Average Error: 29.2 → 0.1
Time: 3.1s
Precision: binary64
\[\log \left(N + 1\right) - \log N\]
\[\begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 1.3125216469234147 \cdot 10^{-06}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.3333333333333333}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\log \left(\sqrt{N + 1}\right) + \log \left(\sqrt{N + 1}\right)\right) - \log N\\ \end{array}\]
\log \left(N + 1\right) - \log N
\begin{array}{l}
\mathbf{if}\;\log \left(N + 1\right) - \log N \leq 1.3125216469234147 \cdot 10^{-06}:\\
\;\;\;\;\frac{1}{N} + \left(\frac{0.3333333333333333}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\

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

\end{array}
(FPCore (N) :precision binary64 (- (log (+ N 1.0)) (log N)))
(FPCore (N)
 :precision binary64
 (if (<= (- (log (+ N 1.0)) (log N)) 1.3125216469234147e-06)
   (+ (/ 1.0 N) (- (/ 0.3333333333333333 (pow N 3.0)) (/ 0.5 (* N N))))
   (- (+ (log (sqrt (+ N 1.0))) (log (sqrt (+ N 1.0)))) (log 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)) <= 1.3125216469234147e-06) {
		tmp = (1.0 / N) + ((0.3333333333333333 / pow(N, 3.0)) - (0.5 / (N * N)));
	} else {
		tmp = (log(sqrt(N + 1.0)) + log(sqrt(N + 1.0))) - log(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)) < 1.312521647e-6

    1. Initial program 59.7

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

      \[\leadsto \color{blue}{\left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}}\]
    4. Using strategy rm
    5. Applied associate--l+_binary64_150.0

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

    if 1.312521647e-6 < (-.f64 (log.f64 (+.f64 N 1)) (log.f64 N))

    1. Initial program 0.2

      \[\log \left(N + 1\right) - \log N\]
    2. Using strategy rm
    3. Applied add-sqr-sqrt_binary64_1000.2

      \[\leadsto \log \color{blue}{\left(\sqrt{N + 1} \cdot \sqrt{N + 1}\right)} - \log N\]
    4. Applied log-prod_binary64_1640.2

      \[\leadsto \color{blue}{\left(\log \left(\sqrt{N + 1}\right) + \log \left(\sqrt{N + 1}\right)\right)} - \log N\]
  3. Recombined 2 regimes into one program.
  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(N + 1\right) - \log N \leq 1.3125216469234147 \cdot 10^{-06}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.3333333333333333}{{N}^{3}} - \frac{0.5}{N \cdot N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\log \left(\sqrt{N + 1}\right) + \log \left(\sqrt{N + 1}\right)\right) - \log N\\ \end{array}\]

Reproduce

herbie shell --seed 2020295 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))