Average Error: 63.0 → 0
Time: 3.9s
Precision: 64
\[n \gt 6.8 \cdot 10^{15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[1 \cdot \log n + \left(0.5 \cdot \frac{1}{n} - \frac{0.16666666666666669}{{n}^{2}}\right)\]
\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1
↓
1 \cdot \log n + \left(0.5 \cdot \frac{1}{n} - \frac{0.16666666666666669}{{n}^{2}}\right)
double f(double n) {
        double r93578 = n;
        double r93579 = 1.0;
        double r93580 = r93578 + r93579;
        double r93581 = log(r93580);
        double r93582 = r93580 * r93581;
        double r93583 = log(r93578);
        double r93584 = r93578 * r93583;
        double r93585 = r93582 - r93584;
        double r93586 = r93585 - r93579;
        return r93586;
}
↓
double f(double n) {
        double r93587 = 1.0;
        double r93588 = n;
        double r93589 = log(r93588);
        double r93590 = r93587 * r93589;
        double r93591 = 0.5;
        double r93592 = 1.0;
        double r93593 = r93592 / r93588;
        double r93594 = r93591 * r93593;
        double r93595 = 0.16666666666666669;
        double r93596 = 2.0;
        double r93597 = pow(r93588, r93596);
        double r93598 = r93595 / r93597;
        double r93599 = r93594 - r93598;
        double r93600 = r93590 + r93599;
        return r93600;
}

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

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

  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. Taylor expanded around 0 0

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

  5. Simplified0

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

  6. Final simplification0

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

Reproduce

herbie shell --seed 2020065 
(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))