Hacker News
points by punty 2 days ago | 0 comments
2^(n/logn) will eventually be smaller than (1+epsilon)^n for any fixed positive epsilon. However, as log(n) grows quite slowly n will have to be exceptionally large. For epsilon=2^-D n will have to be something like 2^(2^D) large.
Aún no hay comentarios.
Hacker News