Average Error: 63.0 → 0.0
Time: 5.7s
Precision: 64
\[n \gt 6.8 \cdot 10^{15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[1 - \left(\left(\left(1 + \frac{0.16666666666666669}{{n}^{2}}\right) + 1 \cdot \log \left(\frac{1}{n}\right)\right) - \frac{0.5}{n}\right)\]
\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1
↓
1 - \left(\left(\left(1 + \frac{0.16666666666666669}{{n}^{2}}\right) + 1 \cdot \log \left(\frac{1}{n}\right)\right) - \frac{0.5}{n}\right)
double f(double n) {
        double r100642 = n;
        double r100643 = 1.0;
        double r100644 = r100642 + r100643;
        double r100645 = log(r100644);
        double r100646 = r100644 * r100645;
        double r100647 = log(r100642);
        double r100648 = r100642 * r100647;
        double r100649 = r100646 - r100648;
        double r100650 = r100649 - r100643;
        return r100650;
}
↓
double f(double n) {
        double r100651 = 1.0;
        double r100652 = 0.16666666666666669;
        double r100653 = n;
        double r100654 = 2.0;
        double r100655 = pow(r100653, r100654);
        double r100656 = r100652 / r100655;
        double r100657 = r100651 + r100656;
        double r100658 = 1.0;
        double r100659 = r100658 / r100653;
        double r100660 = log(r100659);
        double r100661 = r100651 * r100660;
        double r100662 = r100657 + r100661;
        double r100663 = 0.5;
        double r100664 = r100663 / r100653;
        double r100665 = r100662 - r100664;
        double r100666 = r100651 - r100665;
        return r100666;
}

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 associate-+l-0.0

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

  6. Applied associate--l-0.0

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

  7. Simplified0.0

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

  8. Final simplification0.0

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

Reproduce

herbie shell --seed 2020025 
(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))