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Prove the following statements:

  1. $2^n$ is $O(n!)$, and

  2. $n!$ is not $O(2^n)$

not sure where to start with these two... thanks

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Do you mean $n!$ is not $O(2^n)$? – vadim123 Apr 22 '13 at 2:27
ops, yes thanks – Supernatural Apr 22 '13 at 2:30
up vote 2 down vote accepted
  1. Note that $\frac{2^n}{n!}=\frac{2\times 2 \times 2\cdots \times 2}{1\times 2 \times 3\times 4\cdots \times n} \le 2$ for all $n$. Hence $|2^n| \le 2 |n!|$ for all $n$.

  2. Suppose that there is an $M$ such that $|n!| \le M |2^n|$ for all $n$ (sufficiently large). Try to find a contradiction.

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i've checked 0, 0! <= M * 1 this holds – Supernatural Apr 22 '13 at 2:41
since M could be anything I chose shouldn't this always hold? – Supernatural Apr 22 '13 at 2:41
$M$ needs to be chosen, held fixed. It needs to work for all $n$ (except possibly finitely many small ones). – vadim123 Apr 22 '13 at 2:43
thanks! get it now – Supernatural Apr 22 '13 at 2:46

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