Average Error: 29.7 → 0.1
Time: 10.5s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 3632.9677611080442:\\ \;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log N\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{\frac{0.5}{N}}{N}\right)\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 3632.9677611080442:\\
\;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log N\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{\frac{0.5}{N}}{N}\right)\\

\end{array}
double f(double N) {
        double r31025 = N;
        double r31026 = 1.0;
        double r31027 = r31025 + r31026;
        double r31028 = log(r31027);
        double r31029 = log(r31025);
        double r31030 = r31028 - r31029;
        return r31030;
}
↓
double f(double N) {
        double r31031 = N;
        double r31032 = 3632.967761108044;
        bool r31033 = r31031 <= r31032;
        double r31034 = 1.0;
        double r31035 = r31031 + r31034;
        double r31036 = cbrt(r31035);
        double r31037 = r31036 * r31036;
        double r31038 = log(r31037);
        double r31039 = log(r31036);
        double r31040 = log(r31031);
        double r31041 = r31039 - r31040;
        double r31042 = r31038 + r31041;
        double r31043 = r31034 / r31031;
        double r31044 = 0.3333333333333333;
        double r31045 = 3.0;
        double r31046 = pow(r31031, r31045);
        double r31047 = r31044 / r31046;
        double r31048 = 0.5;
        double r31049 = r31048 / r31031;
        double r31050 = r31049 / r31031;
        double r31051 = r31047 - r31050;
        double r31052 = r31043 + r31051;
        double r31053 = r31033 ? r31042 : r31052;
        return r31053;
}

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 < 3632.967761108044

    1. Initial program 0.1

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

    3. Applied add-cube-cbrt0.1

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

    4. Applied log-prod0.1

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

    5. Applied associate--l+0.1

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

    if 3632.967761108044 < 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

      \[\leadsto \color{blue}{\left(\frac{1}{N} + \frac{0.333333333333333315}{{N}^{3}}\right) - \frac{0.5}{N \cdot N}}\]
    4. Using strategy rm

    5. Applied associate--l+0.0

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

    6. Simplified0.0

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 3632.9677611080442:\\ \;\;\;\;\log \left(\sqrt[3]{N + 1} \cdot \sqrt[3]{N + 1}\right) + \left(\log \left(\sqrt[3]{N + 1}\right) - \log N\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{N} + \left(\frac{0.333333333333333315}{{N}^{3}} - \frac{\frac{0.5}{N}}{N}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020043 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))