2log (problem 3.3.6)

Average Error: 29.4 → 0.1

Time: 5.1s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;N \le 8616.08642354138101:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]

C

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

↓

double code(double N) {
	double temp;
	if ((N <= 8616.086423541381)) {
		temp = log(((N + 1.0) / N));
	} else {
		temp = (((1.0 / pow(N, 2.0)) * ((0.3333333333333333 / N) - 0.5)) + (1.0 / N));
	}
	return temp;
}

Error

Bits error versus N

Derivation

1. Split input into 2 regimes

2. if N < 8616.086423541381

   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 8616.086423541381 < 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. Simplified 0.0

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

4. Final simplification 0.1

   \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 8616.08642354138101:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.333333333333333315}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2020057 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))