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Here k>=1 and e >0. This is a homework question and from what i understood is since n^e is a polynomial function it will always grow faster that log^k(n)(log is to the base 2) so ,

log^k(n) = o (n^e)

But when i try to plot the graph at www.wolframalpha.com to confirm my understanding it does not give me the clear picture. Can anyone tell me if there is a better way to analyse this.

Bruno Gelb
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Deeksha
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  • What are `k` and `e` ? Just constants? – Alma Do Apr 10 '14 at 16:45
  • All you need for this is l'Hospitals rule and the fact that if lim n->infty f(n)/g(n) < infty, then f = O(g), and if it's even 0, then f = o(g). Remember that the derivative of log^k(n) is k*log^(k-1)(n)*(1/n). If k is real, just round up and you'll get an upper bound on the limit since log(n) > 1 for almost all n. – G. Bach Apr 10 '14 at 16:49
  • Shouldn't this be on http://cs.stackexchange.com? – The Guy with The Hat Apr 10 '14 at 16:52

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