Re: Re: Re: "Assuming"
- To: mathgroup at smc.vnet.net
- Subject: [mg85905] Re: [mg85868] Re: [mg85858] Re: "Assuming"
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Wed, 27 Feb 2008 04:28:16 -0500 (EST)
- 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>
Andrzej Kozlowski wrote:
>
> On 26 Feb 2008, at 13:43, Daniel Lichtblau wrote:
>
>> [...]
>>
>> I've seen cases where the FullSimplify[something] result differs from
>> something on a finite set of integers. This motivated me several months
>> ago to alter assumptions of integrality, to reality (realness?
>> realhood?), in processing of Integrate.
>>
>> Daniel Lichtblau
>> Wolfram Research
>>
>
> Formally speaking, reasonable measures (e.g. Radon measures) are either
> diffuse or Dirac measures (or linear combinations of these). However,
> for the former, the entire set of integers has measure zero, and for
> the latter certain finite sets will have a non-zero measure. So none of
> these seems to fit the intended meaning of "measure zero".
>
> More seriously; I think the intended meaning is that in
> Simplify[thing1] -> thing2 thing1 and thing2 should both be functions
> of some variable that is defined on an uncountable set, then they may
> be are allowed to differ for a finite number of values. But, if the
> functions are defined only on countable sets (e.g. the set of all
> integers, as is the case with many number theoretic functions) then the
> failure of thing2 to be equal to thing1 on a finite set could be very
> serious. I think in such situations the "set of measure zero" should
> really be the empty set, or perhaps in really exceptional cases a set
> that contains no more than a single point.
>
> Andrzej Kozlowski
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. 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
Daniel Lichtblau
Wolfram Research
- 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"