Average Error: 30.1 → 0.1
Time: 4.1s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 3753.003344783202:\\ \;\;\;\;\left(\log \left({N}^{3} + {1}^{3}\right) - \log \left(N \cdot N + \left(1 \cdot 1 - N \cdot 1\right)\right)\right) - \log N\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(-0.5, \frac{1}{{N}^{2}}, \mathsf{fma}\left(0.33333333333333337, \frac{1}{{N}^{3}}, \frac{1}{N}\right)\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 3753.003344783202:\\
\;\;\;\;\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}:\\
\;\;\;\;\mathsf{fma}\left(-0.5, \frac{1}{{N}^{2}}, \mathsf{fma}\left(0.33333333333333337, \frac{1}{{N}^{3}}, \frac{1}{N}\right)\right)\\

\end{array}
double code(double N) {
	return (log((N + 1.0)) - log(N));
}
↓
double code(double N) {
	double temp;
	if ((N <= 3753.003344783202)) {
		temp = ((log((pow(N, 3.0) + pow(1.0, 3.0))) - log(((N * N) + ((1.0 * 1.0) - (N * 1.0))))) - log(N));
	} else {
		temp = fma(-0.5, (1.0 / pow(N, 2.0)), fma(0.33333333333333337, (1.0 / pow(N, 3.0)), (1.0 / N)));
	}
	return temp;
}

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 < 3753.003344783202

    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 3753.003344783202 < N

    1. Initial program 59.4

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

    3. Applied flip3-+62.2

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

    4. Applied log-div62.1

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

    5. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\left(0.33333333333333337 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}}\]

    6. Simplified0.0

      \[\leadsto \color{blue}{\mathsf{fma}\left(-0.5, \frac{1}{{N}^{2}}, \mathsf{fma}\left(0.33333333333333337, \frac{1}{{N}^{3}}, \frac{1}{N}\right)\right)}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 3753.003344783202:\\ \;\;\;\;\left(\log \left({N}^{3} + {1}^{3}\right) - \log \left(N \cdot N + \left(1 \cdot 1 - N \cdot 1\right)\right)\right) - \log N\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(-0.5, \frac{1}{{N}^{2}}, \mathsf{fma}\left(0.33333333333333337, \frac{1}{{N}^{3}}, \frac{1}{N}\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020060 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))