Timeline for Does ⌊1/𝑛⌋∈Θ(1/𝑛) or to Ω(log𝑛)
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 16, 2021 at 14:16 | comment | added | gnasher729 | At this point we get into hairsplitting. But calculating floor(1/n) requires checking whether n = 1 or not, so it's not O(0), it's only O(1). floor(1/2n) would always be zero. | |
| Jun 15, 2021 at 12:07 | history | edited | nir shahar | CC BY-SA 4.0 |
added 5 characters in body
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| Jun 15, 2021 at 12:06 | comment | added | nir shahar | @integrator yes | |
| Jun 15, 2021 at 11:05 | comment | added | integrator | "Also, you can easily rule out $\Omega(\log(n))$ since $\log$ is an increasing function, while $1/n$ is a decreasing function." As stated, the argument is false, as you can have $f$ increasing and $g$ decreasing such that $g=\Omega(f)$ (e.g. $f(n)=2-1/n$ and $g(n)=2+1/n$). I'm guessing you meant "increasing to infinity", or something of the sort? | |
| Jun 15, 2021 at 10:11 | history | answered | nir shahar | CC BY-SA 4.0 |