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Question is asked with an intent to get/let understand how to verify Asymptotic Θ notation. A Homework question. I am to show that n ≠ Θ(logn)

Solution: Yes, n ≠ Θ(logn).

c1logn ≤ n ≤ c2logn => c2≥n/logn, Ɐ n≥n0 - Impossible

Why c2≥n/logn is not possible?

Dhiraj Aggarwal
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    I'm voting to close this question as off-topic because it's about [cs.se] or [math.se]. – Bernhard Barker Jun 09 '17 at 10:34
  • @Dukeling when u say Off-topic, what do you mean by that? – Dhiraj Aggarwal Jun 09 '17 at 11:06
  • By ["off-topic"](https://stackoverflow.com/help/on-topic) I mean the question is inappropriate for [so] and would be better suited to a different site. But [don't post it on multiple sites](https://meta.stackexchange.com/questions/64068/is-cross-posting-a-question-on-multiple-stack-exchange-sites-permitted-if-the-qu), and probably don't (delete and) reask or migrate it if you already got a satisfactory answer, unless you feel the question might have a lot of future value for others. Some time complexity questions are on topic, but this one's a bit too maths-heavy and code-light for here. – Bernhard Barker Jun 09 '17 at 11:45
  • Guys, it's a complexity based question. I am aware that it is tending towards Maths and less towards code. But it will help beginners to understand. I have started learning things, so please don't demoralise and remove hold. More over if https://stackoverflow.com/questions/2095395/is-logn-%CE%98n-logn can exist, my question is worth being here. – Dhiraj Aggarwal Jun 09 '17 at 15:27

3 Answers3

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Think about it, c2 is a constant if your n gets large to what value will n/log(n) converge?

gue
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Well, 

 f(n) = Θ(g(n))

if and only if there's a finite limit value

 lim f(n)/g(n) = c > 0
   n -> +inf

which means that for any constants c1 < c < c2 we can find n0 such that f(n)/g(n) will be in the [c1..c2] range for all n > n0 (put it differently c1*g(n) < f(n) < c2*g(n) when n > n0). In your case 

 f(n) = n
 g(n) = log(n)

The limit is (let's use L'Hôpital's rule)

lim n/log(n) = lim 1/(1/n) = lim n = +inf
  n -> +inf      n -> +inf     n -> +inf

there's no such a finite constant c (and we can't select any c2 constant such that c < c2).

Dmitry Bychenko
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Simply,

c2 will always be less than n for any value of n which causes a contradiction here. And so c2 ≥ n/logn is never possible.

Dhiraj Aggarwal
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