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Re: A limit bug
*To*: mathgroup at smc.vnet.net
*Subject*: [mg84525] Re: A limit bug
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Wed, 2 Jan 2008 01:18:26 -0500 (EST)
*References*: <20080101110626.081$mK_-_@newsreader.com>
On 2 Jan 2008, at 01:06, David W. Cantrell wrote:
> [Message also posted to: comp.soft-sys.math.mathematica]
>
> Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
>> On 1 Jan 2008, at 11:16, David W.Cantrell wrote:
>>
>>> A recent question in sci.math led to something which should also
>>> interest this group.
>>>
>>> The OP asked about the limit of (p + q)!/(p! q!) as both p and q
>>> increase without bound. And he said later
>>>
>>>> I wasn't sure about it because Mathematica gives me a limit of
>>>> zero.
>>>> Isn't that strange?
>>>
>>> I responded as follows.
>>>
>>> ----------------------------------------------
>>>
>>> Well, it's a bug. I suppose that what you did in Mathematica was
>>> something like
>>>
>>> In[3]:= Limit[Limit[(p + q)!/(p! q!), q -> Infinity], p -> Infinity]
>>>
>>> Out[3]= 0
>>>
>>> But note that there is not any way -- well, at least none known to
>>> me -- in Mathematica to get a true general "two-variable" limit:
>>>
>>> limit f(x,y) as (x,y) -> (x0,y0)
>>>
>>> However, Mathematica can get a correct answer for your limit
>>> problem.
>>>
>>> First, realize that (p + q)!/(p! q!) is Multinomial[p, q]. So you
>>> might try
>>>
>>> In[5]:= Limit[Limit[Multinomial[p, q], q -> Infinity], p ->
>>> Infinity]
>>>
>>> Out[5]= Limit[Limit[Multinomial[p, q], q -> Infinity], p ->
>>> Infinity]
>>>
>>> Since that remains unevaluated (but at least there was now no bug!),
>>> you might consider the possibility that it remained unevaluated for
>>> a good reason, namely, because a little more information had to be
>>> provided:
>>>
>>> In[6]:= Limit[Limit[Multinomial[p, q], q -> Infinity,
>>> Assumptions -> p > 1], p -> Infinity]
>>>
>>> Out[6]= Infinity
>>>
>>> Success! Happy New Year!
>>>
>>> But BTW, note that, curiously, the following fails:
>>>
>>> In[7]:= Limit[Limit[(p + q)!/(p! q!), q -> Infinity,
>>> Assumptions -> p > 1], p -> Infinity]
>>>
>>> Out[7]= Indeterminate
>>>
>>> David
>>>
>>
>> I think using iterated limits above actually obscures what happens.
>> In
>> fact:
>>
>> Limit[Multinomial[p, q], q -> Infinity,
>> Assumptions -> p > 1]
>> Infinity
>>
>> Thus there is not much point in further taking limits as p->Infinity,
>> since the answer, being independent of p will certainly not change.
>> On
>> the other hand
>>
>> Limit[(p + q)!/(p!*q!), q -> Infinity, Assumptions -> p > 1]
>> Infinity/p!
>>
>> We can now see what happened here and the source of the problem.
>
> I think you lost sight of the main problem, the thing I called a bug,
> namely, Out[3] = 0. As best I can tell, we don't know the source of
> that
> problem.
>
> But yes, what you say probably does help to see why Out[7] was
> Indeterminate.
>
>> Mathematica simply computed
>> Limit[(p + q)!/q!, q -> Infinity, Assumptions -> p > 1]
>> Infinity
>>
>> and then divided the answer by p! returning Infinity/p! since
>> Mathematica leaves Infinity/a unevaluated for an undefined symbol a.
>
> But Mathematica can correctly simplify Infinity/a if we give it
> appropriate
> information about a. More about that below.
>
>> Now, of course taking the limit with respect to p as p->Infinity will
>> quite correctly produce Indeterminate.
>
> No, that's not correct. The limit of Infinity/p! as p increases
> without
> bound is determinate. It's simply Infinity.
>
> Perhaps you were thinking of Infinity/Infinity!, but the fact that
> that is
> Indeterminate is irrelevant. (Remember: ...in a _deleted_
> neighborhood...)
>
> BTW, here's something that does work correctly (without using
> Multinomial):
>
> In[13]:= Limit[Assuming[p > 1, FullSimplify[Limit[(p + q)!/(p! q!),
> q -> Infinity]]], p -> Infinity]
>
> Out[13]= Infinity
>
> Note that FullSimplify, rather than just Simplify, must be used.
>
>> I would not call this a bug but rather a bit of a design problem.
>
> FWIW, I hadn't called Out[7] a bug before either. In any event, giving
> Indeterminate for a limit which should actually be Infinity is not as
> serious a problem as giving 0.
>
>> On a related note; although I think Mathematica is right to leave
>> unevaluated expressions such as Infinity/a, I would prefer it to
>> return Infinity in the example below:
>>
>> Assuming[Element[a, Reals] && a != 0,
>> Simplify[Infinity/a]]
>
> I can understand why you'd want that behavior.
> Mathematica does a good job with both
>
> In[15]:= Assuming[a > 0, Simplify[Infinity/a]]
>
> Out[15]= Infinity
>
> and
>
> In[16]:= Assuming[a < 0, Simplify[Infinity/a]]
>
> Out[16]= -Infinity
>
> David W. Cantrell
Touche ;-)
I forgot all about DirectedInfinity etc... I guess I should have gone
easier on the New Year night drinking .
Andrzej Kozlowski
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