Average Error: 29.9 → 0.0
Time: 2.7s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 10135.618383580597:\\ \;\;\;\;-\log \left(\frac{N}{N + 1}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} \cdot \left(\frac{0.333333333333333315}{{N}^{2}} + \left(1 - \frac{0.5}{N}\right)\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 10135.618383580597:\\
\;\;\;\;-\log \left(\frac{N}{N + 1}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{N} \cdot \left(\frac{0.333333333333333315}{{N}^{2}} + \left(1 - \frac{0.5}{N}\right)\right)\\

\end{array}
double code(double N) {
	return (log((N + 1.0)) - log(N));
}
↓
double code(double N) {
	double VAR;
	if ((N <= 10135.618383580597)) {
		VAR = -log((N / (N + 1.0)));
	} else {
		VAR = ((1.0 / N) * ((0.3333333333333333 / pow(N, 2.0)) + (1.0 - (0.5 / N))));
	}
	return VAR;
}

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 < 10135.618383580597

    1. Initial program 0.1

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

    3. Applied diff-log0.1

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

    5. Applied clear-num0.1

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

    6. Applied log-rec0.1

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

    if 10135.618383580597 < N

    1. Initial program 59.5

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

    2. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\left(0.333333333333333315 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}}\]

    3. Simplified0.0

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

  4. Final simplification0.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 10135.618383580597:\\ \;\;\;\;-\log \left(\frac{N}{N + 1}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} \cdot \left(\frac{0.333333333333333315}{{N}^{2}} + \left(1 - \frac{0.5}{N}\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020078 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))