Average Error: 63.0 → 0
Time: 4.5s
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 r83269 = n;
        double r83270 = 1.0;
        double r83271 = r83269 + r83270;
        double r83272 = log(r83271);
        double r83273 = r83271 * r83272;
        double r83274 = log(r83269);
        double r83275 = r83269 * r83274;
        double r83276 = r83273 - r83275;
        double r83277 = r83276 - r83270;
        return r83277;
}
↓
double f(double n) {
        double r83278 = 1.0;
        double r83279 = n;
        double r83280 = log(r83279);
        double r83281 = r83278 * r83280;
        double r83282 = 0.5;
        double r83283 = 1.0;
        double r83284 = r83283 / r83279;
        double r83285 = r83282 * r83284;
        double r83286 = 0.16666666666666669;
        double r83287 = 2.0;
        double r83288 = pow(r83279, r83287);
        double r83289 = r83286 / r83288;
        double r83290 = r83285 - r83289;
        double r83291 = r83281 + r83290;
        return r83291;
}

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 2020036 
(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))