Timeline for Does log(log(n)) grow asymptotically slower than log(n) / log(log(n))?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 10 at 9:57 | comment | added | SilvioM | @JohnKemeny yes but the derivative of $k$ is $1$, so the ratio of the derivatives is $\frac{(2\log k) / k}{1} = \frac{2\log k}{k}$ | |
| Dec 9 at 22:36 | comment | added | John Kemeny | @KellyBundy the derivative of $\log^2 k$ is $2 \log k / k$. | |
| Dec 9 at 21:27 | comment | added | minh quý lê | @KellyBundy I am assuming that the base of $\log$ is $e$. L'hospital method is applied to infer the step. | |
| Dec 9 at 20:09 | comment | added | Kelly Bundy | Can you explain the step $\lim_{k\rightarrow\infty}\frac{\log^2 k}{k}=\lim_{k\rightarrow\infty}\frac{2\log k}{k}$? | |
| Dec 9 at 19:36 | vote | accept | maybesunny | ||
| Dec 9 at 19:31 | history | answered | minh quý lê | CC BY-SA 4.0 |