Average Error: 29.4 → 0.1
Time: 19.3s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 8103.092234919824477401562035083770751953:\\ \;\;\;\;\log \left(\left(N + 1\right) \cdot \frac{1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 8103.092234919824477401562035083770751953:\\
\;\;\;\;\log \left(\left(N + 1\right) \cdot \frac{1}{N}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\

\end{array}
double f(double N) {
        double r43242 = N;
        double r43243 = 1.0;
        double r43244 = r43242 + r43243;
        double r43245 = log(r43244);
        double r43246 = log(r43242);
        double r43247 = r43245 - r43246;
        return r43247;
}
↓
double f(double N) {
        double r43248 = N;
        double r43249 = 8103.0922349198245;
        bool r43250 = r43248 <= r43249;
        double r43251 = 1.0;
        double r43252 = r43248 + r43251;
        double r43253 = 1.0;
        double r43254 = r43253 / r43248;
        double r43255 = r43252 * r43254;
        double r43256 = log(r43255);
        double r43257 = 0.3333333333333333;
        double r43258 = 3.0;
        double r43259 = pow(r43248, r43258);
        double r43260 = r43257 / r43259;
        double r43261 = 0.5;
        double r43262 = r43261 / r43248;
        double r43263 = r43251 - r43262;
        double r43264 = r43263 / r43248;
        double r43265 = r43260 + r43264;
        double r43266 = r43250 ? r43256 : r43265;
        return r43266;
}

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 < 8103.0922349198245

    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 div-inv0.1

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

    if 8103.0922349198245 < N

    1. Initial program 59.3

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

    2. Taylor expanded around inf 0.0

      \[\leadsto \color{blue}{\left(0.3333333333333333148296162562473909929395 \cdot \frac{1}{{N}^{3}} + 1 \cdot \frac{1}{N}\right) - 0.5 \cdot \frac{1}{{N}^{2}}}\]

    3. Simplified0.0

      \[\leadsto \color{blue}{\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 8103.092234919824477401562035083770751953:\\ \;\;\;\;\log \left(\left(N + 1\right) \cdot \frac{1}{N}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{0.3333333333333333148296162562473909929395}{{N}^{3}} + \frac{1 - \frac{0.5}{N}}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019306 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))