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Re: "Assuming"
*To*: mathgroup at smc.vnet.net
*Subject*: [mg85935] Re: "Assuming"
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Thu, 28 Feb 2008 02:48:00 -0500 (EST)
*References*: <20080227092600.592$XT_-_@newsreader.com>
On 27 Feb 2008, at 15:26, David W. Cantrell wrote:
> [Message also posted to: comp.soft-sys.math.mathematica]
>
> 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
Actually, Daniel's In[4] and my In[3] are (essentially) equivalent,
since what happens when my In[3] is evaluated by Mathematica is that
first Daniel's In[4] is evaluated, returning 0 and then the Limit of 0
is taken as m goes to Infinity, returning again 0. The Limit is
therefore irrelevant except for the problem of mathematical
interpretation of what it all actually means, since of course the
assumption that m is an integer and the assumption that m ->1 are
incompatible in Mathematica, which does not consider limits of
sequences. As it was very early in the morning I could not see
anything else in Daniel's message that was problematic and could not
understand what the point of his example was. I think I can see that
now, but I have to confess I am still quite unable to take it at all
seriously.
Andrzej Kozlowski
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