Which function grows faster?

  • Thread starter Dragonfall
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[tex]n^n[/tex]

or

[tex]\prod_{r=0}^{n-1}\left(\begin{array}{c}\sum_{j\leq r}\left(\begin{array}{c}n\\ j\end{array}\right)\\2^r\end{array}\right)[/tex]

Asymptotically, I mean.
 
Last edited:

CRGreathouse

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Since
[tex]\prod_{r=0}^{n-1}{{\sum_{j<r}{n\choose j}}\choose{2^r}}={0\choose1}\prod_{r=1}^{n-1}{{\sum_{j<r}{n\choose j}}\choose{2^r}}=0[/tex]
I'd have to go with [itex]n^n[/itex].
 
1,028
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How about now?
 

CRGreathouse

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It seems that the product grows faster.
 

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