Average Error: 63.0 → 0
Time: 5.2s
Precision: binary64
\[n > 6.8 \cdot 10^{+15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[\log n\]
\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1
↓
\log n
(FPCore (n)
 :precision binary64
 (- (- (* (+ n 1.0) (log (+ n 1.0))) (* n (log n))) 1.0))
↓
(FPCore (n) :precision binary64 (log n))
double code(double n) {
	return (((n + 1.0) * log(n + 1.0)) - (n * log(n))) - 1.0;
}
↓
double code(double n) {
	return log(n);
}

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

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

  3. Simplified0.0

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

  4. Taylor expanded around 0 0

    \[\leadsto \color{blue}{\log n}\]

  5. Final simplification0

    \[\leadsto \log n\]

Reproduce

herbie shell --seed 2021023 
(FPCore (n)
  :name "logs (example 3.8)"
  :precision binary64
  :pre (> n 6.8e+15)

  :herbie-target
  (- (log (+ n 1.0)) (- (/ 1.0 (* 2.0 n)) (- (/ 1.0 (* 3.0 (* n n))) (/ 4.0 (pow n 3.0)))))

  (- (- (* (+ n 1.0) (log (+ n 1.0))) (* n (log n))) 1.0))