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<Jer### [at] enst fr> wrote:
> Moreover your implication goes the wrong way (ie the second inequation
> implies the first, not the other way round).
Hmm, maybe you are forgetting some fundamental things here...
2n lg n > 2n + n lg n | : n lg n
=>
n lg n > lim 2n
because n lg n can never be 0 in this case (so the division is legal). Of
course I am just silently dropping the + 1 here and some other changes to
the function because of the division.
> This doesn't make any mathematical sense since all four limits are
> infinity and therefore can't be compared. What can be said is that for a
> big enough n, you have:
> 2n lg(n) > 2n + n lg(n)
> and
> n lg(n) > 2n
Nope, you can do the following (as what I am up to is not a result, but only
a relative comparison of the growth rate):
lim 2n lg n > lim 2n + n lg n
n->oo n->oo
<=>
lim 2n lg n - 2n + n lg n > 0
n->oo
<=>
lim n lg n - 2n > 0
n->oo
Thorsten
____________________________________________________
Thorsten Froehlich, Duisburg, Germany
e-mail: tho### [at] trf de
Visit POV-Ray on the web: http://mac.povray.org
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