Re: "Assuming"
- To: mathgroup at smc.vnet.net
- Subject: [mg85928] Re: "Assuming"
- From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
- Date: Thu, 28 Feb 2008 02:44:23 -0500 (EST)
- Organization: NewsReader.Com Subscriber
- References: <20080221171506.200$2n_-_@newsreader.com> <200802221221.HAA08545@smc.vnet.net> <200802251237.HAA22859@smc.vnet.net> <200802261243.HAA22523@smc.vnet.net> <3185411B-C98F-476F-9C77-FBC32FE719D2@mimuw.edu.pl> <47C49E87.600@wolfram.com> <fq3ba0$g78$1@smc.vnet.net>
Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
> On 27 Feb 2008, at 00:19, Daniel Lichtblau wrote:
[snip]
> > Here is an example of the behavior in question. I do not pass
> > judgement on whether it should be regarded as a bug or a feature. I
> > simply wanted to give a concrete example where the behavior arises
> > and is difficult to supress.
> >
> > In[2]:= i1 = Integrate[Sin[m*x]*Sin[n*x], {x,0,2*Pi}, Assumptions-
> > >Element[{m,n},Reals]];
> >
> > Check what happens when we assign n->1 and then take limit as m->1.
> >
> > In[3]:= l1 = Limit[i1 /. n->1, m->1]
> > Out[3]= Pi
> >
> > That was fine.
Yes, that limit was fine. But i1 was not valid for all m and n. And indeed
it is that very lack of validity when |m| = |n| which is the common reason
for wishing to find a limit such as l1. This situation can be avoided if we
use an antiderivative for Sin[m*x]*Sin[n*x] which is valid for all m and n:
x/2 (Sinc[(m - n) x] - Sinc[(m + n) x])
Then, applying Newton-Leibniz, we would have gotten
i1alt = Pi (Sinc[2 (m - n) Pi] - Sinc[2 (m + n) Pi])
as an alternative for the definite integral, valid for all m and n.
> > Now see what happens if we assign n->1 and simplify
> > under assumption that m is an arbitrary integer.
> >
> > In[4]:= l2 = Simplify[i1 /. n->1, Element[m,Integers]]
> > Out[4]= 0
In[13]:= Simplify[i1alt /. n -> 1, Element[m, Integers]]
Out[13]= Pi (Sinc[2 (-1 + m) Pi] - Sinc[2 (1 + m) Pi])
In[14]:= FullSimplify[%, Element[m, Integers]]
Out[14]= 0
Thus, Simplify to 0 happened to have been suppressed, but FullSimplify
still gave 0, which I would call a misdemeanor.
> Unless I am missing something obvious (which is possible as I have not
> yet fully woken up) the problem amounts simply to this:
>
> x = (1/2)*(Sin[2*(m - 1)*Pi]/(m - 1) - Sin[2*(m + 1)*Pi]/(m + 1));
>
> In[2]:= Limit[x, m -> 1]
> Out[2]= Pi
>
> In[3]:= Limit[x, m -> 1, Assumptions -> Element[m, Integers]]
> Out[3]= 0
>
> The last answer maybe slightly dubious because it is not perfectly
> clear in what sense the limit is taken here. But it seems to me a very
> minor point and no cause for concern ?
Your In[3] and Daniel's In[4] are asking for different things, a fact which
you probably already noticed, having been awake longer now.
David
- References:
- Re: "Assuming"
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Re: "Assuming"
- From: "Mariano Suárez-Alvarez" <mariano.suarezalvarez@gmail.com>
- Re: Re: "Assuming"
- From: Daniel Lichtblau <danl@wolfram.com>
- Re: "Assuming"