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 r96342 = n;
        double r96343 = 1.0;
        double r96344 = r96342 + r96343;
        double r96345 = log(r96344);
        double r96346 = r96344 * r96345;
        double r96347 = log(r96342);
        double r96348 = r96342 * r96347;
        double r96349 = r96346 - r96348;
        double r96350 = r96349 - r96343;
        return r96350;
}
↓
double f(double n) {
        double r96351 = 1.0;
        double r96352 = n;
        double r96353 = log(r96352);
        double r96354 = r96351 * r96353;
        double r96355 = 0.5;
        double r96356 = 1.0;
        double r96357 = r96356 / r96352;
        double r96358 = r96355 * r96357;
        double r96359 = 0.16666666666666669;
        double r96360 = 2.0;
        double r96361 = pow(r96352, r96360);
        double r96362 = r96359 / r96361;
        double r96363 = r96358 - r96362;
        double r96364 = r96354 + r96363;
        return r96364;
}

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 2020057 
(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))