Re: Limit with restrictions
- To: mathgroup at smc.vnet.net
- Subject: [mg34304] Re: [mg34269] Limit with restrictions
- From: Hannes Egli <hannes.egli at uni-greifswald.de>
- Date: Tue, 14 May 2002 04:11:03 -0400 (EDT)
- References: <d7.1777ef60.2a110117@aol.com>
- Sender: owner-wri-mathgroup at wolfram.com
Thank you very much for your detailed answer.
Unfortunately, it seems that it does not always work as you have proposed.
Consider the following examples:
p = Simplify[m^(-1 + a + b + n)*(a/(a + b + n))^a*((b + n)/(a + b + n))^(b +
n)*(a + b + n), {a > 0, b > 0, n > 0, a + b + n < 1}];
Needs["Calculus`Limit`"];
Limit[p, m -> Infinity]
0
This result is correct. However, if (a + b + n) > 1, the equations should
converge to infinity if m -> infinity, should't it?
Clear[p]
p = Simplify[m^(-1 + a + b + n)*(a/(a + b + n))^a*((b + n)/(a + b + n))^(b +
n)*(a + b + n), {a > 0, b > 0, n > 0, a + b + n > 1}];
Needs["Calculus`Limit`"];
Limit[p, m -> Infinity]
0
In addition, for a simpler equation I do not get any clear result at all:
Clear[p]
p = Simplify[m^(-1 + a + b + n),{a > 0, b > 0, n > 0, a + b + n >1}];
Needs["Calculus`Limit`"];
Limit[p, m -> Infinity]
E^(Infinity*Sign[-1 + a + b + n])
Since I assumed that (a+b+n)>1 the sign of (-1+a+b+n) is unambiguously positive
and the equation should converge to infinity.
Thanks
Hannes Egli
BobHanlon at aol.com schrieb:
> In a message dated 5/13/02 7:00:15 AM, hannes.egli at uni-greifswald.de writes:
>
> >Is there a possibility to combine the command Limit with parameter
> >restrictions?
> >
> >A simple example reads:
> >
> >p= m^(-1 + a + b + n)*(a/(a + b + n))^a*((b + n)/(a + b + n))^(b +
> >n)*(a + b + n)
> >
> >The following restrictions are given:
> >
> >restrictions = {a > 0, b > 0, n > 0, a + b + n < 1}
> >
> >Since Limit does not allow restrictions, I tried to combine Limit with
> >Simplify, since in other examples (e.g. determination of an equation's
> >sign) this has worked very well:
> >
> >Simplify[Limit[p, m -> Infinity],restrictions]
>
> restrictions={a>0,b>0,n>0,a+b+n<1};
>
> p=m^(-1+a+b+n)*(a/(a+b+n))^a*((b+n)/(a+b+n))^(b+n)*(a+b+n);
>
> Needs["Calculus`Limit`"];
>
> Limit[p, m->Infinity]
>
> 0
>
> You would generally want to use Simplify (or FullSimplify) with the
> restrictions before applying Limit. If the resulting limit is
> complicated you might then want to re-use Simplify (or FullSimplify).
>
> Limit[Simplify[p, restrictions], m->Infinity]
>
> 0
>
> However, in general Simplify (or FullSimplify) with the restrictions
> should be applied when defining p so that all subsequent uses of
> p have the simpler form.
>
> p=Simplify[
> m^(-1+a+b+n)*(a/(a+b+n))^a*((b+n)/(a+b+n))^(b+n)*(a+b+n),
> restrictions];
>
> Bob Hanlon
> Chantilly, VA USA