Average Error: 29.3 → 0.1
Time: 3.8s
Precision: binary64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 9410.2070517314387:\\ \;\;\;\;2 \cdot \log \left(\sqrt[3]{N + 1}\right) + \log \left(\frac{\sqrt[3]{N + 1}}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 9410.2070517314387:\\
\;\;\;\;2 \cdot \log \left(\sqrt[3]{N + 1}\right) + \log \left(\frac{\sqrt[3]{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 code(double N) {
	return ((double) (((double) log(((double) (N + 1.0)))) - ((double) log(N))));
}
↓
double code(double N) {
	double VAR;
	if ((N <= 9410.207051731439)) {
		VAR = ((double) (((double) (2.0 * ((double) log(((double) cbrt(((double) (N + 1.0)))))))) + ((double) log(((double) (((double) cbrt(((double) (N + 1.0)))) / 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 < 9410.2070517314387

    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 *-un-lft-identity0.1

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

    6. Applied add-cube-cbrt0.1

      \[\leadsto \log \left(\frac{\color{blue}{\left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) \cdot \sqrt[3]{N + 1}}}{1 \cdot N}\right)\]

    7. Applied times-frac0.1

      \[\leadsto \log \color{blue}{\left(\frac{\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}}{1} \cdot \frac{\sqrt[3]{N + 1}}{N}\right)}\]

    8. Applied log-prod0.1

      \[\leadsto \color{blue}{\log \left(\frac{\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}}{1}\right) + \log \left(\frac{\sqrt[3]{N + 1}}{N}\right)}\]

    9. Simplified0.1

      \[\leadsto \color{blue}{2 \cdot \log \left(\sqrt[3]{N + 1}\right)} + \log \left(\frac{\sqrt[3]{N + 1}}{N}\right)\]

    if 9410.2070517314387 < 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

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 9410.2070517314387:\\ \;\;\;\;2 \cdot \log \left(\sqrt[3]{N + 1}\right) + \log \left(\frac{\sqrt[3]{N + 1}}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2020150 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))