Re: A limit bug

*To*: mathgroup at smc.vnet.net*Subject*: [mg84522] Re: A limit bug*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Wed, 2 Jan 2008 01:16:51 -0500 (EST)*References*: <200801010216.VAA08134@smc.vnet.net> <flctlp$h2s$1@smc.vnet.net>

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

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