Average Error: 29.2 → 0.1
Time: 45.5s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9094.170836439409:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) - \frac{\frac{-1}{3}}{N \cdot \left(N \cdot N\right)}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9094.170836439409:\\
\;\;\;\;\log \left(\frac{1 + N}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) - \frac{\frac{-1}{3}}{N \cdot \left(N \cdot N\right)}\\

\end{array}
double f(double N) {
        double r2562066 = N;
        double r2562067 = 1.0;
        double r2562068 = r2562066 + r2562067;
        double r2562069 = log(r2562068);
        double r2562070 = log(r2562066);
        double r2562071 = r2562069 - r2562070;
        return r2562071;
}
↓
double f(double N) {
        double r2562072 = N;
        double r2562073 = 9094.170836439409;
        bool r2562074 = r2562072 <= r2562073;
        double r2562075 = 1.0;
        double r2562076 = r2562075 + r2562072;
        double r2562077 = r2562076 / r2562072;
        double r2562078 = log(r2562077);
        double r2562079 = r2562075 / r2562072;
        double r2562080 = -0.5;
        double r2562081 = r2562072 * r2562072;
        double r2562082 = r2562080 / r2562081;
        double r2562083 = r2562079 + r2562082;
        double r2562084 = -0.3333333333333333;
        double r2562085 = r2562072 * r2562081;
        double r2562086 = r2562084 / r2562085;
        double r2562087 = r2562083 - r2562086;
        double r2562088 = r2562074 ? r2562078 : r2562087;
        return r2562088;
}

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 < 9094.170836439409

    1. Initial program 0.1

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

    2. Simplified0.1

      \[\leadsto \color{blue}{\log_* (1 + N) - \log N}\]
    3. Using strategy rm

    4. Applied log1p-udef0.1

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

    5. Applied diff-log0.1

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

    if 9094.170836439409 < N

    1. Initial program 59.5

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

    2. Simplified59.5

      \[\leadsto \color{blue}{\log_* (1 + N) - \log N}\]

    3. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\left(\frac{1}{3} \cdot \frac{1}{{N}^{3}} + \frac{1}{N}\right) - \frac{1}{2} \cdot \frac{1}{{N}^{2}}}\]

    4. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9094.170836439409:\\ \;\;\;\;\log \left(\frac{1 + N}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{\frac{-1}{2}}{N \cdot N}\right) - \frac{\frac{-1}{3}}{N \cdot \left(N \cdot N\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2019112 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))