Average Error: 63.0 → 0
Time: 3.8s
Precision: 64
\[n \gt 6.8 \cdot 10^{15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[1 \cdot \log n + \left(0.5 \cdot \frac{1}{n} - \frac{0.16666666666666669}{{n}^{2}}\right)\]
\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 r69250 = n;
        double r69251 = 1.0;
        double r69252 = r69250 + r69251;
        double r69253 = log(r69252);
        double r69254 = r69252 * r69253;
        double r69255 = log(r69250);
        double r69256 = r69250 * r69255;
        double r69257 = r69254 - r69256;
        double r69258 = r69257 - r69251;
        return r69258;
}
↓
double f(double n) {
        double r69259 = 1.0;
        double r69260 = n;
        double r69261 = log(r69260);
        double r69262 = r69259 * r69261;
        double r69263 = 0.5;
        double r69264 = 1.0;
        double r69265 = r69264 / r69260;
        double r69266 = r69263 * r69265;
        double r69267 = 0.16666666666666669;
        double r69268 = 2.0;
        double r69269 = pow(r69260, r69268);
        double r69270 = r69267 / r69269;
        double r69271 = r69266 - r69270;
        double r69272 = r69262 + r69271;
        return r69272;
}

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 2020060 
(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))