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How to prove that (log logn)×(log log log n) = Ω(logn)

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  1. Aug 15, 2015 #1
    Is ##\log \log n \times \log \log \log n = \Omega(\log n)##
    How can we prove it.

    Actually I'm trying to prove that ##f(n) = \lceil(\log \log n)\rceil !## is polynomially bounded. It means

    ##c_1 n^{k_1} \leq f(n) \leq c_2 n^{k_2} \quad \forall n > n_0##

    ##m_1 \log n \leq \log [f(n)] \leq m_2 \log n##

    ##\log [f(n)]=\theta(\log n) \text{ i.e. } \log [f(n)]=\Omega(\log n) \text{ and }\log [f(n)]=O(\log n)##

    I've proved that ##\log [f(n)] = O(\log n)##, But I'm having trouble proving ##\,\log \left[f(n)\right] = \Omega\left(\log n\right)##. Can anybody tell me how can we do it.
     
  2. jcsd
  3. Aug 15, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    For the original problem, isn't it trivial to find c1, k1 to make the left inequality right?
     
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