2log (problem 3.3.6)

Average Error: 29.5 → 0.1

Time: 18.4s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;N \le 4842.291926013775:\\ \;\;\;\;\log \left(\frac{\sqrt{1 + N}}{N}\right) + \log \left(\sqrt{1 + N}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{3}, \frac{\frac{\frac{1}{N}}{N}}{N}, \mathsf{fma}\left(\frac{-1}{2}, \frac{\frac{1}{N}}{N}, \frac{1}{N}\right)\right)\\ \end{array}\]

C

double f(double N) {
        double r2167963 = N;
        double r2167964 = 1.0;
        double r2167965 = r2167963 + r2167964;
        double r2167966 = log(r2167965);
        double r2167967 = log(r2167963);
        double r2167968 = r2167966 - r2167967;
        return r2167968;
}

↓

double f(double N) {
        double r2167969 = N;
        double r2167970 = 4842.291926013775;
        bool r2167971 = r2167969 <= r2167970;
        double r2167972 = 1.0;
        double r2167973 = r2167972 + r2167969;
        double r2167974 = sqrt(r2167973);
        double r2167975 = r2167974 / r2167969;
        double r2167976 = log(r2167975);
        double r2167977 = log(r2167974);
        double r2167978 = r2167976 + r2167977;
        double r2167979 = 0.3333333333333333;
        double r2167980 = r2167972 / r2167969;
        double r2167981 = r2167980 / r2167969;
        double r2167982 = r2167981 / r2167969;
        double r2167983 = -0.5;
        double r2167984 = fma(r2167983, r2167981, r2167980);
        double r2167985 = fma(r2167979, r2167982, r2167984);
        double r2167986 = r2167971 ? r2167978 : r2167985;
        return r2167986;
}

Error

Bits error versus N

Derivation

1. Split input into 2 regimes

2. if N < 4842.291926013775

   1. Initial program 0.1

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

   2. Simplified 0.1

      \[\leadsto \color{blue}{\mathsf{log1p}\left(N\right) - \log N}\]
   3. Using strategy rm

   4. Applied log1p-udef 0.1

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

   5. Applied diff-log 0.1

      \[\leadsto \color{blue}{\log \left(\frac{1 + N}{N}\right)}\]
   6. Using strategy rm

   7. Applied *-un-lft-identity 0.1

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

   8. Applied add-sqr-sqrt 0.1

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

   9. Applied times-frac 0.1

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

   10. Applied log-prod 0.1

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

   11. Simplified 0.1

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

   if 4842.291926013775 < N

   1. Initial program 59.4

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

   2. Simplified 59.4

      \[\leadsto \color{blue}{\mathsf{log1p}\left(N\right) - \log N}\]

   3. Taylor expanded around inf 0.0

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

   4. Simplified 0.0

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

4. Final simplification 0.1

   \[\leadsto \begin{array}{l} \mathbf{if}\;N \le 4842.291926013775:\\ \;\;\;\;\log \left(\frac{\sqrt{1 + N}}{N}\right) + \log \left(\sqrt{1 + N}\right)\\ \mathbf{else}:\\ \;\;\;\;\mathsf{fma}\left(\frac{1}{3}, \frac{\frac{\frac{1}{N}}{N}}{N}, \mathsf{fma}\left(\frac{-1}{2}, \frac{\frac{1}{N}}{N}, \frac{1}{N}\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2019163 +o rules:numerics
(FPCore (N)
  :name "2log (problem 3.3.6)"
  (- (log (+ N 1)) (log N)))