Is it true?

Computer Science Level 2

True or False?

$\large 2^{2n} = O(2^n)$

False True

1 solution

Muhammad Rasel Parvej
Dec 1, 2016

The common error is the subconscious assumption that to verify this relation, we have to find a positive constant $C$ such that $0 \leq 2^{2n} \leq 2^{Cn}$ , for $n\geq n_0$ , where $n_0$ is another positive constant.

But actually we have to find a positive constant $C$ such that $0 \leq 2^{2n} \leq C2^n$ , for $n\geq n_0$ , where $n_0$ is another positive constant.

For the sake of Contradiction, let's assume such $C$ exist so that $0 \leq 2^{2n} \leq C2^n$ .

But $2^{2n} \leq C2^n \implies 2^n \leq C$ , for ALL $n$ greater than some positive constant $n_0$ (Dividing both sides by $2^n$ ). Obviously such a $C$ can't exist.

So, the answer is False .

The common error is the subconscious assumption that to verify this relation

Is this a common misconception that people have?
If yes, why do they do it? Is it that they put the "C" in front of "n"?

Calvin Lin Staff - 4 years, 6 months ago

Yeah, in my experience, I've seen too many people who thinks putting $C$ just before $n$ suffices.

Muhammad Rasel Parvej - 4 years, 6 months ago

And this particular example can be used as counter-example against each of following propositions

$f(n)= \theta (g(n)) \implies 2^{f(n)}= \theta (2^{g(n)})$ .
$f(n)= \theta (g(n)) \implies h(f(n))= \theta (h(g(n))$
$f(cn)= \theta( f(n))$ for any positive constant $C$ .

The first and third ones can be considered as special cases of the second one.

Each of those three propositions may seem intuitively fine and obvious . Most of the people, again in my experience, have found each of those propositions so obvious that they haven't feel any need to verify it mathematically.

And if they use any of those three to solve this problem, they'll get it wrong.

Muhammad Rasel Parvej - 4 years, 6 months ago

Is it true?

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