Average Error: 29.2 → 0.1
Time: 5.0s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 1.113298076 \cdot 10^{-7}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;\log \left(N + 1\right) - \log N \le 1.113298076 \cdot 10^{-7}:\\
\;\;\;\;\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 r62050 = N;
        double r62051 = 1.0;
        double r62052 = r62050 + r62051;
        double r62053 = log(r62052);
        double r62054 = log(r62050);
        double r62055 = r62053 - r62054;
        return r62055;
}
↓
double f(double N) {
        double r62056 = N;
        double r62057 = 1.0;
        double r62058 = r62056 + r62057;
        double r62059 = log(r62058);
        double r62060 = log(r62056);
        double r62061 = r62059 - r62060;
        double r62062 = 1.1132980759498423e-07;
        bool r62063 = r62061 <= r62062;
        double r62064 = 1.0;
        double r62065 = 2.0;
        double r62066 = pow(r62056, r62065);
        double r62067 = r62064 / r62066;
        double r62068 = 0.3333333333333333;
        double r62069 = r62068 / r62056;
        double r62070 = 0.5;
        double r62071 = r62069 - r62070;
        double r62072 = r62067 * r62071;
        double r62073 = r62057 / r62056;
        double r62074 = r62072 + r62073;
        double r62075 = r62058 / r62056;
        double r62076 = log(r62075);
        double r62077 = r62063 ? r62074 : r62076;
        return r62077;
}

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)) < 1.1132980759498423e-07

    1. Initial program 59.9

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

    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 1.1132980759498423e-07 < (- (log (+ N 1.0)) (log N))

    1. Initial program 0.3

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

    3. Applied diff-log0.2

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \le 1.113298076 \cdot 10^{-7}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \mathbf{else}:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020036 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))