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Re: how to explain this weird effect? Integrate

  • To: mathgroup at
  • Subject: [mg46484] Re: how to explain this weird effect? Integrate
  • From: drbob at (Bobby R. Treat)
  • Date: Fri, 20 Feb 2004 00:29:29 -0500 (EST)
  • References: <20040218182324.673$> <c11se0$nkq$>
  • Sender: owner-wri-mathgroup at

I agree with Andrzej; Mathematica can't just substitute limits
everywhere. To do so assumes a context that may not be relevant.

More to the point, I simply don't want Mathematica to spend time
computing limits without asking me. Sometimes these limits would be
very slow to compute and Mathematica would sometimes fail to compute
them, or even get them wrong.

I'd rather have an error message, so that I know there's a formal
discontinuity to deal with in context.


Andrzej Kozlowski <akoz at> wrote in message news:<c11se0$nkq$1 at>...
> On 19 Feb 2004, at 00:23, David W. Cantrell wrote:
> > Andrzej Kozlowski <akoz at> wrote:
> >> On 15 Feb 2004, at 04:19, steve_H wrote:
> >>
> >>> Andrzej Kozlowski <akoz at> wrote in message
> >>> news:<c0kr0g$fo8$1 at>...
> >>>> On 14 Feb 2004, at 03:56, steve_H wrote:
> >>>>
> >>>>> mathematically speaking, 1/a when a=0, is the same as Limit[1/a ,
> >>>>> a->0] So, the final answer should not be different.
> >
> > Right. The only reason that they differ in Mathematica is that, when
> > finding Limit[1/a, a->0], Mathematica assumes a specific path, namely 
> > that
> > a approaches 0 along the positive real axis. But if we're dealing with 
> > a
> > path-independent limit and with, say, C* = C U {oo}, the one-point
> > compactification of C, for the domain and range of the function f(z) = 
> > 1/z,
> > then f is continuous: f(0) and the limit of f(z) as z -> 0 are both oo.
> >
> > [snip]
> >> It is precisely the fact that
> >>
> >> Limit[(f[x+h]-f[x])/h, h->0] is not obtained by "substituting" 0 for 
> >> h,
> >> that is the key point here.
> >
> > That key point is, of course, correct.
> >
> >> What you wrote, that is that 1/x at 0 is the same as Limit[1/x, x->0]
> >> is entirely wrong mathematically.
> >
> > No, it is correct, at least in contexts familiar to me:
> >
> > 1) If we take the codomain of f(x) = 1/x to be R or C, then f(x) is
> > undefined at x = 0 and the corresponding limit does not exist. 
> > Granted, the
> > usual wordings "undefined" and "does not exist" are different, but 
> > they're
> > saying essentially the same thing.
> >
> > 2) If we take the codomain to be C* or R* (the one-point 
> > compactification
> > of R), then f(x) is defined at x = 0, the corresponding limit exists, 
> > and
> > they are the same.
> >
> > Note that, above, I was talking about path-independent limits. But, for
> > better or worse, Mathematica assumes a specific default path when 
> > finding
> > Limit[1/x, x->0]. Due to that, it yields Infinity. But the 
> > corresponding
> > path-independent limit should give ComplexInfinity, which is the same 
> > as
> > the value of 1/x at x = 0 in Mathematica.
> >
> >> It is the kind of thing that is used when people speak informally,
> >> or perhaps by engineers who do not care
> >> about mathematical correctness but only if things work in practice, 
> >> but
> >> it was you who wrote "mathematically".
> >
> > There's nothing inherently informal or incorrect in such usage.
> >
> >> Mathematically, the real valued
> >> function written as 1/x has no value at all at 0.
> >
> > True.
> >
> >> There is no real or complex number called Infinity or ComplexInfinity.
> >
> > True. Those are instead elements of extensions of R and C, resp.
> >
> >> On the other hand
> >> Limit[1/x,x->0] = Infinity has a well defined sense: it means that for
> >> all x which are  small enough  1/x will exceed any chosen number,
> >> however large.
> >
> > More specifically (remembering the default path),
> > ... for all _positive_ x which are small enough...
> >
> > But such a limit statement can have _different_ well defined senses,
> > depending on context. I presume that, for you, the context was strictly
> > that of the reals. In that case, saying "Limit[1/x,x->0] = Infinity" is
> > merely giving a specific way that the limit fails to exist. As you 
> > said,
> > it's because "1/x will exceed any chosen number, however large." But 
