Average Error: 63.0 → 0
Time: 17.2s
Precision: 64
\[n \gt 6.8 \cdot 10^{15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[1 \cdot \log n + \left(\frac{0.5}{n} - \frac{\frac{0.1666666666666666851703837437526090070605}{n}}{n}\right)\]
\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1
↓
1 \cdot \log n + \left(\frac{0.5}{n} - \frac{\frac{0.1666666666666666851703837437526090070605}{n}}{n}\right)
double f(double n) {
        double r76587 = n;
        double r76588 = 1.0;
        double r76589 = r76587 + r76588;
        double r76590 = log(r76589);
        double r76591 = r76589 * r76590;
        double r76592 = log(r76587);
        double r76593 = r76587 * r76592;
        double r76594 = r76591 - r76593;
        double r76595 = r76594 - r76588;
        return r76595;
}
↓
double f(double n) {
        double r76596 = 1.0;
        double r76597 = n;
        double r76598 = log(r76597);
        double r76599 = r76596 * r76598;
        double r76600 = 0.5;
        double r76601 = r76600 / r76597;
        double r76602 = 0.16666666666666669;
        double r76603 = r76602 / r76597;
        double r76604 = r76603 / r76597;
        double r76605 = r76601 - r76604;
        double r76606 = r76599 + r76605;
        return r76606;
}

Error

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

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

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

    \[\leadsto \color{blue}{\left(\left(0.5 \cdot \frac{1}{n} + 1\right) - \left(1 \cdot \log \left(\frac{1}{n}\right) + 0.1666666666666666851703837437526090070605 \cdot \frac{1}{{n}^{2}}\right)\right)} - 1\]

  3. Simplified0.0

    \[\leadsto \color{blue}{\left(\left(1 - \left(1 \cdot \log \left(\frac{1}{n}\right) + 0.1666666666666666851703837437526090070605 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right)} - 1\]
  4. Using strategy rm

  5. Applied inv-pow0.0

    \[\leadsto \left(\left(1 - \left(1 \cdot \log \color{blue}{\left({n}^{-1}\right)} + 0.1666666666666666851703837437526090070605 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1\]

  6. Applied log-pow0.0

    \[\leadsto \left(\left(1 - \left(1 \cdot \color{blue}{\left(-1 \cdot \log n\right)} + 0.1666666666666666851703837437526090070605 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1\]

  7. Taylor expanded around 0 0

    \[\leadsto \color{blue}{\left(0.5 \cdot \frac{1}{n} + 1 \cdot \log n\right) - 0.1666666666666666851703837437526090070605 \cdot \frac{1}{{n}^{2}}}\]

  8. Simplified0

    \[\leadsto \color{blue}{\left(1 \cdot \log n - 0.1666666666666666851703837437526090070605 \cdot \frac{1}{{n}^{2}}\right) + \frac{0.5}{n}}\]

  9. Final simplification0

    \[\leadsto 1 \cdot \log n + \left(\frac{0.5}{n} - \frac{\frac{0.1666666666666666851703837437526090070605}{n}}{n}\right)\]

Reproduce

herbie shell --seed 2019291 
(FPCore (n)
  :name "logs (example 3.8)"
  :precision binary64
  :pre (> n 6.8e15)

  :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))