r50066 : r50076;
        return r50077;
}

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 < 10602.002068345459

    1. Initial program 0.1

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

    3. Applied diff-log0.1

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)}\]
    4. Using strategy rm

    5. Applied div-inv0.1

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

    if 10602.002068345459 < 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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 10602.0020683454586:\\ \;\;\;\;\log \left(\left(N + 1\right) \cdot \frac{1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2020024 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))