Re: A limit bug
- To: mathgroup at smc.vnet.net
- Subject: [mg84532] Re: A limit bug
- From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
- Date: Thu, 3 Jan 2008 05:30:00 -0500 (EST)
- References: <flc7tk$7vo$1@smc.vnet.net> <flfaiq$aac$1@smc.vnet.net>
Norbert Marxer <marxer at mec.li> wrote: > On 1 Jan., 03:18, "David W.Cantrell" <DWCantr... at sigmaxi.net> 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 > > Hello > > And what about telling Mathematica how p and q should approach > Infinity? That has rather limited usefulness for the problem posed by the OP in the parent thread. Recall that he wanted the limit, if it exists, of (p + q)!/(p! q!) as both p and q increase without bound. That's a limit without reference to any specific path -- in other words, a path-independent limit. Telling Mathematica how p and q should approach Infinity (as in your first example below, say), we find that _that_ path-specific limit is Infinity. And that tells us that, if the path-independent limit exists, it must be Infinity. Examining other path-specific limits would be definitive if we ever found one along which the limit was not Infinity, for then we would know that the desired path-independent limit did not exist. But of course, for the question at hand, examining other path-specific limits is futile. Regardless of how many we find to be Infinity, we are still left with the possibility that, along some other path which we didn't consider, the limit might not be Infinity and so the path-independent limit would not exist. > E.g. p and q approach Infinity in the same way: > > Limit[(p + q)!/(p!*q!) /. q -> p, p -> Infinity] > > E.g. q approaches Infinity faster than p: > > Limit[(p + q)!/(p!*q!) /. q -> p^5, p -> Infinity] > > ... and many other ways would give the correct result, i.e. Infinity. Yes, they should give the correct result for path-specific limits, but still leave open the possibility that the desired path-independent limit does not exist. Best regards, David Cantrell