Average Error: 28.6 → 0.0
Time: 2.4s
Precision: binary64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \leq 8069.600434568251:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \leq 8069.600434568251:\\
\;\;\;\;\log \left(\frac{N + 1}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\

\end{array}
(FPCore (N) :precision binary64 (- (log (+ N 1.0)) (log N)))
↓
(FPCore (N)
 :precision binary64
 (if (<= N 8069.600434568251)
   (log (/ (+ N 1.0) N))
   (- (+ (/ 1.0 N) (/ 0.3333333333333333 (pow N 3.0))) (/ 0.5 (* N N)))))
double code(double N) {
	return ((double) (((double) log(((double) (N + 1.0)))) - ((double) log(N))));
}
↓
double code(double N) {
	double tmp;
	if ((N <= 8069.600434568251)) {
		tmp = ((double) log((((double) (N + 1.0)) / N)));
	} else {
		tmp = ((double) (((double) ((1.0 / N) + (0.3333333333333333 / ((double) pow(N, 3.0))))) - (0.5 / ((double) (N * 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 N < 8069.60043456825133

    1. Initial program 0.1

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

    3. Applied diff-log_binary640.1

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

    if 8069.60043456825133 < N

    1. Initial program 59.6

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

    2. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\left(0.3333333333333333 \cdot \frac{1}{{N}^{3}} + \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}}\]

    3. Simplified0.0

      \[\leadsto \color{blue}{\left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \leq 8069.600434568251:\\ \;\;\;\;\log \left(\frac{N + 1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]

Reproduce

herbie shell --seed 2020219 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))