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

  • #1
142
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.
 

Answers and Replies

  • #2
34,474
10,597
For the original problem, isn't it trivial to find c1, k1 to make the left inequality right?
 

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

  • Last Post
Replies
4
Views
684
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
575
  • Last Post
Replies
2
Views
1K
Replies
16
Views
4K
Replies
3
Views
1K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
Top