Average Error: 63.0 → 0.0
Time: 2.9s
Precision: 64
\[n \gt 6.8 \cdot 10^{15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[\left(\left(0.5 \cdot \frac{1}{n} + 1\right) + \left(\log n \cdot 1 - \frac{\frac{0.16666666666666669}{n}}{n}\right)\right) - 1\]
\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1
↓
\left(\left(0.5 \cdot \frac{1}{n} + 1\right) + \left(\log n \cdot 1 - \frac{\frac{0.16666666666666669}{n}}{n}\right)\right) - 1
double code(double n) {
	return ((((n + 1.0) * log((n + 1.0))) - (n * log(n))) - 1.0);
}
↓
double code(double n) {
	return ((((0.5 * (1.0 / n)) + 1.0) + ((log(n) * 1.0) - ((0.16666666666666669 / n) / n))) - 1.0);
}

Error

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original63.0
Target0
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

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

  4. Final simplification0.0

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

Reproduce

herbie shell --seed 2020078 
(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))