Average Error: 29.1 → 0.1
Time: 6.1s
Precision: binary64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 5884.9941336213851:\\ \;\;\;\;\left(\log \left(\sqrt{\sqrt{\frac{N + 1}{N}}}\right) + \log \left(\sqrt{\sqrt{\frac{N + 1}{N}}}\right)\right) + \log \left(\frac{\sqrt{N + 1}}{\sqrt{N}}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 5884.9941336213851:\\
\;\;\;\;\left(\log \left(\sqrt{\sqrt{\frac{N + 1}{N}}}\right) + \log \left(\sqrt{\sqrt{\frac{N + 1}{N}}}\right)\right) + \log \left(\frac{\sqrt{N + 1}}{\sqrt{N}}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\

\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 <= 5884.994133621385)) {
		VAR = ((double) (((double) (((double) log(((double) sqrt(((double) sqrt(((double) (((double) (N + 1.0)) / N)))))))) + ((double) log(((double) sqrt(((double) sqrt(((double) (((double) (N + 1.0)) / N)))))))))) + ((double) log(((double) (((double) sqrt(((double) (N + 1.0)))) / ((double) sqrt(N))))))));
	} else {
		VAR = ((double) (((double) (((double) (1.0 / ((double) pow(N, 2.0)))) * ((double) (((double) (0.3333333333333333 / N)) - 0.5)))) + ((double) (1.0 / 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 < 5884.994133621385

    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 add-sqr-sqrt0.1

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

    6. Applied log-prod0.1

      \[\leadsto \color{blue}{\log \left(\sqrt{\frac{N + 1}{N}}\right) + \log \left(\sqrt{\frac{N + 1}{N}}\right)}\]
    7. Using strategy rm

    8. Applied sqrt-div0.1

      \[\leadsto \log \left(\sqrt{\frac{N + 1}{N}}\right) + \log \color{blue}{\left(\frac{\sqrt{N + 1}}{\sqrt{N}}\right)}\]
    9. Using strategy rm

    10. Applied add-sqr-sqrt0.1

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

    11. Applied sqrt-prod0.1

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

    12. Applied log-prod0.1

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

    if 5884.994133621385 < 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

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 5884.9941336213851:\\ \;\;\;\;\left(\log \left(\sqrt{\sqrt{\frac{N + 1}{N}}}\right) + \log \left(\sqrt{\sqrt{\frac{N + 1}{N}}}\right)\right) + \log \left(\frac{\sqrt{N + 1}}{\sqrt{N}}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2020148 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))