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Is $2^{2n} = O(2^n)$?

My solution is:

$2^n 2^n \leq C_{1}2^n$

$2^n \leq C_{1}$,

TRUE.

Is this correct?

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3  
You're saying all powers of two are bounded by a constant? –  anon Apr 13 '12 at 19:04
    
Please, explain me, I don't understand what is what I must to prove, Thanks. –  Albert Apr 13 '12 at 19:06
6  
Is there really a constant $C_1$ such that $2^n \le C_1$ for all $n$? –  André Nicolas Apr 13 '12 at 19:08
1  
You have to figure out whether or not there exists a constant $C$ such that $2^{2n}\le C2^n$ (or equivalently as you note, $2^n\le C$) for all $n$ beyond a certain, say, $n_0$. In other words, here you have to decide whether or not the integer powers of two have an upper bound. –  anon Apr 13 '12 at 19:08
    
@Albert: En inglés, no existe el signo de interrogación izquierdo. –  Arturo Magidin Apr 13 '12 at 19:14
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2 Answers 2

up vote 6 down vote accepted

If $2^{2n}=O(2^n)$, then there is a constant $C$ and an integer $M$ such that for all $n\ge M$, the inequality $2^{2n}\le C 2^n$ holds.

This would imply that $2^n\cdot 2^n\le C 2^n$ for all $n\ge M$, which in turn implies $$\tag{1} 2^n\le C \quad {\bf for\ all } \quad n\ge M. $$ Can such $C$ and $M$ exist? Note the right hand side of $(1)$ is fixed, and the left hand side...

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Ok, now I understand, thanks a lot. –  Albert Apr 13 '12 at 19:42
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$x^n=o(y^n)$ iff $x<y$, as $(\frac{x}{y})^n\rightarrow 0$. Here 4 and 2, so no.

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