Average Error: 30.1 → 0.1
Time: 3.3s
Precision: binary64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \leq 5723.256773123576:\\ \;\;\;\;\log \left(\frac{N + 1}{\sqrt{N}}\right) - \log \left(\sqrt{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \leq 5723.256773123576:\\
\;\;\;\;\log \left(\frac{N + 1}{\sqrt{N}}\right) - \log \left(\sqrt{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 5723.256773123576)
   (- (log (/ (+ N 1.0) (sqrt N))) (log (sqrt N)))
   (- (+ (/ 1.0 N) (/ 0.3333333333333333 (pow N 3.0))) (/ 0.5 (* N N)))))
double code(double N) {
	return log(N + 1.0) - log(N);
}
↓
double code(double N) {
	double tmp;
	if (N <= 5723.256773123576) {
		tmp = log((N + 1.0) / sqrt(N)) - log(sqrt(N));
	} else {
		tmp = ((1.0 / N) + (0.3333333333333333 / pow(N, 3.0))) - (0.5 / (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 < 5723.256773123576

    1. Initial program 0.1

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

    3. Applied diff-log_binary64_1710.1

      \[\leadsto \color{blue}{\log \left(\frac{N + 1}{N}\right)}\]
    4. Using strategy rm

    5. Applied add-sqr-sqrt_binary64_1010.1

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

    6. Applied *-un-lft-identity_binary64_790.1

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

    7. Applied times-frac_binary64_850.1

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

    8. Applied log-prod_binary64_1650.1

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

    9. Simplified0.1

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

    if 5723.256773123576 < N

    1. Initial program 59.5

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

    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 simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \leq 5723.256773123576:\\ \;\;\;\;\log \left(\frac{N + 1}{\sqrt{N}}\right) - \log \left(\sqrt{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\left(\frac{1}{N} + \frac{0.3333333333333333}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}\\ \end{array}\]

Reproduce

herbie shell --seed 2020292 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))