Average Error: 63.0 → 0.0
Time: 4.2s
Precision: 64
\[n \gt 6.8 \cdot 10^{15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[\left(0.5 \cdot \frac{1}{n} - \frac{0.16666666666666669}{{n}^{2}}\right) - 1 \cdot \log \left(\frac{1}{n}\right)\]
\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1
↓
\left(0.5 \cdot \frac{1}{n} - \frac{0.16666666666666669}{{n}^{2}}\right) - 1 \cdot \log \left(\frac{1}{n}\right)
double f(double n) {
        double r129244 = n;
        double r129245 = 1.0;
        double r129246 = r129244 + r129245;
        double r129247 = log(r129246);
        double r129248 = r129246 * r129247;
        double r129249 = log(r129244);
        double r129250 = r129244 * r129249;
        double r129251 = r129248 - r129250;
        double r129252 = r129251 - r129245;
        return r129252;
}
↓
double f(double n) {
        double r129253 = 0.5;
        double r129254 = 1.0;
        double r129255 = n;
        double r129256 = r129254 / r129255;
        double r129257 = r129253 * r129256;
        double r129258 = 0.16666666666666669;
        double r129259 = 2.0;
        double r129260 = pow(r129255, r129259);
        double r129261 = r129258 / r129260;
        double r129262 = r129257 - r129261;
        double r129263 = 1.0;
        double r129264 = log(r129256);
        double r129265 = r129263 * r129264;
        double r129266 = r129262 - r129265;
        return r129266;
}

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

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

  4. Taylor expanded around inf 0.0

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

  5. Simplified0.0

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

  6. Final simplification0.0

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

Reproduce

herbie shell --seed 2020062 
(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))