Re: how to explain this weird effect? Integrate
- To: mathgroup at smc.vnet.net
- Subject: [mg46457] Re: how to explain this weird effect? Integrate
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Thu, 19 Feb 2004 03:02:10 -0500 (EST)
- Organization: NewsReader.Com Subscriber
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Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
> On 15 Feb 2004, at 04:19, steve_H wrote:
>
> > Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote in message
> > news:<c0kr0g$fo8$1 at smc.vnet.net>...
> >> 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
- References:
- Re: how to explain this weird effect? Integrate
- From: nma124@hotmail.com (steve_H)
- Re: how to explain this weird effect? Integrate
- From: nma124@hotmail.com (steve_H)
- Re: how to explain this weird effect? Integrate