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Does the latter grow faster? I'm assuming that if we have $a^n$ vs $b^n$, if $b>a$ then $a = O(b)$, but if there is a n term in front of a does that change it?

Widawensen
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maregor
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1 Answers1

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Note that for suitably large $n$ we have $n << (4/3)^n$. This shows directly that $4^n$ grows faster.

Rolf Hoyer
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