Thorsten Froehlich wrote:
> 

> <Jer### [at] enstfr>  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.
> 
	All right, then how do you justify the first inequation? The valid
demonstration is:
lg(n) > 2 (for n > 3) | * n
=>
n lg(n) > 2n          | + n lg(n)
=>
2n lg(n) > 2n + n lg(n)

> >  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
> 
> ...
	I wasn't really speaking about the reasoning. My point is that you
can't compare two infinite limits. The valid reasoning is to say that
for n big enough (in this case "big enough" means greater than 3) you
have those equations (without the limits). Or else redefine clearly what
you mean by:
 lim ...
n->oo
since you're not using the standard definition.
	Again it's more a complaint about the form of your reasoning (which is
wrong) than with the content (which for this part is right, I haven't
looked into the rest so I can't speak about the whole)


-- 

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