This is how I would think it n^2 = O(log n)
f(n) <= c*g(n)
should c=1 ?
n0 = 1
n^2 <= log n
0 <= log n - n^2, for all n>=1
we take n=1 then 0 <= log(1) - 1^2
0 <= 0 - 1
0 <= -1
Therefore n^2 = O(logn) false?
Is this correct?
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No, this is false. As for your attempt, you don't get to set $n=1$; you need to show the inequality for all $n \geq n_0$. – Raphael Feb 17 '18 at 17:50
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How exactly? can you give a helping hand ? :) I'm having trouble figuring out proofs – crystyxn Feb 17 '18 at 17:52
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You have to show that no matter how large you choose c, and no matter how large you choose n_0, you will find n ≥ n_0 where n^2 > c log n. You might try to find an n such that n ≥ n_0, n > c, and n > log n. – gnasher729 Feb 17 '18 at 19:09
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@crystyxn You need to look at the definition of O again -- it's all there. As for helping with the task at hand, you'll notice that a lengthy answer at the question I linked you was written by me. – Raphael Feb 17 '18 at 19:31