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I was wondering how people go about showing the proofs with orders of growth? Currently, I have the following functions and I know what order they go in, but I'm not sure how to prove them. I simply put them into a graphing application I saw online and the graph itself obviously showed the growth.

$\begin{align} & 2^{\sqrt{\log n}}\\ & 2^n\\ & n^{4/3}\\ & n \log^3 n\\ & n^{ \log n}\\ & 2^{2^n}\\ & 2^{n^2} \end{align}$

For example, I know that $2^{2^n}$ has the largest growth rate, with $2^{n^2}$ coming in second. I know this because if I plug the value of $5$ it turns into $2^{32}$ vs $2^{25}$, $6$ is $2^{64}$ vs $2^{36}$. But, is that the proof itself?

As far as the odd ones out, I have no idea how I could show a proof comparison of $n^{4/3}$ vs $n\log^3 $n vs $2^n$, etc.

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    @ChrisTaylor, I'm not *as* lost, but still a bit. I guess I know the order, but the whole part of showing a proof to it outside of just plugging in values gets me. And, the mixed bag of types makes it harder as well. If it were just$2^n$vs 2^n^2 vs 2^lg(n) that would be easy...2011-09-19

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General Rules of Thumb which hold for sufficiently large $n$:

$\begin{align} &\log^c(n) \leq n^{\epsilon}, \quad \text{ for any } \epsilon > 0, \quad \text { and any constant } c \\ \\ &n^k \leq k^n, \qquad \text{where } k \text{ is any constant } k > 1 \end{align} $

As Chris noted above, we can write this as

$ A \ll \log^c (n) \ll n^{\epsilon} \ll k^n $

To actually prove these rules, you just compute limits (assuming you know calculus). To see that $\log (n) \ll n^{\varepsilon}$ for example, just note that

$ \lim_{n \to \infty} \frac{\log n}{n^{\epsilon}} = \lim_{n \to \infty} \frac{n^{-1}}{\epsilon n^{\epsilon - 1}} = 0 = \lim_{n \to \infty} \frac{1}{\epsilon n^{\epsilon}} = 0. $

where $A$ and $c$ are constants, $\epsilon > 0$, and $k$ is a real number such that $k > 1$.

If you want to show that $\log^c(n) \ll n^{\epsilon}$ for any positive integer $c$, then just note that

$ \lim_{n \to \infty} \frac{\log^c (n)}{n^\epsilon} \leq \left( \lim_{n \to \infty} \frac{\log (n)}{n^{\epsilon/c}} \right) \dots \left( \lim_{n \to \infty} \frac{\log(n)}{n^{\epsilon/ c}}\right) = 0 $ by what we showed above. We can actually take $c$ be to be any arbitrary real number and the above still holds. This is easy to see since every real number is less than or equal to some integer (infinitely many in fact!). I will leave it to you to show that $n \log^3 (n) \ll n^{4/3}$, for example. The more complicated functions work the same way but we might need to be slightly creative when evaluating the limits.

Hope this helps.