Big O relation between $2^n$ and $2^{2n}$

Question

I know that:

If $f(n) = O(g(n))$ , then there are constants $M$ and $x_0$ , such that

$f(n) <= M*g(n), \forall n > n_0$

The other, plain English way of defining it is,

If $f(n)=O(g(n))$ then for large $n$ , $f(n)$ would grow as fast as $g(n)$.

I got confused when comparing $2^n$ with $2^{2n}$. Here , $f(n) = 2^n$ and $g(n) = 2^{2n}$. Clearly , $f(n)$ is smaller than $g(n)$ by a factor of $2^n$. So there will be constants $A$ and $x_0$ such that the first definition above is met.

However, for large $n$ , $2^{2n}$ would grow much faster than $2^n$, leaving $2^n$ far behind. That is $2^{2n}$ won't be an asymptotic/tight bound for $2^n$ .

So, is $2^n = O(2^{2n})$ or not? (or did I just create a confusing situation out of nothing)

Answer 1

The more strong estimate is correct:

$2^n = o(2^{2n})$

Just look at definition of the little-o. For any constant $c \gt 0$ you need to find a constant $n_0$, such that for all $n \ge n_0$ you get $2^n \le c \cdot 2^{2n}$. Apparently, the following inequality will give you this $n_0$:

$n_0 \ge ln(1 / c)$

Answer 2

Remember that $O$ is an upper bound.

$2^n = O(2^{2n})$, but $4^n = 2^{2n} \ne O(2^n)$. I.e., $2^{2n}$ is not within a constant factor of $2^n$.

Answer 3

The statement:

If $f(n)=O(g(n))$ then for large $n$ , $f(n)$ would grow as fast as $g(n)$.

is incorrect, and that's probably where your confusion is. Taking from Raphael's answer in How does one know which notation of time complexity analysis to use? :

$f \in \cal{O}(g)$ means that $f$ grows at most as fast as $g$...

Note "at most as fast," not "as fast." If you want "as fast," you're probably looking for $\Theta$; taking from Raphael's answer again:

$f \in \Theta(g)$ means that $f$ grows about as fast as $g$...

but as you've already probably guessed, $2^n \notin \Theta(2^{2n})$. For $2^n \in \Theta(2^{2n})$ to be true, $2^{2n}$ would itself have to be $\cal{O}$$(2^n)$, which I think you can see is clearly not the case.