Average Error: 29.6 → 0.1
Time: 3.4s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 5358.03365705759552:\\ \;\;\;\;\log \left(N + 1\right) + \log \left(\frac{1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 5358.03365705759552:\\
\;\;\;\;\log \left(N + 1\right) + \log \left(\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 r47184 = N;
        double r47185 = 1.0;
        double r47186 = r47184 + r47185;
        double r47187 = log(r47186);
        double r47188 = log(r47184);
        double r47189 = r47187 - r47188;
        return r47189;
}
↓
double f(double N) {
        double r47190 = N;
        double r47191 = 5358.0336570575955;
        bool r47192 = r47190 <= r47191;
        double r47193 = 1.0;
        double r47194 = r47190 + r47193;
        double r47195 = log(r47194);
        double r47196 = 1.0;
        double r47197 = r47196 / r47190;
        double r47198 = log(r47197);
        double r47199 = r47195 + r47198;
        double r47200 = 2.0;
        double r47201 = pow(r47190, r47200);
        double r47202 = r47196 / r47201;
        double r47203 = 0.3333333333333333;
        double r47204 = r47203 / r47190;
        double r47205 = 0.5;
        double r47206 = r47204 - r47205;
        double r47207 = r47202 * r47206;
        double r47208 = r47193 / r47190;
        double r47209 = r47207 + r47208;
        double r47210 = r47192 ? r47199 : r47209;
        return r47210;
}

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 < 5358.0336570575955

    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

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

    6. Applied log-prod0.1

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

    if 5358.0336570575955 < 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

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 5358.03365705759552:\\ \;\;\;\;\log \left(N + 1\right) + \log \left(\frac{1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

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