Average Error: 29.6 → 0.1
Time: 5.5s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 3335.45546527291162:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\log \left(N + 1\right)\right)\right) - \log N\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(1, \frac{1}{N}, \mathsf{fma}\left(0.333333333333333315, \frac{1}{\left(1 - \log \left(\frac{1}{N}\right)\right) \cdot {N}^{3}}, 5.55112 \cdot 10^{-17} \cdot \frac{\log \left(\frac{1}{N}\right)}{{\left(1 - \log \left(\frac{1}{N}\right)\right)}^{3} \cdot {N}^{3}}\right) - \mathsf{fma}\left(5.55112 \cdot 10^{-17}, \frac{1}{{\left(1 - \log \left(\frac{1}{N}\right)\right)}^{3} \cdot {N}^{3}}, \mathsf{fma}\left(0.333333333333333315, \frac{\log \left(\frac{1}{N}\right)}{\left(1 - \log \left(\frac{1}{N}\right)\right) \cdot {N}^{3}}, 0.5 \cdot \frac{1}{{N}^{2}}\right)\right)\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 3335.45546527291162:\\
\;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\log \left(N + 1\right)\right)\right) - \log N\\

\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(1, \frac{1}{N}, \mathsf{fma}\left(0.333333333333333315, \frac{1}{\left(1 - \log \left(\frac{1}{N}\right)\right) \cdot {N}^{3}}, 5.55112 \cdot 10^{-17} \cdot \frac{\log \left(\frac{1}{N}\right)}{{\left(1 - \log \left(\frac{1}{N}\right)\right)}^{3} \cdot {N}^{3}}\right) - \mathsf{fma}\left(5.55112 \cdot 10^{-17}, \frac{1}{{\left(1 - \log \left(\frac{1}{N}\right)\right)}^{3} \cdot {N}^{3}}, \mathsf{fma}\left(0.333333333333333315, \frac{\log \left(\frac{1}{N}\right)}{\left(1 - \log \left(\frac{1}{N}\right)\right) \cdot {N}^{3}}, 0.5 \cdot \frac{1}{{N}^{2}}\right)\right)\right)\\

\end{array}
double code(double N) {
	return (log((N + 1.0)) - log(N));
}
↓
double code(double N) {
	double temp;
	if ((N <= 3335.4554652729116)) {
		temp = (expm1(log1p(log((N + 1.0)))) - log(N));
	} else {
		temp = fma(1.0, (1.0 / N), (fma(0.3333333333333333, (1.0 / ((1.0 - log((1.0 / N))) * pow(N, 3.0))), (5.551115123125783e-17 * (log((1.0 / N)) / (pow((1.0 - log((1.0 / N))), 3.0) * pow(N, 3.0))))) - fma(5.551115123125783e-17, (1.0 / (pow((1.0 - log((1.0 / N))), 3.0) * pow(N, 3.0))), fma(0.3333333333333333, (log((1.0 / N)) / ((1.0 - log((1.0 / N))) * pow(N, 3.0))), (0.5 * (1.0 / pow(N, 2.0)))))));
	}
	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 < 3335.4554652729116

    1. Initial program 0.1

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

    3. Applied expm1-log1p-u0.1

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\log \left(N + 1\right)\right)\right)} - \log N\]

    if 3335.4554652729116 < N

    1. Initial program 59.4

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

    3. Applied expm1-log1p-u60.3

      \[\leadsto \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\log \left(N + 1\right)\right)\right)} - \log N\]

    4. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\left(1 \cdot \frac{1}{N} + \left(0.333333333333333315 \cdot \frac{1}{\left(1 - \log \left(\frac{1}{N}\right)\right) \cdot {N}^{3}} + 5.55112 \cdot 10^{-17} \cdot \frac{\log \left(\frac{1}{N}\right)}{{\left(1 - \log \left(\frac{1}{N}\right)\right)}^{3} \cdot {N}^{3}}\right)\right) - \left(5.55112 \cdot 10^{-17} \cdot \frac{1}{{\left(1 - \log \left(\frac{1}{N}\right)\right)}^{3} \cdot {N}^{3}} + \left(0.333333333333333315 \cdot \frac{\log \left(\frac{1}{N}\right)}{\left(1 - \log \left(\frac{1}{N}\right)\right) \cdot {N}^{3}} + 0.5 \cdot \frac{1}{{N}^{2}}\right)\right)}\]

    5. Simplified0.0

      \[\leadsto \color{blue}{\mathsf{fma}\left(1, \frac{1}{N}, \mathsf{fma}\left(0.333333333333333315, \frac{1}{\left(1 - \log \left(\frac{1}{N}\right)\right) \cdot {N}^{3}}, 5.55112 \cdot 10^{-17} \cdot \frac{\log \left(\frac{1}{N}\right)}{{\left(1 - \log \left(\frac{1}{N}\right)\right)}^{3} \cdot {N}^{3}}\right) - \mathsf{fma}\left(5.55112 \cdot 10^{-17}, \frac{1}{{\left(1 - \log \left(\frac{1}{N}\right)\right)}^{3} \cdot {N}^{3}}, \mathsf{fma}\left(0.333333333333333315, \frac{\log \left(\frac{1}{N}\right)}{\left(1 - \log \left(\frac{1}{N}\right)\right) \cdot {N}^{3}}, 0.5 \cdot \frac{1}{{N}^{2}}\right)\right)\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 3335.45546527291162:\\ \;\;\;\;\mathsf{expm1}\left(\mathsf{log1p}\left(\log \left(N + 1\right)\right)\right) - \log N\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(1, \frac{1}{N}, \mathsf{fma}\left(0.333333333333333315, \frac{1}{\left(1 - \log \left(\frac{1}{N}\right)\right) \cdot {N}^{3}}, 5.55112 \cdot 10^{-17} \cdot \frac{\log \left(\frac{1}{N}\right)}{{\left(1 - \log \left(\frac{1}{N}\right)\right)}^{3} \cdot {N}^{3}}\right) - \mathsf{fma}\left(5.55112 \cdot 10^{-17}, \frac{1}{{\left(1 - \log \left(\frac{1}{N}\right)\right)}^{3} \cdot {N}^{3}}, \mathsf{fma}\left(0.333333333333333315, \frac{\log \left(\frac{1}{N}\right)}{\left(1 - \log \left(\frac{1}{N}\right)\right) \cdot {N}^{3}}, 0.5 \cdot \frac{1}{{N}^{2}}\right)\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020065 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))