Average Error: 29.2 → 0.1
Time: 5.5s
Precision: binary64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 4139.10814493664566:\\ \;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 4139.10814493664566:\\
\;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\

\end{array}
double code(double N) {
	return ((double) (((double) log(((double) (N + 1.0)))) - ((double) log(N))));
}
↓
double code(double N) {
	double VAR;
	if ((N <= 4139.108144936646)) {
		VAR = ((double) (((double) exp(((double) log(((double) log(((double) (N + 1.0)))))))) - ((double) log(N))));
	} else {
		VAR = ((double) ((1.0 / N) + ((double) ((0.3333333333333333 / ((double) pow(N, 3.0))) - (0.5 / ((double) (N * N)))))));
	}
	return VAR;
}

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 < 4139.10814493664566

    1. Initial program 0.1

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

    3. Applied add-exp-log0.1

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

    if 4139.10814493664566 < 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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 4139.10814493664566:\\ \;\;\;\;e^{\log \left(\log \left(N + 1\right)\right)} - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020182 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))