Average Error: 63.0 → 0.0
Time: 3.8s
Precision: 64
\[n \gt 6.8 \cdot 10^{15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[\left(\left(1 - \left(1 \cdot \left(\left(-1\right) \cdot \log n\right) + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1\]
\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1
↓
\left(\left(1 - \left(1 \cdot \left(\left(-1\right) \cdot \log n\right) + 0.16666666666666669 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right) - 1
double f(double n) {
        double r93523 = n;
        double r93524 = 1.0;
        double r93525 = r93523 + r93524;
        double r93526 = log(r93525);
        double r93527 = r93525 * r93526;
        double r93528 = log(r93523);
        double r93529 = r93523 * r93528;
        double r93530 = r93527 - r93529;
        double r93531 = r93530 - r93524;
        return r93531;
}
↓
double f(double n) {
        double r93532 = 1.0;
        double r93533 = 1.0;
        double r93534 = -r93533;
        double r93535 = n;
        double r93536 = log(r93535);
        double r93537 = r93534 * r93536;
        double r93538 = r93532 * r93537;
        double r93539 = 0.16666666666666669;
        double r93540 = 2.0;
        double r93541 = pow(r93535, r93540);
        double r93542 = r93533 / r93541;
        double r93543 = r93539 * r93542;
        double r93544 = r93538 + r93543;
        double r93545 = r93532 - r93544;
        double r93546 = 0.5;
        double r93547 = r93546 / r93535;
        double r93548 = r93545 + r93547;
        double r93549 = r93548 - r93532;
        return r93549;
}

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 pow10.0

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

  6. Applied pow-flip0.0

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

  7. Applied log-pow0.0

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

  8. Final simplification0.0

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

Reproduce

herbie shell --seed 2020059 
(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))