Average Error: 63.0 → 0.0
Time: 7.5s
Precision: 64
\[n \gt 6.8 \cdot 10^{15}\]
\[\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1\]
↓
\[1 - \left(\left(\left(1 + \frac{0.16666666666666669}{{n}^{2}}\right) + 1 \cdot \log \left(\frac{1}{n}\right)\right) - \frac{0.5}{n}\right)\]
\left(\left(n + 1\right) \cdot \log \left(n + 1\right) - n \cdot \log n\right) - 1
↓
1 - \left(\left(\left(1 + \frac{0.16666666666666669}{{n}^{2}}\right) + 1 \cdot \log \left(\frac{1}{n}\right)\right) - \frac{0.5}{n}\right)
double f(double n) {
        double r316 = n;
        double r317 = 1.0;
        double r318 = r316 + r317;
        double r319 = log(r318);
        double r320 = r318 * r319;
        double r321 = log(r316);
        double r322 = r316 * r321;
        double r323 = r320 - r322;
        double r324 = r323 - r317;
        return r324;
}
↓
double f(double n) {
        double r325 = 1.0;
        double r326 = 0.16666666666666669;
        double r327 = n;
        double r328 = 2.0;
        double r329 = pow(r327, r328);
        double r330 = r326 / r329;
        double r331 = r325 + r330;
        double r332 = 1.0;
        double r333 = r332 / r327;
        double r334 = log(r333);
        double r335 = r325 * r334;
        double r336 = r331 + r335;
        double r337 = 0.5;
        double r338 = r337 / r327;
        double r339 = r336 - r338;
        double r340 = r325 - r339;
        return r340;
}

Error

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original63.0
Target0.0
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. Using strategy rm

  5. Applied associate-+l-0.0

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

  6. Applied associate--l-0.0

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

  7. Simplified0.0

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

  8. Final simplification0.0

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

Reproduce

herbie shell --seed 2020025 
(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))