Average Error: 29.0 → 0.1
Time: 2.6s
Precision: binary64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 54049.172079891905:\\ \;\;\;\;\left(\log \left({N}^{3} + {1}^{3}\right) - \log \left(N \cdot N + \left(1 \cdot 1 - N \cdot 1\right)\right)\right) - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} \cdot \left(1 - \frac{0.5}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 54049.172079891905:\\
\;\;\;\;\left(\log \left({N}^{3} + {1}^{3}\right) - \log \left(N \cdot N + \left(1 \cdot 1 - N \cdot 1\right)\right)\right) - \log N\\

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

\end{array}
double code(double N) {
	return ((double) (((double) log(((double) (N + 1.0)))) - ((double) log(N))));
}
↓
double code(double N) {
	double VAR;
	if ((N <= 54049.172079891905)) {
		VAR = ((double) (((double) (((double) log(((double) (((double) pow(N, 3.0)) + ((double) pow(1.0, 3.0)))))) - ((double) log(((double) (((double) (N * N)) + ((double) (((double) (1.0 * 1.0)) - ((double) (N * 1.0)))))))))) - ((double) log(N))));
	} else {
		VAR = ((double) (((double) (1.0 / N)) * ((double) (1.0 - ((double) (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 < 54049.172079891905

    1. Initial program 0.1

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

    3. Applied flip3-+0.1

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

    4. Applied log-div0.1

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

    if 54049.172079891905 < N

    1. Initial program 59.5

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

    2. Taylor expanded around -inf 64.0

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

    3. Simplified0.1

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 54049.172079891905:\\ \;\;\;\;\left(\log \left({N}^{3} + {1}^{3}\right) - \log \left(N \cdot N + \left(1 \cdot 1 - N \cdot 1\right)\right)\right) - \log N\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} \cdot \left(1 - \frac{0.5}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020164 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1.0)) (log N)))