Average Error: 29.6 → 0.1
Time: 4.5s
Precision: 64
\[\log \left(N + 1\right) - \log N\]
↓
\[\begin{array}{l} \mathbf{if}\;N \le 8086.425104129199098679237067699432373047:\\ \;\;\;\;\log \left(1 \cdot \frac{1}{N} + 1\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.3333333333333333148296162562473909929395}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]
\log \left(N + 1\right) - \log N
↓
\begin{array}{l}
\mathbf{if}\;N \le 8086.425104129199098679237067699432373047:\\
\;\;\;\;\log \left(1 \cdot \frac{1}{N} + 1\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.3333333333333333148296162562473909929395}{N} - 0.5\right) + \frac{1}{N}\\

\end{array}
double f(double N) {
        double r92298 = N;
        double r92299 = 1.0;
        double r92300 = r92298 + r92299;
        double r92301 = log(r92300);
        double r92302 = log(r92298);
        double r92303 = r92301 - r92302;
        return r92303;
}
↓
double f(double N) {
        double r92304 = N;
        double r92305 = 8086.425104129199;
        bool r92306 = r92304 <= r92305;
        double r92307 = 1.0;
        double r92308 = 1.0;
        double r92309 = r92308 / r92304;
        double r92310 = r92307 * r92309;
        double r92311 = r92310 + r92308;
        double r92312 = log(r92311);
        double r92313 = 2.0;
        double r92314 = pow(r92304, r92313);
        double r92315 = r92308 / r92314;
        double r92316 = 0.3333333333333333;
        double r92317 = r92316 / r92304;
        double r92318 = 0.5;
        double r92319 = r92317 - r92318;
        double r92320 = r92315 * r92319;
        double r92321 = r92307 / r92304;
        double r92322 = r92320 + r92321;
        double r92323 = r92306 ? r92312 : r92322;
        return r92323;
}

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 < 8086.425104129199

    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. Taylor expanded around 0 0.1

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

    if 8086.425104129199 < N

    1. Initial program 59.7

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

    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

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

  4. Final simplification0.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 8086.425104129199098679237067699432373047:\\ \;\;\;\;\log \left(1 \cdot \frac{1}{N} + 1\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{{N}^{2}} \cdot \left(\frac{0.3333333333333333148296162562473909929395}{N} - 0.5\right) + \frac{1}{N}\\ \end{array}\]

Reproduce

herbie shell --seed 2019318 
(FPCore (N)
  :name "2log (problem 3.3.6)"
  :precision binary64
  (- (log (+ N 1)) (log N)))