Average Error: 29.4 → 0.1
Time: 4.2s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 5578.40769447651019:\\ \;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log N\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{{N}^{2}}, \frac{0.333333333333333315}{N} - 0.5, \frac{1}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 5578.40769447651019:\\
\;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log N\right)\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(\frac{1}{{N}^{2}}, \frac{0.333333333333333315}{N} - 0.5, \frac{1}{N}\right)\\

\end{array}
double code(double N) {
	return (log((N + 1.0)) - log(N));
}
↓
double code(double N) {
	double temp;
	if ((N <= 5578.40769447651)) {
		temp = (log((cbrt((N + 1.0)) * cbrt((N + 1.0)))) + (log(cbrt((N + 1.0))) - log(N)));
	} else {
		temp = fma((1.0 / pow(N, 2.0)), ((0.3333333333333333 / N) - 0.5), (1.0 / N));
	}
	return temp;
}

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 < 5578.40769447651

    1. Initial program 0.1

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

    3. Applied add-cube-cbrt0.1

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

    4. Applied log-prod0.1

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

    5. Applied associate--l+0.1

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

    if 5578.40769447651 < 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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 5578.40769447651019:\\ \;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log N\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{{N}^{2}}, \frac{0.333333333333333315}{N} - 0.5, \frac{1}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020058 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))