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Given the following functions i need to arrange them in increasing order of growth

a) $2^{2^n}$

b) $2^{n^2}$

c) $n^2 \log n$

d) $n$

e) $n^{2^n}$

My first attempt was to plot the graphs but it didn’t gave the correct answer so I took a look on How do I determine the increasing order of growth of a set of functions?

and calculated the log for all the functions listed above and got

a) $2^{2^n} \Rightarrow 2^n + \log 2$

b) $2^{n^2} \Rightarrow n^2 + \log 2$

c) $n^2 \log n \Rightarrow 2\log n + \log \log n$

d) $n \Rightarrow \log n$

e) $n^{2^n} \Rightarrow 2^n + \log n$

and then i plotted them on graph and got the answer : dcbea

but when i submitted the answer it seems to be incorrect.

What i am doing wrong?

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To answer @sol4me *If you know some better way then please share, i will be glad to know.*

First, never trust a plot. You just saw it may hurt, so even if it *can* help, it’s not a proof.

Then, you must know some basic comparison scales, for example, as $n\rightarrow \infty$, and fixed $a>0, b>0, c>1$ and any real $d$,

$$d \ll \log \log n \ll \log^a n \ll n^b \ll c^n \ll n! \ll n^n$$

Where I write $a_n \ll b_n$ iff $\frac{a_n}{b_n} \rightarrow 0$ (this is **not** a standard notation !)

Of course, there are many other possibles asymptotic comparisons, these are just the most frequent.

You have also some allowed operations, for example,

- if $\xi>1$ is a fixed real and $1 \ll a_n \ll b_n$, then $\xi^{a_n} \ll \xi^{b_n}$.
- if $s_n \ll a_n$ and $s_n \ll b_n$, and if $a_n \ll b_n$, then $a_n + s_n \ll b_n + s_n$.

You prove such things by computing the limit. Taking $\log$ as you did may be very useful (for example in the first case above).

Finally, you have to apply these comparisons to your case, and sometimes it’s a bit tricky, but honestly here it’s not.

- $n^2 \ll 2^n$ so $2^{n^2} \ll 2^{2^n}$
- $2 \ll n$, so $2^{2^n} \ll

n^{2^n}$. If in doubt, write the quotient

$a_n=\left(\frac{2}{n}\right)^{2^n}$, and since

$\frac{2}{n}<\frac{1}{2}$ as soon as $n>4$, you have $a_n \rightarrow

0$ - $1 \ll \log n $, so $n^2 \ll n^2 \log n$, and since $n \ll n^2$, you have $n \ll n^2\log n$.
- $n \ll n^2$ so $2^n \ll 2^{n^2}$, and also $\log n \ll n$ so $n^2 \log n \ll n^3$. Then $n^3 \ll 2^n$, hence $n^2\log n \ll n^3 \ll 2^n \ll 2^{n^2}$, and especially $n^2 \log n \ll 2^{n^2}$.

When you have a doubt, write what the comparison means as a limit, and compute the limit.

And remember, these comparisons are *asymptotic*. Sometimes the smallest $n$ such that the inequality really holds may be *very* large, but it’s nevertheless only a fixed, finite number. You may have inequality for $n> 10^{10^{10}}$ for example, so trying to plot things is often hopeless.

If you want to know more about such methods, there are good readings, such as *Concrete Mathematics*, by Graham, Knuth and Patashnik.

As @Jean-Claude Arbaut mentioned, the problem was in calculating log.

To summarize we need to first calculate the log for all the functions and then plot them for **wide range of number**. I plotted it here

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