Average Error: 29.4 → 0.1
Time: 5.1s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 8616.08642354138101:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 8616.08642354138101:\\
\;\;\;\;\log \left(\frac{N + 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 r53566 = N;
        double r53567 = 1.0;
        double r53568 = r53566 + r53567;
        double r53569 = log(r53568);
        double r53570 = log(r53566);
        double r53571 = r53569 - r53570;
        return r53571;
}
↓
double f(double N) {
        double r53572 = N;
        double r53573 = 8616.086423541381;
        bool r53574 = r53572 <= r53573;
        double r53575 = 1.0;
        double r53576 = r53572 + r53575;
        double r53577 = r53576 / r53572;
        double r53578 = log(r53577);
        double r53579 = 1.0;
        double r53580 = 2.0;
        double r53581 = pow(r53572, r53580);
        double r53582 = r53579 / r53581;
        double r53583 = 0.3333333333333333;
        double r53584 = r53583 / r53572;
        double r53585 = 0.5;
        double r53586 = r53584 - r53585;
        double r53587 = r53582 * r53586;
        double r53588 = r53575 / r53572;
        double r53589 = r53587 + r53588;
        double r53590 = r53574 ? r53578 : r53589;
        return r53590;
}

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 < 8616.086423541381

    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 8616.086423541381 < N

    1. Initial program 59.4

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

    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

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 8616.08642354138101:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2020057 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))