Average Error: 63.0 → 0
Time: 14.4s
Precision: 64
\[n \gt 6.8 \cdot 10^{15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[\left(1 \cdot \log n - \frac{0.1666666666666666851703837437526090070605}{{n}^{2}}\right) + \frac{0.5}{n}\]
\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1
↓
\left(1 \cdot \log n - \frac{0.1666666666666666851703837437526090070605}{{n}^{2}}\right) + \frac{0.5}{n}
double f(double n) {
        double r51215 = n;
        double r51216 = 1.0;
        double r51217 = r51215 + r51216;
        double r51218 = log(r51217);
        double r51219 = r51217 * r51218;
        double r51220 = log(r51215);
        double r51221 = r51215 * r51220;
        double r51222 = r51219 - r51221;
        double r51223 = r51222 - r51216;
        return r51223;
}
↓
double f(double n) {
        double r51224 = 1.0;
        double r51225 = n;
        double r51226 = log(r51225);
        double r51227 = r51224 * r51226;
        double r51228 = 0.16666666666666669;
        double r51229 = 2.0;
        double r51230 = pow(r51225, r51229);
        double r51231 = r51228 / r51230;
        double r51232 = r51227 - r51231;
        double r51233 = 0.5;
        double r51234 = r51233 / r51225;
        double r51235 = r51232 + r51234;
        return r51235;
}

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

    \[\leadsto \color{blue}{\left(\left(0.5 \cdot \frac{1}{n} + 1\right) - \left(1 \cdot \log \left(\frac{1}{n}\right) + 0.1666666666666666851703837437526090070605 \cdot \frac{1}{{n}^{2}}\right)\right)} - 1\]

  3. Simplified0.0

    \[\leadsto \color{blue}{\left(\left(1 - \left(1 \cdot \log \left(\frac{1}{n}\right) + 0.1666666666666666851703837437526090070605 \cdot \frac{1}{{n}^{2}}\right)\right) + \frac{0.5}{n}\right)} - 1\]

  4. Taylor expanded around 0 0

    \[\leadsto \color{blue}{\left(0.5 \cdot \frac{1}{n} + 1 \cdot \log n\right) - 0.1666666666666666851703837437526090070605 \cdot \frac{1}{{n}^{2}}}\]

  5. Simplified0

    \[\leadsto \color{blue}{1 \cdot \log n + \left(0.5 \cdot \frac{1}{n} - \frac{0.1666666666666666851703837437526090070605}{{n}^{2}}\right)}\]

  6. Final simplification0

    \[\leadsto \left(1 \cdot \log n - \frac{0.1666666666666666851703837437526090070605}{{n}^{2}}\right) + \frac{0.5}{n}\]

Reproduce

herbie shell --seed 2019294 
(FPCore (n)
  :name "logs (example 3.8)"
  :precision binary64
  :pre (> n 6.8e15)

  :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))