- #1
evinda
Gold Member
MHB
- 3,836
- 0
Hello! (Wave)
I want to prove the following sentences:
That's what I have tried :
I want to prove the following sentences:
- $ \displaystyle{ \lg^2 n=o(2^{\sqrt{2 \lg n}})}$
- $ \displaystyle{ 2^{2^{n+1}}=\omega(2^{2^n})}$
That's what I have tried :
- We suppose that: $\lg^2 n=o(2^{\sqrt{2 \lg n}})$.
$$$$
So, $\forall c>0, \exists n_0 \geq 1$ such that $\forall n \geq n_0$:
$$\lg^2 n<c \cdot 2^{\sqrt{2 \lg n}}$$
$$$$
Do I have to find now a $n_0$ ? If so,how can I do this? (Thinking)
$$$$ - We suppose that $2^{2^{n+1}}=\omega(2^{2^n})$.
$$$$
Then, $\forall c>0, \exists n_0 \geq 1$ such that $\forall n \geq n_0$:
$$c \cdot 2^{2^{2^n}}<2^{2^{n+1}} \\ \Rightarrow c \cdot 2^{2^{2^n}}<2^{2^n \cdot 2} \\ \Rightarrow c \cdot 2^{2^{2^n}}<(2^{2^n})^2 \\ \Rightarrow c<2^{2^n}\\ \Rightarrow \lg c<2^n \\ \Rightarrow n>\lg (\lg c), \text{ if c>1}$$
$$$$
Now I found a restriction for $c$, but the relation should stand for each $c$. So, have I done something wrong? (Sweating)
Last edited: