- #1
22990atinesh
- 143
- 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.
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.