Re: 4^n > n^4
From: J Silverman (jhs_at_math.brown.edu)
Date: 10/26/04
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Date: 26 Oct 2004 15:09:19 -0700
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote in message news:<cll0d1$go7$3@south.jnrs.ja.net>...
> tsmith wrote:
>
> > 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?
>
> Try showing that 4^{n+1}/(n+1)^4 > 4^n/n^4.
That's a nice elementary solution. Or, slightly fancier, take logs of
both sides, so need to show
n/log(n) > 4/log(4).
Now use elementary calculus to show that
f(x) = x/log(x)
is strictly increasing for x >= 5 (actually, for x > e). So in
particular,
f(x) > f(4) for all x > 4.
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