Average Error: 29.6 → 0.2
Time: 3.7s
Precision: binary64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 1.9252119898283127 \cdot 10^{-09}:\\ \;\;\;\;\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + \frac{1}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;\log \left(N + 1\right) - \log N \leq 1.9252119898283127 \cdot 10^{-09}:\\
\;\;\;\;\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\log \left(1 + \frac{1}{N}\right)\\

\end{array}
(FPCore (N) :precision binary64 (- (log (+ N 1.0)) (log N)))
↓
(FPCore (N)
 :precision binary64
 (if (<= (- (log (+ N 1.0)) (log N)) 1.9252119898283127e-09)
   (- (/ 1.0 N) (* 0.5 (/ 1.0 (pow N 2.0))))
   (log (+ 1.0 (/ 1.0 N)))))
double code(double N) {
	return log(N + 1.0) - log(N);
}
↓
double code(double N) {
	double tmp;
	if ((log(N + 1.0) - log(N)) <= 1.9252119898283127e-09) {
		tmp = (1.0 / N) - (0.5 * (1.0 / pow(N, 2.0)));
	} else {
		tmp = log(1.0 + (1.0 / N));
	}
	return tmp;
}

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 (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N)) < 1.925212e-9

    1. Initial program 60.1

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

    2. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}}\]

    if 1.925212e-9 < (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N))

    1. Initial program 0.4

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

    3. Applied diff-log_binary64_1700.4

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

    4. Taylor expanded around 0 0.4

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

  4. Final simplification0.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 1.9252119898283127 \cdot 10^{-09}:\\ \;\;\;\;\frac{1}{N} - 0.5 \cdot \frac{1}{{N}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + \frac{1}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2021045 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))