2log (problem 3.3.6)

Average Error: 29.1 → 0.1

Time: 4.6s

Precision: binary64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;N \le 236675.39451860223:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.5}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]

C

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

↓

double code(double N) {
	double VAR;
	if ((N <= 236675.39451860223)) {
		VAR = ((double) log(((double) (((double) (N + 1.0)) / N))));
	} else {
		VAR = ((double) (((double) (((double) (1.0 / ((double) pow(N, 2.0)))) * ((double) (((double) (0.5 / N)) - 0.5)))) + ((double) (1.0 / N))));
	}
	return VAR;
}

Error

Bits error versus N

Derivation

1. Split input into 2 regimes

2. if N < 236675.39451860223

   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 236675.39451860223 < N

   1. Initial program 59.7

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

   2. Taylor expanded around inf 0.1

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

   3. Simplified 0.1

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

4. Final simplification 0.1

   \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 236675.39451860223:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.5}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2020147 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))