Average Error: 63.0 → 0.0
Time: 4.6s
Precision: 64
\[n \gt 6.8 \cdot 10^{15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[\left(\left(1 - \left(1 \cdot \left(\log 1 - \log n\right) + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1\]
\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1
↓
\left(\left(1 - \left(1 \cdot \left(\log 1 - \log n\right) + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1
double f(double n) {
        double r65858 = n;
        double r65859 = 1.0;
        double r65860 = r65858 + r65859;
        double r65861 = log(r65860);
        double r65862 = r65860 * r65861;
        double r65863 = log(r65858);
        double r65864 = r65858 * r65863;
        double r65865 = r65862 - r65864;
        double r65866 = r65865 - r65859;
        return r65866;
}
↓
double f(double n) {
        double r65867 = 1.0;
        double r65868 = 1.0;
        double r65869 = log(r65868);
        double r65870 = n;
        double r65871 = log(r65870);
        double r65872 = r65869 - r65871;
        double r65873 = r65867 * r65872;
        double r65874 = 0.16666666666666669;
        double r65875 = 2.0;
        double r65876 = pow(r65870, r65875);
        double r65877 = r65868 / r65876;
        double r65878 = r65874 * r65877;
        double r65879 = r65873 + r65878;
        double r65880 = r65867 - r65879;
        double r65881 = 0.5;
        double r65882 = r65881 / r65870;
        double r65883 = r65880 + r65882;
        double r65884 = r65883 - r65867;
        return r65884;
}

Error

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original63.0
Target0.0
Herbie0.0
\[\log \left(n + 1\right) - \left(\frac{1}{2 \cdot n} - \left(\frac{1}{3 \cdot \left(n \cdot n\right)} - \frac{4}{{n}^{3}}\right)\right)\]

Derivation

  1. Initial program 63.0

    \[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]

  2. Taylor expanded around inf 0.0

    \[\leadsto \color{blue}{\left(\left(0.5 \cdot \frac{1}{n} + 1\right) - \left(1 \cdot \log \left(\frac{1}{n}\right) + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right)} - 1\]

  3. Simplified0.0

    \[\leadsto \color{blue}{\left(\left(1 - \left(1 \cdot \log \left(\frac{1}{n}\right) + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right)} - 1\]
  4. Using strategy rm

  5. Applied add-exp-log0.0

    \[\leadsto \left(\left(1 - \left(1 \cdot \log \left(\frac{1}{\color{blue}{e^{\log n}}}\right) + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1\]

  6. Applied add-exp-log0.0

    \[\leadsto \left(\left(1 - \left(1 \cdot \log \left(\frac{\color{blue}{e^{\log 1}}}{e^{\log n}}\right) + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1\]

  7. Applied div-exp0.0

    \[\leadsto \left(\left(1 - \left(1 \cdot \log \color{blue}{\left(e^{\log 1 - \log n}\right)} + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1\]

  8. Applied rem-log-exp0.0

    \[\leadsto \left(\left(1 - \left(1 \cdot \color{blue}{\left(\log 1 - \log n\right)} + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1\]

  9. Final simplification0.0

    \[\leadsto \left(\left(1 - \left(1 \cdot \left(\log 1 - \log n\right) + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1\]

Reproduce

herbie shell --seed 2020034 
(FPCore (n)
  :name "logs (example 3.8)"
  :precision binary64
  :pre (> n 6.8e+15)

  :herbie-target
  (- (log (+ n 1)) (- (/ 1 (* 2 n)) (- (/ 1 (* 3 (* n n))) (/ 4 (pow n 3)))))

  (- (- (* (+ n 1) (log (+ n 1))) (* n (log n))) 1))