Re: -- Wrong limits do not commute

G. Rodrigues wrote:

On 3 okt, 15:06, "G. Rodrigues"
<sorlak...@xxxxxxxxxxx> wrote:

G. Rodrigues wrote:

<snip>

Let me ask you one thing: the non-equality of
iterated limits means
geometrically that if you have a function f(x,
y)
say, then examining
the behaviour of f along two different parallel
paths (meaning, they
start and end at the same points) leads to
different results. Is your
contention that every such limiting procedure
along parallel paths
should yield the same results and only fails
todo so in cunningly
contrived cases that never appear in the
ptractical applications of
mathematics?

How can two different parallel paths start and
end at
the same points ?

This is standard terminology in some fields of
mathematics. If it bothers you, replace "parallel
paths" by "two paths that start and end at the
same
point". Now, will you or will you not please
answer the question?

Yes. As soon as I understand what it _means_.

What do you not understand? To fix ideas, consider
the equality of an iterated limit of the form

lim_{x->0} lim_{y->0} f(x, y)= lim_{y->0}
lim_{x->0} f(x, y)

? Don't you mean:

lim_{x->oo} lim_{y->oo} f(x, y) = lim_{y->oo}
lim_{x->oo} f(x, y)


No, I meant exactly what I wrote. Use a suitable transformation of variables to reduce one case to the other. Or is that "wrong", "blind" and "unstable"?

Look, let us assume that the double limit of f when (x, y) -> oo exists and is equal to the iterated limits. Call it l. By defining g = f - l you obtain a function that *no matter how you approach oo* decays off to 0. Contending that such functions (minus a constant) are the only ones of interest to applied mathematics is so trivially wrong that I do not know where to begin.

Geometrically, it means that starting at some point
near (0, 0), and going first parallel to the x-axis
then parallel to the y-axis, is the same (as far as
taking the limit) as first going parallel to the
y-axis then going parallel to the x-axis. Needless to
say, I am not being rigorous here, just trying to
convey the idea. I can do that if you want me to, at
the cost of considerable obfuscation.

Of course, there is nothing special about "first
move parallel to the x-axis then move parallel to the
y-axis", etc., so we can generalize to any pair of
parallel paths (as per the above definition). And
there is nothing special about the limit taken; you
can conceivably replace it by other limiting
processes such as taking the derivative or the (line)
integral. Thus my question above.

But anyway, both limits are approximated by f(X,Y) ,
where X and Y are
large (iff you mean oo instead of 0 )and thus should
_not_ be different
in practical applications. And be equal to
lim_{(x,y)->(oo,oo)} f(x,y) ,
i.e. the double limit of the same function.


See above for (one) reply.

I take it that your answer to my question above is yes. Congratulations, you have just trivialized General Relativity. The curvature tensor of spacetime allows us to associate to every path c starting at x_0 and ending at x_1 and to every tangent vector v_0 at x_0, a tangent vector v_1 at x_1. This vector is called parallel transport along c of v_0 and can be given by a limiting process. If you have another path d starting and ending at the same points, the parallel transport maybe different from the one along c, and this difference is measured precisely by the curvature tensor. If the curvature is 0 then they are indeed the same (technical note: actually not even this is true, the paths must be homotopic). General relativity with 0 curvature is not particularly interesting from the point of view of modelling reality: no second order effects, no local geometry, no gravitational waves...

Regards,
G. Rodrigues
.

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