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I was asked to find the tightest upper bound for (logn)^2

What I think should be the right way of doing this is as follows:

Since we know log X < X, we can let X be n^(1/4)

log n^(1/4) < n^(1/4)

log n < 4 * n^(1/4)

(logn)^2 < 4 * n^(1/2)

Thus, (logn)^2 = O(n^(1/2))

Plotting the graph out for it seems to suggest it as well. On a side note, I tried substituting n^(1/8) and realised perhaps an even tighter bound would be O(n^(1/8))

Javier Tan
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  • It is known that $n^\varepsilon$, with any $\varepsilon>0$, grows faster than any postitive power of $\log n$. See https://math.stackexchange.com/q/2857391 – Gary Mar 11 '24 at 01:50
  • Ahh that answers my doubt! Thank you @Gary – Javier Tan Mar 11 '24 at 02:02

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