Re: 4^n > n^4
From: Marcel Martin (mm_at_ellipsa.no.sp.am.net)
Date: 10/26/04
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Date: Tue, 26 Oct 2004 15:35:52 +0200
tsmith a écrit :
>
> Show for eact natural number n >= 5, 4^n > n^4.
>
> I've tried to do this using the techniques similar to the book examples and
> the previous exercises I worked. But I cannot get very far. It is late and
> may be missing something obvious. But is there a trick to doing this?
Notice that 4^n = (2^n)^2 and that n^4 = (n^2)^2, so, proving
(4^n > n^4) and proving (2^n > n^2) are equivalent.
Now, you could show that
1) (2^n > n^2) is true for n = 5
2) (2^n > n^2) --> (2^(n+1) > (n+1)^2)
-- mm http://www.ellipsa.net/ mm@ellipsa.no.sp.am.net ( suppress no.sp.am. )
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