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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?

  • 0
    @Mark: True... but there are some LaTeX artifacts that produce them, so best to not include them.2012-04-13

2 Answers 2

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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.2012-04-13
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$x^n=o(y^n)$ iff x, as $(\frac{x}{y})^n\rightarrow 0$. Here 4 and 2, so no.