Re: Cantor Confusion

David Marcus writes
Of course, I agree formally sin(pi/x) is undefined at x=0.

What does "formally" mean?

Because pi/0 is not defined, but thought that were good reasons for assigning sin(pi/x) a value of 0 at 0, as opposed to 0.3 , -.25 or any other possible number. (Actually, having read the intelligent and helpful post on the value of 0^0 I am suddenly not so sure about this, but anyway, that was what I meant by the "formally")

But, if something is undefined it is usually because
the coordinate system describing the underlying thing
is inappropriate at that point. e.g. y = mx + c is
not a good basis for describing all lines in a plane
(a better set of coordinates are the minimum distance from
the origin and a unit vector).

That is not what "undefined" means.

I think that you miss my point, or discount it. A line parrallel to the y axis has an infinite gradient; such a line viewed in (m,c) coordinates it has an "undefined" value (inf, inf). But the line is perfectly sensible at e.g. x = k. So, the same thing, I think (maybe), applies to functions. asin(y) is multivalued, difficult to comprehend, but a shift of perspective gives y = sin(x), which is readily comprehensible. Ditto, possibly, for situations involving infinities large or small - a change in variables potentially makes things more comprehensible.

I don't know about sin(pi/x) since it is performing an
infinite number of oscillations around x = 0. But a value
of anything other than 0 at 0 would make the function
discontinuous at 0 because it is antisymmetric (or odd, if
you prefer).

We say what function we are talking about. Let

f(x) = sin(1/x), if x <> 0.

g(x) = sin(1/x), if x <> 0,
0, if x = 0.

h(x) = sin(1/x), if x <> 0,
1, if x = 0.

f is not defined at zero. g and h are defined at zero. Using Wikipedia's
definition of "odd", f and h are not odd, g is odd. It is meaningful to
ask whether f can be extended to all of R so that its extension is odd.
The answer to this question is yes.

I don't know about Wikipedia - I used the term "antisymmetric". f,g,h are all antisymmetric. g is "obviously" correct - because if you invert the universe left to right, there is no reason to suppose that sin(1/x) changes sign at x =0. Possibly not a killer argument.

exp(-1/(x^2)) is also "not defined" at x = 0, but
it would be a mistake to define the function and all
its derivatives as e.g. 42.0 at x=0, and then say
that the function and all its derivatives can be made less
than any value d by choosing some 0<x<eta.

It is not a "mistake" to define a function as follows.

k(x) = e^(-1/x^2), x <> 0,
42, x = 0.

It is true that k is not continuous, but it is a perfectly valid
function. Do you know the definition of "function"?

Doubtless you lot have some legally watertight definition. As far as I am concerned, a function is a formula that provides a value for a given input (or inputs, in a multidimensional situation).

However, we can't simply define derivatives to be whatever we want.
Derivatives are determined by the values of the function. The function k
is not continuous at zero, so it is not differntiable at zero.

(incidentally
how does exp(-1/x^2) get off the ground anyway? OK, it's not
defined at x = 0, but you CAN define a function f by
e.g. Taylor expansion in x about some x0, such that f(x) =
exp(1/x^2) for all x!=0, and that IS defined at x =0,
and that has f and all its derivatives = 0 at x =0 ?)

What does "get off the ground" mean? I don't understand why you want to
use a Taylor expansion to define the function (and I'm not sure if the
Taylor expansion you want to use is zero at zero; I think it isn't).

Well maybe that would explain how it "gets off the ground". Excuse the loose talk, but I thought that would be clear - if a function and all its derivatives are zero at a point, then if you move an infinitessimal distance delta, to order (delta)**2 the function and all its derivatives are still all zero. And in a Taylor expansion to a finite offset, the function is still zero, because all its derivatives are zero.

Why
not just do it directly? I.e., let

F(x) = e^(-1/x^2), x <> 0,
0, x = 0.

This is a perfectly good function. It is infinitely differentiable, but
not analytic. I.e., all of its derivatives exist, but its Taylor series
at zero does not equal the function in any neighborhood of zero.

Well that is undoubtedly the answer to the question. Why doesn't the Taylor expansion work might be a better way of phrasing the question?

Re example of 1/x, the discontinuity of 1/x at x = 0 suggests strongly
that in some sense + and -inf are the same location,
but I don't expect that suggestion to be viewed very sympathetically here.

The function 1/x isn't defined at zero. We can't extend it to be
continuous because the limits as we approach zero from the left and the
right don't exist. Infinity is not a number. Although, in real analysis,
we do consider the extended real numbers where we include +inf and
-inf as numbers. However, not all of the usual arithmetic properties
extend to +/-inf, so you have to be careful. There are other contexts
where a point at infinity is sometimes used, for example in complex
analysis or geometry.

Yes.

> You seem to have a funny idea about math.
> Mathematical objects are as we
> define them. You seem to want to argue in reverse.

I think that you lot discover meaningful mathematical
objects; I don't think you define them. If they are
universal truths, you could have an intelligent discussion
with the aliens when they land...

I think you don't understand how the word "define" is used in
mathematics. What math courses have you had?

I am definitely not conversant with your technical definitions of words. Me, as far as Maths courses go, if you want to talk about anything dirty, like applied math, Fourier series, signal processing, vectors and differential equations, I am pretty clued up. But as far as set theory, number theory, pure maths in general, de nada. So from your perspective I am a Klutz. But I am not trying to engage/compete ... just curious, no ego here, happy to admit ignorance. Yes, I should read a book ( I am, actually) but that would just result in more questions...
--
Andy Smith

.

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