MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: how to explain this weird effect? Integrate

To avoid further pedantic objections I should have written in my first 
response to the message below that if one chooses to "extend 
continuously"  1/x to 0 the range of the new function or mapping will 
no longer be contained in a "complete ordered field". And then to avoid 
further irrelevant pedantry I should explain what I mean by 
"contained", e.g by means of an injective continuous field homomorphism 
or something like that. But to what puropse?
In fact the whole argument sounds to me not that different form saying 
that every mapping is continuous since one can always give the domain 
the discrete topology. Quite true and equally irrelevant.

Andrzej Kozlowski
Chiba, Japan

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

  • Prev by Date: Re: Re: Help Browser issue in 5.0.1 on Mac OS X
  • Next by Date: Re: corrected RE: Re: Computing sets of equivalences
  • Previous by thread: Re: how to explain this weird effect? Integrate
  • Next by thread: Re: how to explain this weird effect? Integrate