2log (problem 3.3.6)

Average Error: 29.3 → 0.1

Time: 4.1s

Precision: 64

Math

\[\log \left(N + 1\right) - \log N\]

↓

\[\begin{array}{l} \mathbf{if}\;N \le 10104.7750084909585:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{N}, 1 - \frac{0.5}{N}, 0.333333333333333315 \cdot \frac{1}{{N}^{3}}\right)\\ \end{array}\]

C

double code(double N) {
	return ((double) (((double) log(((double) (N + 1.0)))) - ((double) log(N))));
}

↓

double code(double N) {
	double VAR;
	if ((N <= 10104.775008490958)) {
		VAR = ((double) log(((double) (((double) (N + 1.0)) / N))));
	} else {
		VAR = ((double) fma(((double) (1.0 / N)), ((double) (1.0 - ((double) (0.5 / N)))), ((double) (0.3333333333333333 * ((double) (1.0 / ((double) pow(N, 3.0))))))));
	}
	return VAR;
}

Error

Bits error versus N

Derivation

1. Split input into 2 regimes

2. if N < 10104.775008490958

   1. Initial program 0.1

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

   3. Applied diff-log 0.1

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

   if 10104.775008490958 < N

   1. Initial program 59.5

      \[\log \left(N + 1\right) - \log N\]

   2. Taylor expanded around inf 0.1

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

   3. Simplified 0.1

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

4. Final simplification 0.1

   \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 10104.7750084909585:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{N}, 1 - \frac{0.5}{N}, 0.333333333333333315 \cdot \frac{1}{{N}^{3}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020113 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))