> > if the
> > context is instead R* or C*, then it's literally (and rigourously) 
> > correct
> > to say that the value of the path-independent limit _is_ oo.
> >
> > [snip]
> >> In fact the situation is
> >> exactly the opposite of what you wrote, this kind of thing may be
> >> acceptable in a practically oriented mathematical program like
> >> Mathematica, would not be in a serious mathematics book (or at least
> >> the author would make excuses about the informality assuming that
> >> everybody knows the true state of affairs).
> >
> > No, "this kind of thing" can be perfectly acceptable "in a serious
> > mathematics book"!
> >
> >>> Do you know of a function that is well behaved and smooth at a point
> >>> 'a', where  the limit of f(x) as x approaches 'a' will not be f(a) ? 
> >>> if
> >>> so, please tell us about this function.
> >>
> >> Were you joking? This is the definiton of continuous at a. 1/x is not
> >> defined at 0 and there is not way to define it so as to make it
> >> continuous at 0.
> >
> > Sure there is a way: Take its codomain to be R* or C*, for example.
> >
> > David Cantrell
> >
> >
> You are entering into a pointless dispute which is about definitions at 
> best and has nothing to do the aims of this list. But is you like ....
>   One can of course consider functions with values in the Riemann 
> sphere. (I certainly know about that since I am a professor of 
> mathematics, a topologist, and most of my comparatively recent 
> published work has been, in fact,  on the topology of the space of 
> holomorphic functions form complex projective spaces to the Riemann 
> sphere and related stuff). But when you compactify the complex plane 
> (and of course you can do the same thing with the real line though you 
> need two extra points, which you can call Infinity and -Infinity  and 
> topologically you get the interval) you are no longer dealing with real 
> valued or complex valued functions. The range of your functions is no 
> longer a field, either of real numbers or complex numbers. You can no 
> longer perform ordinary algebraic operations on your functions and you 
> have to introduce lot's of additional care. All of this needs a lot of 
> additional justification and one would expect a serious mathematician 
> to know that. Of course if one is talking with people to whom this sort 
> of thing is bread and butter one need not go into the details each time 
> but this list is not such a place.
> Moreover, even to assume that when 1/x is a function in the complex 
> plane not to mention on the Riemann sphere is actually choosing one 
> context for Mathematica, which is not the only context in which 
> Mathematica is used. Actually x can equally well be n element of an 
> abstract abelian group written multiplicatively, or, say an element of 
> the ring Z/p for a prime p, etc. Mathematica deals with such things 
> too. To say that 1/x is not defined at 0 as Infinity is nonsense, 
> though of course one could actually always define some sort of new 
> mathematical object that allows one to do that. But that is completely 
> beside the point, mathematics is very flexible and if you do not have 
> an actual contradiction one can usually invent knew theories or new 
> conventions to make sense of what was nonsense before. That is what 
> happened with Leibniz, Abraham Robinson and non-standard analysis. But 
> bringing up these issues here seems to me only like trying to show off 
> ones mathematical knowledge without any relevance to Mathematica, which 
> is not intended for this sort of theoretical and specialised purposes, 
> though of course one can always adapt it oneself if one needs it. The 
> reason not do do so in Mathematica is of course that buy forcing on it 
> a particular interpretation you are making it harder for other peoples 
> to use theirs, even your proposal to use automatically continuous 
> extensions of certain functions might make things harder for people who 
> needed to work with discontinuous functions, and forcing on certain 
> expressions interpretations taken form complex analysis will make it 
> harder for people who want to use them to work in abstract ring theory, 
> without a topology and nay infinities.
> Andrzej Kozlowski
> Chiba, Japan

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